Atomic StructureI. Atomic structure A. Different model of atomic structure B. Isotopic mass and relative atomic mass 1. Mass spectrometer a) Determination of atomic mass b) Determination of molecular mass C. Radioactivity of nuclide 1. Nature of radioactivity a) Penetrating power of radiation 2. Radioactive decay a) Half life b) Use of radioactive isotope II. Electronic structure of atoms A. Emission spectrum of hydrogen 1. Origin of emission and absorption spectrum 2. Pattern of energy levels of hydrogen 3. Determination of the ionization energy a) Through the convergence limit of Lyman series b) Through Rydberg constant B. Detail electronic configuration 1. Definition of ionization energy 2. Successive ionization energy 3. Periodicity of First ionization energy a) First I.E. across periods b) First I.E. down a group 4. Quantum numbers a) Energy of electrons in different orbitals 5. Building up of electrons in an atom a) Aufbau principle b) Pauli exclusion principle c) Hund's rule d) Building up of electrons e) Detail electronic configuration of some atoms 6. Electronic configuration of first 36 elements a) Extra stability of half-filled and full-filled subshell 7. Position of s-, p-, d- and f-block elements III. Atomic orbitals A. Shape of s, p and d orbitals IV. Periodic Table A. Variation in atomic radius B. Variation in electronegativity Energetics I. Enthalpy Change (∆H) A. Definition of a system B. Definition of enthalpy (H) and enthalpy change (∆H) 1. Constant pressure process a) Work done by a system C. Standard enthalpy change of formation (∆H f o) 1. Definition of ∆H f o a) Standard enthalpy change (∆Ho) b) Standard enthalpy change of formation (∆H f o) c) Standard enthalpy change of formation of an element 2. Direct determination of ∆H f o a) Use of simple calorimeter D. Other enthalpy change (∆H) 1. Examples of enthalpy change a) Enthalpy of combustion b) Enthalpy of atomization c) Enthalpy of neutralization d) Enthalpy of hydrogenation e) Enthalpy of solution f) Enthalpy of reaction E. Hess’s Law 1. Indirect determination of ∆H f o a) Hess’s Law (1) Use of energy level diagram / energy cycle (2) Use of equation II. Bonding Energy A. Bond energy 1. Definition of bond energy 2. Determination of bond energy term (bond enthalpy) B. Strenght of covalent bond 1. Relationship between bond length and bond energy 2. Factors affecting the bond strength C. Estimation of enthalpy change by bond energy III. Energetics of formation of ionic compounds A. Electron affinity B. Lattice energy 1. Born-Haber cycle a) Determination of lattice energy by Born-Haber cycle 2. Factors affecting the value of lattice energy C. Stoichiometry of ionic compounds IV. Gibbs free energy change (∆G) A. Limitation of enthalpy change (∆H) B. Entropy (S) and entropy change (∆S) C. Gibbs free energy change (∆G) Chemical Bonding Page 1 Chemical Bonding I. Shape of molecule A. Lewis structure of molecule 1. Oxidation no. of individual atoms in a molecule 2. Formal charge of individual atoms in a molecule 3. Limitation of octet rule B. Valence Shell Electron Pair Repulsion Theory (VSEPRT) 1. Octet expansion of elements in Period 3 2. Position of lone pair 3. 3-dimensional representation of molecular shape C. Hybridization Theory 1. Overlapping of atomic orbital 2. Hybridization theory 3. Examples a) sp hybridizaion b) sp 2 hybridizaion c) sp 3 hybridizaion d) sp 3 d hybridizaion e) sp 3 d 2 hybridizaion 4. Structure and shape of hydrocarbons a) Characteristic of Carbon-Carbon bond b) Shape of hydrocarbons (1) Saturated hydrocarbons (2) Unsaturated hydrocarbons (3) Aromatic hydrocarbons (a) Delocalization of π-electrons (b) Stability of benzene D. Molecular Orbital Theory II. Ionic bonding A. Strength of ionic bond - lattice energy B. X-ray diffraction 1. Electron density map of sodium chloride C. Periodicity of ionic radius 1. Definition of radius 2. Periodicity of ionic radius 3. Size of isoelectronic particles Chemical Bonding Page 2 III. Covalent bonding A. H 2 + ion B. Electron diffraction C. Covalent radius 1. Definition of covalent radius 2. Addivity of covalent radius 3. Breaking down of addivity in covalent radius and bond energy a) Resonance (1) Resonance structure of benzene (2) Resonance structure of nitrate ion (3) Rules in writing resonance structures b) Breaking down of addivity in bond enthalpies (1) Delocalization energy D. Dative covalent bond 1. Examples of H 3 N→BF 3 and Al 2 Cl 6 a) H 3 N→BF 3 b) Al 2 Cl 6 IV. Bonding intermediate between ionic and covalent A. Differences between ionic bond and covalent bond B. Incomplete electron transfer in ionic compound 1. Electron density map of LiF comparing with those of NaCl and H 2 2. Difference among lattice energies of Na, Ag and Zn compounds a) Lattice energies of sodium halide, silver halide and zinc sulphate b) Bonding intermediate between covalent and ionic 3. Polarization of ionic bond a) Fajans’ Rules in polarization of ionic bond C. Electronegativity 1. Definition of electronegativity 2. Pauling scale of electronegativity D. Polarity in covalent bond 1. Deflection of a liquid jet by an electric field 2. Dipole moment a) Vector quantity of dipole moment b) Polarity of molecule c) Factors affecting dipole moment (1) Inductive effect (I) (2) Mesomeric effect / resonance effect (R) (3) Presence of lone pair V. Metallic bonding A. Electron sea model of metal B. Strength of metallic bond C. Melting and boiling of metal D. Strength of ionic bond, covalent bond and metallic bond Intermolecular forces Intermolecular Forces I. Van der Waals’ forces A. Discovery of van der Waals' forces B. Origin of van der Waals' forces 1. Induced dipole-induced dipole attractions 2. Dipole-dipole interactions 3. Dipole-induced dipole attractions C. Relative strength of different origins of van der Waals' forces II. Hydrogen bond A. Nature B. Strength C. Solubility and hydrogen bond D. Intramolecular hydrogen bond E. Examples of hydrogen bonding 1. Ethanoic acid dimer 2. Simple hydrides - CH 4 , NH 3 , H 2 O and HF 3. Structure and physical properties of ice 4. Biochemical importance of hydrogen bond a) DNA b) Protein F. Strength of van der Waals' forces and hydrogen bond III. Relationship between structures and properties States of Matter I. Solid state A. Metallic structure 1. Hexagonal close packing 2. Cubic close packing - face centred cubic 3. Tetrahedral holes and octahedral holes 4. Body-centred cubic structure B. Giant ionic structure C. Covalent giant structure 1. Allotrope of carbon a) Diamond b) Graphite 2. Silicon(IV) oxide D. Molecular crystal 1. Iodine – face centred cubic 2. Carbon dioxide (Dry ice) – face centred cubic II. Gaseous state A. Gas laws 1. Charles' law 2. Boyle's law 3. Avogadro's law 4. PVT surface B. Ideal gas equation (Ideal gas law) 1. Molar volume C. Determination of molecular mass 1. Experimental determination of molar mass of a gas 2. Experimental determination of molar mass of a volatile liquid D. Dalton's law of partial pressure 1. Mole fraction 2. Relationship between partial pressure and mole fraction Chemical Kinetics Page 1 Chemical Kinetics I. Rates of chemical reactions A. Definition of rate of reaction 1. Unit of rate of reaction B. Measuring the rate of reaction 1. Different approaches a) Constant amount approach b) Constant time approach 2. Interpretation of physical measurements made in following a reaction a) Volume of gas formed b) Colorimetric measurement II. Order of reaction A. Rate Law or rate equation 1. Differential form and integrated form of rate law 2. Graphical presentation of reaction rate 3. Order of reaction a) Experimental determination of order of reaction (1) By measuring the rates of reaction at different reactant concentration (2) By plotting graph of ln rate versus ln [X] B. Examples of different reaction 1. First order reaction a) Half life (t ½ ) b) Carbon-14 dating c) Examples of calculation 2. Second order reaction 3. Zeroth order reaction III. Collision theory and activation energy A. Collision theory 1. Distribution of molecular speed (Maxwell-Boltzman distribution) 2. Effect of temperature change on molecular speeds 3. Collision theory and activation energy a) Arrhenius equation B. Determination of activation energy 1. By measuring the rate of reaction at different temperature 2. By plotting graph of ln k versus 1 T 3. By plotting graph of ln rate versus 1 T Chemical Kinetics Page 2 IV. Energy profile of reaction A. Transition state 1. Order of reaction leads to interpretation of reaction mechanism at molecular level a) Multi-stages reaction (1) Alkaline hydrolysis of 2-chloro-2-methylpropane (2) Reaction between bromide and bromate(V) in acidic medium b) Rate determining step 2. Energetic stability and kinetic stability B. Effect of catalyst 1. Characteristics of catalyst 2. Theory of catalyst a) Homogeneous catalysis b) Heterogeneous catalysis 3. Application of catalyst a) Use of catalyst in Haber process and hydrogenation b) Catalytic converter c) Enzymes V. Factors influencing the rate of reaction A. Concentration B. Temperature C. Pressure D. Surface area E. Catalyst F. Light Chemical Equilibria Page 1 Chemical Equilibria I. Nature of equilibrium A. Dynamic nature B. Examples of equilibrium 1. Bromine water in acidic and alkaline medium 2. Potassium dichromate in acidic and alkaline medium 3. Hydrolysis of bismuth chloride C. Equilibrium law 1. Introduction to different equilibrium constant (K) – K c , K p , K w , K a , K b , K In , K d 2. Equilibrium constant (K c and K p ) 3. Degree of dissociation (α) 4. Concentration of solid 5. Determination of equilibrium constant (K c ) a) Esterification b) Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) 6. Relationship between rate constant and equilibrium constant D. Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) 1. Concentration 2. Pressure 3. Temperature a) Equation : ln K = constant - ∆H T 4. Effect of catalyst on equilibria 5. Examples of calculation Chemical Equilibria Page 2 II. Acid-base Equilibria A. Acid-base Theory 1. Arrhenius definition 2. Bronsted-Lowry definition 3. Lewis definition B. Strength of acid 1. Leveling effect C. Dissociation of water 1. Ionic product of water D. pH and its measurement 1. Definition of pH 2. Temperature dependence of pH 3. Measurement of pH a) Use of pH meter (1) Calibration of pH meter b) Use of indicator (1) Colour of indicator E. Strong and weak acids/bases 1. Measuring of pH or conductivity of acid / base 2. Dissociation constant (K a and K b ) a) Dissociation of polybasic acid (1) Charge effect 3. Calculation involving pH, K a and K b a) Relationship between K a and K b (pK a and pK b ) b) Relationship between pH, pOH and pK w c) Some basic assumptions applied 4. Experimental determination of K a F. Buffer 1. Principle of buffer action a) pH of an acidic buffer b) pH of an alkaline buffer 2. Calculation of buffer solution G. Theory of Indicator H. Acid-base titration 1. Difference between equivalence point and end point 2. Titration using pH meter 3. Titration using indicator a) Choosing of indicator 4. Thermometric titration 5. Conductimetric titration a) Strong acid vs Strong base b) Weak acid vs Strong base c) Weak acid vs Weak base Chemical Equilibria Page 3 III. Redox Equilibria A. Redox reaction 1. Balancing of redox reaction a) By combining balanced half-ionic equation (1) Steps of writing balanced half-ionic equation (2) Combining half-ionic equations b) By the change in oxidation no. 2. Faraday and mole 3. Calculation of mass liberated in electrolysis B. Electrochemical cells 1. Measurement of e.m.f. a) Potentiometer b) High impedance voltmeter / Digital multimeter 2. Use of salt bridge a) Requirement of a salt bridge b) An electrochemical cell does not need salt bridge 3. Cell diagrams (IUPAC conventions) C. Electrode potentials 1. Standard hydrogen electrode 2. Relative electrode potential (Standard reduction potential) a) Meaning of sign and value 3. Electrochemical series 4. Use of electrode potential a) Comparing the oxidizing and reducing power b) Calculation of e.m.f. of a cell c) Prediction of the feasibility of redox reaction (1) Disproportionation reaction D. Secondary cell and fuel cell 1. Lead-acid accumulator 2. Hydrogen-oxygen fuel cell E. Corrosion of iron and its prevention 1. Electrochemical process of rusting 2. Prevention of rusting a) Coating b) Sacrificial protection c) Alloying Phase Equilibria I. One component system A. Water 1. Phase diagram of water 2. Isotherm of water 3. PVT surface of water B. Carbon dioxide 1. Phase diagram of carbon dioxide 2. Isotherm of carbon dioxide II. Two components system A. Ideal system 1. Raoult's law 2. Composition of the liquid mixture comparing with composition of the vapour 3. Boiling point-composition diagram a) Definition of boiling point b) Boiling point-composition diagram c) Fractional distillation B. Non-ideal solution 1. Deviation from Raoult's law a) Positive deviation b) Negative deviation c) Enthalpy change of mixing 2. Elevation of boiling point and depression of freezing point by an involatile solute 3. Boiling point of two immiscible liquid a) Steam distillation III. Three components system A. Partition of a solute between two phases 1. Solvent extraction a) Calculation 2. Paper chromatography a) R f value b) Separation of amino acids Atomic Structure I. Atomic structure A. Different model of atomic structure B. Isotopic mass and relative atomic mass 1. Mass spectrometer a) Determination of atomic mass b) Determination of molecular mass C. Radioactivity of nuclide 1. Nature of radioactivity a) Penetrating power of radiation 2. Radioactive decay a) Half life b) Use of radioactive isotope II. Electronic structure of atoms A. Emission spectrum of hydrogen 1. Origin of emission and absorption spectrum 2. Pattern of energy levels of hydrogen 3. Determination of the ionization energy a) Through the convergence limit of Lyman series b) Through Rydberg constant B. Detail electronic configuration 1. Definition of ionization energy 2. Successive ionization energy 3. Periodicity of First ionization energy a) First I.E. across periods b) First I.E. down a group 4. Quantum numbers a) Energy of electrons in different orbitals 5. Building up of electrons in an atom a) Aufbau principle b) Pauli exclusion principle c) Hund's rule d) Building up of electrons e) Detail electronic configuration of some atoms 6. Electronic configuration of first 36 elements a) Extra stability of half-filled and full-filled subshell 7. Position of s-, p-, d- and f-block elements III. Atomic orbitals A. Shape of s, p and d orbitals IV. Periodic Table A. Variation in atomic radius B. Variation in electronegativity I. Atomic structure Unit 1 Page 1 Topic I. Atomic Structure Unit 1 Reference Reading 1.1, 1.3 Chemistry in Context, 3rd Edition, ELBS pg. 2–4, 57–58 Physical Chemistry, 4th Edition, Fillans pg. 3–5, 10–17 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 6–12 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 1, 2,–4, 54–62 Syllabus Different model of atomic structure Mass spectrometer Notes I. Atomic Structure A. Different model of atomic structure Dalton’s model Thomson’s model Rutherford’s model Bohr’s model Wave model Orbital model Dalton’s model (1808) Atom is a solid, indivisible sphere. Thomson’s model (1899) Atom is made of electrons evenly spread throughout positive mass. Rutherford’s model (1911) Atom is made of electrons and nucleus where most of the space is empty. Bohr’s model (1913) The electron is moving around the nucleus in certain circular orbits. Wave model (1924) The electron is moving around the nucleus in a wave like motion. Orbital model (1927) The exact position of the electron cannot be predicted or measured, an electron density cloud is used to describe the probability of finding an electron in certain region. N.B. Model is just as good as what you want it to explain. B. Isotopic mass and relative atomic mass Isotopic mass – The relative isotopic mass of a particular isotope of an element is the mass of one atom of that isotope on the Carbon-12 scale. (a mass has no unit). Relative atomic mass – The relative atomic mass of an element is the weighted average of the isotopic masses of its natural isotopes on the Carbon-12 scale. Carbon-12 scale – A scale in which Carbon-12 isotope is chosen as the standard and 12 units is assigned to it by definition. The masses of other atoms are determined by comparing with this standard. I. Atomic structure Unit 1 Page 2 1. Mass Spectrometer A device in which particles are ionized and the accelerated ions are separated according to their mass to charge (m/e) ratio. Construction of Mass Spectrometer Heater – to vaporize the sample, either element or molecule. Ionization chamber – gas molecules are bombarded with fast speed electrons to form positive ions. M (g) + e - (fast) → M + (g) + 2e - Accelerating plate / Negatively charged plate – to accelerate the beam of positive ions across the electric field. Magnet / Magnetic field – to deflect the ion beam, the magnetic field is adjusted so that ions of a particular mass are focused onto the ion-detector Detector – detect the signal and pass it on to a recorder. Vacuum pump – to evacuate the apparatus. This allow the positive ions to travel in straight lines without collision and prevents presence of other substances in the spectrometer that will ionize and affect the result Knowing the magnetic field strength(B), Accelerating voltage(V) and the radius of the path(r), the mass of the ion can be calculated. (Calculation not required.) Important uses of Mass Spectrometer 1. Measurement of isotopic masses of an element. 2. Measurement of molecular mass. 3. Identification of compounds - the various m/e signals and peak heights are processed by computer. 4. Determination of ionization energies. Mass Spectrum A spectrum is obtained with a mass spectrometer, in which a beam of ion is arranged in order of increasing mass to charge (m/e) ratio. The detector current is directly proportional to the relative abundance of the ions reaching the detector. So by using a mass spectrum of a sample of an element, the relative atomic mass of the element can be determined. I. Atomic structure Unit 1 Page 3 a) Determination of relative atomic mass from mass spectrum (Not to scale) The four isotopes of lead are lead-204, lead-206, lead-207 and lead-208. The height of the peak at 204 = 0.07" The height of the peak at 206 = 1.81" The height of the peak at 207 = 1.66" The height of the peak at 208 = 2.81" Total height of the four peaks = 0.07" +1.81" + 1.66" + 2.81" = 6.35" Since, the height of the peak ∝ the relative abundance The relative atomic mass of lead = 204 × 0.07" 6.35" + 206 × 1.81" 6.35" + 207 × 1.66" 6.35" + 208 × 2.81" 6.35" = 207(207.19) b) Determination of molecular mass Another important use of mass spectrometer is to determine the mass of a molecule. Similar to an atom, a molecule will be ionized upon bombardment with fast moving electrons. Once a molecule is ionized, it becomes energetic and unstable. It tends to break up into several fragments. Since the fragment always has a smaller mass than the original molecule, the peak on the mass spectrum with the highest (m/e) ratio corresponds to the original molecule. For example, the following is the spectrum of a sample of methane CH 4 . Ionization CH 4 + e - → CH 4 + + 2e - Fragmentation CH 4 + → CH 3 + + H and CH 3 + → CH 2 + + H etc. Fragment 12 CH 4 + 12 CH 3 + 12 CH 2 + 12 CH + 12 C + m/e 16 15 14 13 12 Besides the peak with m/e ratio 16, a very small peak (about 1% of that of 16) with m/e ratio 17 is found. It is caused by the existence of trace amount of C-13 isotope in the nature. I. Atomic structure Unit 1 Page 4 Carbon-12 scale After the invention of mass spectrometer, structure of numerous number of organic molecules have been studied. Because all organic compound contains carbon, it would be more convenient if carbon is used as the standard of reference in mass spectrum. As a result, in 1961, C-12 was chosen as the standard for atomic mass and carbon-12 scale is established. In carbon-12 scale, by definition, 12 g of C-12 contains 1 mole of C-12 atoms. Since the definition is solely arbitrary, there is no wonder why Avogadro number is not a whole number. Indeed the experimental determination of the more precise value of Avogadro number is still the interest of many scientists. Glossary orbital – Loosely speaking, it is used to describe the geometrical figure which describes the most probable location of an electron. More accurately, it is used to denote the allowed energy level for electrons. (relative) isotopic mass (relative) atomic mass Carbon–12 scale Past Paper Question 90 2A 3 a 96 1A 1 a i ii iii 98 2A 4 c 99 2A 1 a i 90 2A 3 a 3a Give an account of the use of a mass spectrometer for determining the relative masses of particles. 6 The modern method of determining atomic masses uses the mass spectrometer. 2 marks - the sample (an element) is bombarded by electrons to form positively charged ions. - the ions are accelerated by the electric field between plates A and B. - the ions travel along until deflected by the magnetic field - the magnetic field is adjusted so that ions of a particular mass are focused into the ion-detector - knowing the magnetic field strength, accelerating voltage and the radius of the path, the atomic mass can be calculated. - the mass spectrometer responds to the mass-to-charge (m/e) of the ionized species 4 marks C Many candidates did not point out that the mass spectrometer responds to the mass to charge ratio. Some candidates did not know the positively charged ions are accelerated by the electric field. I. Atomic structure Unit 1 Page 5 96 1A 1 a i ii iii 1a i Write down the number of neutrons, protons and electrons in one atom of carbon-12, 12 C, and in one atom of carbon-13, 13 C . 1 12 C 6n, 6p, 6e ½ mark 13 C 7n, 6p, 6e ½ mark ii The isotopic mass of 12 C is 12.000 atomic mass unit (a.m.u.). Calculate the mass, in kg, of 1 mole of 12 C atoms. (1 a.m.u. = 1.6605 × 10 -27 kg; Avogadro constant, L = 6.0221 × 10 23 mol -1 ) 2 Mass of 1 mole of 12 C = 12.000 × 1.6605 × 10 -27 × 6.0221 × 10 23 1 mark = (0.0119996) = 0.0120 kg 1 mark (Accept answers which could round off to 0.012) iii The following data were obtained from the mass spectrum of a carbon-containing compound : Ion Mass / a.m.u. Relative intensity 12 C + 12.000 100.00 13 C + 13.003 1.12 Using the above data, calculate the relative atomic mass of carbon. 2 relative atomic mass = 12.000 × 100.00 + 13.003 × 1.12 (100.00 + 1.12) 1 mark = 1214.57 101.12 = 12.011 1 mark (Accept answers which could round off to 12.01) (Deduct ½ mark if unit is given) C Well answered in general, though some candidates did not give the unit, kg, as required. Many candidates wrongly assigned the unit a.m.u. to the relative atomic mass of carbon. 98 2A 4 c 4c Sketch the expected mass spectrum for a gas sample having the composition : N 2 78%, O 2 21% and CO 2 1%. (You only need to consider the major isotope of each element.) 3 99 2A 1 a i 1a i Rubidium occurs naturally in two isotopic forms. The table below lists the mass and relative abundance of each isotope. Isotope Mass / a.m.u. Relative abundance 85 Rb 84.939 72.15% 87 Rb 86.937 27.85% (I) Suggest an experimental method to detect the isotopes of rubidium and state how the relative abundance of each isotope can be obtained. (II) Calculate the relative atomic mass of rubidium. I. Atomic structure Unit 2 Page 1 Topic I. Atomic Structure Unit 2 Reference Reading 1.2 Chemistry in Context, 3rd Edition ELBS pg. 83–89, 93–95 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 13–20 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 76–81 Syllabus Nature of radioactivity Radioactive decay Notes C. Radioactivity of nuclide 1. Nature of radioactivity There are totally 109 elements, but owing to the presence of the isotope, there are over 1600 different nuclei. Only 238 nuclei are stable, all others exhibit some kind of radioactivity. They eject proton, neutron and electron or emit strong electromagnetic radiation. The process is called radioactive decay and the elements are said to be radioactive. a) Penetrating power of radiation There are 3 types of radioactivity. Emission Nature Relative velocity Range in air Relative penetrating power α particle He nuclei (2p + 2n) 1/20 velocity of light a few cm 1 β particle fast moving electrons 3 – 99% velocity of light a few m 100 γ radiation electromagnetic wave 100% velocity of light a few km 10000 Penetrating power is ∝ 1 mass ; ∝ 1 charge ; ∝ velocity Emission Relative mass Relative charge Relative effect of electric and magnetic field α particle 4 units + 2 small deflection β particle 1 / 1840 unit - 1 large deflection γ radiation – 0 no deflection Effect of electric field Effect of magnetic field (The magnetic field is going into the paper.) The deflection is ∝ charge and ∝ 1 mass All of the three kinds of radioactivity have the ability to ionize the air, so a Geiger-counter can be used to measure the activity of the radioactive material. I. Atomic structure Unit 2 Page 2 2. Radioactive decay In a nuclear reaction, unlike ordinary chemical reaction, nuclei are involved instead of electrons. In the reaction, 1. one element can be changed to another element by radioactive decay, atomic fission or atomic fusion. 2. a large amount of energy is released in the process, the energy is from the binding energy in the nucleus. The process in which an element changes to another element in nuclear reaction is known as transmutation. Alpha Decay This is a decay common to most of the radioactive isotopes of elements with Z > 82 (Pb). (high p: n ratio) e.g. a) 92 238 U → 2 4 He + 90 234 Th b) 88 226 Ra → 2 4 He + 86 222 Rn c) 88 224 Ra → 2 4 He + 86 220 Rn When an element undergoes alpha decay, its mass no. will be reduced by 4 and the atomic no. will be reduced by 2. Note: i) the charge of the particle, e.g. the charge on He nucleus, is usually omitted for simplification. ii) the mass number and the atomic number in all nuclear reaction must be conserved. Beta Decay This is a decay common to most radioactive isotopes of elements with Z < 82. (low p: n ratio) e.g. a) 6 14 C→ −1 0 e + 7 14 N b) 47 108 Ag → −1 0 e + 48 108 Cd c) 0 1 n → −1 0 e + 1 1 H When an element undergoes beta decay, its mass no. will remain unchanged and the atomic no. will increase by 1. Gamma Radiation Unlike alpha and beta decay, an element emitting gamma radiation will not change to another element. It will only change from a high energy(excited) state to an low(stable) energy state. e.g. a) 90 234 Th * → γ–ray + 90 234 Th Most of the heavy isotope will undergo multiple decays to form a series. The U-238 isotope can decay through multiple stages to a stable Pb-206 isotope I. Atomic structure Unit 2 Page 3 a) Half life of radioactive decay Half life - the time taken by a given amount of radioactive isotope to decay to half of the original amount. e.g. Radioactive Isotopes Half-life Uranium-238 4.5 × 10 9 years Carbon-14 5.7 × 10 3 years Polonium-214 1.5 × 10 -4 seconds The stability of an isotope is directly proportional to its half life. For example, if there are 100 Uranium-238 isotopes, it will take 4.5 x 10 9 years to decay to half of the original amount (50 Uranium-238 isotopes). The rate of radioactive decay, unlike rate of chemical reaction, only depends on the amount of isotopes present. The changes in external factors such as temperature and pressure have no effect on it. Rate of radioactive decay The rate of the decay decreases as the decay proceeds because the rate of decay is directly proportional to the amount of isotope remains. This forms an exponential decay curve. After 5 half-lives, the fraction remaining will be = ½ × ½ × ½ × ½ × ½ = 0.03125 = 3.125% I. Atomic structure Unit 2 Page 4 Artificial Transmutation of elements Artificial transmutation reactions are usually achieved by bombarding an element with a stream of particles, especially neutrons, protons, alpha particles and other small nuclei. e.g. Uranium-238 can be transmutated to Uranium-239 by bombarded with an neutron. i.e. 92 238 U + 0 1 n → 92 239 U The unstable Uranium-239 then decays in two stages by emitting β-particles. i.e. a) 92 239 U → 93 239 Np + −1 0 e b) 93 239 Np → 94 239 Pu + −1 0 e In nature the, 7 14 N in air, is transmutated to 6 14 C by the neutron from the cosmic ray (fast moving neutron). i.e. 7 14 N + 0 1 n → 6 14 C+ 1 1 H In some cases, if the speeds of intruding neutrons are different, it will cause different modes of nuclear reaction. e.g. a) 13 27 Al + 0 1 n → 11 24 Na + 2 4 He b) 13 27 Al + 0 1 n → 12 27 Mg + 1 1 p c) 13 27 Al + 0 1 n → 13 28 Al + γ-ray b) Use of radioactive isotope 1. Treatment of cancer 2. Study of metabolic pathways. 3. Thickness gauges and empty packet detection. 4. Archaeological and geological dating. 5. Sterilizing surgical instruments. 6. Detection of leakage. (e.g. water pipes) Glossary nuclide - A type of atom specified by its atomic number, atomic mass, and energy state, such as carbon 14. penetrating power radioactive decay half-life I. Atomic structure Unit 2 Page 5 Past Paper Question 91 2A 1 a i ii 93 1A 1 a i ii 95 2A 2 a i 99 1A 2 b i ii 91 2A 1 a i ii 1a The radioactive decay of 92 238 U may be represented by 92 238 U → 82 206 Pb + α + β. i State the nature of α and β particles and compare their penetrating power. 3 α particles are bare helium nucleus with two positive charge and four units of mass. 1 mark β particles are fast electrons, which is negatively charged. 1 mark The penetrating power of β particles is 100 times stronger than α particles. 1 mark ii Balance the above equation. 2 92 238 U → 82 206 Pb + 8 2 4 He + 6 −1 0 e 2 marks C Many candidates incorrectly stated that α particles are helium atoms. Some candidates did not emphasize that the electrons in β particles are moving very fast. 93 1A 1 a i ii 1a Name the particles X and Y in the following nuclear reactions, and gives their charge and mass. Compare the action of a magnetic field upon the paths of X and Y 4 i 90 229 88 225 Th X Ra → + X is _______________ X is / helium nucleus / α particle 1 mark ii 88 225 89 225 Ra Y Ac → + Y is _______________ Y is / fast moving electron / β particle. 1 mark The path of α and β particles are deflected/bent by a magnetic field 1 mark but they are deflected in opposite direction. 1 mark OR by a diagram The magnetic field is going into the paper. Bent motion 1 mark Correct direction 1 mark 95 2A 2 a i 2a 88 228 Ra decays by the emission of β particles. The half-life for the decay is 6.67 years. i Write a balanced equation to represent the decay of 88 228 Ra . 1 88 228 Ra → 89 228 Ac + −1 0 e 1 mark C Many candidates did not make use of the Periodic Table given in the question paper to find the symbol (Ac) for the product of the decay, and hence they arrived at incorrect answers. 99 1A 2 b i ii 2b Radioactive decay of radium (Ra) in rocks and soil produces radioisotopes of radon (Rn). One of the isotopes 86 222 Rn decays to give 84 218 Po with a half-life of 3.82 days. i Write the nuclear equation for the decay of 86 222 Rn . ii Explain why 86 222 Rn is considered hazardous to health. II. Electronic Structure of atoms Unit 1 Page 1 Topic II. Electronic Structure of atoms Unit 1 Reference Reading 2.1 Chemistry in Context, 3rd Edition ELBS pg. 69–71 Physical Chemistry, 4th Edition, Fillans pg. 63–74 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 31–36 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 64–68 Syllabus Emission spectrum of hydrogen Determination of ionization energy Notes II. Electronic Structure of atoms A. Emission spectrum of hydrogen Emission Spectroscopy White light from tungsten lamp consists of a continuous range of wavelengths of visible light. The white light can be separated into different colour light by a prism and forms a continuous spectrum of visible light, like a rainbow. But some lights, for example, the light from the sodium lamp or green light from the flame test of copper, are not composed of a continuous range of wavelengths, instead they are only mixtures of a limited number of wavelengths. In this case, the spectrum is known as a line spectrum. 1. Origin of emission Emission spectrum i. At room temperature, nearly all the atoms in a given sample of substance are in the ground state, the electrons occupy the orbital of lowest energy. The atom is in the ground state because all electrons in the atoms are having the lowest possible energy. ii. Energy is given to the atoms by heating the element with Bunsen flame or hitting with electrons in discharge tube. The electrons takes up the energy and are promoted to a higher energy state, excited state. iii. Since the electrons in excited state are not stable, they tends to fall back to a lower energy state to dissipate their extra energy in form of radiation. Each transition involves a definite amount of energy, called a quantum. iv. Different transition has different energy and each transition corresponds to a single line in the emission spectrum. The greater the number of electrons making a particular transition, the more intense the spectral line is. This gives coloured emission lines on dark background. v. The relationship between the wavelength(λ) and frequency(υ) of any electromagnetic radiation is given by λ = c υ where c is the velocity of light vi. And, the relationship between energy of radiation is given by the equation ∆E = hυ where h is Planck’s constant and υ is frequency II. Electronic Structure of atoms Unit 1 Page 2 2. Pattern of energy levels of hydrogen Balmer series i. The emission line spectrum of hydrogen atom consists of many lines, some of them fall in the visible region and some of them fall in the invisible region. ii. The visible part of the spectrum was first studied by the scientist Balmer. This series of spectrum lines was named after him as Balmer series. iii. The red line, the line with longest wavelength in the visible region, corresponds to the transition of an electron from the third energy level (n=3) to the second energy level (n=2). It gives out a quantum of light, called photon. With increasing frequencies, the lines get closer and closer until they merge, this is because the difference between energy levels gets smaller as the frequency increases. Relationship between emission spectrum and energy levels iv. The point, where the spectral lines merge, corresponds to the electron transition between energy levels with n = ∞ and n = 2. v. At n = ∞, the electron has effectively left the atom. That means the atom has been ionized. II. Electronic Structure of atoms Unit 1 Page 3 Complete Hydrogen Emission Spectrum i. Besides Balmer series, the complete spectrum is composed of many series. The origin of other series is similar to that of Balmer series, the only difference is the destination of the electrons. ii. Bohr obtained an expression for the frequencies of the lines in the spectrum. υ = − cR n n H ( ) 1 1 1 2 2 2 where R H is Rydberg constant c is velocity of light n 1 and n 2 are integers, called principal quantum number, with n 2 greater than n 1 . n 1 : the principal quantum number of the energy level to which the electron returns. n 2 : the principal quantum number of the energy level from which the electron leaves. II. Electronic Structure of atoms Unit 1 Page 4 3. Determination of the ionization energy The ionization energy of hydrogen can be determined by measuring the frequency the convergence limit (υ ∞→1 ) of Lyman series or by determining the Rydberg constant through the frequency of a specific line. a) Through the convergence limit of Lyman series Ionization of hydrogen = ∆E = h υ ∞→1 However, it is quite difficult to determine the convergence limit accurately as the lines are getting close together when approaching the convergence limit and the spectrum becomes a bright continum beyond the limit. Therefore, the limit is usually determined graphically. i. In Lyman series, there is a convergence limit in the line spectrum. It represents the point where the difference in energy levels diminished. ii. The frequencies of the first lines in Lyman series are measured. The differences in frequencies between two adjacent lines are calculated. e.g. ∆υ 1 = (υ 2 - υ 1 ), ∆υ 2 = (υ 3 - υ 2 ) ...iii. A graph of υ against ∆υ is plotted. iv. The curve is extrapolated, and the υ at y-intercept, where ∆υ = 0, will correspond to the transition of electron from n = ∞ to n = 1. v. The Ionization energy can be calculated by I.E. = hυ b) Through Rydberg constant For example, the Rydberg constant can be determined by measuring the frequency of the first line of Lyman series (υ 2→1 ). Thereafter, by applying the Ryberg equation, the ionization energy of hydrogen can be calculated. i. The Rydberg constant is calculated from the first line in the Lyman series. υ 2 1 2 2 1 1 1 2 → = − cR H ( ) ii. The frequency corresponding to the ionization process is calculated by υ ∞→ = − ∞ 1 2 2 1 1 1 cR H ( ) iii. The ionization energy(I.E.), the energy transition from the ground state and the continuum for one electron is Ionization energy(I.E.) = h υ ∞→1 = hcR H The ionization energy per mole = LhcR H where L is the Avogadro's constant Glossary emission spectrum absorption spectrum Planck’s constant Balmer series Lyman series Paschen series Rydberg equation II. Electronic Structure of atoms Unit 1 Page 5 Past Paper Question 90 2A 2 c 95 1A 1 b i ii 97 2A 2 a 99 1A 2 a i ii 90 2A 2 c 2c Give a brief account for the Balmer series in the spectrum of the hydrogen atom. 4 It is a series of lines resulting from photons emitted or absorbed by the hydrogen atom with wavelength in the visible region. These spectral lines are the result of transitions of the electron from one energy level to another with the lower energy level with quantum number n 1 = 2. 2 marks The wavelengths of this series satisfies the formula, 1 λ = R H ( 1 2 2 - 1 n 2 2) 1 mark where R H is the Rydberg constant n 2 = 3,4,5 ... λ = wavelength 1 mark C Many candidates gave the correct formula for the relationship between wavelength and quantum numbers, but only a few defined their symbols. 95 1A 1 b i ii 1b i What can you deduce from the fact that the spectral lines in the atomic emission spectrum of hydrogen are not equally spaced ? 1 The energy differences between electron shells in a hydrogen atom are not the same (converge). 1 mark ii In the atomic emission spectrum of hydrogen, the convergence limit for the Lyman series occurs at 3.275 x 10 15 Hz. Calculate the ionization energy of hydrogen, in kJ mol -1 (Planck constant, h = 6.626 x 10 -34 J s; Avogadro constant, L = 6.023 x 10 23 mol -1 ) 2 I.E. = hυ = (6.26 × 10 -34 ) × (3.275 × 10 15 × 6.023 x 10 23 ) 1 mark = 1.307 × 10 6 Jmol -1 / 1307 kJmol -1 1 mark C Many candidates were unable to point out that the energy differences between two energy levels are not the same and hence the spectral lines in the emission spectrum are not equally spaced. Many candidates neglected to give the answer to (ii) in kJmol -1 . 97 2A 2 a 2a Describe and account for the characteristics of the emission spectrum of atomic hydrogen. 7 99 1A 2 a i ii 2a i Outline the origin of atomic emission. ii Briefly explain why each series of atomic emission lines of hydrogen converges at the lower wavelength end. (3 marks) II. Electronic Structure of atoms Unit 2 Page 1 Topic II. Electronic Structure of atoms Unit 2 Reference Reading 2.2–2.3 Modern Physical Chemistry ELBS pg. 26–33, 39–41 Chemistry in Context, 3rd Edition ELBS pg. 71–78 Physical Chemistry, 4th Edition, Fillans pg. 74–77, 80–84, 88–95 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 37–42, 49–53 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 38–42, 64, 68–71, 73–74, 170, 173–175 Syllabus Detail electronic configuration Notes B. Detail electronic configuration 1. Definition of first ionization energy X (g) → X + (g) + e - First ionization energy – Energy required to remove an electron from a free atom in gaseous state. Normally measured in electron volts (eV) or kJmol -1 . (1eV = charge of electron × 1V = 1.60 × 10 -19 C × 1V = 1.60 × 10 -19 J) 2. Successive ionization energy (presence of electron shells) The information from successive ionization energy can be used to deduce the electronic configuration of an atom. The ionization of Potassium (K) can take place in steps 1st ionization K (g) → K + (g) + e - ∆H I1 = 419 kJmol -1 2nd ionization K + (g) → K 2+ (g) + e - ∆H I2 = 3051 kJmol -1 3rd ionization K 2+ (g) → K 3+ (g) + e - ∆H I3 = 4412 kJmol -1 i. The value of successive ionization energy increases as more electrons are removed from the atom. This is because extra energy is needed to removed an electron from an ion of increasing positive charge. ii. The 2nd I.E. is much higher than 1st I.E. It can be observed from the graph that the I.E.s of K atom can be divided into 4 groups. 1 in the first group, 8 in the second group, 8 in the third group and remaining 2 in the fourth. iii. This indicates that a K atom has four electron shells and electronic configuration of 2,8,8,1. II. Electronic Structure of atoms Unit 2 Page 2 3. Periodicity of First ionization energy (presence of electron sub-shell) a) First I.E. across periods First ionisation Energy of the first 38 elements i. The 1st I.E. increases roughly across a period due to the increase in nuclear charge. ii. Besides the main trends, some irregularities are found within a period. This suggests the presence of sub- shells (s, p, d and f sub-shells) within an electron shell (n = 1, 2, 3 ...) iii. In period 2 and 3, the increasing trends can further be broken up into 3 parts, in a 2-3-3 manner. This suggest the presence of subshells. iv. In period 4, from K to Kr, the trend can be broken into 4 parts, in a 2-10-3-3 manner. This suggest the presence of another set of subshells. b) First I.E. down a group Li Na K Rb Cs 350 370 390 410 430 450 470 490 510 530 F i r s t I . E . / k J m o l - 1 F Cl Br I 900 1000 1100 1200 1300 1400 1500 1600 1700 F i r s t I . E . / k J m o l - 1 He Ne Ar Kr Xe 1000 1200 1400 1600 1800 2000 2200 2400 F i r s t I . E . / k J m o l - 1 1st I.E. of group I 1st I.E. of group VI 1st I.E. of group 0 i. On down a group, the 1st I.E. decreases as the number of electron shells or the size of the atom increases. The attraction between the outermost electron and the nucleus decreases with increasing atomic size. II. Electronic Structure of atoms Unit 2 Page 3 A prevailing explanation for these variation in ionization energy is that different electrons occupy different orbitals in an atom. Since different orbitals have different energies, the electron in different orbital shows difference in energy. A more energetic electron would be removed more readily and has a lower ionization energy. 4. Quantum numbers To facilitate the discussion of the detail electronic configuration of an atom, a set of four quantum numbers is assigned to each electron in an atom. Quantum numbers is the I.D. no of the electron in an atom. The four quantum numbers are Allowed value Principal quantum number (n) specify the principal energy level. Any natural number Subsidiary quantum number (l) specify the shape of the orbital. Depends on n, in the range of 0, 1, 2 .. (n-1) Magnetic quantum number (m) specify the direction of the orbital. Depends on l, in the range of -l to +l Spin quantum number (s) specify the spin of the electron. Either +½ or -½ Electrons with different subsidiary quantum numbers can be denoted by the letters s, p, d and f, or s orbital, p orbital, d orbital and f orbital. l = 0, 1, 2, 3 s p d f a) Energy of electrons in different orbitals i. The principal quantum no. (n) determines the principal energy level of the orbital. ii. With the same principle quantum no. (n), the energy of orbitals increases in following order, s < p < d < f. iii. The energy of orbitals is independent of magnetic quantum number (m) and spin quantum number (s). iv. The orbitals are labelled according to the values of n, l and m. e.g. The orbitals with n = 2, l = 1 and m = 1, 0 and -1 are called 2p x , 2p y and 2p z orbitals. II. Electronic Structure of atoms Unit 2 Page 4 5. Building up of electrons in an atom a) Aufbau principle In building up the electronic configuration of an atom in its ground state, the electrons are placed in the orbitals in order of increasing energy b) Pauli exclusion principle – In any atom, no two electrons can have all four quantum numbers the same. c) Hund’s rule (Rule of maximum multiplicity) Electrons in degenerate orbitals will occupy each orbital singly before electron pairing takes place. To generalize the three rules, the following procedure of filling the orbitals shown be observed i. Of the available orbitals, the added electron will always occupy the one with the lowest energy. (Aufbau principle) ii. Each orbital may hold only two electrons, and they must have opposite spin. (Pauli exclusion principle) iii. Where a number of orbitals of equal energy are available, the added electron will go into empty orbital, keeping electron spins the same, before spin-pairing occurs. (Hund's rule) According to the Aufbau principle, the electrons will fill the available orbitals with lowest energy first. The order of energy levels need not to be exactly parallel to the principal quantum. When the atom becomes heavy, the high energy levels tend to overlap each other. For example, the energy of 4s orbital is lower than that of 3d orbital. II. Electronic Structure of atoms Unit 2 Page 5 The order of energy of the orbitals can be predicted by the pattern shown on the left. The order of the energy levels mainly depends on the attraction between the orbital and the nucleus. The stronger the attraction, the lower will be the energy of the orbital. (because of lower potential energy) This once again depends on the spatial distribution of the orbital. Spatial distribution of 3s and 3p orbitals Moreover, from the fact that 4s orbital is less energetic than 3p orbital. The effect of penetration of penetration must be more significant than the effect of distance and shielding effect. For example, 3s orbital has lower energy than 3p orbital because i. Distance from the nucleus 3s orbital is closer to the nucleus than 3p orbital, thus the 3s orbital will experience more attraction. ii. Penetration of 3s orbital Spatial distribution of 3s orbital has 2 small peaks close to the nucleus and make it experience more attraction. iii. Shielding / Screening effect Besides attraction with the nucleus, 3p electrons also experience repulsion will K- shell, L-shell and 3s electrons. This is known as shielding / screening effect of inner electrons. Note : The completely filled shell can be denoted by K, L, M, N according to the principal quantum no Shielding effect and effective nuclear charge Owing to shielding effect, the electrons in an atom experience attraction from the nucleus less than the actual nuclear charge. The effective nuclear charge is always less than actual nuclear charge. There is two kinds of shielding effect Primary shielding effect This is stronger than the secondary shielding effect. This is the shielding effect of the inner electrons on the outer electrons from the nucleus. i.e. electrons in shells of lower principal quantum number are more effective shields than electrons in shells of higher quantum number and s-orbital electrons is more effective in shielding than p orbital electrons. Secondary shielding effect The shielding effect of the electrons/repulsion between electrons in the same orbital. This is much weaker than primary shielding effect. II. Electronic Structure of atoms Unit 2 Page 6 d) Building up of electrons i. The quantum no. of the electron in hydrogen atom in ground state n = 1, l = 0, m = 0, s = +½ 1s 1 electron in box 1s 1 Electron with spin quantum no. +½ is represented by an upward arrow. ii. The quantum no. of the two electrons in helium atom in ground state n = 1, l = 0, m = 0, s = +½ n = 1, l = 0, m = 0, s = -½ 1s 2 electron in box 1s 2 iii. The quantum no. of the three electrons in lithium atom in ground state n = 1, l = 0, m = 0, s = +½ n = 1, l = 0, m = 0, s = -½ n = 2, l = 0, m = 0, s = +½ 1s 2s 0 0 1s 2 2s 1 iv. The quantum no. of the six electrons in carbon atom in ground state n = 1, l = 0, m = 0, s = +½ n = 1, l = 0, m = 0, s = -½ n = 2, l = 0, m = 0, s = +½ n = 2, l = 0, m = 0, s = -½ n = 2, l = 1, m = 1, s = +½ n = 2, l = 1, m = 0, s = +½ 1s 2s 2p 0 0 000 1s 2 2s 2 2p 2 e) Detail electronic configuration of some atoms i. The electronic configuration of argon atom in ground state 1s 2s 2p 3s 3p 0 0 000 0 000 1s 2 2s 2 2p 6 3s 2 3p 6 II. Electronic Structure of atoms Unit 2 Page 7 ii. The electronic configuration of potassium atom in ground state 1s 2s 2p 3s 3p 4s 0 0 000 0 000 0 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 iii. The electronic configuration of scandium atom (Z=21) in ground state 1s 2s 2p 3s 3p 3d 4s 0 0 000 0 000 00000 0 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 or [Ar]3d 1 4s 2 or K L 3s 2 3p 6 3d 1 4s 2 Note : Writing of electronic configuration follows the principal quantum no., not the order of placing electrons iv. The electronic configuration of germanium atom (Z=32) in ground state 1s 2s 2p 3s 3p 3d 4s 4p 0 0 000 0 000 00000 0 000 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 2 or [Ar]3d 10 4s 2 4p 2 or K L M 4s 2 4p 2 6. Electronic configuration of first 36 elements 1 H 1s 1 2 He 1s 2 3 Li 1s 2 2s 1 4 Be 1s 2 2s 2 5 B 1s 2 2s 2 2p 1 6 C 1s 2 2s 2 2p 2 7 N 1s 2 2s 2 2p 3 8 O 1s 2 2s 2 2p 4 9 F 1s 2 2s 2 2p 5 10 Ne 1s 2 2s 2 2p 6 11 Na 1s 2 2s 2 2p 6 3s 1 12 Mg 1s 2 2s 2 2p 6 3s 2 13 Al 1s 2 2s 2 2p 6 3s 2 3p 1 14 Si 1s 2 2s 2 2p 6 3s 2 3p 2 15 P 1s 2 2s 2 2p 6 3s 2 3p 3 16 S 1s 2 2s 2 2p 6 3s 2 3p 4 17 Cl 1s 2 2s 2 2p 6 3s 2 3p 5 18 Ar 1s 2 2s 2 2p 6 3s 2 3p 6 19 K 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 20 Ca 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 21 Sc 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 22 Ti 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2 23 V 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2 24 Cr 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 25 Mn 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2 26 Fe 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 27 Co 1s 2 2s 2 2p 6 3s 2 3p 6 3d 7 4s 2 28 Ni 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 4s 2 29 Cu 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1 30 Zn 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 31 Ga 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 1 32 Ge 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 2 33 As 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 3 34 Se 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 4 35 Br 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 36 Kr 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 a) Extra stability of half-filled and full-filled subshell It can be observed that Cr (Z=24) and Cu (Z=29) do not follow the general trend. Instead of [Ar]3d 4 4s 2 , Cr has the electronic configuration [Ar]3d 5 4s 1 Instead of [Ar]3d 9 4s 2 , Cu has the electronic configuration [Ar]3d 10 4s 1 Extra stability is associated with i. half-filled orbital and ii. completely/full-filled orbital. This is because i. a half-filled or full-filled d orbitals mean an even distributed electrons in an atom which make the electron repulsions weaker. ii.4s 2 is less favorable than 4s 1 as pairing up of the 4s electrons creates greater repulsion between the 2 electrons. This also accounts for the anomaly of first I.E. of Be and N in period 2. Be has a full-filled 2s orbital and N has a half-filled 2p orbital. II. Electronic Structure of atoms Unit 2 Page 8 7. Position of s-, p-, d- and f- block elements According to the electronic structure of elements, the periodic table can be divided into different blocks. s-block elements : Group IA to Group IIA e.g. Na 1s 2 2s 2 2p 6 3s 1 p-block elements : Group IIIA to Group VIIA e.g. N 1s 2 2s 2 2p 3 d-block elements : Transition metal e.g. Mn 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2 f-block elements : Lanthanide and Actinide e.g. Pr 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 4f 3 5s 2 5p 6 Across the s-block, electron is being added into the s sub-shell. While moving across, p-, d- and f- block, the electron is added into the p, d and f sub-shell. Glossary First ionization energy electron shell /orbital electron sub-shell/sub-orbital quantum number Aufbau principle Pauli exclusion principle Hund's rule degenerate orbitals – orbitals with the same energy penetration shielding / screening effect effective nuclear charge electron in box half-filled orbital full-filled orbital s-block p-block d-block f-block transition metal Past Paper Question 90 2B 4 a 92 2A 1 a ii iii 93 1A 2 a v 95 1A 1 a i 95 2B 5 a iv 95 2B 6 a 96 1A 1 c i ii 97 2A 2 b i ii iii 98 1A 1 a 98 2B 8 a 90 2B 4 a 4 Account for the following observations. 4a The first ionization energies of the second row elements are in the order Li < B < Be < C < O < N < F < Ne. 4 Increase in effective nuclear charge with full explanation, i.e. no. of protons increase 1 mark From Li to Ne, electrons are added in the same principal shell, the shielding effect involved is only a secondary one 1 mark Extra stability, Be > B 1s 2 2s 2 filled 2s orbital; stability 1 mark N > O 1s 2 2s 2 2p 3 half-filled electronic configurations; half-filled stability 1 mark C Some candidates failed to mention or explain the abnormally high ionization potentials of Be and N. II. Electronic Structure of atoms Unit 2 Page 9 92 2A 1 a ii iii 1a ii Arrange the following elements in order of increasing first ionization energy : F, Ne and Na. Explain your order. 2½ Na < F < Ne 1 mark Na has the least I.E. because there is a single 3s-electron which is not strongly attracted towards the nucleus and losing it would accomplish the completed K- and L- shell status. It is for this reason that an electron in Ne requires the highest I.E. To lose an electron means that the F atom would move further away from the K- and L- shell status and this requires a relatively high I.E. 1½ mark iii Describe and explain the variation of the successive ionization energies of potassium. 3 There is a gradual increase in the successive ionization energies in general because the nucleus becomes effectively more positively-charged as each electron is being lost. The increase is greatest when the 2nd electron are being ionized, suggesting the 2nd electron is closer to the nucleus than the first. Similar greater but less abrupt I.E.’s occur when the 10th and 18th electrons are being ionized. 3 marks 93 1A 2 a v 2a Answer the following questions by choosing in each case one of the species listed below, putting it in the box and giving the relevant equation(s). Al (s) , AlCl 3(s) , AlO 2 - (aq) , Na (s) , CO 3 2- (aq) , Cu 2+ (aq) , P 4 O 10(s) , S (s) , S 2 O 3 2- (aq) , Zn 2+ (aq) v Which species has the following ionization energies: 1st 500; 2nd 4600; 3rd 6900; 4th 9500 kJmol -1 ? 1 Na (s) 1 mark 95 1A 1 a i 1a i The first four successive ionization energies of an element A are 578, 1817, 2746 and 10813 kJ mol -1 respectively. To which group in the Periodic Table does A belong ? 1 Group III / IIIA / 3 / 3A 1 mark 95 2B 5 a iv 5a The table below lists some properties of the alkali metals. Element Atomic radius / nm Ionic radius / nm First ionization energy / kJmol -1 Standard electrode potential / V Melting point / ºC Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 iv Explain the trend in the first ionization energy from Li to Cs. 2 The atomic radius increases as the group is descended hence the outer-shell electrons become less strongly held and it is easier to remove an e - from the gaseous atom of the element. 2 marks C iv the effective attractive forces on the electron is greatest in the case of the smallest first atom. 95 2B 6 a 6a Write the electronic configuration of a titanium atom in its ground state. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2 3d 4s [Ar] 11000 2 1 mark (accept any correct form of electronic configuration) II. Electronic Structure of atoms Unit 2 Page 10 96 1A 1 c i ii 1c Explain why i the first ionization energy of oxygen is greater than that of sulphur; 1 S has 1 more e - shell / has a large atomic size than O, ½ mark the attraction experienced by the outermost electron is smaller hence S has a smaller I.E. ½ mark (DO NOT accept explanations in terms of effective nuclear charge; deduct ½ mark for any misconception) C The question was poorly answered. Many candidates incorrectly attributed the difference in ionization energies to effective nuclear charge and/or screening effect. In fact, the effective nuclear charge experienced by the outermost shell electrons of sulphur was greater than that of oxygen. Such answers reflected the fact that many candidates did not fully understand the concepts associated with screening effect and effective nuclear charge. ii the first ionization energy of oxygen is smaller than that of fluorine. 1 F has 1 more proton in the nucleus, ½ mark the effect of increase in nuclear charge has outweighed the screening effect of the additional electron hence F has a greater I.E. ½ mark Or, Effective nuclear charge experienced by outermost e - of atom increases across a period. ∴ outermost e - of O is more easily removed than that of F. 1 mark C Many candidates pointed out that the outermost shell electron of fluorine experiences a greater effective nuclear charge. But they explained their answers rather vaguely. This again showed that they did not grasp these concepts well but just produced their answers by citing some terms learnt in their lessons. Some candidates incorrectly explained the difference in ionization energies between O and F in terms of electronegativity. 97 2A 2 b i ii iii 2b i What is an electron shell in an atom ? 6 ii For each of the electron shells with principal quantum numbers 1, 2 and 3 list the subshells. iii Deduce the maximum number of electrons in each of the electron shells in (ii). 98 1A 1 a 1a Arrange the following chemical species in the order of increasing ionization enthalpy. Explain your arrangement. N + , N and O 3 98 2B 8 a 8a Which of the following, (I) or (II), is the ground state electronic configuration of chromium atom ? Explain your answer. 3d 4s (I) [Ar] 11111 1 (II) [Ar] 11110 2 2 III. Atomic orbitals Page 1 Topic III. Atomic orbitals Reference Reading Modern Physical Chemistry ELBS pg. 41–42 Physical Chemistry, 4th Edition, Fillans pg. 78–80 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 39–40 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 72 Syllabus 2.4 Shape of atomic orbitals Notes III. Atomic orbitals In the orbital model of atom, the electrons are moving in a wave like motion. Because of its extremely small mass, the position of individual electron cannot be located precisely by any phyical mean, an electron density diagram is therefore used to described the distribution of electrons. Atomic orbital is the region within which there is a high probability, i.e 90%, of finding an electron. Different orbital have different shape, size and energy. A. Shape of s, p and d orbitals s orbital s-orbital is spherical in shape. The electron density at the nucleus is zero. For 1s orbital of hydrogen, the electron density (probability density of finding a electron) increases as the distance from the nucleus increases and reaches a maximum at a distance equal to Bohr’s radius. And then, the electron density decreases as the distance from the nucleus increases. p orbital p-orbital is dump bell in shape. The three p orbitals are perpendicular to each other. The three p- orbitals have the same energy (degenerate). d orbital There are 5 degenerate d-orbitals. d xy , d xz and d yz orbitals lie on xy, xz and yz plate respectively. d x 2 - y 2 orbital lies on xy plate with the lobes are directing to the x and y axes. d z 2 orbital looks like p z orbital but with a ring at the centre. Glossary probability density degenerate s orbital p orbital d orbital III. Atomic orbitals Page 2 Past Paper Question 91 2A 3 c 92 1A 2 c 96 1A 1d 91 2A 3 c 3c Give a brief account of the electron density of the 1s orbital of a hydrogen atom. 4 1s orbital is spherical in shape, the variation is exactly the same in any direction from the nucleus. At a certain distance from nucleus, there is a maximium probability of electron density. (The most probable distance of the electron from the nucleus of the hydrogen atom is equal to the Bohr’s radius.) The electron density is zero at the nucleus and at infinite distance from the nucleus. C Most candidates were able to give the correct shape of the 1s orbital, however, they failed to describe the variation of electron density as a function of distance from the nucleus. 92 1A 2 c 2c Give the shape of a 1s-orbital and comment on its physical significance. 2 1s-orbital is spherical. ½ mark Physical Significance - The probability of finding an electron at various points in space is called an orbital. ½ mark - For a give 1s-orbital, the region of highest probability of finding the electron, say 90% of the charge distribution is pictorially represented by a single surface along which the probability of finding the electron is constant. ½ mark - They should be thought of as shapes with fuzzy and indistinct edges, like a cloud. ½ mark C Only a handful of candidates were able to explain clearly the meaning of the spherical shape of the 1s-orbital and its physical significance, i.e. the most simple description of the surface of the sphere represents the probability of finding, say 90 %, of the 1s-electron. A majority of the candidates did not make the important point that orbitals should be thought of as shapes with fuzzy and indistinct edges, like clouds in reality. 96 1A 1d 1d Sketch the pictorial representation of a p orbital and indicate the location of the nucleus in your diagram. 1 nucleus (or along any axis) 1 mark C Many candidates did not do what the question asked for and drew three p-orbitals instead of one. IV. Periodic Table Page 1 Topic IV. Periodic Table Reference Reading Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. 37-41 Physical Chemistry, 4th Edition, Fillans pg. 85–88 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 111–112, 334–335 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 170–171, 175–176 Syllabus 2.5 Variation in atomic radii Variation in electronegativity Notes IV. Periodic Table A. Variation in atomic radius Covalent radius – Half of the distance between two nuclei in an element where the atoms are chemically combined together. Van der Waals’ radius – Half of the distance between two nuclei in an element where the atoms are not chemically combined together. If not specified, atomic radius is normally referring to covalent radius. However, for those atoms where formation of chemical bond is impossible, atomic radius will be referring to van der Waals’ radius. a) Factor affect the size of atomic radius proton to electron ratio (p/e) – An increase in this ratio causes an increase in effective nuclear charge, resulting a contraction of the electron cloud of the ion. The ionic radius of a cation is, of course, smaller than the atom itself. Across a period – Atomic size decreases as atomic number increases. All electrons go into the same quantum shell and the increase in nuclear charge is not effectively shielded. Effective nuclear charge increases across a period and has a greater attraction on the outermost electrons drawing them closer to the nucleus. Down a group – Atomic size increases as no. of electron shells increases. The electron in the outermost shell is effectively shielded from the nucleus by the inner electrons. IV. Periodic Table Page 2 Variation of atomic radii B. Variation in electronegativity Electronegativity is a measure of tendency of an atom in a stable molecule to attract electrons within a bonds. Therefore, it is closely related to the effective nuclear charge experienced by the bonding electrons. The scientist Pauling computed a scale of electronegativity according to bond dissociation energies, ionization energy and electron affinity values of an atom. In Pauling scale, 4 is assigned to the most electronegative atom, fluorine. Pauling electronegativity values of some elements (The shaded ones are more electronegative than C) H 2.1 Li 1.0 Be 1.5 B 2.0 C 2.5 N 3.0 O 3.5 F 4.0 Na 0.9 Mg 1.2 Al 1.5 Si 1.8 P 2.1 S 2.5 Cl 3.0 K 0.8 Ca 1.0 _ Ge 1.8 As 2.0 Se 2.4 Br 2.8 Rb 0.8 Sr 1.0 _ Sn 1.8 Sb 1.9 Te 2.1 I 2.5 Cs 0.7 Ba 0.9 In general, electronegativity values increase from left to right across each period and decrease down each group. Noble gases do not has any value of electronegativity because they do not form any stable molecule at all. Glossary atomic radius covalent radius van der Waals radius proton to electron ratio electronegativity IV. Periodic Table Page 3 Past Paper Question 89 2B 4 d ii 92 2A 3 b i ii 95 2B 5 a I 89 2B 4 d ii 4d Comment on the relative oxidizing properties of ii the non-metal C, N, O and F. 3 C < N < O < F 1 mark - increase in electron affinity / electronegativity from C to F - decrease in atomic size - 2p electrons are less effectively shielded - increase in nuclear charge - F is particularly reactive because of the weak F–F bond due to lone pair-lone pair repulsion 2 marks 92 2A 3 b i ii 3b i Define the covalent radius of an atom. 1 The covalent radius is defined as on-half the distance between two atoms of the same kind held together by covalent bond. 1 mark ii State and explain the trends in the covalent radius on going down any group and going across a short period of the periodic table. 2 Going down any group of the periodic table the covalent radius increases ½ mark because the elements become larger as the no. of electron increases. ½ mark Going across any period the covalent radius decreases ½ mark because the nuclear charge increases. ½ mark 95 2B 5 a i 5a The table below lists some properties of the alkali metals. Element Atomic radius / nm Ionic radius / nm First ionization energy / kJmol -1 Standard electrode potential / V Melting point / ºC Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 i Explain why the atomic radii increase from Li to Cs. 1 As the group is descended, the next atom has another shell of electrons around the nucleus, therefore larger the atomic radius. 1 mark C as the group is descended, the 'next' atom has one more shell of electrons around the nucleus; Energetics I. Enthalpy Change (∆H) A. Definition of a system B. Definition of enthalpy (H) and enthalpy change (∆H) 1. Constant pressure process a) Work done by a system C. Standard enthalpy change of formation (∆H f o) 1. Definition of ∆H f o a) Standard enthalpy change (∆Ho) b) Standard enthalpy change of formation (∆H f o) c) Standard enthalpy change of formation of an element 2. Direct determination of ∆H f o a) Use of simple calorimeter D. Other enthalpy change (∆H) 1. Examples of enthalpy change a) Enthalpy of combustion b) Enthalpy of atomization c) Enthalpy of neutralization d) Enthalpy of hydrogenation e) Enthalpy of solution f) Enthalpy of reaction E. Hess’s Law 1. Indirect determination of ∆H f o a) Hess’s Law (1) Use of energy level diagram / energy cycle (2) Use of equation II. Bonding Energy A. Bond energy 1. Definition of bond energy 2. Determination of bond energy term (bond enthalpy) B. Strenght of covalent bond 1. Relationship between bond length and bond energy 2. Factors affecting the bond strength C. Estimation of enthalpy change by bond energy III. Energetics of formation of ionic compounds A. Electron affinity B. Lattice energy 1. Born-Haber cycle a) Determination of lattice energy by Born-Haber cycle 2. Factors affecting the value of lattice energy C. Stoichiometry of ionic compounds IV. Gibbs free energy change (∆G) A. Limitation of enthalpy change (∆H) B. Entropy (S) and entropy change (∆S) C. Gibbs free energy change (∆G) I. Enthalpy Change (∆H) Unit 1 Page 1 Topic I. Enthalpy Change (∆H) Unit 1 Reference Reading Chemistry in Context, 3rd Edition ELBS pg. 165–166 Physical Chemistry, 4th Edition, Fillans pg. 99 – 101 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 204 – 206 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 146–147 Syllabus 3.0–3.1 Definition of a system Enthalpy and enthalpy change Notes I. Enthalpy Change (∆H) The discovery of detailed electronic structure of atom serves as a very good example in how does a chemist work. Nobody have seen the atom directly. The electronic structure is only the result of deduction from some macroscopic observations (ionization energies and successive ionization energies). The energy change associated with a reaction provides another useful tools to investigate the chemistry involved. A. Definition of a system In the study of energetics, the region or particular quantity of matter of interest is defined as the system. The surroundings are defined as everything other than the system. For example, the flask, gas syringe and everything inside them is defined as the system. The atmosphere and everything outside the flask and syringe is defined as the surroundings. B. Definition of enthalpy (H) and enthalpy change (∆H) The enthalpy of a substance, sometimes called its heat content, is an indication of the total energy content of the substance. The absolute enthalpy (H) of a substance depends on i. potential energy inherent in the electrical and nuclear interactions of the constituent particles (chemical energy), ii. kinetic energy possessed by the atoms and sub-atomic particles (kinetic energy) The absolute enthalpy (H) of a substance is not measurable since it is still not possible to measure all the interaction between the sub-atomic particles in an atom. Instead, we can measure the enthalpy change (∆H) of a system. Enthalpy change (∆H) can be arbitrarily defined as the heat change (q) at constant pressure. I. Enthalpy Change (∆H) Unit 1 Page 2 1. Constant pressure process Consider the reaction between magnesium and hydrochloric acid : Mg HCl MgCl H s aq aq g ( ) ( ) ( ) ( ) + → + 2 2 2 Initial stage Immediate after reaction Final stage 1. Starting from room temperature (25ºC), we allow the magnesium and hydrochloric acid to react. 2. Immediately after the reaction, the temperature has increased due to heat energy released in the reaction. 3. We let the system cool back to room temperature (25ºC). We measure all the heat transferred between system and surroundings in order to return the system to 25ºC, this is the enthalpy change for the reaction. By law of conservation of energy, H 1 and H 2 must be equal. When the system is cooled down to 25 ºC from 50 ºC, some energy is lost to the surrounding in form of heat energy(q). By definition, ∆H = H final - H initial = H 3 - H 1 = q Since H 1 is larger than H 3 in this exothermic reaction, it can be seen that ∆H of an exothermic reaction is always negative. Strictly speaking, according to the definition, in order to determine the enthalpy change, the amount of heat exchanged (q) between the system and the surrounding have to be determined. Since the quantity of the heat change (q) is difficult to be measured, it would be more convenient if we assume that all the heat energy evolved in the above example is absorbed by the system. Then the amount of the heat evolved can be calculated from the heat capacity of the system. ∆H = heat change (q) = - Heat capacity of the system × temperature change of the system (∆T) N.B. In an exothermic reaction, ∆T is positive but q and ∆H are negative, so a negative sign is added in the expression. I. Enthalpy Change (∆H) Unit 1 Page 3 a) Work done by a system Upon mixing of magnesium and dilute hydrochloric acid in the flask, hydrogen gas evolves. The volume of the system increase and the system has done a work against the atmospheric pressure. Work done is defined as Force(F) × Distance(d) = [Pressure(P) × Area(A)] × [Volume change(∆V) ÷ Area(A)] = Pressure(P) × Volume change(∆V) = P∆V ∆V = V final - V initial which is positive if the volume of the system is increasing, and is negative if the volume of the system is decreasing. If the volume is kept constant in this case, the pressure will no longer be a constant. More heat will evolve in this reaction since the work done against atmospheric pressure is now exhibited in form of heat. Therefore, if the pressure of the system is not kept constant, according to the definition of ∆H, the heat change (q) measured would not be the same as the value of ∆H. Glossary system surrounding enthalpy change (∆H) heat change / transfer work done I. Enthalpy Change (∆H) Unit 2 Page 1 Topic I. Enthalpy Change (∆H) Unit 2 Reference Reading 3.2 Modern Physical Chemistry ELBS pg. 215–222 Chemistry in Context, 3rd Edition ELBS pg. 165 – 169 Physical Chemistry, 4th Edition, Fillans pg. 102 – 104 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 208 – 210 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 148–150 Syllabus Standard enthalpy change of formation (∆H f o) Determination of enthalpy change Other enthalpy change Notes C. Standard enthalpy change of formation (∆H f o) i. Standard state For the basis of comparison and tabulation of data, a standard state has been arbitrarily defined. All the reactants and products i. in the pure state if they are crystalline substances or liquids ii. at 1 atmosphere pressure if they are gases iii. at a concentration of 1M if they are dissolved in a solvent iv. at a specified temperature, usually, but not necessarily, 298K (25ºC) The condition of 1 atm and a specified temperature is called standard condition. ii. Meaning of superscripts o = standard º = pure substance * = excited electronic state t = transition state, activated complex ∞ = infinite dilution (with respect to all solutes present) 1. Definition of ∆H f o a) Standard enthalpy change (∆Ho) The enthalpy change (∆H) for a reaction which occurs under standard conditions is called the standard enthalpy change. It is represented by the symbol ∆Ho. Note : The superscript o does not contain the information of the temperature. To be specific, the standard enthalpy change at 298K should be written as ∆Ho(298K). If the temperature is not specified, by convention, it is assumed to be 298K. e.g. i. 2H 2(g) + O 2(g) → 2H 2 O (g) ∆H = -484 kJmol -1 ii. 2H 2(g) + O 2(g) → 2H 2 O (l) ∆H = -572 kJmol -1 It can be seen that the value of enthalpy change depends on the physical state of the reactant and product. Therefore, in writing of equation, the physical state of individual species must be specified. I. Enthalpy Change (∆H) Unit 2 Page 2 b) Standard enthalpy change of formation (∆H f o) Standard enthalpy change of formation (∆H f o)– The enthalpy change which occurs when 1 mole of a compound is formed, under standard conditions, from its elements in their standard state. e.g. i. Standard enthalpy of formation of carbon dioxide C (graphite) + O 2(g) → CO 2(g) ∆H f o(298K) = -393.5 kJmol -1 ii. Standard enthalpy of formation of ethane 2C (graphite) + 2H 2(g) → C 2 H 4(g) ∆H f o(298K) = -52.3 kJmol -1 Standard enthalpy of formation give us a rough ideal on the energetic stability of the compound respect to its constituent elements. A more negative standard enthalpy formation means more energetically stable. c) Standard enthalpy change of formation of an element According to the definition, standard enthalpy change of formation of an element, e.g. oxygen, is the enthalpy change associated with the following change, O 2(g) → O 2(g) . Actually, there is no change, the standard enthalpy change of formation of an element is always zero. 2. Direct determination of ∆H f o Some of the enthalpy of formation can be determined directly by measuring the heat change in direct synthesis of the compound, e.g. ∆H f o[CO 2(g) ], ∆H f o[SO 2(g) ]. a) Use of simple calorimeter A simple calorimeter can be used to determine the amount of heat absorbed or released in a reaction. For an exothermic reaction, the temperature will raise. For an endothermic reaction, the temperature will fall. If the heat capacity of the system is known. The heat energy evolved or absorbed can be calculated from the temperature change. I. Enthalpy Change (∆H) Unit 2 Page 3 e.g. Determination of enthalpy of reaction of Zn (s) + 2Ag + (aq) → Zn 2+ + 2Ag (s) An excess of zinc powder was added to 50.0 cm 3 of 0.100M AgNO 3(aq) in a polystyrene cup. Initially, the temperature was 21.10ºC and it rose to 25.40ºC. Calculate the enthalpy change for the reaction : Zn (s) + 2Ag + (aq) → Zn 2+ + 2Ag (s) Assume that the density of the solution is 1.00 gcm -3 and its specific heat capacity is 4.18 kJkg -1 K -1 . Ignore the heat capacity of the metals. i. Since the polystyrene cup is an insulator, its heat capacity is assumed to be zero and there is no energy exchanged between system and surroundings. ii. All the chemical energy released in the reactions transformed into heat energy which raises the temperature of the solution. iii. As there is no lost of energy to the surroundings, the total energy change in the system is zero. ∆H, due to reaction (at constant T) + change in heat energy of solution = 0 iv. ∆H + mc p ∆T = 0 where m is mass of the solution c p is the specific heat capacity of the solution under constant pressure ∆T is the temperature change in K v. ∴ ∆H = -mc p ∆T = - 50.0 1000 kg × 4.18 kJkg -1 K -1 × 4.30 K = -0.899 kJ vi. The value -0.899 kJ corresponds to 0.00500 mole of silver ions. In order to obtain the value for 2 mole of silver ions as specified by the equation, The enthalpy change using 2 mole of Ag + ions = -0.899 kJ × 2 mole 0.00500 mole = -360 kJ vii. Zn (s) + 2Ag + (aq) → Zn 2+ + 2Ag (s) ∆H = -360kJmol -1 viii. Strictly speaking, ∆H cannot be written as ∆Ho in this question because the conditions of the experiment were not standard, but the values of ∆H and ∆Ho would be very close. ix. Usually, a graph is plotted and extrapolated to determined the change in temperature, ∆T. I. Enthalpy Change (∆H) Unit 2 Page 4 D. Other standard enthalpy change (∆Ho) 1. Examples of enthalpy change a) Standard enthalpy of combustion ∆H c o It refers to the complete combustion of 1 mole of the substance under standard conditions. e.g. C (graphite) + O 2(g) → CO 2(g) ∆H c o[C (graphite) ] = -393.5 kJmol -1 b) Standard enthalpy of atomization ∆Ho at It refers to the formation of 1 mole of gaseous atoms from the element in the defined physical state under standard conditions. e.g. i. C (graphite) → C (g) ∆Ho at [C (graphite) ]= +715 kJmol -1 ii. ½H 2(g) → H (g) ∆Ho at [½H 2(g) ]= +218 kJmol -1 For a compound, the value would be the enthalpy change involved in atomization of 1 mole of the compound being concerned. e.g. CO 2(g) → C (g) + 2O (g) ∆Ho at [CO 2(g) ] = +1606 kJmol -1 c) Standard enthalpy of neutralization ∆Ho neutralization It is the heat change when an acid and a base react to form 1 mole of water under standard condition. e.g. HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (l) ∆Ho neutralization = -57.1 kJmol -1 CH 3 COOH (aq) + NaOH (aq) → CH 3 COONa (aq) + H 2 O (l) ∆Ho neutralization = -55.2 kJmol -1 Neutralization of a weak acid is found to be less exothermic than that of a strong acid because energy is required to ionize the molecules of weak acid first. d) Standard enthalpy of hydrogenation ∆Ho hydro It is the heat change when 1 mole of an unsaturated compound is converted to saturated compound by reaction with gaseous hydrogen at 1 atm. e.g. CH 2 CH=CHCH 2(g) + 2H 2(g) → CH 3 CH 2 CH 2 CH 3(g) ∆Ho hydro = -238.8 kJmol -1 e) Standard enthalpy of solution ∆Ho sol’n It is the heat change when 1 mole of a substance is dissolved at 1 atm in a stated amount of solvent or infinite amount of solvent. e.g. i. LiCl 100H O LiCl (s) 2 (l) (aq,100H O) 2 + → ∆Ho sol’n = -35 kJmol -1 ii. LiCl LiCl s H O aq l ( ) ( ) ( ) 2 → ∆Ho sol’n = -37.2 kJmol -1 f) Standard enthalpy of reaction ∆Ho rxn It is the heat change in a reaction at 1 atm between the number of moles of reactants shown in the equation for the reaction. e.g. 2Al (s) + Fe 2 O 3 (s) → 2Fe (s) + Al 2 O 3(s) ∆Ho rxn = -845.6 kJmol -1 Glossary standard enthalpy of formation standard state calorimeter enthalpy of combustion enthalpy of atomization enthalpy of neutralization I. Enthalpy Change (∆H) Unit 2 Page 5 Past Paper Question 91 2A 1 b ii 92 2A 1 b i 99 1B 7 b 91 2A 1 b ii 1b ii The enthalpy of neutralization of ethanoic acid with aqueous sodium hydroxide is -55.2 kJ mol -1 while that of hydrochloric acid is -57.3 kJ mol -1 . Account for the difference in these two values. 2 The enthalpy of neutralization of a strong acid (HCl) by a strong base (NaOH) is -57.3 kJ mol -1 . The enthalpy of neutralization of ethanoic acid with NaOH is -55.2 kJ mol -1 because ethanoic acid is a weak acid and is only slightly ionized, and the difference is due to the enthalpy of ionization. 2 marks 92 2A 1 b i 1b i Define the standard enthalpy of formation of a compound, using CH 3 OH (l) as an illustration. 1 The standard enthalpy of formation of a compound is the standard enthalpy change that occurs when one mole of the compound is made from its constituent elements under standard conditions (298 K and 1 atmospheric pressure). e.g. C (s) + 2H 2(g) + ½O 2(g) → CH 3 OH (l) 1 mark C i Many candidates did not include temperature and pressure in their definition of standard enthalpy of formation. 99 1B 7 b 7b In an experiment to determine the enthalpy change of neutralization, a polystyrene foam cup was used as a calorimeter. When a solution of an acid was poured into a solution of an alkali in the calorimeter, the temperature rise was recorded by a thermometer which also served as a stirrer. State THREE sources of error in the result obtained in such an experiment. I. Enthalpy Change (∆H) Unit 3 Page 1 Topic I. Enthalpy Change (∆H) Unit 3 Reference Reading 3.3 Modern Physical Chemistry ELBS pg. 222–226 Chemistry in Context, 3rd Edition ELBS pg. 170–174 Physical Chemistry, Fillans pg. 62–75 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 210 – 213 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 150–157 Syllabus Hess’s Law Applications of Hess’s Law Notes E. Hess’s Law 1. Indirect determination of ∆H f o Not ∆H f o of all compounds can be determined directly. This is because i. Not all compounds can be synthesized directly from its constituent elements at standard state. e.g. CH 4(g) ii. In some reactions, the extent of the reaction is difficult to control. e.g. NH 3(g) iii. Presence of side products / side reactions. 4Na (s) + O 2(g) → 2Na 2 O (s) , 2Na (s) + O 2(g) → Na 2 O 2(s) For the ∆H f o of the compound which cannot be determined directly, law of conservation of energy (Hess’s Law) can be used to determine its value. a) Hess’s Law Hess's Law - The change in enthalpy accompanying a chemical reaction is independent of the pathway between the initial and final states. The enthalpy change for a reaction is constant and is the sum of enthalpy changes for the intermediate steps. The standard enthalpy change of the reaction, C (s) + ½O 2(g) → CO (g) , cannot be determined directly since the extent of the reaction is difficult to control, inevitably some C (s) will remain and some CO 2(g) will be present. But, the following 2 data can be determined experimentally by calorimetry, C (s) + O 2(g) → CO 2(g) ∆Ho = -394 kJmol -1 CO (g) + ½O 2(g) → CO 2(g) ∆Ho = -283 kJmol -1 I. Enthalpy Change (∆H) Unit 3 Page 2 (1) Use of energy level diagram / energy cycle i. Energy Cycle By Hess's Law, the heat change accompanying Route 1 should equal to that of Route 2. ∴ ∆H 1 = ∆H 2 + ∆H 3 ∆H 2 = ∆H 1 - ∆H 3 ∴ ∆H f o[CO (g) ] = ∆H 2 = -394 kJmol -1 - (-283 kJmol -1 ) = -111 kJmol -1 ii. Energy Level Diagram To represent the relative enthalpy of the substance involved, an energy level diagram may be used. Steps of drawing an energy level diagram 1. Draw a line to represent the enthalpy of 1 mole of graphite and 1 mole of oxygen at standard condition. 2. The combustion of 1 mole of graphite produces CO 2(g) and releases 394 kJ. 3. The combustion of CO (g) gives the same product, 1 mole of CO 2(g) . The energy released is 283 kJ. 4. The gap represents the heat of formation of CO (g) . Steps of constructing energy cycle / energy level diagram 1. Write down an equation according to the definition of the problem. 2. Write down the common reference to which both reactant and product are related. In the above example, it is CO 2(g) . 3. Complete the cycle / diagram. N.B. In general, there are three popular common reference used in construction of energy cycles. They are i. combustion products ii. constituent elements in their standard states iii. constituent atoms I. Enthalpy Change (∆H) Unit 3 Page 3 (2) Use of equation The standard heat of reaction can also be determined if the standard heat of formation of the reactants and products are given. The generalized statement of this approach in the calculation of ∆Ho rxn is ∆Ho rxn = Σ∆H f o(products) - Σ∆H f o(reactants) e.g. C (s) + O 2(g) → CO 2(g) ∆H f o = -394 kJmol -1 CO (g) + ½O 2(g) → CO 2(g) ∆H f o = -283 kJmol -1 If you want to determine the heat of formation of CO (g) , C (s) + ½O 2(g) → CO (g) , following method may be used. For the reaction, CO (g) + ½O 2(g) → CO 2(g) , ∆Ho rxn = -283 kJmol -1 ∆Ho rxn = Σ∆H f o(products) - Σ∆H f o(reactants) -283 kJmol -1 = {∆H f o[CO 2(g) ]} - {∆H f o[CO (g) ] + ∆H f o[½O 2(g) ]} -283 kJmol -1 = (-394 kJmol -1 ) - (∆H f o[CO (g) ] + 0 kJmol -1 ) ∆H f o[CO (g) ] = (-394 kJmol -1 ) - (-283 kJmol -1 ) = -111 kJmol -1 Note : By definition, ∆H f o of any element, e.g. O 2 , is zero. Or simply, by doing some manipulations on the equations. C (s) + O 2(g) → CO 2(g) ∆H f o = -394 kJmol -1 CO (g) + ½O 2(g) → CO 2(g) ∆H f o = -283 kJmol -1 i. Reverse the second equation and the sign of ∆H f o. ii. Add the two equations together. C (s) + O 2(g) → CO 2(g) ∆H f o = -394 kJmol -1 CO 2(g) → CO (g) + ½O 2(g) ∆H f o = +283 kJmol -1 --------------------------------------------------------------------------------------------- C (s) + O 2(g) + CO 2(g) → CO 2(g) + CO (g) + ½O 2(g) --------------------------------------------------------------------------------------------- C (s) + ½O 2(g) → CO (g) ∆H f o = -111 kJmol -1 Glossary Hess’s Law energy level diagram energy cycle I. Enthalpy Change (∆H) Unit 3 Page 4 Past Paper Question 90 1A 2 b 91 1B 4 b i ii iii 91 2A 1 b i 92 2A 1 b ii 93 2A 3 b i ii 94 2A 2 a i ii 95 2A 1 b 97 2A 2 c 98 2A 2 c i ii 90 1A 2 b 2b Show how ∆Ho f for CuSO 4 ·5H 2 O (s) can be determined using the following data: ∆Ho soln CuSO 4 ·5H 2 O (s) = + 8 kJ mol -1 ; ∆Ho f CuSO 4(s) = - 773 kJ mol -1 ∆Ho soln CuSO 4(s) = - 66 kJ mol -1 ; ∆Ho f H 2 O (l) = - 286 kJ mol -1 3 aq + Cu (s) + S (s) + 9/2 O 2(g) + 5H 2(g) CuSO 4 ?H 2 O (s) + aq aq + CuSO 4(s) + 5H 2 O (l) CuSO 4(aq) + 5H 2 O (l) ∆H f CuSO 4 ?H 2 O (s) ∆H f CuSO 4(s) + 5 ? ∆H f H 2 O (l) ∆H soln CuSO 4(s) ∆H soln CuSO 4 ?H 2 O (s) Cu (s) + S (s) + 2O 2(g) → CuSO 4(s) - 773 kJ 5H 2(g) + 2½O 2(g) → 5H 2 O (l) 5 × - 286 kJ aq + CuSO 4(s) → CuSO 4(aq) - 66 kJ CuSO 4(aq) + 5H 2 O (l) → CuSO 4 ·5H 2 O (s) + aq - 8 kJ ------------------------------------------------------------------------------------------- Cu (s) + S (s) + 5H 2(g) + 4½O 2(g) → CuSO 4 ·5H 2 O (s) ∴∆H f o for CuSO 4 ·5H 2 O (s) = (-773) + 5(-286) + (-66) - 8 = -2277 kJ mol -1 2 marks for working (cycle/equations and expressions) 1 mark for correct answer (sign, numeric and units) Deduct ½ mark for wrong units. C Only about half of the candidates did this numerical problem correctly. Many errors arose as a result of a poorly constructed Born-Haber cycle. Very few candidates used the more analytical approach of setting up equations. 91 1B 4 b i ii iii 4b You are required to determine indirectly the enthalpy of formation of calcium carbonate(s). You are give the reagents: calcium, calcium carbonate, a strong acid, and water; the enthalpies of formation of carbon dioxide (g) and water (l) are available. i Give equations of suitable reactions for this determination. 2 CaCO 3(s) + 2HCl (aq) → CaCl 2(aq) + CO 2(g) + H 2 O (l) Ca (s) + 2HCl (aq) → CaCl 2(aq) + H 2(g) 2 marks (missing or incorrect physical state -½; wrong coefficient -½; inappropriate acid e.g. H 2 SO 4 and HNO 3 -½) ii Name the major pieces of apparatus needed for this determination. 2 Calorimeter (beaker or plastic cup, stirrer, temperature bath, insulation, electric heater) ½ mark thermometer or thermocouple ½ mark measuring cylinder, burette or pipette ½ mark balance ½ mark iii Suggest one experimental difficulty in the direct determination of the enthalpy of formation of calcium carbonate from its elements. 2 Ca (s) + C (s) + 3 2 O 2(g) → CaCO 3(s) Side reactions: 2Ca (s) + O 2(g) → 2CaO (s) ; C (s) + O 2(g) → CO 2(g) 2 marks for any one explanation: - extent of reaction cannot be controlled - side reactions - directly combustion of Ca can be violent - cannot react or cannot react to form CaCO 3 (1 mark only) C This section showed that the candidates were weak in applying general chemical knowledge to experiments. Most candidates scored marks by pointing out a very common piece of equipment and giving a partial explanation of the experimental difficulty. Of those who gave the equations with correct species, many failed to give balanced coefficients and states of the species. Many incorrect explanations like "heat loss to surroundings" and "large heat of formation" were given for (iii). I. Enthalpy Change (∆H) Unit 3 Page 5 91 2A 1 b ii 1b i Calculate the enthalpy of formation of NaCl (s) from the following data : Reaction ∆Ho/kJmol -1 NaOH (aq) + HCl (aq) → NaCl (aq) + H 2 O (l) -57.3 H 2(g) + ½O 2(g) → H 2 O (l) -285.9 ½H 2(g) + ½Cl 2(g) → HCl (g) -92.3 HCl (g) + aq → HCl (aq) -71.9 Na (s) + ½O 2(g) + ½H 2(g) + aq → NaOH (aq) -425.6 NaCl (s) + aq → NaCl (aq) +3.9 4 Na (s) + ½O 2(g) + ½H 2(g) + aq → NaOH (aq) ∆Ho = -425.6 kJ mol -1 NaOH (aq) + HCl (aq) → NaCl (aq) + H 2 O (l) ∆Ho = -57.3 kJ mol -1 H 2 O (l) → H 2(g) + ½O 2(g) ∆Ho = +285.9 kJ mol -1 ½H 2(g) + ½Cl 2(g) → HCl (aq) ∆Ho = -92.3 + (-71.9) = -164.2 kJ mol -1 NaCl (aq) → NaCl (s) + aq ∆Ho = -3.9 kJ mol -1 1 mark ∆Ho = (-425.6) + (-57.3) + (+285.9) + (-164.2) + (-3.9) = -365.1 kJ mol -1 (no unit -½) 3 marks C Some candidates were not able to give a correct energy cycle and hence failed to obtain the correct answer. Some weaker candidates did not even give state symbols in the chemical equations. 92 2A 1 b ii 1b ii Given the following thermochemical data at 298K : Standard enthalpy of combustion of CH 3 OH (l) -726.6 kJmol -1 Standard enthalpy of formation of CO 2(g) -393.5 kJmol -1 Standard enthalpy of formation of H 2 O (l) -285.8 kJmol -1 Calculate the standard enthalpy of formation of CH 3 OH (l) at 298K. 3 CH 3 OH (l) + 3 2 O 2(g) → CO 2(g) + 2H 2 O (l) ∆Ho = -726.6 kJmol -1 ∆Ho = Σn p (∆H f o) p - Σn r (∆H f o) r ∆Ho = (∆H f o(CO 2 ) + 2∆H f o(H 2 O) - (∆H f o(CH 3 OH) + 3 2 ∆H f o(O 2 )) Since ∆H f o(O 2 ) is zero. ∴ -726.6 = (-393.5 + 2 × (-285.8)) - (∆H f o(CH 3 OH) + 0) correct steps 2 marks ∴ ∆H f o(CH 3 OH) = -238.5 kJ mol -1 missing sign(-1), missing unit(-½) correct answer 1 mark 93 2A 3 b i ii 3b Given the following thermochemical data at 298 K: Compound ∆ H o combustion /kJ mol -1 ∆ H o formation /kJ mol -1 cyclopropane (g) -2091 — propene (g) -2058 — propane (g) -2220 — water (l) — -285.8 i Calculate the enthalpy change involved in the hydrogenation of cyclopropane to propane. 4 (1) ∆ (g) + 9 2 O 2(g) → 3H 2 O (l) + 3CO 2(g) ∆H = -2091 kJmol -1 (2) CH 3 CH 2 CH 3(g) + 5O 2(g) → 4H 2 O (l) + 3CO 2(g) ∆H = -2220 kJmol -1 (3) H 2(g) + ½O 2(g) → H 2 O (l) ∆H = -285.8 kJmol -1 1 mark (1) + (3) - (2) ∆ (g) + H 2(g) → CH 3 CH 2 CH 3(g) 1 mark ∆H = -2091 + (-285.8) - (-2220) = -156.8 kJmol -1 2 marks ii Calculate the enthalpy change involved in the conversion of cyclopropane to propene. Comment on the relative stabilities of cyclopropane and propene. 4 (4) H 2 C=CH–CH 3(g) + 9 2 O 2(g) → 3CO 2(g) + 3H 2 O (l) ∆H = -2058 kJmol -1 (1) - (4) ∆ (g) → H 2 C=CH–CH 3(g) 1 mark ∆H = -2091 - (-2058) = -33 kJmol -1 1 mark The conversion from cyclopropane to propene is exothermic, hence propene is more stable 1 mark because in the structure of cyclopropane there is squeezing of bond angles / ring is highly strained / poor overlapping of atomic orbital. 1 mark C Weaker candidates produced wrong answers because they used the wrong signs and some could not correlate the fact that the cyclopropane molecule was highly strained, and that it would release the excess energy upon conversion to propene, resulting in an exothermic reaction. I. Enthalpy Change (∆H) Unit 3 Page 6 94 2A 2 a i ii 2a Given the following thermochemical data. Reaction ∆Ho 298 / kJ mol -1 C (graphite) + 2H 2(g) → CH 4(g) - 75.0 C (graphite) + O 2(g) → CO 2(g) - 393.5 H 2(g) + ½O 2(g) →H 2 O (l) - 285.9 i Calculate the enthalpy change ∆Ho 298 for the reaction CH 4(g) + 2O 2(g) → 2H 2 O (l) + CO 2(g) . 2 ∆Ho 298 = -393.5 + 2(-285.9) - (-75.0) = - 890.3 kJmol -1 2 marks ii The enthalpy change ∆Ho 298 is - 801.7 kJ mol -1 for the following reaction. CH 4(g) + 2O 2(g) → 2H 2 O (g) + CO 2(g) Calculate the enthalpy change of vaporization of water at 298K. 2 CH 4(g) + 2O 2(g) → 2H 2 O (g) + CO 2(g) ∆H = - 801.7 kJ mol -1 - CH 4(g) + 2O 2(g) → 2H 2 O (l) + CO 2(g) ∆H = - 890.3 kJ mol -1 –––––––––––––––––––––––––––––––––––––––––––––––––––––– 2H 2 O (l) → 2H 2 O (g) ∆H = + 88.6 kJ mol -1 1 mark ∆Ho vaporization (H 2 O) = + 44.3 kJ mol -1 1 mark C Many candidates gave +88.6 kJ, instead of + 44.3 kJ mol -1 for the enthalpy change of vaporization of water. 95 2A 1 b 1b You are provided with the following thermochemical data: Reaction ∆Ho 298 /kJmol -1 Ag (s) + aq → Ag + (aq) + e - +105.56 1 2 N 2(g) + 3 2 O 2(g) + aq + e - → NO 3 - (aq) -207.36 1 2 Cl 2(g) + aq + e - → Cl - (aq) -167.15 Ag (s) + 1 2 Cl 2(g) → AgCl (s) -127.07 Calculate the standard enthalpy change for the reaction AgNO 3(aq) + HCl (aq) → AgCl (s) + HNO 3(aq) . 4 The reaction AgNO 3(aq) + HCl (aq) → AgCl (s) + HNO 3(aq) is equivalent to Ag + (aq) + Cl - (aq) → AgCl (s) 1 mark ∆H = Σ∆H f (products) - Σ∆H f (reactants) 1 mark = -127.07 - (+105.56) - (-167.15) = -65.48 kJmol -1 2 mark (Accept any correct method using an energy cycle.) C Some candidates could not construct the energy cycle correctly. Mistakes in the numerical calculation were common. 97 2A 2 c 2c Given the following thermochemical data at 298 K: Standard enthalpy change of formation of CO 2(g) = -393.5 kJ mol -1 Standard enthalpy change of formation of H 2 O (l) = -285.8 kJ mol -1 Standard enthalpy change of combustion of CH 3 CH 2 OH (l) = -1336.9 kJ mol -1 calculate the standard enthalpy change of formation of CH 3 CH 2 OH (l) at 298 K. 4 I. Enthalpy Change (∆H) Unit 3 Page 7 98 2A 2 c i ii 2c Both H 2(g) and CH 3 OH (l) are possible fuels for powering rockets. Their combustion reactions are shown below. H 2(g) + ½O 2(g) → H 2 O (g) CH 3 OH (l) + 1½O 2(g) → CO 2(g) + 2H 2 O (g) Compound molar mass / g ∆H f o / kJmol -1 CO 2(g) 44 -394 H 2 O (g) 18 -242 CH 3 OH (l) 32 -239 i For each of the above reactions, calculate the enthalpy change at 298 K per kg of the fuel-oxygen mixture in the mole ratio as indicated in the stoichiometric equation ii The effectiveness of a fuel can be estimated by dividing the enthalpy change per kg of the fuel-oxygen mixture in its combustion reaction by the average molar mass of the product(s) in g. Deduce which of the above two fuels is more effective in powering rockets. II. Bonding Energy Unit 1 Page 1 Topic II. Bonding Energy Unit 1 Reference Reading 3.4.1 Modern Physical Chemistry ELBS pg. 229–233 Chemistry in Context, 3rd Edition ELBS pg. 174–180 Physical Chemistry, Fillans pg. 76–83 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 214 – 215 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 157–159 Syllabus Definition of bond enthalpies Determination of bond enthalpies Notes II. Bonding Energy This is the principles of energetics and bonding energy again. i. Energy change always involves with bond formation and bond breaking. Usually, the bond formation process evolves energy, i.e. exothermic and bond breaking process absorbs energy, i.e. endothermic. ii. Bonding is just a balance between attractions and repulsions. If the attractions between atoms are stronger that the repulsions, the bond will form. Otherwise, the bond will be weak or will not form at all. A. Bond energy 1. Definition of bond energy In discussion of bond energy, two definitions have to be made clear. Bond Dissociation Energy, D: The energy required to break one mole of a particular covalent bonds in molecules. e.g. Cl–Cl bond in Cl 2 molecule In contract, enthalpy of atomization is the energy required to produce one mole of gaseous atom, for a diatomic molecule, it is half of D. Bond Energy Terms, E(X–Y): Also known as average bond energy or bond enthalpy, is an average value of bond dissociation energies for a given type of bond. e.g. E(C–H) is the average value of bond dissociation energies of C-H bond in all kinds of hydrocarbon. 2 1 C C H H H H H OH Not all C–H bonds are equivalent, even in the same molecule. For example, C–H bond at position 1 is stronger than the bond at position 2. The presence of the electronegative oxygen makes the C–H bonds become more polar and stronger. By convention, if not specified, bond energy is referring to bond energy term (average bond energy). Enthalpy changes for four successive dissociation reactions : Reaction ∆Ho(298K) kJmol -1 CH 4(g) → CH 3(g) + H (g) +425 CH 3(g) → CH 2(g) + H (g) +470 CH 2(g) → CH (g) + H (g) +416 CH (g) → C (g) + H (g) +335 Since dissociation energy refers to a particular bond, it is important to specify the process involved. The standard enthalpy change for the reaction, CH 4(g) → C (g) + 4H (g) , ∆Ho at [CH 4(g) ] = (+425) + (+470) + (416) + (+335) = +1646 kJmol -1 The bond energy terms, E(C–H), in CH 4(g) = +1646 kJmol -1 ÷ 4 = 413 kJmol -1 In contrast with the bond dissociation energy, bond energy term does not refer to one particular bond but it is the average bond energy of a particular type of bond. II. Bonding Energy Unit 1 Page 2 After the collection of a number of bond dissociation energies for a particular type of bond, the bond energy term can be determined. Bond E(X-Y) kJmol -1 C–H 413 C–C 347 C–Cl 346 2. Determination of bond energy term (bond enthalpy) Bond energy term can be determined by i. thermochemical methods (by Hess’s Law) ii. spectroscopic methods (not required) iii. electron impact methods (not required) Determination of E(C=O) in CO 2 by using an energy level diagram E(C=O) should be half of the enthalpy of reaction of CO 2(g) → C (g) + 2O (g) , ∆Ho at [CO 2(g) ] The enthalpy changes of the first three stages can be determined experimentally. Stage 1 Atomization of graphite C (graphite) → C (g) ∆Ho at [C (graphite) ] = +715 kJmol -1 Stage 2 Atomization of oxygen O 2(g) → 2O (g) ∆Ho at [O 2(g) ] = +498 kJmol -1 Stage 3 Enthalpy of formation of carbon dioxide C (graphite) + O 2(g) → CO 2(g) ∆H f o[CO 2(g) ] = -393 kJmol -1 Stage 4 Enthalpy of atomization of carbon dioxide CO 2(g) → C (g) + 2O (g) ∆Ho at [CO 2(g) ] By Hess’s Law ∆Ho at [CO 2(g) ] = - ∆H f o[CO 2(g) ] + ∆Ho at [C (graphite) ] + ∆Ho atm [O 2(g) ] = - (-393 kJmol -1 ) + 715 kJmol -1 + 498 kJmol -1 = +1606 kJmol -1 E(C=O) = ∆Ho at [CO 2(g) ] ÷ 2 = +1606 kJmol -1 ÷ 2 = +803 kJmol -1 Glossary bond energy bond energy terms (average bond energy) bond dissociation energy enthalpy of atomization Past Paper Question II. Bonding Energy Unit 2 Page 1 Topic II. Bonding Energy Unit 2 Reference Reading 3.4.2–3.4.3 Modern Physical Chemistry ELBS pg. 229–233 Chemistry in Context, 3rd Edition ELBS pg. 174–180 Physical Chemistry, Fillans pg. 76–83 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. Syllabus Strength of covalent bond Factors affecting the bond strength Notes B. Strength of covalent bond 1. Relationship between bond length and bond energy The bond strength of a bond can be measured by the bond energy. The stronger the bond, the more energy will be required to break the bond. For identical atoms, the bond strength is inversely proportional to the bond length. The stronger the bond, the closer will be the nuclei drawn together. Therefore, bond strength can be served as a indicator of bond strength. Bond Bond length / nm Bond energy C–C 0.154 +347 kJmol -1 C=C 0.134 +612 kJmol -1 2. Factors affecting the bond strength i. atomic size ii. number of bonding electrons / bond order iii. bond polarity i. atomic size Bond Bond length/nm Bond energy F–F 0.142 +158 kJmol -1 Cl–Cl 0.199 +242 kJmol -1 Br–Br 0.228 +193 kJmol -1 I–I 0.266 +151 kJmol -1 Bond Bond length/nm Bond energy F–H 0.092 +568 kJmol -1 Cl–H 0.127 +432 kJmol -1 Br–H 0.141 +366 kJmol -1 I–H 0.161 +298 kJmol -1 As the size of the atoms gets bigger, the bonding electrons are further from the nucleus and the shielding effect of the electron is getting more significant. Therefore, the strength of the bond decreases with increasing atomic size. Fluorine shows anomaly in group VII, this is because of the repulsion between lone pairs on the two atoms which weakens the bond. However if F is bonded to an atom with no lone pair, owing to the small size of F, the bond would be very strong. This explains why F four a larger variety of compounds that the other elements because the compound formed is energetically more stable than the others. ii. number of bonding electrons / bond order Bond Bond energy / kJmol -1 C–C +346 C=C +610 C≡C +837 N–N +163 N=N +410 N≡N +945 For the same kind of bond, e.g. C–C bond, the bond energy increases with increasing no. of bonding electron. II. Bonding Energy Unit 2 Page 2 iii. bond polarity Bond Difference in E.N. Bond energy N–H 0.9 +388 O–H 1.4 +463 F–H 1.9 +562 Polarization of the bond gives the bond certain ionic character. The gives the bond certain ionic attraction on top of covalent bond. Glossary bond length bond order molecular orbital theory bond polarity Past Paper Question 90 2B 4 c 91 2B 4 c i 95 2B 4 a ii 95 2B 4 c I 98 1A 1 b ii 90 2B 4 c 4 Account for the following observations. 4c Dinitrogen tetraoxide and hydrogen peroxide are both unstable to heat. 3 N 2 O 4 d 2NO 2 H 2 O 2 → 2HO· or homolytic bond fission/state the products 1 mark weak N–N and O–O bond strengths 1 mark due to lone pair-lone pair repulsions in O 2 and long N–N bond in N 2 1 mark 91 2B 4 c i 4c i Why is the Cl-Cl bond stronger than the F-F or Br-Br bonds? 2 We expect that the X-X bond energy decreases on descending the group because the distance between the nuclear protons and the bond-pair electron increases, so the attractive force decreases. This is so for Cl 2 and Br 2 . 1 mark The F-F bond is weaker than that of Cl 2 due to non-bonding electron repulsion between lone pairs, on the F nuclei. 1 mark C i The abnormally weak F–F bond strength was attributed to 'internuclear repulsion', to 'the small size of the fluorine atom', or to 'electron repulsion' and 'lone-pair repulsion', and few answers cited non-bonding electron repulsion between the two fluorine atoms. Some candidates thought that the Cl–Cl bond is stronger than the Br–Br bond because 'chlorine is more electronegative' or because 'the effective nuclear charge of Br is smaller'. Two points were required for the second part of the answer : a mention of the change in distance between nuclear protons and shared (bond-pair) electrons, and the consequent change in attractive force. 95 2B 4 a ii 4a Explain the following facts: ii The bond dissociation energy of F 2 is less than that of Cl 2 . 2 Unusually short F–F distance leads to high repulsion due to lone pair electrons of fluorine atoms. Therefore F–F bond is weaker than expected. 2 marks C Many candidates did not know that the small bond dissociation energy of F 2 is due to the strong repulsion of lone electron pairs of the fluorine atoms in the F 2 molecule. 95 2B 4 c i 4c Consider the data given below for the hydrogen halides and answer the questions that follow. Standard enthalpy change of formation / kJmol -1 Bond dissociation energy / kJmol -1 H–F -269.4 +562 H–Cl -92.8 +430 H–Br -36.8 +367 H–I +26.1 +298 i Explain briefly the trend in the bond dissociation energy of the hydrogen halides. 3 Trend : Bond dissociation energy decreases with increasing molecular mass of hydrogen halides. 1 mark Explanation : The H–X bond lengths increase as the atomic radii of the halogens increase. The longer the bond, the weaker it is and the bond dissociation energy is smaller. 2 marks C Some candidates did not know that the H–X bond lengths increase as the atomic radii of the halogens increase from F to I and that the longer the bond, the weaker it is and the smaller is the bond dissociation energy. 98 1A 1 b ii 1b ii Explain why the carbon-oxygen bond lengths in CO and CO 2 are different. 2 II. Bonding Energy Unit 3 Page 1 Topic II. Bonding Energy Unit 3 Reference Reading 3.4.4 Modern Physical Chemistry ELBS pg. 229–233 Chemistry in Context, 3rd Edition ELBS pg. 174–180 Physical Chemistry, Fillans pg. 76–83 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 157–158 Syllabus Estimation of bond enthalpies and bond length Notes C. Estimation of enthalpy change by bond energy With the atomization energies and Hess's Law, bond energy terms can be used for estimation of enthalpy changes. Since energy is required in bond breaking and energy is released in bond formation, the overall enthalpy can be estimated by the equation, ∆H = energy required - energy released = ΣE(bond broken) - ΣE(bond formed). i. Estimation of Enthalpy of formation of methane, CH 4(g) Value estimation by bond energy terms E(C–H) +413 kJmol -1 ∆Ho at [C (graphite) ] +715 kJmol -1 ∆Ho at [½H 2(g) ] +218 kJmol -1 ∆H f o[CH 4(g) ] = energy required - energy released = (∆Ho at [C (graphite) ] + 4 × ∆Ho at [½H 2(g) ]) - (4 × E(C–H)) = (715 + 4 × 218) - (4 × 413) = -65 kJmol -1 Value determination by Hess’s Law ∆H c o[C (graphite) ] -393.5 kJmol -1 ∆H c o[H 2(g) ] -285.8 kJmol -1 ∆H c o[CH 4(g) ] -890.3 kJmol -1 ∆H f o[CH 4(g) ] = (∆H c o[C (graphite) ] + 2 × ∆H c o[H 2(g) ]) - ∆H c o[CH 4(g) ] = (-393.5 + 2 × (-285.8)) - (-890.3) = -74.8 kJmol -1 It can be seen that the value estimated by bond energy terms differs from the value determined by Hess’s Law. This is because bond energy terms is only the average bond energy and not the actual value of the bond being concerned. Therefore, the value from Hess’s Law is always preferred. II. Bonding Energy Unit 3 Page 2 ii. Estimation of Enthalpy of formation of 1-chloroethane, CH 3 CH 2 Cl (g) , by bond energy terms E(C–H) +413 kJmol -1 E(C–C) +347 kJmol -1 E(C–Cl) +346 kJmol -1 ∆Ho at [C (graphite) ] +715 kJmol -1 ∆Ho at [½H 2(g) ] +218 kJmol -1 ∆Ho at [½Cl 2(g) ] +122 kJmol -1 ∆H f o[CH 3 CH 2 Cl (g) ] = energy required - energy released = (2 × ∆Ho at [C (graphite) ] + 5 × ∆Ho at [½H 2(g) ] + ∆Ho at [½Cl 2(g) ]) - (5 × E(C–H) + E(C–C) + E(C–Cl)) = (2 × 715 + 5 × 218 + 122) - (5 × 413 + 347 + 346) = -116 kJmol -1 iii. Use of bond energy term to predict the heat of reaction For the reaction, H–H(g) + Cl–Cl(g) → 2H–Cl(g) E(H–H) +435.9 kJmol -1 E(Cl–Cl) +243.4 kJmol -1 E(H–Cl) +432.0 kJmol -1 ∆H = energy required - energy released = ΣE(bond broken) - ΣE(bond formed) = (E(H–H) + E(Cl–Cl)) - (2 × E(H–Cl)) = (435.9 + 243.4) - (2 × 432.0) = -184.7 kJmol -1 (experimental value 184.6 kJmol -1 ) Glossary Past Paper Question 96 2A 1 a i 96 2A 1 a i ii 1a For the hydrogenation of buta-1,3-diene, H 2 C=CH–CH=CH 2(g) + 2H 2(g) → CH 3 CH 2 CH 2 CH 3(g) the experimental molar enthalpy change is -239 kJmol -1 . i Estimate the molar enthalpy change for the above hydrogenation using the bond energy terms below : Bond Bond energy term / kJmol -1 H–H 436 C–H 413 C–C 346 C=C 611 2 ∆H = -2E C–C - 4E C–H + 2E C–C + 2E H–H 1 mark or = -2(346) - 4(413) + 2(611) + 2(436) (1 mark) = -250 kJmol -1 1 mark (0 marks for omitting the eve sign; deduct ½ mark for no units) III. Energetics of formation of ionic compounds Page 1 Topic III. Energetics of formation of ionic compounds Reference Reading 3.5.1–3.5.2 Modern Physical Chemistry ELBS pg. 69–70, 74, 226–228 Chemistry in Context, 3rd Edition ELBS pg. 180–182 Physical Chemistry, Fillans pg. 84–90 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 216 – 218 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 85–86, 160–162 Syllabus Lattice energy Electron affinity and magnitude of electron affinity Determination of lattice energy by Born-Haber cycle Notes III. Energetics of formation of ionic compounds A. Electron affinity Electron affinity, ∆H e o or EA – the enthalpy change which occurs when one mole of electrons are added to one mole of neutral atom in the gaseous state to form one mole of gaseous ions under standard conditions., i.e. A (g) + e - → A - (g) . The value of the electron affinity is depending on the attraction between the incoming electron and the nucleus and the shielding effect offered by the existing electrons. i.e. the electron affinity of O, O (g) + e - → O - (g) ∆H e o = -142 kJmol -1 i.e. the electron affinity of O - (or second electron affinity of O) O - (g) + e - → O 2- (g) ∆H e o = +791 kJmol -1 The first EA of O is exothermic because the attraction between the incoming electron and the nucleus is atronger than the repulsion between the incoming electron and the existing electrons. The second EA of O is endothermic because in the presence of the extra electron, the extra electron charge will offer a stronger repulsion with the incoming electron. Furthermore, the electron cloud of O - will expand and the incoming electron will occupy a position further from the nucleus which weakens the attraction with the nucleus. B. Lattice energy Lattice energy, ∆Ho lat – the enthalpy change which occurs when one mole of the ionic compound is formed, as a crystal lattice, from its constituent gaseous ions. The value of lattice energy is directly proportional to the strength of the ionic bond. e.g. Na + (g) + Cl - (g) → NaCl (s) ∆Ho lat = -788.5 kJmol -1 Since lattice energy always associates with bond formation, it must have a negative value, i.e. exothermic. The lattice energy cannot be determined directly by experiment, we construct an energy diagram called Born- Haber cycle to determine the lattice energy. 1. Born-Haber cycle Born-Harber cycle – A thermodynamic cycle derived by application of Hess’s Law. Commonly used to calculate lattice energies of ionic solids and average bond energies of covalent compound. III. Energetics of formation of ionic compounds Page 2 a) Determination of lattice energy by Born-Haber cycle Generalized Born-Haber cycle for an ionic compound Transformation of elements in their standard states to gaseous ions, ∆H 1 ∆Ho at [Na (s) ] = +108 kJmol -1 ∆H i o[Na (g) ] = +500 kJmol -1 ∆Ho at [½Cl 2(g) ] = +121 kJmol -1 ∆H e o[Cl (g) ] = -364 kJmol -1 ∆H 1 = ∆Ho at [Na (s) ] + ∆H i o[Na (g) ] + ∆Ho at [½Cl 2(g) ] + ∆H e o[Cl (g) ] = 108 + 500 + 121 + (-364) = 365 kJmol -1 Standard heat of formation of the ionic crystal lattice, ∆H 2 ∆H 2 = ∆Ho f [NaCl (s) ] = -411 kJmol -1 Lattice energy, ∆Ho lat = ∆Ho lat [NaCl (s) ] Finally, by Hess’s Law ∆H 1 + ∆Ho lat [NaCl (s) ] = ∆H 2 ∆Ho lat [NaCl (s) ] = ∆H 2 − ∆H 1 = (-411) - (+365) = -776 kJmol -1 Detail Born Haber cycle of sodium chloride Note: ∆H i o – ionization energy, I.E. ∆H e o – electron affinity, E.A. 3. Factors affecting the value of lattice energy The magnitude of the lattice energy depends on i. The packing efficiency of the crystal lattice – small and similarity in size of ions result in better packing and more negative lattice energy ii. The charge density of the constituent ions – higher the charge density of the ion, stronger will be the electrostatic attraction and the lattice energy will be more negative. Lattice energy ∝ 1 r + + r - r + , r - : radii of cation and anion III. Energetics of formation of ionic compounds Page 3 C. Stoichiometry of ionic compounds The calculation of the lattice energies and use of Born-Haber cycle can also be used to explain why group I, II and III metal always form M + , M 2+ and M 3+ ion respectively. e.g. Born-Haber cycle for Ca + Cl - , Ca 2+ (Cl - ) 2 and Ca 3+ (Cl - ) 3 The lattice energies are calculated based on the mathematical model. By summing up the terms, the enthalpy of formation of the 3 hypothetical ionic crystal can be calculated. Ca (s) + ½Cl 2(g) → Ca + Cl - (s) ∆H f o = -155 kJmol -1 Ca (s) + Cl 2(g) → Ca 2+ (Cl - ) 2(s) ∆H f o = -763 kJmol -1 Ca (s) + 3 2 Cl 2(g) → Ca 3+ (Cl - ) 3(s) ∆H f o = +1356 kJmol -1 The enthalpy of formation of CaCl 2(s) is most exothermic among the three. This explain why calcium chloride has the empirical formula CaCl 2 because CaCl 2(s) is the most stable one among the three. The octet rule only helps to predict the formation of Ca 2+ ion but it doesn't offer explanation. The rationale behind the formation of Ca 2+ ion is solely from an energetic point of view. Glossary lattice energy electron affinity Born-Haber cycle intermediate type of bond III. Energetics of formation of ionic compounds Page 4 Past Paper Question 92 1A 3 d 94 1A 1 a i ii iii 96 2B 5 a i ii 98 1A 1 c i 99 2A 1 a ii 92 1A 3 d 3d Briefly describe and explain the change in the lattice energy of Group I chlorides on descending the group. 3 The lattice energy decreases in magnitude on descending the group. NaCl 771; CsCl 645 kJ mol -1 1 mark This is due to increasing size of cation results in smaller attraction between cation and anion. 2 marks 94 1A 1 a i ii iii 1a i Write an equation to represent the change related to the second electron affinity of oxygen. 1 O - (g) + e - → O 2- (g) 1 mark ii The first and second affinities of oxygen are -142 kJmol -1 and +791 kJmol -1 respectively. Explain why they have opposite signs. 2 Addition of an e - to the O (g) is exothermic Q attraction between the nucleus and the incoming e - outweighs the repulsion between the e - in O atom and the incoming e - . 1 mark After addition of an e - , the size of O expands, hence the attraction for another electron is weakened, and the repulsion between the O - (g) ion and the e - makes the second electron affinity an endothermic process. 1 mark iii Explain why all the inert gases have positive first electron affinities. 1 Screening effect of the stable p 6 configuration of inert gases (s 2 in the case of He) makes the addition of an extra e - an unfavourable process ∴ inert gases have +ve first E.A. 1 mark C It was surprising to find that many candidates could not distinguish between electron affinity and electronegativity and hence they performed badly in this part. 96 2B 5 a i ii 5a The first electron affinities of chlorine and bromine are -364 kJmol -1 and -295 kJmol -1 respectively. i What is the meaning of 'electron affinity' ? 2 Electron affinity is the molar enthalpy change ½ mark for the following process at standard state/conditions ½ mark X (g) + e - → X - (g) Or electron affinity of an element X is the enthalpy change ½ mark when 1 mole of X - (g) is formed from gaseous atoms of X 1 mark under standard conditions. ½ mark C Most candidates did not give a precise definition for 'electron affinity'. They did not point out that the electron affinity of an element X is the molar enthalpy change when atoms of X in the gaseous state take up electrons to form X - (g) ions under standard conditions. ii Explain why the first electron affinity of chlorine is more negative than that of bromine. 2 Cl has a smaller size / a smaller no. of electron shells 1 mark ∴ Cl can exert a stronger attraction on an extra electron 1 mark 98 1A 1 c i 1c The theoretical lattice enthalpies of NaCl (s) and AgCl (s) , and the experimental lattice enthalpy of AgCl (s) are given in the table below. Compound Theoretical lattice enthalpy Experimental enthalpy NaCl (s) - 770 ? AgCl (s) - 833 - 905 i Calculate the experimental lattice enthalpy of NaCl (s) using the following thermochemical data. ∆Ho/kJ mol -1 Atomization enthalpy of Na (s) 108 First ionization enthalpy of Na (g) 495 Bond dissociation enthalpy of Cl 2(g) 239 Electron affinity of Cl (g) - 349 Enthalpy change of formation of NaCl (s) - 411 3 III. Energetics of formation of ionic compounds Page 5 99 2A 1 a ii 1a ii Using the thermochemical data given below, calculate the lattice enthalpy of rubidium chloride. ∆Ho 298 / kJ mol -1 Rb (s) → Rb (g) 82 Rb (g) → Rb + (g) + e - 403 Cl 2(g) → 2Cl (g) 242 Cl (g) + e - → Cl - (g) -348 2Rb (s) + Cl 2(g) → 2RbCl (s) -861 IV. Gibbs free energy change (∆G) Page 1 Topic IV. Gibbs free energy change (∆G) Reference Reading 3.0 Modern Physical Chemistry ELBS pg. Chemistry in Context, 5th Edition ELBS pg. 339–350 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. Assignment Reading Syllabus Gibbs free energy change Notes IV. Gibbs free energy change (∆G) (not required in A-Level) So far, we have applied the concept of energetic to different chemical systems such as ionization energy, bond energy and lattice energy. However, in certain instances, changes in enthalpy are only part of the driving force for the change. Some processes take place spontaneously even though they are endothermic. i.e. ∆H > 0. The observation seems to be contradictory, the products of an endothermic change is less stable than the reactants. A. Limitation of enthalpy change (∆H) For example, melting is an endothermic change. Ice melts readily at room temperature. Although the heat content of water is higher than ice, does that mean ice is more stable than water ? If ice is more stable than water, why does it change to water ? Therefore, enthalpy change ∆H, is not a very good description of the relatively stability of systems. ∆H Ice Water melting process B. Entropy (S) and entropy change (∆S) Consider another example, the free expansion of a balloon in vacuum. Since no work is done in the expansion, the temperature will remain constant, i.e. ∆H = 0. The size of the balloon increases and more spaces are available for the gaseous molecules to travel. Therefore, the arrangement and the movement of the molecules are getting more random. free ansion in vacuum exp → The expansion is spontaneous which means that the bigger balloon is more stable than the smaller balloon though the ∆H is 0. It can be generalized that the state with a high degree of disorderness is always favourable in nature. The scientist called the disorderness in arrangement of a system – entropy S and the change in the disorderness – entropy change ∆S. In the above example, the arrangement of the molecules becomes more random and the entropy of the system increases. i.e. ∆S > 0. IV. Gibbs free energy change (∆G) Page 2 C. Gibbs free energy change (∆G) In order to address the limitation of enthalpy change ∆H in describing the relative stability of different systems. The scientist J. Willard Gibbs defined another function – Gibbs free energy change ∆G. Gibbs free energy change is defined as ∆G = ∆H - T∆S where ∆H is enthalpy change T is temperature in Kelvin ∆S is entropy change For example, ice has a more ordered packing that water. Thus, melting is a process associated with an increase in disorderness i.e. ∆S > 0. Enthalpy change (∆H) of melting = 6010 kJmol -1 Entropy change (∆S) of melting = 22.0 kJmol -1 Temperature ∆H of melting -T∆S of melting ∆G of melting -10ºC (263.15K) 6010 kJmol -1 -5790 kJmol -1 220 kJmol -1 0ºC (273.15K) 6010 kJmol -1 -6010 kJmol -1 0 kJmol -1 10ºC (283.15K) 6010 kJmol -1 -6230 kJmol -1 -220 kJmol -1 Obviously, whether ice will melt or freeze is depending on temperature. At any temperature above 0ºC, ice tends to melt. At any temperature below 0ºC, water tends to freeze. Moreover, the value of ∆G is always smaller than zero. All spontaneous changes take place in the direction of decrease in free energy. i.e. ∆G < 0. However, free energy change ∆G only indicates the energetic stability of a system. Whether the change will take place is also depending on the kinetic stability of the system which will be discussed in the chapter of rate of reaction. Glossary spontaneous reacton disorderness entropy (S) entropy change (∆S) Gibbs free energy change (∆G) Past Paper Question Chemical Bonding Page 1 Chemical Bonding I. Shape of molecule A. Lewis structure of molecule 1. Oxidation no. of individual atoms in a molecule 2. Formal charge of individual atoms in a molecule 3. Limitation of octet rule B. Valence Shell Electron Pair Repulsion Theory (VSEPRT) 1. Octet expansion of elements in Period 3 2. Position of lone pair 3. 3-dimensional representation of molecular shape C. Hybridization Theory 1. Overlapping of atomic orbital 2. Hybridization theory 3. Examples a) sp hybridizaion b) sp 2 hybridizaion c) sp 3 hybridizaion d) sp 3 d hybridizaion e) sp 3 d 2 hybridizaion 4. Structure and shape of hydrocarbons a) Characteristic of Carbon-Carbon bond b) Shape of hydrocarbons (1) Saturated hydrocarbons (2) Unsaturated hydrocarbons (3) Aromatic hydrocarbons (a) Delocalization of π-electrons (b) Stability of benzene D. Molecular Orbital Theory II. Ionic bonding A. Strength of ionic bond - lattice energy B. X-ray diffraction 1. Electron density map of sodium chloride C. Periodicity of ionic radius 1. Definition of radius 2. Periodicity of ionic radius 3. Size of isoelectronic particles Chemical Bonding Page 2 III. Covalent bonding A. H 2 + ion B. Electron diffraction C. Covalent radius 1. Definition of covalent radius 2. Addivity of covalent radius 3. Breaking down of addivity in covalent radius and bond energy a) Resonance (1) Resonance structure of benzene (2) Resonance structure of nitrate ion (3) Rules in writing resonance structures b) Breaking down of addivity in bond enthalpies (1) Delocalization energy D. Dative covalent bond 1. Examples of H 3 N→BF 3 and Al 2 Cl 6 a) H 3 N→BF 3 b) Al 2 Cl 6 IV. Bonding intermediate between ionic and covalent A. Differences between ionic bond and covalent bond B. Incomplete electron transfer in ionic compound 1. Electron density map of LiF comparing with those of NaCl and H 2 2. Difference among lattice energies of Na, Ag and Zn compounds a) Lattice energies of sodium halide, silver halide and zinc sulphate b) Bonding intermediate between covalent and ionic 3. Polarization of ionic bond a) Fajans’ Rules in polarization of ionic bond C. Electronegativity 1. Definition of electronegativity 2. Pauling scale of electronegativity D. Polarity in covalent bond 1. Deflection of a liquid jet by an electric field 2. Dipole moment a) Vector quantity of dipole moment b) Polarity of molecule c) Factors affecting dipole moment (1) Inductive effect (I) (2) Mesomeric effect / resonance effect (R) (3) Presence of lone pair V. Metallic bonding A. Electron sea model of metal B. Strength of metallic bond C. Melting and boiling of metal D. Strength of ionic bond, covalent bond and metallic bond I. Shape of molecule Unit 1 Page 1 Topic I. Shape of molecule Unit 1 Reference Reading 4.0–4.1 Modern Physical Chemistry ELBS pg. 84–86 Organic Chemistry, Solomons, 6th Edition pg. 8–14 Syllabus Lewis structure Notes Chemical Bonding I. Shape of molecule The shape of a molecule can be predicted and explained by several different theory i. Valence Shell Electron Pair Repulsion Theory – repulsions among electron centres (bond pair and lone pair electrons) have to be minimized. ii. Hybridization Theory – overlapping of hybridized atomic orbitals iii. Molecular Orbital Theory – overlapping of atomic orbitals to form molecular orbital A. Lewis structure of molecule Before the shape of a molecule can be derived, it would be easier to draw the Lewis structure (cross-dot diagram) of the molecule first. Steps of drawing Lewis structure i. Determination of central atom Not all atoms can be the central atom in a molecule. e.g. in H 2 O, H can only be the peripheral atom since it can form only 1 bond. ii. Determination the no. of the bonds have to be formed. Count the electron available Count the number of valence electrons provided by each atom. If the molecule is negatively charged, add extra no. of electrons accordingly. If the molecule is positively charged, subtract the no. of electrons accordingly. Count the electron required Count the total no. of electrons required if the outermost shells of all the atoms have to be filled. e.g. in NH 3 , there should be 8 electrons for N and 2 electrons for each H atom. Determine the no. of bonds The total no. of bond pairs can be calculated by the following method. Assuming each atom in the molecule needs the max. no. of e - to fill its valence shell e.g. 2 for H, 8 for O or N. Formation of each bond can save the use of 2 electrons. no. of bond pair = (no. of electron required - no. of electron available) ÷ 2 e.g. NH 3 no. of bond pair = ((2 × 3 + 8) - 8) ÷ 2 = 3 e.g. NO 3 - no. of bond pair = ((8 × 4) - 24) ÷ 2 = 4 iii. Drawing the skeleton of the molecule Within a covalent molecule, there should be at least 1 bond pair between 2 atoms. iv. Put the rest of the electrons into the diagram until octet is fulfilled and include the charge if any. I. Shape of molecule Unit 1 Page 2 Examples NH 3 i. Central atom – N ii. N 5 3H 1 × 3 e - available 8 N 8 3H 2 × 3 e - required 14 no. of bond = (14 - 8) ÷ 2 = 3 iii. N H H H Total 8 3 N–H 6 Remaining 2 iv. N H H H 3 N–H 6 1 lone pair 2 Total 8 NH 4 + i. Central atom – N ii. N 5 4H 1 × 4 +1 (charge) -1 e- available 8 N 8 4H 2 × 4 e - required 16 no. of bond = (16 - 8) ÷ 2 = 4 iii. N H H H H Total 8 4 N–H 8 Remaining 0 iv. N H H H H + 4 N–H 8 Total 8 NO 3 - i. Central atom – N ii. N 5 3O 6 × 3 -1 1 e- available 24 N 8 3O 8 × 3 e- required 32 no. of bond = (32-24)÷2 = 4 iii. N O O O Total 24 3 N–O 6 Remaining 18 iv. - N O O O 1 N=O 4 2 N–O 4 8 lone pairs 16 Total 24 1. Oxidation no. of individual atoms in a molecule Oxidation no. – The difference between the number of electrons associated with an atom in a compound as compared with an atom of the element. Oxidation no. is related to the difference in electronegativity of the atoms in the molecule. It is assumed that the bonding electrons will be attracted to the more electronegative atom from the less electronegative. i.e. the more electronegative atom will has more electrons than if neutral, results in negative oxidation no. e.g. CO 2 C O O Oxygen is more electronegative than carbon, the 4 electrons in the double bond is attracted towards the oxygen. Oxidation no. of oxygen = no. of electron in neutral O atom - no. of electrons on O in CO 2 = 6 - 8 = -2 Oxidation no. of carbon = no. of electron in neutral C atom - no. of electrons on C in CO 2 = 4 - 0 = +4 e.g. NO 3 - - N O O O Oxygen is more electronegative than nitrogen O.N. of the doubly bonded oxygen = no. of e - in neutral O atom - no. of e - on O in NO 3 - = 6 - 8 = -2 O.N. of the singly bonded oxygen = no. of e - in neutral O atom - no. of e - on O in NO 3 - = 6 - 8 = -2 O.N. of nitrogen = no. of e - in neutral N atom - no. of e - of N in NO 3 - = 5 - 0 = +5 2. Formal charge of individual atoms in a molecule I. Shape of molecule Unit 1 Page 3 Unlike oxidation no., formal charge is independent of difference in electronegativity. Formal charge of an atom is calculated by assuming that the bond pair electrons are shared equally between the two atoms bonded together. e.g. CO 2 C O O Formal charge of oxygen = no. of e- in neutral O atom - no. of e- associated with O in CO 2 = 6 - 6 = 0 Formal charge of carbon = no. of e- in neutral C atom - no. of e- associated with C in CO 2 = 4 - 4 = 0 e.g. NO 3 - N O O O - - + Formal charge of the doubly bonded oxygen = no. of e - in neutral O atom - no. of e - associated with O in NO 3 - = 6 - 6 = 0 Formal charge of the singly bonded oxygen = no. of e - in neutral O atom - no. of e - associated with O in NO 3 - = 6 - 7 = -1 Formal charge of nitrogen = no. of e - in neutral N atom - no. of e - associated with N in NO 3 - = 5 - 4 = +1 Non-zero formal charge is usually indicated on the Lewis structure. Similar to oxidation no., the sum of formal charges equals the overall charge carried by the molecule. To minimize the fuzziness of calculation, some formal charges of some atoms can be memorized. N.B. The positive sign and negative sign must be accompanied for all oxidation no. and formal charges. I. Shape of molecule Unit 1 Page 4 3. Limitation of octet rule The forth-mentioned method of drawing Lewis structure is only limited to the elements in Period 2. This is because, for principal quantum number 2, there are only 2s, 2p x , 2p y and 2 z orbitals available. Each orbital can accommodate 2 electrons and makes up to a total of 8. Though the largest no. of valence electron would be 8, the no. can be less than 8. For example, in NO 2 , N O O - + there are totally ( 5 + 6 + 6 ) = 15 valence electrons which is an odd number. It is not possible for all atoms to have an octet structure. Actually, there are 8 electrons about each O atom and only 7 eletrons about N. However, for period 3 elements, there are 3d orbitals available which is capable to accommodate 10 more electrons. This is why P can form both PCl 3 and PCl 5 but N can only form NCl 3 . This phenomenon is called expansion of octet (or expansion of coordination sphere) and will be explained in the chapter of hybridization theory. For those which does not obey octet rule, e.g. PCl 5 , SO 2 , another method have to be used to draw the Lewis structure. It is assumed that all valence electrons are capable to form bond with peripheral atoms. And the peripheral will form the required number of bond to achieve octet. e.g. in PCl 5 , P has 5 valence electrons and each Cl is capable to form 1 single bond. Therefore, the Lewis structure will become P Cl Cl Cl Cl Cl . e.g. in SO 2 , S has 6 valence electrons and each O is capable to form 1 double bond. Therefore, the Lewis structure will become S O O . e.g. in SO 4 2- ion, S has 6 valence electrons ; 2 O atoms with 0 formal charge each forming 2 bond ; 2 O atoms with -1 formal charge each forming 1 bond. Therefore, the Lewis structure will become S O O O - O - . e.g. XeF 4 is a noble gas compound not obeying octet rule. Xe has 8 outermost electrons, each F is capable to form 1 bond. As a result, 4 electrons about Xe will remains as the lone pairs. Therefore, the Lewis structure will become Xe F F F F . However, for some structures do not obey octet rule, sometimes memorization is required. Glossary Lewis structure Oxidation no. formal charge Past Paper Question 92 2B 6 Aa i 92 2B 6 Ba i 92 2B 6 Aa i 6Aa i Write the formula in each case of ONE compound or ion in which nitrogen has the oxidation state +5, +3, -1 and -3. 2 NO 3 - , NO 2 - , NH 2 OH, NH 3 ½ mark each 92 2B 6 Ba i 6Ba i Write the formula in each case of ONE compound or ion in which sulphur has the oxidation state -2, +2, +4 and +6. 2 S 2- , S 2 O 3 2- , SO 3 2- , SO 4 2- ½ mark each I. Shape of molecule Unit 2 Page 1 Topic I. Shape of molecule Unit 2 Reference Reading Modern Physical Chemistry ELBS pg. 84–86 Physical Chemistry, Fillans pg. 126–143 Organic Chemistry, Solomons, 6th Edition pg. 9–14 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 125–128, 131–132 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 90–92 Syllabus 4.2 Shape of molecule Valence Shell Electron Pair Repulsion Theory Notes B. Valence Shell Electron Pair Repulsion Theory (VSEPRT) The shape of a molecule was related to the number of electrons in the outer shell of the central atom. 1. Electron-pairs, whether in bonding orbitals or lone-pair orbitals, arrange themselves in space so as to minimize their mutual repulsions. 2. The repulsion between lone-pair/lone-pair is greater than that between lone-pair/bond-pair which is, greater than that between bond-pair/bond-pair. 3. Where there are more than four pairs of electrons to consider. the interactions where the electron-pairs make an angle greater than 90° at the central atom are ignored. No. of electron centre in the outer shell of the central atom Shape of the framework Bond angle Example 2 linear 180º BeCl 2 3 trigonal planar 120º BF 3 4 tetrahedral 109º28' CH 4 NH 3 H 2 O 5 trigonal bipyramidal 120º 90º PF 5 SF 4 ClF 3 6 octahedral 90º SF 6 IF 5 7 pentagonal bipyramidal 72º 90º IF 7 Note: An electron centre can either be a lone pair, single bond, double bond or triple bond. I. Shape of molecule Unit 2 Page 2 1. Octet expansion of elements in Period 3 Elements in period 3 and onwards are capable to use the low lying energy d-orbitals available to form more bonds. Therefore, the element in period 3 can exceed the octet by expansion of its coordination sphere. This would be further explained in hybridization theory. 2. Position of lone pair For the molecule with 5 electron centres, the lone pair always occupies the equatorial position to minimize the repulsion. Since lone pair electrons, comparing with bonding pair, is closer to the central atom and has greatest repulsion with other electron centres, only repulsions with the lone pair need to be considered. e.g. in SF 4 If the lone pair occupies the equatorial position, there would be only 2 90º lone pair / bond pair repulsions. If the lone pair occupies the axial position, there would be 3 90º lone pair / bond pair repulsions. The 120º repulsion is neglected as it is much weaker than 90º repulsion. Steps of using VSEPR theory to determine the shape of a molecule. 1. Draw the Lewis structure. 2. Count the number of electron centre around the central atom and choose the corresponding framework. 3. Put the electron centres (with the atom if applicable) on the vertex of the framework so that the repulsion among them are minimized. 4. Determine the bond angle. In the determination of the shape of a molecule, only the geometry of the atoms are considered. The position of the lone pair is not counted. 3. 3-dimensional representation of molecular shape a lone pair a bond on the paper a bond going into the paper a bond coming out of the paper Usually two bonds are used to defined the plane of the paper and then the other bonds are added with respect to this plane. I. Shape of molecule Unit 2 Page 3 Example Lewis structure no. of electron centre Framework chosen Shape of molecule Bond angle BF 3 B F F F 3 single bonds Total : 3 trigonal planar B F F F ∠ F–B–F : 120 º CH 4 C H H H H 4 single bonds Total : 4 tetrahedral C H H H H ∠ H–C–H : 109½º NH 3 N H H H 3 single bonds 1 lone pair Total : 4 trigonal pyramidal N H H H ∠ H–N–H : ≈ 107º H 2 O O H H 2 single bonds 2 lone pair Total : 4 angular / bent / V-shaped O H H ∠ H–O–H : ≈ 105º Although all CH 4 , NH 3 and H 2 O has 4 electron centres, they possess slightly different bond angle, 109½º, 107º and 105º respectively. 109½º is the angle for a perfect tetrahedral. However, the repulsion with lone pair is greater than the repulsion with bond pair. The presence of a lone pair on N squeezes the bond angle to 107º, the two lone pairs on O squeezes the bond angle even further to 105º Example Lewis structure no. of electron centre Framework chosen Shape of molecule Bond angle PCl 5 P Cl Cl Cl Cl Cl 5 single bonds Total : 5 trigonal bipyramidal P Cl Cl Cl Cl Cl ∠ Cl–P–Cl : 120º × 3 ∠ Cl–P–Cl : 90º × 6 SF 6 S F F F F F F 6 single bonds Total : 6 octahedral S F F F F F F ∠ F–S–F : 90º NH 4 + N H H H H + 4 single bonds Total : 4 tetrahedral N H H H H + ∠ H–N–H : 109½º NH 2 - N H H - 2 single bonds 2 lone pairs Total : 4 angular / bent / V-shaped N H H - ∠ H–N–H : ≈ 105º CO 2 O C O 2 double bonds Total : 2 linear C O O ∠ O=C=O : 180º SO 2 S O O 2 double bonds 1 lone pair Total : 3 angular / bent / V-shaped S O O ∠ O=S=O : ≈ 120º I. Shape of molecule Unit 2 Page 4 Glossary Lewis structure Oxidation no. formal charge Valence shell electron pair repulsion theory octet expansion Past Paper Question 91 1A 3 d i ii 91 2B 5 e 92 1A 3 g i ii 93 1A 1 b i ii 93 1A 3 a ii 94 1A 1 c i ii 95 1A 2 e i ii 96 1A 2 b i ii 97 1A 3 b i ii iii 98 1A 3 a i ii iii 98 1A 4 a i 98 2A 1 c i 99 2A 3 c i 91 1A 3 d i ii 3d Draw the molecular shapes of i PCl 5(g) 1 P Cl Cl Cl Cl Cl 1 mark ii SF 4(g) 1 S F F F F 1 mark 91 2B 5 e 5e Why is the bond angle in NF 3 smaller than that in NH 3 ? 2 (Valence Shell Electron Pair Repulsion approach) Bond pair electrons nearer to F than H ½ mark since F is more electronegative ½ mark ∴ bond-pair/bond pair repulsion is smaller in NF 3 than in NH 3 . 1 mark N H H H N F F F or (Hybridization approach) The primary scheme of hybridization of N is sp 3 the p orbital is more directional than s orbital and have been attracted strongly towards the electronegative F as a result the N in NF 3 has higher s-character than N in NH 3 Bond angle ∝ s-character e.g. Ideal bond energy of following hybridization: sp = 180º, sp 2 = 120º, sp 3 = 109½º 92 1A 3 g i ii 3g Draw diagrams showing the shapes of the following molecules. Indicate the lone pairs (if any) on each central atom. i ICl 2 1½ I Cl Cl 1 e - only Linear shape 1 mark Lone pairs ½ mark ii XeOF 4 1½ Xe O F F F F note: Xe is [Kr]4d 10 5s 2 5p 6 Square-pyramidal 1 mark Lone pair ½ mark I. Shape of molecule Unit 2 Page 5 93 1A 1 b i ii 1b For each of the following molecules, draw a three-dimensional structure and state the molecular geometry. i SiF 4 2 Si F F F F tetrahedral 2 marks ii OF 2 2 O F F bent / V-shaped 2 marks 93 1A 3 a ii 3a Consider the following compound, X: ii Suggest expected values for the bond angles α, β and γ. 1½ α: 108º-110º β: 178º-182º γ: 118º-122º ½ mark each 94 1A 1 c i ii 1c For each of the following molecules, draw a three-dimensional structure showing the positions of the bond electron pairs and lone electron pairs (if any). In each case, state the molecular geometry and whether the molecule possesses a non-zero dipole moment. i BF 3 2 B F F F 1 mark trigonal / triangular planar ½ mark no dipole moment ½ mark ii ClF 3 2 Cl F F F 1 mark T-shaped ½ mark Possesses a net dipole ½ mark [0 mark if shape of ClF 3 is described as triangular planar] 95 1A 2 e i ii 2e For each of the following species, draw a three-dimensional structure showing the bond electron pairs and lone electron pairs of the central atom. State the shape of the species in each case. i ICl 4 - 1½ I Cl Cl Cl Cl - (1 mark, no charge -½ mark, no lone pair -½ mark) Square planar ½ mark ii SCl 2 S Cl Cl 1 mark V-shaped / bent / angular ½ mark 1½ 96 1A 2 b i ii I. Shape of molecule Unit 2 Page 6 2b For each of the following chemical species, draw a three-dimensional structure showing the bond electron pairs and lone electron pair(s) of the central atom underlined. State the shape of the species in each case. i ClO 3 - 1½ - Cl O O O pyramidal 1½ mark (Deduct ½ mark for omitting the negative charge) ii NOF 1½ O N F angular or V-shaped 1½ mark (Deduct ½ mark for missing out the lone pair) (1 mark for 3-D structure; ½ mark for the shape) 97 1A 3 b i ii iii 3b For each of the following sulphur-containing chemical species, state its shape and the oxidation state of sulphur. 3 i H 2 S ii SO 2 iii SO 4 2- 98 1A 3 a i ii iii 3a For each of the nitrogen-containing chemical species below, state its shape and the oxidation state of nitrogen. i NO 2 + 1 ii NH 3 1 iii NO 3 - 1 98 1A 4 a i 4a Alcohol E has the structure CH 3 CH(OH)C 2 H 5 . i Draw a three-dimensional representation of E. 1 98 2A 1 c i 1c i Draw the three-dimensional structure of BF 3 . 4 99 2A 3 c i 3c Consider the hydrides of three Period 3 elements : SiH 4 , PH 3 and H 2 S i For each hydride, draw a three-dimensional structure showing the bond electron pairs and lone electron pair(s), if any, of the central atom. I. Shape of molecule Unit 3 Page 1 Topic I. Shape of molecule Unit 3 Reference Reading Modern Physical Chemistry ELBS pg. 84–86 Physical Chemistry, Fillans pg. 126–143 Organic Chemistry, Solomons, 6th Edition pg. 26–37 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 133–140 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 419 Syllabus 4.3 Hydridization Theory Notes C. Hybridization Theory Actually, the electrons in the molecule are not stationary, they undergo wave like motion. The valence shell electron pair repulsion theory is not good enough to explain this. Therefore, the scientist developed another theory, hybridization theory. Questions that cannot be answered by VSEPRT 1. electrons are not stationary 2. the strength of the 2 bonds in a double bond are not the same. 3. No. of unpaired e - available for bond formation. 4. shape of some molecules 5. strength of some bonds 1. Overlapping of atomic orbital Instead of fixed electron centre, covalent bond is now viewed as an overlapping of atomic orbitals. In hydrogen, two hydrogen 1s atomic orbitals are overlapped to form a covalent bond. The two 1s orbitals interfere with each other constructively. This causes a higher electron density along the internuclear axis and create the attraction to bring the two atoms together. There are two kinds of overlaps i. head-on overlap ii. sideway overlap Head-on overlap (End to End overlap) of the orbital give σ (sigma) bond. Sideway overlap (Laterally overlap) of the orbital give π (pi) bond. Since the bond strength of a covalent bond is directly proportional to the area of overlap, in most of the case, the σ bond will be stronger than the π bond. Also, the π electrons, unlike σ electrons, are not concentrated along the internuclear axis. In large atoms, π bonds are not favorable because, being removed from the line between the centres of the atoms, the π bond rapidly weakens as the size of the atom increases. e.g. C=C double exists but Si=Si double bond doesn't exist. I. Shape of molecule Unit 3 Page 2 σ bond and π bond can be distinguished by the shape of their cross section. σ bond has a cylindrical cross section about the bond axis but π bond does not has a cylindrical cross section. Orbital representation of one σ bond and two π bonds Cross section ` 2. Hybridization theory Hybridization means mixing. Hybridization theory addresses many questions that cannot be explained by Valence Shell Electron Pair Repulsion Theory. i. electrons are not stationery, they undergo wave like motion. ii. The strength of C=C (612 kJmol -1 ) is not twice of the strength of C–C (347 kJmol -1 ) bond. i.e. the two bonds in double bond are not equvalent. iii. There are not enough unpaired electrons in a C atom in ground state (1s 2 2s 2 2p 2 ) for the formation of 4 single bonds. iv. The four single bonds in CH 4 are equivalent and the shape is tetrahedral. Consider CH 4 molecule, experimental findings tell us that the 4 C–H bonds are equivalent and the molecule is tetrahedral A C atom in ground state has only 2 unpaired electron in 2 different 2p orbitals which can only form 2 single bond. 1s 2s 2p ground state C 2 2 110 If one electron is promoted from 2s orbital to 2p orbital, you will get 4 unpaired electrons for bond formation. However, if the 4 unpaired electrons overlap with 1s electron of the 4 H atoms independently, the bond formed will not have equivalent strength. This is because the 4 unpaired electrons are not equivalent. One is a 2s electron and 3 are 2p electrons. Furthermore, the bond angle will not be 109½º any more. The angle between different p orbitals are only 90º. 1s 2s 2p excited state C * 2 1 111 Therefore, scientist proposed that the 4 unpaired electrons interfere with each other first (i.e. mixed with each other) to form 4 equivalent orbitals before any bond is formed. Because the new orbitals are formed from one s orbital and three p orbital. They are called sp 3 hybridized atomic orbitals or sp 3 hybrid orbitals. 1s sp 3 hybridized C * 2 1111 It can be seen that the main purpose of developing hybridization theory, or any new theory, is to explain the experiemental findings. I. Shape of molecule Unit 3 Page 3 The process of combining the orbitals is called hybridization. Note : Hybridized atomic orbitals = atomic orbitals ∑ no. of hybridized atomic orbitals = no. of atomic orbitals used 4 s -o rb i ta l hybridization excited state hybridized atomic orbital overlapping of orbitals (bond formation) ground state (more stable product) promotion ∆H In some of the hybridization processes, electron is promoted to the higher orbital first. e.g. from 2s to 2p orbital in C. This requires energy. Since bond formed will be stronger with hybridization, this will be paid off by the energy released during bond formation. Since hybridization is just a kind of interference, only the orbitals with similar energy can be hybridized. i.e. 2s orbital cannot hybridize with 3s orbital Hybridized orbital No. of electron pairs (bond & lone pairs) Geometrical arrangement Typical examples sp 2 Linear BeCl 2 , C 2 H 2 , CO 2 sp 2 3 Trigonal planar BeF 3 , C 2 H 4 sp 3 4 Tetrahedral CH 4 , NH 3 , H 2 O sp 3 d or dsp 3 (not required) 5 Trigonal bipyramidal PCl 5 sp 3 d 2 or d 2 sp 3 (not required) 6 Octahedral Fe(CN) 6 3+ ,SF 6 * Table needed to be remembered. 3. Examples a) sp hybridization e.g. C 2 H 2 (linear) At first, one 2s electron is promoted to 2p orbital. One 2s orbital and one 2p orbital undergo hybridization to form two hybridized orbitals. sp hybridized orbitals consists of two lobes with linear geometry. 1s 2s 2p ground state C 2 2 110 1s 2s 2p promotion → excited state C * 2 1 111 1s sp 2p hybridization → hybridized C * 2 11 11 One sp hybridized orbital overlaps with the 1s orbital of H to form a C–H σ bond. Another sp hybridized orbital overlaps with the sp hybridized orbital of another C to form a C–C σ bond. The two p orbitals on one C overlap with the two p orbitals of another C laterally to form 2 C–C π bond. It can be seen that not all the three C–C bond are equivalent. Because of the cylindrical symmetry of the σ bond, the C–H bond is free to rotate. Rotation of the C≡C bond will break the two π bonds, therefore C≡C bond is not free to rotate. Orbital representation of ethyne (H–C≡C–H) I. Shape of molecule Unit 3 Page 4 b) sp 2 hybridization e.g. BF 3 (trigonal planar) 1s 2s 2p 1s 2s 2p ground state B 2 2 100 promotion → excited state B* 2 1 110 1s sp 2 2p hybridization → hybridized B* 2 111 0 The three sp 2 hybridized orbitals overlap with 2p orbitals of the three fluorine to form 3 σ bonds. sp 2 hybridized orbitals consists of three lobes with trigonal planar geometry. On B, there is still an empty 2p orbital and it does not fulfill octet. BF 3 is an electron deficiency species and ready to accept electron from other molecule. one 2s and two 2p orbitals sp 2 hybridized orbitals e.g. C 2 H 4 (trigonal planar) 1s 2s 2p 1s 2s 2p ground state C 2 2 110 promotion → excited state C * 2 1 111 1s sp 2 2p hybridization → hybridized C * 2 111 1 Unlike BF 3 , the unhybridized p orbital of the C at right angle to the plane has an unpaired electron. The p orbitals from the two C overlap laterally to form a π bond. Since the π bond is not free to rotate, ethene has a planar structure. I. Shape of molecule Unit 3 Page 5 c) sp 3 hybridization e.g. CH 4 (tetrahedral) 1s 2s 2p 1s 2s 2p ground state C 2 2 110 promotion → excited state C * 2 1 111 1s sp 3 hybridization → hybridized C * 2 1111 sp 3 hybridized orbitals consists of four lobes with tetrahedral geometry directed to the four vertices of a tetrahedron. CH 4 e.g. NH 3 (trigonal pyramidal) 1s 2s 2p 1s sp 3 ground state N 2 2 111 hybridization → hybridized N 2 1112 One of the sp 3 hybridized orbital is occupied by an electron pair. This is the lone pair electrons of the molecule. NH 4 e.g. H 2 O (angular) 1s 2s 2p 1s sp 3 ground state O 2 2 211 hybridization → hybridized O 2 1122 In CH 4 molecule, being symmetrical, adopts the bond angle of regular tetrahedron 109º29'. In NH 3 molecule, the bond angle is decreased by 2.2º (106º45'). In H 2 O molecule, the bond angle is decreased by 5.0º (104º31') from the tetrahedral value. The tend can be predicted by the valence shell electron pair repulsion theory which state that the lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion and a lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion. H 2 O It can be observed that the bond angle is related to the percentage of s character in the hybridized orbital. sp – 180º (50% s character), sp 2 – 120º (33% s character), sp 3 – 109½º (25% s character) The higher the s character, the larger will be the bond angle. I. Shape of molecule Unit 3 Page 6 d) sp 3 d hybridization (not required in A-Level) e.g. PF 5 (trigonal bipyramidal) 3s 3p 3d 3s 3p 3d P [Ne] 2 111 00000 promotion → P * [Ne] 1 111 10000 sp 3 d 3d hybridization → P * [Ne] 11111 0000 The hybridization process involves the promotion of an electron from 3s orbital to 3d orbital. The central atom, P, does not obey octet rule, this is known as expansion of the octet. This is impossible for elements in period 2, as there is no low lying energy d orbital available. i.e. Energy required to promote 2p electrons to 3s orbital will be large that cannot be paid off in the bond formation process. e) sp 3 d 2 hybridization (not required in A-Level) e.g. SF 6 (octahedral) [Ne] 3s 2 3p 4 3s 3p 3d 3s 3p 3d S [Ne] 2 211 00000 promotion → S * [Ne] 1 111 11000 sp 3 d 2 3d hybridization → S * [Ne] 111111 000 The six sp 3 d 2 hybrid orbitals are identical. They overlap with the p orbitals of the six fluorine atoms to form a perfect octahedral. e.g. XeF 4 (square planar) [Kr] 4d 10 5s 2 5p 6 5s 5p 5d 5s 5p 5d Xe [Kr] 4d 10 2 222 00000 promotion → Xe [Kr] 4d 10 2 211 11000 sp 3 d 2 5d hybridization → Xe [Kr] 4d 10 221111 000 Not all noble gases can form compound. As the principle quantum number increases, the energy gap between the orbitals gets smaller. For heavy noble gas atom, electron from lower energy level may be promoted to higher energy level to yield unpaired electrons for bonding. Xe F F F F I. Shape of molecule Unit 3 Page 7 Glossary Lewis structure Oxidation no. formal charge Valence shell electron pair repulsion theory octet expansion Hybridization theory σ(sigma) bond π(pi) bond ground state excited state I. Shape of molecule Unit 3 Page 8 Past Paper Question 90 2B 4 d 90 2B 6 a i iii v 94 2B 4 c 95 1A 2 a 98 2B 5 a i 90 2B 4 d 4 Account for the following observations. 4d Oxygen usually forms compounds with oxidation numbers -2, -1 and 0, whereas sulphur can form compounds with many more oxidation numbers ranging from -2 to +6. 3 Sulphur has empty low lying energy 3d orbital - to expand its coordination sphere, thus capable of forming compounds with a wide range of oxidation number. 1 mark Oxygen has a much higher electronegativity than S and therefore it usually acts as an oxidant. 1 mark Oxygen cannot expand its octet/has no low lying energy orbital for bonding. 1 mark 90 2B 6 a i iii v 6a The following species are either impossible to prepare or very unstable. Explain, in each case, why this is so. i NCl 4 1½ Nitrogen cannot have 9e - in its valence shell and cannot expand its octet because N has no low lying energy orbitals for bonding. 1½ mark iii ArCl 2 1½ Ar does not form compounds because of the stability of close-shell electronic configuration. 1½ mark v [PO 4 ] 2- 1½ P cannot have oxidation state higher than +5. 1½ mark 94 2B 4 c 4c What is the highest oxidation state of iodine? Give an example of an iodine-containing compound in which iodine is in this oxidation state. 2 Hightest O.S. of I = +7 1 mark Example : IF 7 , IO 4 - 1 mark 95 1A 2 a 2a Explain why phosphorus can form PCl 3 and PCl 5 , while nitrogen can form only NCl 3 . 2 P has low-lying empty 3d orbital / energy level, promotion of e - to 3d, gives 5 unpaired electrons. It is capable to form 5 covalent bond. 1½ mark In N, there is no low-lying d orbitals, excitation of e - is not possible. ∴ It can form only 3 covalent bonds. ½ mark 98 2B 5 a i 5a Consider the following compound F. CH CH CH 3 C CH a b c d F 5 i Give the hybridization states of the carbon atoms, a, b, c and d. I. Shape of molecule Unit 4 Page 1 Topic I. Shape of molecule Unit 4 Reference Reading 4.4 Principles of Organic Chemistry, Peter R.S. Murray, 2 nd Edition pg. 14–23, 33–34, 122–129 Organic Chemistry, Solomons, 6 th Edition pg. 52–59, 61–65 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 110–111, 141–143 Syllabus The unique nature of carbon Structure and Shape of hydrocarbons Notes 4. Structure and shape of hydrocarbons a) Characteristic of Carbon–Carbon bond i. Bond strength Comparing with other single bond, C–C bond is the one with the highest bond strength. Therefore, it is capable to form a stable long chain molecule. This characteristic is called catenation of C atom. Bond Bond energy C–C +346 kJmol -1 N–N +163 kJmol -1 F–F +158 kJmol -1 Cl–Cl +242 kJmol -1 Br–Br +193 kJmol -1 I–I +151 kJmol -1 ii. Ability to form multiple bond Furthermore, C is capable to form relative stable double bond and triple bond. Bond Bond energy / kJmol -1 C–C +346 C=C +610 C≡C +837 This make a lot of different kinds of connection possible in the organic molecule. There is over millions of different organic compounds identified and may be billions not identified. b) Shape of hydrocarbons (1) Saturated hydrocarbons In saturated hydrocarbons, all C atoms are tetrahedrally bonded. The arrangement can be viewed as the repulsion of the four bonding electron pairs or the result of sp 3 hybridization. Furthermore, since σ bond has cylindrical (circular) cross section and the area doesn't change when the bond rotates, all σ bond are free to rotate. Shape of methane Shape of ethane I. Shape of molecule Unit 4 Page 2 (2) Unsaturated hydrocarbons Rigidity of double and triple bond – Rotation of the C=C and C≡C bonds will break the one π bond and two π bonds respectively, therefore C=C and C≡C bond are not free to rotate. Shape of ethene (H 2 C=CH 2 ) Shape of ethyne (H–C≡C–H) (3) Aromatic hydrocarbons By X-ray diffraction, it was found that the six C–C bonds in benzene are equivalent and with the bond length intermediate between single C–C bond and double C=C bond. Kekulé structure Type of bond Bond length / nm Single C–C 0.154 Benzene C–C 0.139 Double C=C 0.134 Triple C≡C 0.120 According to energetic studies, the actual benzene is more stable that the Kekulé structure. This is due to the delocalization of the π electrons / resonance of π electrons (a) Delocalization of π-electrons 2s 2p C ground state [He] 2 110 2s 2p promotion → C excited state [He] 1 111 sp 2 p hybridization → C sp 2 hybrid atomic orbital [He] 111 1 The six carbon atoms undergo sp 2 hybridization. One lobe of each C sp 2 hybrid orbital overlaps with lobe of another C to form C–C σ bond. Another lobe overlap with 1s orbital of H to form C–H σ bond. The p orbitals overlap above and below the plane laterally to form an extensive π orbital. This gives benzene a planar structure which guarantees maximium p orbital overlapping. I. Shape of molecule Unit 4 Page 3 (b) Stability of benzene Chemical properties of benzene, hexane, cyclohexane and cyclohexene H H H H H H H H H H H H H H H H H H H H H H H H H H H H Benzene Hexane Cyclohexane Cyclohexene Reaction Benzene Hexane Cyclohexane Cyclohexene bromine in CCl 4 (in dark) No reaction. Br 2 decolorizes slowly only in the presence of Fe catalyst with evolution of HBr. (Electrophilic substitution) No reaction. Upon heating, Br 2 decolorizes slowly with evolution of HBr. (Radical substitution) Similar to hexane. (Radical substitution) Decolorize immediately. (Electrophilic addition) acidified potassium manganate(VII) KMnO 4(aq) /H + (aq) No reaction. No reaction. No reaction. Purple manganate(VII) decolorize immediately. (Oxidation) neutral potassium manganate(VII) KMnO 4(aq) No reaction. No reaction. No reaction. Purple manganate(VII) turns to black ppt. of manganese(VI) oxide, MnO 2(s) . (Oxidation) alkaline potassium manganate(VII) KMnO 4(aq) /OH - (aq) No reaction. No reaction. No reaction. Purple manganate(VII) turns to green manganate(VI), MnO 4 2- (aq) , immediately. (Oxidation) The chemical properties of benzene is so different from ordinary alkene e.g.(cyclohexene). It resembles the properties of alkane instead. This is due to the extra stability from delocalization of π electrons. For example, its reaction with bromine is a substitution reaction (electrophilic substitution) instead of addition reaction (electrophilic addition) in other typical alkenes. This will be explained with more details in the section of reaction mechanism. Glossary catenation electrophilic substitution electrophilic addition reaction mechanism I. Shape of molecule Unit 4 Page 4 Past Paper Question 93 1A 3 a i 94 2C 9 b i 96 1A 3 c i ii 97 1A 4 c I 99 2A 3 b 93 1A 3 a i 3a Consider the following compound, X: i Give the hybridization state of the carbon atoms indicated by a, b and c. 1½ a: sp 3 b: sp c: sp 2 ½ mark each 94 2C 9 b i 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem. i Describe the bonding and shape of the ethene molecule in terms of the type and spatial arrangement of the orbitals involved. 3 There are 4 C–H bonds and 1 C=C bond in an ethene molecule. The C’s in ethene are sp 2 hybdrized The C–H bond is formed by overlapping of a sp 2 hybrid orbital with an atomic orbital of H. 1 mark The C=C bond consisted of a s bond and a p bond. These bonds are formed by the overlapping of two sp 2 hybride orbitals and of the two unhybridized p orbitals on C’s respectively. 1 mark C C H H H H π bond σ bond Since for sp2 hybridization, the hybrid orbitals are arranged on a triangular plane. ∴ Ethene molecule is planar and the C–C–H bond angles are 120º 1 mark 96 1A 3 c i ii 3c Consider the following structure : i Give the hybridization states of the carbon atoms a, b and c. 1½ a: sp 3 , b sp, c: sp 2 3 × ½ mark ii Suggest expected values for the bond angles α, β and γ. 1½ a : 108-110º, b : 180º, g : 119-121º 3 × ½ mark (Deduct ½ mark for missing out the degree notation) 97 1A 4 c i 4c i Give the hybridization state of the carbon atoms and the bond angles in ethene. 3 99 2A 3 b 3b Briefly describe the bonding in benzene and give TWO pieces of supportive evidence. I. Shape of molecule Unit 5 Page 1 Topic I. Shape of molecule Unit 5 Reference Reading Modern Physical Chemistry ELBS pg. 84–86 Organic Chemistry, Solomons, 6th Edition pg. 23–26 Syllabus 4.5 Molecular Orbital Theory Notes D. Molecular Orbital Theory (Not required in A-Level) According to hybridization theory, all electrons in O 2 should be paired up. 1s 2s 2p 1s sp 2 2p ground state O 2 2 211 hybridization → hybridized O 2 221 1 O undergoes sp 2 hybridization. The hybridized sp 2 orbital of the two O overlaps with each other to form a σ bond. The unhybridized 2p orbitals of the overlaps laterally to form a π bond. This approach fails to explain why liquid O 2 is paramagnetic, a phenomenon arisen from unpaired electron. Scientist developed another theory called molecular orbital theory to explain this. For example, in H 2 molecule, the two 1s orbitals can interfere constructively to create a bonding molecular orbital or destructively to create a anti-bonding molecular orbital. In bonding molecular orbital, the electron density is concentrated along the internuclear axis, thus tends to bring the two hydrogen atoms together. In anti-bonding molecular orbital, the electron density is concentrated on the two sides of the molecule, also there is a node(an area with zero electron density) between the two hydrogen atoms. This tends to break the two hydrogen atoms apart. As a result, bonding orbital has lower energy than anti- bonding orbital. Similar to Aufbau principle in building up of electrons, the molecular orbital with lower energy will be filled first. Each molecular orbital can accommodate 2 electrons. Electron pair in bonding orbital will have opposite spin and electron pair in anti-bonding orbital will have similar spin. N.B. atomic orbital = molecular orbital ∑ no. of atomic orbital = no. of molecular orbital I. Shape of molecule Unit 5 Page 2 Molecular orbitals of oxygen When the two oxygen atoms is approaching each other, the atomic orbitals start to overlap. 1s orbital is not involved in the overlapping since it is inner orbital. The overlappings of 2s, 2p x , 2p y and 2p z give rise to 6 different energy levels. The 12 electrons fill up the molecular orbitals according to the energy. 2 unpaired electron can be found in the 2 degenerate anti-bonding π * 2p molecular orbital. This explain why oxygen is paramagnetic. No. of electrons in bonding orbital = 10 No. of electrons in anti-bonding orbital = 6 Bond order = (no. of e - in bonding orbital - no. of e - in anti-bonding orbital) ÷ 2 = (10 - 6) ÷ 2 = 2 Glossary Molecular orbital theory paramagnetism Past Paper Question II. Ionic bonding Page 1 Topic II. Ionic bonding Reference Reading 4.7 Modern Physical Chemistry ELBS pg. 64–66 Chemistry in Context, 3rd Edition ELBS pg. 133, 192–195 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 99–104, 145–146, 149–151 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 83–86, 115–118, 125–127, 171–173 Syllabus Strength of ionic bond X-ray diffraction Atomic and ionic size Notes II. Ionic bonding Like all other kinds of bonding, ionic bond is a result of balance between attraction and repulsion. There are electrostatic attractions among oppositely charged particle and repulsion among similarly charged particle. The ions will come to equilibrium when the attractive and repulsive forces balance one another an the ions will stay in positions in the crystal lattice. The structure of ionic compound is called giant ionic structure or giant ionic lattice. II. Ionic bonding Page 2 Different ionic lattices CsCl - Simple cubic NaCl - Face-centered cubic (Octahedral) r + r - = 0.933 r + r - = 0.524 Two factors govern the packing of ions to form giant lattice: i. Each ion tends to have the highest no. of neighbours of opposite charge, as this would increase the lattice stability. The number of nearest neighbours is called the coordination umber of the central ion. ii.. Relative sizes of the ions. The arrangement of the ions in an ionic crystal is determined by the ionic radii ratio (r + /r - ): Coordination no. Shape r + /r - Example 8 Simple cubic 0.73 and above CsCl 6 Face centered cubic (Octahedral) 0.41 - 0.73 NaCl 4 Face-centered cubic (Tetrahedral) 0.22 - 0.41 ZnS Ionic bonding is non-directional in nature. An ion is surrounded by many oppositely charged ions in all directions, i.e. no particular orientation is favoured. Characteristics of ionic compounds 1. Good electrolytes. 2. Hard solids because the inter-ionic forces within an ionic crystal are usually strong. 3. High melting point. 4. Soluble in water and insoluble in organic solvents. A. Strength of ionic bond - lattice energy By definition, mM n+ (g) + nX m- (g) → M m X n(s) ∆Ho lat Lattice energy can be served as a good estimate for the strength of the ionic bond. It has a typical value range from 600–1000 kJmol -1 . II. Ionic bonding Page 3 B. X-ray diffraction The structure of an ionic crystal can be determined by a technique called X-ray diffraction. When X-rays strike a crystal, they are diffracted by the electrons in the atoms or ions. The larger the atom, the more electrons it possesses and the brighter the spot will be on the diffraction pattern. From the pattern of spots on the X-ray film, the distribution of electron in the crystal can be calculated. Points of equal density in the crystal are joined by contours giving an electron density map. 1. Electron density map of sodium chloride From the electron density map, it can be seen that electrons are concentrated near the nuclei in ionic compounds. In NaCl, ions are not perfectly spherical but tend to expand to fill up any empty space present, until their charge clouds experience repulsion from that of the neighbouring ions. The electron density between the ions fall almost to zero. Electron density map of sodium chloride C. Periodicity of ionic radius 1. Definition of ionic radius Ionic radius – the effective radius of ions in crystal lattices. It is the distance measured from the centre of the ion to the region where the electron density is zero. 2. Periodicity of ionic radius Similar to atomic radius, the size of an ion is mainly governed by the strength of the effective nuclear charge. On top of this, it is also affected by the presence of other ions. If an ion is present in a surrounding with few ions around, its electron cloud would tend to expand to fill up the space, until it is repelled by the electron clouds of other ions Size of cation It is always smaller than the size of the corresponding atom as electron is removed. The proton to electron ratio increases. The nuclear charge will have a greater effect on the fewer remaining electrons, which makes the electron cloud contracts. Size of anion It is always bigger than the size of the corresponding atom as electron is added. The additional electron enters the highest energy level without an increase in nuclear charge. This causes greater repulsion among the electrons. The above discussion applies to simple ion only. Certainly, a polyatomic ion is much larger than a simple ion because it contains a larger no. of atoms. II. Ionic bonding Page 4 3. Size of isoelectronic particles O 2- ion, F - ion, Na + ion, Mg 2+ ion are said to be isoelectronic because they contain the same no. of electron and have the same electronic configuration, – 2,8. Therefore, their sizes show the order of O 2- > F - > Na + > Mg 2+ . Glossary electrovalent bond coordination no. X-ray diffraction electron density map ionic radius isoelectronic Past Paper Question 92 1A 3 b 95 2B 5 a ii iii 97 1A 1 c 92 1A 3 b 3b The size of three anions is in the order: Br - < H - < I - . Briefly explain why H - is smaller than I - , but larger than Br - . 2 I - has more e - shells ([Kr]5s 2 5p 6 ) than H - (1s 2 ). 1 mark H - is large due to electron repulsion between 2 electrons in 1s orbital. 1 mark 95 2B 5 a ii iii 5a The table below lists some properties of the alkali metals. Element Atomic radius / nm Ionic radius / nm First ionization energy / kJmol -1 Standard electrode potential / V Melting point / ºC Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 ii Explain why the atomic radius is larger than the ionic radius in each element and why the ratio, atomic radius : ionic radius is greatest in the case of Li. 3 The atom has one more electron shell than the cation. 1 mark In M + , the nuclear charge outweighs the screening effect of the electrons and the electron cloud of M + experiences a stronger attraction. Therefore, the ionic radii are smaller than the atomic radii. 1 marks Atomic radius : ionic radius ratio is highest in the case of Li, because it has the smallest size. In Li + , the e - experience the strongest attraction ∴ contraction in size is greatest. 1 mark iii Explain why the ionic radius of K + is larger than that of Ca 2+ , although they have the same electronic configuration. 2 Electronic configuration of K + and Ca 2+ are both 1s 2 2s 2 2p 6 3s 2 3p 6 . 1 mark Both number and arrangement of electrons are the same in two cations. Ca 2+ is doubly charged, while K + is singly charged, the outermost e - in Ca 2+ experiences a stronger attraction (effective nuclear charge). Hence the radius of Ca 2+ is smaller. 1 mark 97 1A 1 c 1c Arrange, with explanation, the following chemical species in the order of decreasing size. F, O and O - 3 III. Covalent bonding Page 1 Topic III. Covalent bonding Reference Reading 4.6.0–4.6.2 Modern Physical Chemistry ELBS pg. 78–82, 112–113 Chemistry in Context, 3rd Edition ELBS pg. 101–105 Organic Chemistry, Solomons, 5th Edition pg. 15–18, 506–509 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 114–115 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 86–90, 92–94 Syllabus H 2 + ion Additivity of covalent radius Breaking down of additivity due to resonance Dative covalent bond Notes III. Covalent bonding A. H 2 + ion For simplicity, H 2 + ion can be used as the simplest model of covalent bond. In H 2 + ion, only 1 electrons and 2 protons need to be considered. There are only 3 forces involved. 1 repulsion term between the 2 protons and 2 attraction terms between the electron and 2 protons. The bond length (equilibrium distance between two nuclei) depends on the balance of the repulsion and the attraction. Electron density map for the H 2 + ion The two positive nuclei are bond together by sharing the negative electron cloud. This arrangement leads to a lower potential energy than if the electron cloud were not shared. And a lower P.E. results in greater stability. Electron density is not uniform on the electron density map, but is greater along the internuclear axis. The high concentration of electron between two positive nuclei serves to bind the nuclei together - covalent bond. The attraction is even greater with the neutral H 2 molecule, where two electrons are available for bonding. This is revealed in the decrease in the internuclear distance and increase in bond dissociation energy. Internuclear distance (nm) Bond Dissociation energy (kJmol -1 ) H 2 + 0.104 257 H 2 0.074 435 III. Covalent bonding Page 2 B. Electron diffraction The electron density map of hydrogen cannot be obtained by X-ray diffraction / X-ray crystallography because hydrogen is volatile and cannot be crystallized easily. And also, the electron density of the hydrogen atom is too low to diffract the X-ray. Another similar method called electron diffraction is used to determine the structure of a gaseous molecule. When a beam of electrons is passed through a gas or vapour at low pressure. The nuclei of atoms in gas molecule scatter (diffract) the electrons. From the diffraction pattern, the distance between the nuclei and the structure of the molecule can be determined. C. Covalent radius 1 Definition of covalent radius Covalent radius – half the distance between two atoms of the same kind held together by covalent bond. 2. Additivity of covalent radius If there is a molecule A–B, the interatomic distance A–B is approximately equal to the arithmetic mean of the A–A and B–B distance. A B A A B B − = − − + 2 and because of this simple relationship, it is possible to obtain interatomic distance of certain compound by simple addition of covalent radii. e.g. Average bond length of C–C bond : 0.154 nm Bond length of H–H bond in H 2 : 0.074 nm By calculation, bond length of C–H bond = 0.154 nm + 0.074 nm 2 = 0.114 nm Average bond length of C–H bond determined by experiment : 0.108 nm The two values rather agree with each other. 3. Breaking down of additivity in covalent radius and bond energy This simple additive relationship of covalent radii was found to fail for certain molecules. e.g. benzene. The reason is the molecules are actually resonance hybrid of two or more possible structures. There are compounds whose bond length differ from calculated values. e.g. benzene and nitrate ion. Compound Bond lengths (nm) Calculated Experimental benzene C=C 0.134 nm C–C 0.154 nm 0.139 nm NO 3 − N–O 0.136 nm N=O 0.115 nm 0.121 nm All bond lengths are found to be equivalent. This is due to the delocalization of electrons / reasonance of electrons. III. Covalent bonding Page 3 a) Resonance / Delocalization of electrons (1) Resonance structure of benzene In benzene molecule, six carbon atoms are joined together in a ring by overlap of carbon sp 2 hybrid orbitals, forming C–C σ bonds. C–H σ bonds are formed by further overlap with hydrogen 1s orbitals. the p-orbitals of each carbon atoms overlap with each other to formal a new molecular orbital. The electrons are delocalized in a system of π orbitals above and below the ring. This make all the six C–C bonds equivalent. Delocalization of electron is also called resonance of electrons. Two Lewis structures are used to represent the real structure. Resonance structures of benzene Representation of hybrid or Since one π bond is shared between two C–C bonds, the bond order of the C C bond is said to be 1½ . (2) Resonance structure of nitrate ion Usually, formal charge is included in the resonance structure and the hybrid structure to see the distribution of the electron. Since one π bond is shared among three N–O bonds, the bond order of the N O bond is said to be 12 . N.B. Resonance structures are only representations on paper. They do not exist. The real structure of the molecule cannot be represented by any individual resonance structure. Furthermore, the electrons are not shifting forth and back in the structure. Instead, they are evenly distributed in the structure. Orbital representation is always the better way to represent the actual structure. N O O O sp 2 sp 2 sp 2 sp 2 (3) Rules in writing resonance structures i. Resonance structures exist only on paper. ii. In writing resonance structures, only electrons are allowed to move. iii. The actual molecule or ion will be better represented by a hybrid of these structures. iv. The energy of the actual molecule is lower than the energy that might be estimated for any contributing structure. III. Covalent bonding Page 4 b) Breaking down of additivity in bond enthalpies The presence of delocalization of π electrons is further confirmed by the breaking down of addivitiy in bond enthalpies. C C C C C C H H H H H H or H H H H H H or Kekulé structure of benzene German scientist Kekulé proposed that benzene is a ring compound with alternating C–C single bond and C=C double bond joining the carbon atoms together. To verify the validity of the structure, some reactions of benzene are studied. e.g hydrogenation Hydrogenation of benzene 3H 2(g) + H H H H H H H H H H H H H H H H H H (l) Benzene Cyclohexane Estimation of enthalpy of hydrogenation by bond energy terms Bond broken Bond formed 3 C=C 3 C–C 3 H–H 6 C–H E(C–H) +413 kJmol -1 E(H–H) +436 kJmol -1 E(C–C) +347 kJmol -1 E(C=C) +612 kJmol -1 ∆H = energy required - energy released = ΣE(bond broken) - ΣE(bond formed) = (3 × E(C=C) + 3 × E(H–H)) - (3 × E(C–C) + 6 × E(C–H)) = (3 × 612 + 3 × 436) - (3 × 347 + 6 × 413) = -375 kJmol -1 (experimental value -210 kJmol -1 ) The calculated value is so differed from the experimental value, less heat is released than the calculated value. The difference between these values indicates that the actual structure of benzene is more stable than Kekulé structure by (-210) - (-375) = 165 kJmol -1 III. Covalent bonding Page 5 (1) Delocalization energy This extra stability is attributed to the delocalization of the bonding electrons over all 6 carbon atoms. The double bond electrons are not localized at fixed positions, instead, they delocalized around the ring and make the 6 C–C bonds equivalent. This distribution minimizes the repulsion among the electrons and give a more stable structure. H H H H H H or Resonance structure of benzene 165 kJmol -1 (delocalization energy) -210 kJmol -1 -375 kJmol -1 H The breakdown of the additivity rule demonstrates the importance of bond energy terms in understanding the chemical bonding in benzene. III. Covalent bonding Page 6 D. Dative covalent bond Dative covalent bond – The linkage of two atoms by a pair of electrons, both electrons being provided by one of the atoms. The dative covalent bond is also known as coordinate bond. The formation of a dative bond requires an unbonded(lone) pair of electrons in one atom (donor) and an empty orbital in another atom (acceptor). 1 Examples of H 3 N→BF 3 and Al 2 Cl 6 a) H 3 N→BF 3 Boron trifluoride does not possess an octet of electrons; this is a suitable compound with which to introduce dative covalency. BF 3 gas reacts with NH 3 gas to give a white solid with the composition NH 3 BF 3 . 1s 2s 2p 1s 2s 2p ground state B 2 2 100 promotion → excited state B* 2 1 110 1s sp 2 2p hybridization → hybridized B* 2 111 0 B undergoes sp 2 hybridization and uses the three sp 2 hybridized orbitals to overlap with 2p orbitals of fluorine. BF 3 has a planar structure and a vacant 2p orbital. The 2p orbital accepts the lone pair from NH 3 and the hybridization of B changes from sp 2 to sp 3 . b) Al 2 Cl 6 In gaseous state, two AlCl 3(g) molecule dimerizes to form Al 2 Cl 6(g) molecule. 3s 3p 3s 3p Al [Ne] 2 100 promotion → Al * [Ne] 1 110 sp 2 3p hybridization → hybridized Al * [Ne] 111 0 AlCl 3(g) has a planar structure similar to BF 3(g) . The vacant 3p orbital accepts lone pair electrons on chlorine atom to form dative bond. Al Cl Cl Cl Cl Al Cl Cl Glossary Octet rule electron diffraction additivity of covalent radius resonance hybrid structure delocalization delocalization energy dative covalent bond / coordinate bond III. Covalent bonding Page 7 Past Paper Question 92 2A 3 b iii 92 2B 5 a iii 95 1A 1 c 96 2A 1 a i ii 98 2A 1 c ii 92 2A 3 b iii 3b iii The covalent radius of carbon is 0.077 nm. The measured carbon-carbon bond length in benzene is 0.139 nm. Estimate the carbon-carbon bond length in ethane. Explain any difference in the carbon-carbon bond lengths in these two molecules. 3 C–C bond length of CH 3 –CH 3 = 2 × 0.077 = 0.154 nm 1 mark C–C bond in ethane is a single bond which is longer than the carbon-carbon bond in benzene. In benzene, it is because of delocalization electrons in the ring. The average bond order is larger than 1. 1 + 1 mark 92 2B 5 a iii 5a iii Draw resonance electronic structures of CO and CO 2 , showing lone pair electrons and charges where appropriate. 2 C O C O C O Any 2 structures C O O C O O C O O Any 2 structures ½ mark each C Very few candidates were able to draw resonance structures for CO and CO 2 . The charges on atoms in these structures are neglected in many textbooks, but the students should be able to deduce these by "book-keeping" of the lone pair and bond pair electrons. 95 1A 1 c 1c Account for the fact that the carbon-oxygen bond lengths in CO, CO 2 and CO 3 2- are 0.113, 0.116 and 0.129 nm respectively. 3 CO C O - + ½ mark Bond order = 3 / the bond formed between C & O is triple bond. It is the shortest. ½ mark CO 2 C O O ½ mark Bond order = 2 / the bond formed between C & O is double bond. ½ mark CO 3 2- It is a resonance hybrid of the following : C O O O - - C O O O - - C O O O - - ½ mark Bond order = 1 1 3 (intermediate between C–C and C=C bond). It is the longest. ½ mark C This question was not well answered. The answer required descriptions or diagrams of the electronic structures of the three compounds and the deduction of the bond orders. It is more meaningful to show lone pairs of electrons, and formal charges in these electronic structure diagrams. Many candidates could not give the correct structure for CO, and few described the resonance of the CO 3 2- structures. III. Covalent bonding Page 8 96 2A 1 a i ii 1a For the hydrogenation of buta-1,3-diene, H 2 C=CH–CH=CH 2(g) + 2H 2(g) → CH 3 CH 2 CH 2 CH 3(g) the experimental molar enthalpy change is -239 kJmol -1 . i Estimate the molar enthalpy change for the above hydrogenation using the bond energy terms below : Bond Bond energy term / kJmol -1 H–H 436 C–H 413 C–C 346 C=C 611 2 ∆H = -2E C–C - 4E C–H + 2E C–C + 2E H–H 1 mark or = -2(346) - 4(413) + 2(611) + 2(436) (1 mark) = -250 kJmol -1 1 mark (0 marks for omitting the eve sign; deduct ½ mark for no units) ii Explain why the estimated value differs from the experimental value. 2 Buta-1,3-diene is a conjugated system / delocalization of π electrons / resonance / mesomerism occurs in buta-1,3-diene. 1 mark or C C C C H H H H H H (1 mark) or H 2 C CH CH CH 2 H 2 C CH CH CH 2 etc. + - (1 mark) The molecule is thus stabilized. ∴ energy released in hydrogenation is less than that in non-conjugated system. 1 mark (Award 1 mark for the explanation: the bond energy given is only the average bond energy / bond energy of a covalent bond depends on its bonding environment) C Very few candidates related the difference in ∆H values to the delocalization of π electrons in the conjugated system. 98 2A 1 c ii 1c ii BF 3 reacts with NH 3 to form an adduct, BF 3 ·NH 3 . Account for the formation of the adduct and draw its three- dimensional structure. IV. Bonding intermediate between ionic anc covalent Page 1 Topic IV. Bonding intermediate between ionic and covalent Reference Reading 4.8 Chemistry in Context, 3rd Edition ELBS pg. 111–113 Organic Chemistry, Solomons, 6th Edition pg. 38–41 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 111–114 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 96–99, 162 Syllabus Intermediate type of bond Incomplete electron transfer in ionic compound Electronegativity Polarization of ionic bond Polarity in covalent bond Notes IV. Bonding intermediate between ionic and covalent A. Differences between ionic bond and covalent bond The similarity in nature of ionic and covalent bonds is reflected by the same order of magnitude of lattice energies (600-1000 kJmol -1 ) and covalent bond energies (300-1000 kJmol -1 ). i. Distribution of electrons In case of ionic solid, in certain region, the electron density between the ions fall almost to zero. In case of covalent molecule, the electron density is highest between the bonding atoms and decreases outwards. The electron density is concentrated along a line between 2 nuclei. ii. Ionic compounds are formed by complete transfer of electron. Covalent molecules are formed by sharing of electron among the bonding nuclei. iii. In ionic compounds, oppositely charged ions attract electrostatically to one another. In covalent compounds, the positive charged nuclei also attract electrostatically the electrons shared by the atoms involved. Moreover, both of them are electrostatic in nature. B. Incomplete electron transfer in ionic compound 1. Electron density map of LiF comparing with those of NaCl and H 2 NaCl (nearly purely ionic) LiF (intermediate between ionic and covalent) H 2 (purely covalent) All the bonding arise from the interaction between the electrons and the nuclei. Different arrangements of the electrons and atoms will give different kind of bonding. From the electron density map, one can see that electrons are concentrated near the nuclei in ionic compounds while there are substantial electron density between the nuclei of hydrogen atoms. In the LiF, its electron density map shows that it is somewhere between a typical ionic compound and covalent compound. IV. Bonding intermediate between ionic anc covalent Page 2 2. Difference among lattice energies of Na, Ag and Zn compounds Besides the Born-Haber cycle, the lattice energies can also be calculated in terms of electrostatic interactions among ions. The mathematical treatment of the calculation is beyond the scope of the syllabus, but it is worthy to note the basic assumptions and the limitation of the model. Further reading about the calculation of lattice energy can be found on pg. 71–73 of Modern Physical Chemistry ELBS. Assumptions of the mathematical model for calculation: i. The geometrical arrangement of the ions within the crystal and the interionic distances are known. These can be determined by a method called X-ray diffraction. ii. The bonding within the crystal is completely ionic and all ions are spherical in shape, i.e. no sharing of electron. a) Lattice energies of sodium halide, silver halide and Zinc sulphide Experimental lattice Compound Theoretical lattice energy energy from Born-Haber cycle Percentage error NaCl -777 kJmol -1 -781 kJmol -1 0.5 % NaBr -731 kJmol -1 -742 kJmol -1 1.5 % NaI -686 kJmol -1 -699 kJmol -1 1.9 % AgCl -768 kJmol -1 -890 kJmol -1 15.9% AgBr -759 kJmol -1 -877 kJmol -1 15.5% AgI -736 kJmol -1 -867 kJmol -1 17.8% ZnS -3427 kJmol -1 -3615 kJmol -1 5.5% b) Bonding intermediate between covalent and ionic In sodium halide, the theoretical values calculated from the mathematical model close to the values determined by the Born-Haber cycle. This is because the difference in electronegativity between sodium and halogen is very large, thus the electron transfer is almost complete and shows no evidence of electron sharing. This agrees with the basic assumption of the model. The great discrepancy of between the theoretical values and the experimental values of silver halides and zinc sulphide implies that our model for calculation is not realistic. Owing to the small size of the silver ion which polarizes the large halide ion easily, the bond possesses some covalent characters. The experimental values also suggest that the bonds are stronger than the expected. This is also an evidence of the presence of bonding intermediate between covalent and ionic. IV. Bonding intermediate between ionic anc covalent Page 3 The discrepancies implies the presence of an appreciable proportion of covalent bonding. Partial covalent properties is due to incomplete electron transfer in ionic compound and this is called polarization of ions. From spectroscopic studies of the vapours of alkali metal halides, the internuclear distance in these molecule is less than in the corresponding ionic bond. Halide Internuclear separation (nm) Crystal Vapour LiBr 0.275 0.217 LiI 0.300 0.239 This implies a stronger bond than in the crystal. This is caused by a higher concentration of electron density between the nuclei than in the crystal. This arises from distortion of the electron cloud of one ion or both from a spherical distribution. Polarization of an ion representing a transition from ionic bonding to covalent. 3. Polarization of ionic bond When an ionic bond is formed between the cation X + and anion Y - , owing to its charge, the X + ion would tend to attract the outer electrons in the charge cloud of Y - . This causes a displacement of the outer electrons of Y - back to X + , so that the outer electron of X is not completely transferred to Y. The ionic bond formed is said to have some covalent character. a) Fajans’ Rules in polarization of ionic bond i. The polarizing power of a cation is increased as (a) the size is reduced and (b) the charge is increased. ii. The polarizability of anion increases as (a) the size increases and (b) the charge is increased. iii. Covalent character is stronger for cations which do not have a noble gas electron arrangement (d electrons have poor shielding properties). Thus Na + is less polarizing than Cu + . i.e. Ions of transition metal are usually much smaller than the ions of main group metals) IV. Bonding intermediate between ionic anc covalent Page 4 C. Electronegativity 1. Definition of electronegativity Electronegativity – A measure of tendency of an atom in a stable molecule to attract electrons within a bond. In H–Cl molecule, the bonding electrons are much closer to the chlorine than to the hydrogen; the chlorine being the more electronegative atom. 2. Pauling scale of electronegativity There are several method to measure the electronegativity of an element. One of the commonly used method is Pauling scale of electronegativity. Pauling defined the electronegativity of an atom as the power of that atom in a molecule to attract electrons. In contrast with electron affinity, it is a measure of the power of an single gaseous atom to attract electrons. In Pauling scale, 4 is assigned to the most electronegative atom, fluorine. Pauling electronegativity values of some elements (The shaded ones are more electronegative than C) H 2.1 Li 1.0 Be 1.5 B 2.0 C 2.5 N 3.0 O 3.5 F 4.0 Na 0.9 Mg 1.2 Al 1.5 Si 1.8 P 2.1 S 2.5 Cl 3.0 K 0.8 Ca 1.0 _ Ge 1.8 As 2.0 Se 2.4 Br 2.8 Rb 0.8 Sr 1.0 _ Sn 1.8 Sb 1.9 Te 2.1 I 2.5 Cs 0.7 Ba 0.9 In general, electronegativity values increase from left to right across each period and decrease down each group. D. Polarity in covalent bond 1. Deflection of a liquid jet by an electric field A dry glass rod rubbed with polythene develops a positive charge. An ebonite rod rubbed with fur develops a negative charge A number of liquids are tested Liquid showing Liquid showing a marked deflection no or small deflection Water Trichloromethane Tetrachloromethane Propanone Ethoxyethane Nitrobenzene Benzene Cyclohexene Cyclohexane Ethanol IV. Bonding intermediate between ionic anc covalent Page 5 The zero or very small deflection with tetrachloromethane is due to the symmetry of the molecule in which the four dipoles counteract each other exactly. Molecules which do not possess a permanent dipole moment are nevertheless polarised to some extent when placed in an electrical field. 2. Dipole moment If two charges of equal magnitude but opposite in sign, +q and -q, are separated by a distance d, then the system is said to have a dipole moment (µ) of magnitude give by: Dipole moment (µ) = charge × distance between the +ve and -ve centres = q × d This dipole moment is usually expressed in terms of Debye (D). 1 D ≈ 3.34 × 10 -30 Cm. For example, in HCl, Cl attracts electrons more strongly than H does, the dipole moment of the HCl is 1.1 D. Cl is said to exert a negative inductive effect. a) Vector quantity of dipole moment Dipole moment is a vector quantity. For a molecule with more than one polar bond, the dipole moment of the molecule is given by the vector sum of the dipole moments caused by the various polar bonds. In asymmetric molecules (e.g. HCl, CH 3 CH 2 Cl), the center of the positive charge and the center of the negative charge do no coincide and a permanent dipole moment results. b) Polarity of molecule The overall distortion of charge in molecules, which results from unequal sharing of electrons, is known as polarization. It represents the departure of the bond from purely covalent, it introduces some ionic character into the bond. Polarization of ions represent the existence of some covalent character in the ionic bonding. Polarization of a covalent bond represents the existence of some ionic character in the covalent bond. IV. Bonding intermediate between ionic anc covalent Page 6 c) Factors affecting dipole moment The dipole moment of a molecule is affected by i. inductive effect – a substituent effect on an organic compound due to the permanent polarity or polarizability of groups. This is transmitted through σ bond. ii. mesomeric effect / resonance effect – delocalization of electron via π bond. iii. presence of lone pair. (1) Inductive effect (I) This refers to the displacement of electron cloud mainly due to difference in electronegativity. -I effect : electron withdrawing +I effect : electron donating (2) Mesomeric effect / resonance effect (R) This refers to the stabilization due to delocalization of electron through π bond.. -R effect : electron withdrawing +R effect : electron donating For example, in benzenamine, i. N is more electronegative than C and N exerts a -I effect. ii. by resonance, lone pair on N is donated to the benzene ring and N exerts a +R effect. NH 2 NH 2 NH 2 N H H The final dipole moment would be the sum of the two effects. N.B. Benzene ring can serves as a reservoir of electrons in resonance. If it is attached to a +R group, it will behave as a -R group. On another hand, if it is attached to a -R group, it will behave as a +R group. (3) Presence of lone pair Lone pair can be considered as a negative centre and contributes a dipole moment. For example, experimentally, NH 3 has a stronger dipole moment (1.46 D) than NF 3 (0.24D). But 1. The difference in electronegativity between N(3.0) and F(4.0) is greater than that between N(3.0) and H(2.1). 2. The NF 3 has a bond angle smaller than NH 3 has. According to these two factors only, NF 3 should has a stronger dipole moment than NH 3 . But in NF 3 , the lone pair imposes a dipole moment opposing the dipole moment of N–F, this results in a much weaker resultant dipole moment. Furthermore, the direction of the dipole moment cannot be determined because we do not known which is stronger. IV. Bonding intermediate between ionic anc covalent Page 7 Glossary polarization of ion polarizing power polarizability Fajans rules electronegativity Pauling scale dipole moment vector inductive effect mesomeric effect / resonance effect Past Paper Question 97 2A 1 a i ii iii 98 1A 1 c ii 97 2A 1 a i ii iii 1a i Explain the terms 'dipole' and 'dipole moment', using HBr as an example. 7 ii Explain why the dipole moment of HF is greater than that of HI. iii State the effect of an electric field on molecules of the following compounds and explain the effect in terms of dipole moment. Cl Cl Cl Cl 98 1A 1 c ii 1c The theoretical lattice enthalpies of NaCl (s) and AgCl (s) , and the experimental lattice enthalpy of AgCl (s) are given in the table below. Compound Theoretical lattice enthalpy Experimental enthalpy NaCl (s) - 770 ? AgCl (s) - 833 - 905 ii Why is there such a large difference between the theoretical and experimental lattice enthalpies of AgCl (s) ? 1 V. Metallic bonding Page 1 Topic V. Metallic Bonding Reference Reading 4.9 Modern Physical Chemistry ELBS pg. 74–76 Chemistry in Context, 3rd Edition ELBS pg. 137 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 146–147 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 121–123 Syllabus Metallic bonding Electron Sea model Delocalization model Notes V. Metallic bonding A. Electron Sea model of metal The metal atoms are closely packed (high coordination number) to form a crystal lattice. Valence electrons of individual atoms are contributed to form a sea of electrons. The lattice of positive ions is then immersed in the sea of electrons. The attraction arises from the electrostatic attraction among the positive ions, mobile electrons and positive ions. General properties of metal i. Thermal conductivity – the presence of mobile electrons ii. Electrical conductivity – the presence of mobile electrons iii. Malleability and ductility – non-directional electrostatic attraction between nuclei and mobile electrons iv. Shiny lustre – The mobile electron is excited and re-emits the energy in form of lustre. From another point of view, metallic bond may be considered an delocalization of valence electrons from each atom over the entire metallic crystal. (Further reading about band theory of metallic bonding can be found in Modern Physical Chemistry ELBS pg. 158–159) B. Strength of metallic bond Strength of a metallic bond depends on i. packing efficiency of the crystal Transition metal atoms are much smaller than the atoms in Group 1A and 2A. ii. availability of the valence electron Group 1A metal has 1 valence e - . Group 2A metal has 2 valence e - . Transition metal use 3d e - on top of 2 4s e - . iii. effective nuclear charge In general, the metallic bond strength increases with increasing effective nuclear charge on the valence electrons. However, if the effective nuclear charge is too high, this will lower the availability of valence electrons and the bond strength will drop. e.g. Hg has a first I.E. 10.44 V comparing with 5.14 V of Na. V. Metallic bonding Page 2 C. Melting and boiling of metal Upon melting, only small amount of the metallic bonds are broken. Majority of the metallic bond have to be broken upon boiling. This causes a big gap between the melting point and boiling point of metal. e.g. Na (371K/1156K), Fe (1809K/3135K) In molecular substance, most of the attractions among the molecules have been broken upon melting. This results in a smaller gap between the melting point and boiling point. e.g. N 2 (63K/77K), I 2 (387K/458K) D. Strength of ionic bond, covalent bond and metallic bond Estimated by Range (kJmol -1 ) Ionic bond (non-directional) Lattice energy -780 (NaCl) – -3791(MgO) Covalent bond (directional) Bond energy terms 158 (E(F–F)) – 945.4 (E(N≡N)) Metallic bond (non-directional) Atomization energy 107.3 (Na) – 514.2 (V) Ionic bond Covalent Metallic 0 1000 2000 3000 4000 kJmol -1 Glossary malleability ductility lustre Past Paper Question 92 1A 2 d ii iii 98 1A 3 b i ii 92 1A 2 d ii iii 2d ii Describe the bonding in metallic crystals. 1½ ‘Free electron’ model or ‘Mobile electron’ description. 1½ mark iii Of the three energy ranges in kJmol -1 given below: 5 - 100 200 - 700 800 - 1500, which is the most likely energy range for the change M (s) → M (g) , where M is a metal? ½ The best answer out of the three is 200-700 kJ mol -1 . ½ mark 98 1A 3 b i ii 3b Sketch the trends for the properties mentioned in (i) and (ii) below, and account for the trend in each case. i melting point of the alkali metals, Li, Na and K 2 ii boiling point of the Period 3 elements, Na, Mg and Al 2 V. Metallic bonding Page 3 Intermolecular forces Intermolecular Forces I. Van der Waals’ forces A. Discovery of van der Waals' forces B. Origin of van der Waals' forces 1. Induced dipole-induced dipole attractions 2. Dipole-dipole interactions 3. Dipole-induced dipole attractions C. Relative strength of different origins of van der Waals' forces II. Hydrogen bond A. Nature B. Strength C. Solubility and hydrogen bond D. Intramolecular hydrogen bond E. Examples of hydrogen bonding 1. Ethanoic acid dimer 2. Simple hydrides - CH 4 , NH 3 , H 2 O and HF 3. Structure and physical properties of ice 4. Biochemical importance of hydrogen bond a) DNA b) Protein F. Strength of van der Waals' forces and hydrogen bond III. Relationship between structures and properties I. Van der Waals' forces Page 1 Topic I. Van der Waals’ forces Reference Reading 4.10.1 Modern Physical Chemistry ELBS pg. 93–95 Chemistry in Context, 3rd Edition ELBS pg. 114–117 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 116–118 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 99–103 Syllabus van der Waals' forces Notes Introduction to intermolecular forces and intramolecular forces Study of chemistry can be divided into 2 main categories : 1. Chemical properties of substances (e.g. stability and reactivity) These are depending on the intramolecular forces i.e. forces within a molecule – covalent bond. 2. Physical properties of substances (e.g. melting point, boiling point and solubility) These are depending on the intermolecular forces i.e. forces among the molecule. e.g. van der Waals'' forces and hydrogen bond. There are two type of intermolecular forces : van der Waals'' forces and hydrogen bond. van der Waals' forces induced dipole-induced dipole attractions dipole-dipole interactions dipole-induced dipole attractions Hydrogen bond Intermolecular forces I. Van der Waals’ forces A. Discovery of van der Waals' forces van der Waals, Johannes Diderik (van der vals), 1837–1923, was a Dutch physicist. He discovered that real gas doesn't obey ideal gas equation perfectly. In ideal gas equation, it is assumed that there is no interaction among the gas molecules and the gas molecules occupy no volume. Ideal gas law : PV = nRT where P is the pressure measured V is the volume measured n is the number of mole of gas molecule R is the universal gas constant T is the temperature in Kelvin Instead, the real gas follows another equation named after him, van der Waals' equation or real gas equation. van der Waals' equation : ( P + an 2 V 2 ) (V - nb) = nRT where a and b are called van der Waals' constant. a is a measurement of the attraction among the particles and b is a measurement of the volume of the gas molecule. I. Van der Waals' forces Page 2 The values of the empirical constants a and b are depending on the nature of the gas. Gas a / atmdm 6 mol -2 b / dm 3 mol -1 He 0.0341 0.0237 N 2 1.39 0.0391 CO 2 3.63 0.0427 CH 3 CH 2 CH 2 CH 3 147 0.123 NH 3 4.19 0.0372 It can be seen that the attractions among the molecules and the size of the molecules increase from He to butane. The weak forces of mutual attraction is called van der Waals forces. B. Origin of van der Waals' forces van der Waals' forces have 3 origins i. induced dipole-induced dipole attractions ii. dipole-dipole (permanent dipole-permanent dipole) interactions iii. dipole-induced dipole attractions 1. Induced dipole-induced dipole attractions induced dipole-induced dipole attractions is a kind of electrostatic attraction arises from fluctuation of electron cloud. Even at 0 K, electrons in a molecule are in constant motion. Slight relative displacement of electrons in a molecule will give rise to a temporary electrical dipole. It will induce another temporary dipole on a neighbouring molecule and an attractive force will be established between the two. Induced dipole-induced dipole attractions also has many other names i. London force – it was first explained by scientist London in 1930. ii. Dispersion force – it is caused by the unsymmetrical dispersion of electron cloud. iv. Temporary induced dipole-temporary induced dipole attractions – dipole is not permanent. Strength of induced dipole-induced dipole attractions depends on the polarizability of the electrons involved. i. larger molecular size ⇒ larger electron cloud ⇒ higher polarizability ⇒ stronger induced dipole- induced dipole attractions A larger molecule means a larger electron cloud and higher polarizability, this gives strong induced dipole- induced dipole attractions. In general, an unbranched molecule has a larger size than branched molecule. Butane (b.p. -0.5ºC) has a higher boiling point than 2-methylpropane (b.p. -12ºC). ii. small effective nuclear charge ⇒ more diffuse electron cloud ⇒ higher polarizability ⇒ stronger induced dipole-induced dipole attractions Fluorine is highly electronegative with high effective nuclear charge on the valence electrons. CF 4 (m.m. 88) has a greater molecular mass than benzene (m.m. 78) but its boiling point (-128ºC) is lower than that of benzene (80ºC). Teflon, poly(tetrafluoroethene), is used to make the non-sticky surface of frying pan. The high electronegativity, hence low polarizability of fluorine atom results in very small induced dipole-induced dipole attractions. I. Van der Waals' forces Page 3 2. Dipole-dipole interactions (permanent dipole-permanent dipole interactions) There is dipole-dipole interactions among all polar molecules. The oppositely charged ends tend to attract each other and the similarly charged ends tend to repel each other. The molecules will arrange in a way with lowest potential energy for the system. This means maximum amount of attractions. 3. Dipole-induced dipole attractions Like a temporary dipole, a polar molecule can also induce another temporary dipole on another molecule. The attractions caused by this mean is called dipole-induced dipole attractions. C. Relative strength of different origins of van der Waals' forces Molecule induced dipole-induced dipole attractions dipole-dipole interactions (permanent dipole-permanent dipole interactions) dipole-induced dipole attractions Ar (non-polar) 100 % nil nil N 2 (non-polar) 100 % nil nil CH 4 (non-polar) 100 % nil nil CO (polar) ≈ 100 % 0.005 % 0.08 % HCl (polar) 81 % 15 4 % Although dipole-dipole interactions and dipole-induced dipole attractions are very weak comparing with induced dipole-induced dipole attractions, polar molecule always has a higher m.p. and b.p. than a non-polar molecule with comparable molecular mass. This is because on top of dipole-dipole interactions and dipole-induced dipole attractions, there is always induced dipole-induced dipole attractions among polar molecules. Glossary van der Waals’ forces temporary dipole induced dipole-induced dipole attractions polarizability dipole-dipole interactions dipole-induced dipole attractions Past Paper Question 91 2A 2 a 95 2B 4 a i 98 1A 1 b i 91 2A 2 a 2a Arrange the following substances in order of increasing boiling points: C 2 H 5 Cl, CH 4 and C 2 H 6 . Explain your order by comparing the relative magnitudes and nature of the intermolecular forces. 4 CH 4 < C 2 H 6 < C 2 H 5 Cl 1 mark C 2 H 5 Cl has one highly polarizable Cl atom and has a permanent dipole moment, while C 2 H 6 is less polarizable than C 2 H 5 Cl and has no permanent dipole moment. Thus C 2 H 5 Cl is expected to have stronger intermolecular forces than C 2 H 6 , and a higher boiling point. 1 mark Both CH 4 and C 2 H 6 are non-polar but the molecular size and surface area of C 2 H 6 is larger than those of CH 4 , so C 2 H 6 will have a stronger van der Waals’ forces. 2 marks I. Van der Waals' forces Page 4 95 2B 4 a i 4a Explain the following facts: i The boiling points of the halogens increase as the group is descended. 2 Boiling points of halogens depend on the strength of their intermolecular forces (van der Waal’s forces) which is related to the strength of the instantaneous dipole of the molecule. 1 mark Descending the group, with the increase in relative molecular size, the strength of the instantaneous dipole increases. Hence more energy is needed for the boiling of the higher halogens. 1 mark 98 1A 1 b i 1b i An iodine molecule can be represented by the diagram below, with each dot '·' representing an atomic nucleus. (I) Using one or more diagrams of this kind, illustrate your understanding of the two terms 'covalent radius' and 'van der Waals' radius'. (II) Account for the difference between the covalent radius and van der Waals' radius for iodine. 2 II. Hydrogen bond Page 1 Topic II. Hydrogen bond Reference Reading 4.10.2 Modern Physical Chemistry ELBS pg. 90–92 Chemistry in Context, 3rd Edition ELBS pg. 117–126 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 118–123 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 103–112 Syllabus Hydrogen bond Strength of hydrogen bond Intramolecular hydrogen bond Notes II. Hydrogen bond A. Nature Hydrogen bond can be considered as a kind of exceptionally strong dipole-dipole interactions. When a hydrogen is bonded to a very electronegative element e.g. N, O and F, the bond pair electrons would be attracted towards the latter one. The hydrogen atom without the bonding electron will become a naked proton. It is then attracted by the lone pair of another very electronegative element. H F H F bond pair very close to the electronegative atom naked proton intermolecular hydrogen bond δ - δ - δ + δ + Since the charge separation is large, this will give a strong dipole. The electrostatic attraction between the naked proton and the lone pair is called hydrogen bond. The size of proton, N, O and F are small, this facilitates close approach of two dipoles. Necessary condition for the formation of hydrogen bond. 1. A hydrogen atom attaching to a very electronegative atom i.e. N, O, F. 2. A lone pair on a very electronegative atom i.e. N, O, F. II. Hydrogen bond Page 2 B. Strength i. Loneliness of the proton If the hydrogen is bonded to a more electronegative atom, the covalent bond will be more polarized. This causes the proton to be more isolated and the hydrogen bond would be stronger. ii. Availability of the lone pair on the electronegative atom C CH 3 CH 3 CH 3 OH CH 3 CH 2 CH 2 CH 2 OH 2-methylpropan-2-ol pentan-1-ol (b.p. 82.4 °C, m.p. 25.6 °C) (b.p. 138.1 °C, m.p. -78.9 °C) Steric effect – If the lone pair is sterically hindered by bulky groups around it, it would be less available to form hydrogen bond. Though both melting point and boiling point are depending on the attraction among the particles, their value may or may not parallel to each other. e.g. 2-methylpropan-2-ol (b.p. 82.4 °C, m.p. 25.6 °C), pentan-1-ol (b.p. 138.1 °C, m.p. -78.9 °C) Arrangement of particles Ordered Random Random Attraction among the particles High Medium Low C. Solubility and hydrogen bond The solubility of the molecule is parallel to its ability to attract with the water molecule. In certain extent, this is related to its ability to form hydrogen bond with water molecules. N.B. The solubility is depending on the polarity rather than the size of the molecule. e.g. glucose (C 6 H 12 O 6 ) is soluble in water but methane (CH 4 ) is insoluble in water. II. Hydrogen bond Page 3 D. Intramolecular hydrogen bond Besides intermolecular hydrogen bond, hydrogen bond may be formed within a single molecule, i.e. intramolecular hydrogen bond. Boiling point of nitrophenol : O H N O O - + N O O O H - + N O O O H - + 2-nitrophenol (214ºC) 3-nitrophenol (290ºC) 4-nitrophenol (279ºC) 2-nitrophenol has lower boiling point and solubility than the other two isomer. Intramolecular hydrogen bond is formed within the molecule. The intramolecular hydrogen bonding prevents formation of intermolecular bonding hydrogen bond between molecules. Intramolecular hydrogen bonding is impossible in 3- and 4-nitrophenol. E. Examples of hydrogen bonding 1. Ethanoic acid In gas phase or organic solvent, ethanoic acid dimerizes to form a dimer. The presence of an eight-membered ring is confirmed by electron diffraction studies. H 3 C C O O H H O O C CH 3 In aqueous solution, the molecules of a carboxylic acid link up with water molecules rather than form dimers. 2. Simple hydrides - CH 4 , NH 3 , H 2 O and HF CH 4 NH 3 H 2 O HF melting point (K) 91 195 273 190 boiling point (K) 109 240 373 293 There is no hydrogen bond among CH 4 molecule. 1 per molecule in NH 3 and HF. 2 per molecule in H 2 O. Furthermore, boiling point is a better indicator of strength of hydrogen bond than melting point because on top of attractions, melting process involves breaking of crystal lattice. Therefore, melting point is also depending on the packing of the crystal. Consider the boiling point of group V, VI and VII hydrides. The boiling point of the first member is much higher than the expected because of the presence of hydrogen bond on top of van der Waals' forces. The boiling points of group IV hydrides follow the general trend of increasing boiling point with increasing molecular size. This is because of the presence of van der Waals' forces only. Boiling point of Group IV, V, VI and VII hydrides II. Hydrogen bond Page 4 3. Structure and physical properties of ice Each water is capable to form two hydrogen bonds. Including the two covalent bond, oxygen is tetrahedrally coordinated. In ice, the tetrahedrally coordinated oxygen atoms fit into a nonplanar, six-membered ring of water molecules associated through hydrogen bonding. The ring structure extends in a 3-dimensional way. This gives ice an open structure and lower density than water. When ice melts, the energy absorbed breaks some of the intermolecular associations and the hexagonal crystal lattice collapses. The density of water increases with increasing temperature up to 4ºC as the hexagonal structure continues to collapse. 4. Biochemical importance of hydrogen bond a) DNA (Deoxyribonucleic acids) Hydrogen bonding plays a crucial role in the structures of many biochemical substances, such as deoxyribonucleic acids (DNA). The two helical nucleic acid chains are held together by hydrogen bonds. These hydrogen bonds are formed between specific pairs of bases on the DNA chains, such as between the adenine unit in one chain and a thymine unit in the other; or between a guanine unit in one chain and a cytosine unit in the other. II. Hydrogen bond Page 5 b) Protein Protein are made up from α-amino acid molecules. The –NH 2 group condenses with –COOH group to form a peptide link, ( C N O H ). The polypeptide forms a helix structure with the amide groups joined together by hydrogen bonds. R C H 2 N H C O OH C N C C N C C N C O O O H H H H H H R R R α-amino acid chain of polypeptide F. Strength of van der Waals’ forces and hydrogen bond H-bonds have the strength in the range 5–40 kJmol -1 . van der Waals’ forces have roughly the same strength. For a H-bonded molecule, there are both hydrogen bond and van der Waals’ force. Therefore, a H-bonded substance always has a higher boiling point than those not H-bonded. polar molecule non-polar molecule molecule with H-bond induced-induced induced-induced induced-induced dipole-dipole dipole-dipole H-bond Estimated by Range (kJmol -1 ) Ionic bond (non-directional) Lattice energy -780 (NaCl) – -3791(MgO) Covalent bond (directional) Bond energy terms 158 (E(F–F)) – 945.4 (E(N≡N)) Metallic bond (non-directional) Atomization energy 107.3 (Na) – 514.2 (V) Comparing with ionic bond, covalent bond and metallic bond, both H-bond and van der Waals' forces are very weak. This explains why only 100ºC is required to boil water but 8000ºC is required to decompose water into hydrogen and oxygen atoms. O H H O H H O H H strong covalent bond weak hydrogen bond and van der Waals' forces Glossary hydrogen bond steric effect intramolecular hydrogen bond open structure DNA double helix structure amino acid peptide link amide group polypeptide II. Hydrogen bond Page 6 Past Paper Question 92 2B 6 Ac i 92 2B 6 Bc i 92 2C 7 d 93 1A 3 d 94 1A 2 d i ii iii 94 2A 2 b i 94 2B 4 b 97 1A 1 d i ii 98 1A 2 b 98 2B 6 a i 99 1A 1 b iii 92 2B 6 Ac i 6Ac i Account for the difference in boiling points between NH 3 and PH 3 . 2 PH 3 molecules - van der Waals intermolecular forces 1 mark NH 3 molecules - additional hydrogen bonding 1 mark 92 2B 6 Bc i 6Bc i Account for the difference in boiling points between H 2 O and H 2 S. 2 Intermolecular hydrogen bonding in H 2 O but not H 2 S 1 mark H 2 S van der Waals forces and permanent dipole-permanent dipole interactions only 1 mark 92 2C 7 d 7 A carboxylic acid P, with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H 2 SO 4 gave the desired ester Q but with much of the starting material P unchanged. (Relative atomic masses : H 1.0; C 12.0; O 16.0) 7d Which has a higher boiling point, the carboxylic acid or its methyl ester? Explain your answer. 3 Carboxylic acid has higher b.p. than its ester ½ mark Hydrogen bonding 1 mark R C O O H H R C O O or lone pair on oxygen bonded to acidic proton 1 mark Need more energy to break off H-bonding. ∴higher b.p. ½ mark 93 1A 3 d 3d The boiling point and density of butan-1-ol are 117.9 ºC and 0.81 gcm -3 respectively, and those of its isomer 2- methylpropan-2-ol are 82.2 ºC and 0.79 gcm -3 respectively. Account for these differences. 2 Alcohol molecules are held together by hydrogen bond. ½ mark Methylpropan-2-ol is a tertiary alcohol and butan-1-ol is a primary alcohol. ½ mark Due to steric interaction, 3º alcohol forms weaker intermolecular hydrogen bonds so its boiling point is lower. 1 mark Further, the molecules are less closely packed and therefore results in lower density. ½ mark 94 1A 2 d i ii iii 2d i Explain the term “hydrogen bonding”. 2 H atom is situated between 2 small / electronegative atoms. 1 mark One of these has a lone pair of electrons / bond energy in the range of 2 – 10 kJmol -1 . 1 mark ii Draw a diagram of the structure of a compound which has hydrogen bonds. Indicate the hydrogen bond(s) clearly. 1 A actual example of intra / intermolecular hydrogen bond. The example must show X–HLY or XLHLX 1 mark iii Explain why (I) the boiling point of CH 4 is lower than that of SiH 4 , and (II) the boiling point of NH 3 is higher than that of PH 3 . 2 (I) Boiling points is expected to increase with number of electrons in the molecule (CH 4 → SiH 4 ) due to increasing van der Waal’s forces. 1 mark (II) Boiling point of NH 3 higher than the expected (i.e. >PH 3 ) due to increased intermolecular attraction (hydrogen bonding) 1 mark II. Hydrogen bond Page 7 94 2A 2 b i 2b Account for each of the following: i Concentrated H 3 PO 4 has a high viscosity. 2 In H 3 PO 4 , the intermolecular force is H-bond, ½ mark and each H 3 PO 4 molecule is capable of forming 3 H-bonds with its neighbours. With strong intermolecular forces, H 3 PO 4 has a high viscosity. 1 mark P O O O O H H H ½ mark 94 2B 4 b 4b Explain why hydrogen fluoride is a liquid at room temperature and atmospheric pressure while the other hydrogen halides are gases. 2 The anomalous boiling point of HF is due to extensive H-bonding between HF molecules and other hydrogen halides have no hydrogen bondings among the molecules. 1 mark H F H F H F H F H F H F H F (diagram showing H-bonds in HF) 1 mark 97 1A 1 d i ii 1d Explain why 2 i the boiling point of HF is higher than that of HCl; ii the boiling point of HI is higher than that of HBr. 98 1A 2 b 2b Which compound, H 2 O or F 2 O, would you expect to have a higher boiling point ? Explain your answer. 2 98 2B 6 a i 6a Consider the structures of the two synthetic polymers shown below. HN(CH 2 ) 6 NHCO(CH 2 ) 4 CO n nylon-6.6 CH 2 CH 2 n poly(ethene) i Suggest an explanation for the fact that nylon-6,6 has a higher tensile strength than poly(ethene). 99 1A 1 b iii 1b Account for each of the following : iii At 273 K, ice has a smaller density than water. III. Relationship between structures and properties Page 1 Topic III. Relationship between structures and properties Reference Reading 4.0.1.0 4.11 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 109 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 402–403, 468–469 Syllabus The relationship between structures and properties of materials Notes In chemistry, chemists intend to explain the chemical and physical properties of substances in term of particles and the interaction among the particles. All properties can be explained according to the following framework. particles → attractions among particles → arrangement of particles → structure → properties Example Particles Attractions among particles Arrangement of particles Structure Physical properties Chemical properties Zn (s) Zn atom strong metallic bond due to delocalized electron hexagonal close packing giant metallic structure high thermal and electrical conductivities; hard; malleable; ductile; dense; high m.p. & b.p.; insoluble in any solvent except Hg (l) reducing agent Hg (l) Hg atom weak metallic bond due to less delocalized electron random arrangement in liquid liquid at room temperature high thermal and electrical conductivities; relatively low m.p. & b.p. very weak reducing agent NaCl (s) Na + ion & Cl - ion strong ionic bond face centered cubic giant ionic structure low thermal and electrical conductivities; only conducts electricity in molten state or aqueous state; soluble in water quite inert AgCl (s) Ag + ion & Cl - ion ionic bond with very strong covalent character cubic structure giant ionic structure low thermal and electrical conductivities; insoluble in water quite inert H 2 O (l) H 2 O molecule strong O–H covalent bond and relatively weak intermolecular forces among molecules (Hydrogen bond and van der Waals' forces) H 2 O is angular in shape; random arrangement in liquid; open structure in ice simple molecular structure / molecular crystal in ice low thermal and electrical conductivities; relatively low m.p. and b.p.; high heat capacity quite inert; only decomposes into H and O atom at 8000K H 2(g) H 2 molecule strong H–H covalent bond; extremely weak intermolecular forces among molecules (van der Waals' forces only) H–H is linear in shape; random arrangement in gas simple molecular structure low thermal and electrical conductivities; extremely low m.p. and b.p.; low heat capacity reducing agent III. Relationship between structures and properties Page 2 Example Particles Attractions among particles Arrangement of particles Structure Physical properties Chemical properties Graphite C atom strong covalent bond with delocalized electrons between C atoms; weak van der Waals' forces between hexagonal layers C atoms are arranged in hexagonal layers with large gap between layers giant covalent structure high thermal and electrical conductivities; extremely high m.p. and b.p.; brittle; insoluble in any solvent reducing agent Diamond C atom strong covalent bond between C atoms C atoms are arranged tetrahedral network giant covalent structure low thermal and electrical conductivities; high m.p. and b.p.; extremely hard; insoluble in any solvent reducing agent Glucose C 6 H 12 O 6 molecule strong covalent bond between C, H and O; weak intermolecular forces among molecules glucose molecule has a ring structure simple molecule structure low thermal and electrical conductivities; moderate m.p. and b.p.; soluble in water chars upon heating; a weak reducing agent. Glossary particle bonding / attraction arrangement structure properties III. Relationship between structures and properties Page 3 Past Paper Question 90 2A 3 c 91 1A 2 d iii 93 2B 4 b iv 94 2C 9 b iv 99 1A 1 b i 90 2A 3 c 3c Arrange the following substances in order of increasing melting point: NaF, F 2 , HF. Explain your order in terms of the bonding involved. 4 Melting involves the overcoming of forces between the units of which the solid is made. Two of the given solid HF, F 2 would be expected to form molecular solids; the other, NaF would be expected to form ionic solid. The coulombic forces holding an ionic solid together are strong, hence NaF would be expected to have high melting point. 1 mark The attraction between F 2 molecules is quite weak and due only to van der Waals forces would be expected to have a low melting point. 1 mark Even though HF forms molecular solid but because of the presence of hydrogen bonding between H and neighbouring F atom, which can hold the molecules together more strongly and therefore, expected to have much higher melting point than F 2 . 1 mark F 2 (-219.6ºC) < HF(-89.1ºC) < NaF(993ºC) 1 mark 91 1A 2 d iii 2d iii Explain why the electrical conductivities of iron and of caesium chloride are different in the solid state. 1 In CsCl, the ions are held together by strong coulombic forces and cannot be easily displaced. Because of lack of mobility of ions, conductivity is low. In iron, the iron atoms occupy the BCC sites and electrons are delocalized. Conductivity is high because electrons can move. 1 mark 93 2B 4 b iv 4b Account for each of the following facts: iv There is a significant difference in the melting point between silicon (m.p. 1410 ºC) and lead (m.p. 328 ºC). 2 Silicon has a giant covalent network structure. C.N. = 4 Lead has a close packed metallic structure. C.N. = 12 1 mark The Si–Si in silicon is strong compared with the metallic bond in lead and in the liquid state of lead is residual metallic bond remained. ∴ silicon has a higher melting point than lead. 1 mark 94 2C 9 b iv 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. PVC is more rigid and durable than PE, but incineration of PVC causes a more serious pollution problem. iv Explain why PVC is more rigid than PE. 1 The structure of PVC can be represented by The attraction between polymer chains of PVC is dipolar attraction while that between PE polymer is van der Waal's force which is weaker when compared with the former. Hence PVC is more rigid and more durable than PE. 1 mark 99 1A 1 b i 1b Account for each of the following : i At 298 K and 1 atm pressure, carbon dioxide is a gas whereas silicon dioxide is a solid. States of Matter I. Solid state A. Metallic structure 1. Hexagonal close packing 2. Cubic close packing - face centred cubic 3. Tetrahedral holes and octahedral holes 4. Body-centred cubic structure B. Giant ionic structure C. Covalent giant structure 1. Allotrope of carbon a) Diamond b) Graphite 2. Silicon(IV) oxide D. Molecular crystal 1. Iodine – face centred cubic 2. Carbon dioxide (Dry ice) – face centred cubic II. Gaseous state A. Gas laws 1. Charles' law 2. Boyle's law 3. Avogadro's law 4. PVT surface B. Ideal gas equation (Ideal gas law) 1. Molar volume C. Determination of molecular mass 1. Experimental determination of molar mass of a gas 2. Experimental determination of molar mass of a volatile liquid D. Dalton's law of partial pressure 1. Mole fraction 2. Relationship between partial pressure and mole fraction I. Solid State Unit 1 Page 1 Topic I. Solid state Unit 1 Reference Reading 4.12.1 Modern Physical Chemistry ELBS pg. 154–155 Chemistry in Context, 3rd Edition ELBS pg. 134–136 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 147–149 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 45, 119–121 Syllabus Close packing structure Body centred cubic structure Notes I. Solid state A. Metallic structure In metal crystal, the metal atoms tends to pack with each other closely and has a high coordination no. Coordination no. – the number of atoms closest to a particular atom. There are two kinds of close packing - hexagonal close packing and cubic close packing. 1. Hexagonal close packing (74% of the space is filled) Some of the metal have hexagonal close packing (h.c.p.). e.g. Cobalt, Titanium and Zinc In hexagonal close packing, each atom is surrounded by 12 atoms, i.e. coordination is 12. 6 on the same layer, 3 on the top and 3 at the bottom. Since the orientation of the first and third layer are the same, this is also called abab... pattern. Unit cell – smallest portion of the crystal which contains all the fundamentals (information) without repetition. No. of atoms in an unit cell 12 at the corner 12 × 1 6 = 2 2 on the face 2 × 1 2 = 1 3 in the cell 3 × 1 = 3 Total no. of atoms in an unit cell = 6 I. Solid State Unit 1 Page 2 2. Cubic close packing - face centred cubic (74% of the space is filled) Cubic close packing (c.c.p.) is also called face centre cubic packing (f.c.c.). Examples are Copper, Nickel and Calcium The coordination no. of any atom in the structure is also 12. The first layer has a different orientation with the third layer of atoms. This is called abcabc... pattern. By looking at a four layers unit cell, 1 atom at the first layer, 6 at the second layer, 6 at the third layer and 1 at the fourth layer, a face centred cubic unit cell can be constructed. No. of atoms in an unit cell 8 at the corner 8 × 1 8 = 1 6 on the face 6 × 1 2 = 3 Total no. of atoms in an unit cell = 4 I. Solid State Unit 1 Page 3 3. Tetrahedral holes and octahedral holes Although the crystal is closely packed, there is still some spaces between the atoms. They are called holes. There are two types of holes – tetrahedral hole and octahedral hole Octahedral hole Octahedral hole is surrounded by 6 atoms. It is the space between two layers of triangularly arranged atoms. From another angle, it can be seen that the six atoms are arranged in a form of octahedron. Tetrahedral hole Tetrahedral hole is surround by 4 atoms. The four atoms are arranged in a form of tetrahedron. It has a smaller size than octahedral hole. 4. Body-centred cubic structure (68% of the space is filled) e.g. Sodium, Potassium and Chromium Body-centred cubic (b.c.c.) structure is not a close packing structure. Only 68% of the space is filled comparing with 74% in the two close packing structures. In body-centred cubic structure, the coordination no. of any atom is 8. No. of atoms in an unit cell 8 at the corner 8 × 1 8 = 1 1 at the centre 1 × 1 = 1 Total no. of atoms in an unit cell = 2 Glossary coordination no. hexagonal close packing cubic close packing face centred cubic tetrahedral hole octahedral hole body centred cubic unit cell I. Solid State Unit 1 Page 4 Past Paper Question 90 1A 2 c 91 1A 2 d i 92 1A 2 d i 97 1A 1 a i ii 90 1A 2 c 2c Compare and contrast the two close packing atomic arrangements in metals. 3 Metals having the hexagonal close packing has the abab... arrangement for atoms in different layers. Metals having the face-centred cubic close packing has the abcabc... arrangement for atoms in different layers. 2 marks Both arrangements have the same packing efficiency. or 1 mark Atoms in both close packing arrangements all have the co-ordination number of 12. 91 1A 2 d i 2d i Iron has a body-centred structure. Draw a unit cell representation of iron. 1 1 mark 92 1A 2 d i 2d The arrangement of atoms in metals can be described by the close-packing of spheres. i Which close-packed structure does abcabcabc... describe? Indicate on the diagram below one tetrahedral hole (marking it T) and one octahedral hole (marking it O). 1½ Cubic closest packed structure. ½ mark ½ mark for O and ½ mark for T 97 1A 1 a i ii 1a At room temperature, iron has a body-centred cubic structure. 2 i Draw the unit cell representation of iron. ii Deduce the number of atoms in one unit cell of iron. I. Solid state Unit 2 Page 1 Topic I. Solid state Unit 2 Reference Reading 4.12.2 Modern Physical Chemistry ELBS pg. 159–162, 165–166 Chemistry in Context, 3rd Edition ELBS pg. 138–146 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 149–151 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 124–126 Syllabus Giant ionic structure Notes B. Giant ionic structure Ionic crystal is the result of a balance between attractions and repulsions. If the attraction forces and the repulsion forces are not balanced, the structure will be unstable. The ions will be rearranged and another more stable structure will be resulted. This depends on the relative size of the ions in the ionic crystal. Stable Limiting radius Unstable Stable The arrangement of the ions in an ionic crystal is determined by the ionic radii ratio (r + /r - ): Coordination no. Shape r + /r - Example 8 Simple cubic 0.73 and above CsCl 6 Face centered cubic (Octahedral) 0.41 - 0.73 NaCl 4 Face-centered cubic (Tetrahedral) 0.22 - 0.41 ZnS Caesium chloride Simple cubic Sodium chloride face-centred cubic (octahedral) Zinc blende (Zinc sulphide) face centred cubic (tetrahedral) Cs + ions occupy the centre of the cubes formed by 8 Cl - ions Na + ions occupy the octahedral holes formed by 6 Cl - ions Zn 2+ ions occupy half of the tetrahedral holes formed by 4 S 2- ion no. of Cs + ions in the unit cell 1 at the centre 1 × 1 = 1 Total no. of Cs + ions = 1 no. of Cl - ions in the unit cell 4 at the corner 4 × 1 4 = 1 Total no. of Cl - ions = 1 no. of Na + ions in the unit cell 1 at the centre 1 × 1 = 1 12 on the edge 12 × 1 4 = 3 Total no. of Na + ions = 4 no. of Cl - ions in the unit cell 6 on the face 6 × 1 2 = 3 8 at the corner 8 × 1 8 = 1 Total no. of Cl - ions = 4 no. of Zn 2+ ions in the unit cell 4 in the cell 4 × 1 = 4 Total no. of Zn 2+ ions = 4 no. of S 2- ions in the unit cell 6 on the face 6 × 1 2 = 3 8 at the corner 8 × 1 8 = 1 Total no. of S 2- ions = 4 Coordination no. of Cs + ion = 8 Coordination no. of Cl - ion = 8 Coordination no. of Na + ion = 6 Coordination no. of Cl - ion = 6 Coordination no. of Zn 2+ ion = 4 Coordination no. of S 2- ion = 4 I. Solid state Unit 2 Page 2 Both the no. of ions in the unit cell and ratio of coordination no. can be used to determine the empirical formula of the ionic compound. The structure of an ionic crystal is labeled according to the positions of the formula units, not the position of individual ions. The position of each formula unit is called a lattice point. Take CsCl as the example, consider each pair of Cs + ion and Cl - ion as a lattice point. The lattice points are arranged in a simple cubic manner i.e. 8 lattice point occupy the 8 corners of a cube. Glossary giant ionic structure zinc blende Past Paper Question 94 1A 1 b i ii 99 2A 2 c i ii 94 1A 1 b i ii 1b The crystal structure of a compound A x B y can be described as a simple cubic lattice of A atoms with B atoms at the middle of all the edges. i What is the empirical formula of this compound? 1 No. of atom A per unit-cell = 8 × 1 8 = 1 No. of atom B per unit-cell = 12 × 1 4 = 3 ∴ Empirical formula of the compound is AB 3 . 1 mark ii What are the coordination numbers of an A atom and a B atom respectively? 1 Coordination number of A = 6 ½ mark Coordination number of B = 2 ½ mark 99 2A 2 c i ii 2c i Consider the unit cell of calcium fluoride shown below: (I) State the respective coordination numbers of each calcium ion and each fluoride ion. (II) Describe the lattice of calcium ions and that of fluoride ions. ii (I) Draw the unit cell of caesium chloride. (II) Describe the lattice of caesium ions and that of chloride ions in caesium chloride. I. Solid state Unit 3 Page 1 Topic I. Solid state Unit 3 Reference Reading 4.12.3 Modern Physical Chemistry ELBS pg. 159–162, 165–166 Chemistry in Context, 3rd Edition ELBS pg. 138–146 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 152–154 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 46, 123–124 Syllabus Giant covalent structure Notes C. Giant covalent structure 1. Allotrope of carbon Allotropy – The existence of more than one physical form of the element in the same state. e.g. graphite and diamond, red phosphorus and white phosphorus, oxygen and ozone. Carbon has two common allotropes – graphite and diamond. The two allotropes have different physical properties but similar chemical properties. This can be explained by the structures of the two allotropes. Properties Diamond Graphite appearance colourless, transparent crystal black, shiny solid hardness hardest natural substance known brittle density 3.5 gcm -3 2.2 gcm -3 electrical conductivity does not conduct conducts in one direction melting point 3550 ºC 3730 ºC a) Diamond Each carbon is sp 3 hybridized. The sp 3 hybrid orbitals are overlapped to form strong σ covalent bond ⇒ hard, high density, high melting point and boiling point. The bonding is extended to form a 3-dimensional network. b) Graphite Each carbon is sp 2 hybridized. The sp 2 hybrid orbitals are overlapped to form strong σ covalent bond. The p orbitals perpendicular to the carbon layer are overlapped to form π bond. ⇒ higher melting point than diamond. This gives a hexagonal layer structure. There are only weak van der Waals’ forces between the layers. ⇒ brittle and low density The π electrons are delocalization above and below the layer in a direction parallel to the layer. ⇒ conducts electricity in one direction, shiny appearance I. Solid state Unit 3 Page 2 2. Silicon(IV) oxide Quartz / Sand / silicon(IV) oxide / silicon oxide / silicon dioxide (SiO 2 ) The structure of quartz is similar to that of diamond. Each silicon atom is joined to 4 oxygen atoms and each oxygen atom is then connected to another silicon atom. The silicon and oxygen are joined together by strong covalent bond. Upon melting, a lot of energy is required to break the Si–O covalent bond. Therefore, silicon(IV) oxide has a relatively high melting point 1610ºC. Glossary allotrope hexagonal layer structure Past Paper Question 90 2B 4 b 92 2A 3 a i ii 98 2A 2 a i ii 90 2B 4 b 4 Account for the following observations. 4b SiO 2 is a solid with a high melting point, whereas CO 2 is a gas at room temperature. 4 SiO 2 - three dimensional polymeric solid with very strong Si-O bonds. (Si-O bond must be stated; with strong bonds only - 0 mark) 2 marks CO 2 - exists as discrete molecules with only weak intermolecular van der Waals’ forces. 2 marks 92 2A 3 a i ii 3a i Describe the bonding and intermolecular forces in ice and in SiO 2 solid. 2 Ice consists of covalent H 2 O molecules held together by hydrogen bonding. Quartz, SiO 2 solid, is an infinite three dimensional network(giant structure) solid in which each Si atom is covalently bonded in a tetrahedral arrangement to four O atoms. 2 marks ii What type of interactions must be overcome to melt these solid? 2 Hydrogen bonding must be overcome in order to melt ice. In order for SiO 2(s) to melt, some of the strong Si-O covalent bonds have to be broken. 2 marks 98 2A 2 a i ii 2a The structures of the two allotropes of carbon, diamond and graphite, are shown below. 6 i Comment on the three different carbon-carbon distances as indicated in the above structures. ii With reference to the above structures, explain why diamond is hard whereas graphite is soft enough to be used as lubricant. I. Solid state Unit 4 Page 1 Topic I. Solid state Unit 4 Reference Reading 4.12.4 Modern Physical Chemistry ELBS pg. 159–162, 165–166 Chemistry in Context, 3rd Edition ELBS pg. 138–146 Physical Chemistry, Fillans pg. 212–220 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 151–152 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 47, 130–132 Syllabus Molecular crystal Notes D. Molecular crystal Molecular crystal consists of molecules held in lattice sites by weak intermolecular forces such as van der Waals’ forces or hydrogen bonds. The structure of molecular crystals are determined by the shape of the basic molecular units which tend to pack together as efficient as possible. 1. Iodine – face centred cubic 2. Carbon dioxide (Dry ice) - face centred cubic Glossary molecular crystal Past Paper Question II. Gaseous state Unit 1 Page 1 Topic II. Gaseous state Unit 1 Reference Reading 0.5.3.1–0.5.3.2 ILPAC 9 (2nd ed.), John Murray, pg 7–19 Modern Physical Chemistry (4th ed.), Bell & Hyman Ltd., pg 143–144 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., pg 155 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., pg 163–165, 167 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 137–139 Syllabus Charles' law Molar volume Boyle's law Avogadro's law Ideal gas equation (Ideal gas law) Molar volume of gases Notes II. Gaseous state A. Gas Laws 1. Charles’ Law For a given mass of gas, the volume is directly proportional to the absolute temperature (Kelvin scale) at a given pressure. V ∝ T 2. Boyle’s Law For a given mass of gas, the volume is inversely proportional to the pressure at a given temperature. V ∝ 1 P 3. Avogadro’s Law At constant pressure and temperature, the volume of a gas is directly proportional to the number of gaseous particle. V ∝ N or V ∝ n N : no. of particle, n : no. of mole of particle II. Gaseous state Unit 1 Page 2 4 PVT surface PVT relationship of an ideal gas can be represented by a 2-dimensional surface. PVT surface for an ideal gas B. Ideal gas equation (Ideal gas law) By combining the three gas laws, an ideal gas equation can be derived. By combining V ∝ T, V ∝ 1 P and V ∝ n ⇒ V ∝ nT P Therefore, PV nT = constant The constant is called gas constant, R and the relationship becomes PV = nRT Unit P : 1 atm = 101325 Nm -2 (Pa) = 760 mmHg V : 1 m 3 = 1000 dm 3 = 1000000 cm 3 n : no. of mole T : must be in Kelvin (K) R : 8.314 JK -1 mol -1 = 82.05 cm 3 atmK -1 mol -1 II. Gaseous state Unit 1 Page 3 1. Molar volume Since the volume occupied by a gas is mainly space whose size is depending on the motion of the particles only, all gases have the similar molar volume. i.e. the molar volume is independent of the nature of the particles. It is found that the molar volume of any gas at 0 ºC (273 K) and 1 atm is about 22.4 dm 3 . The condition 0 ºC and 1 atm is called standard temperature and pressure (stp) or standard condition. Gas Formula Molar volume at stp, V m / dm 3 mol -1 Helium He 22.396 Hydrogen H 2 22.432 Carbon dioxide CO 2 22.262 By using the ideal gas equation, the molar volume of a gas at different condition can also be determined. For example, the condition 25ºC (273K) and 1 atm is called room temperature and pressure (rtp) or room condition. The molar volume at stp can be determined by using ideal gas equation. PV = nRT P 1 V 1 n 1 T 1 = P 2 V 2 n 2 T 2 1 atm × 22.4 dm 3 1 mole × 273 K = 1 atm × V 2 1 mole × 298 K V 2 = 24.5 dm 3 Since the value is only an approximate value, usually the molar volume of a gas is assumed to be 24 dm 3 at rtp. Glossary Charles' law absolute temperature (Kelvin scale) Boyle's law Avogadro's law PVT surface ideal gas equation gas constant / universal gas constant molar volume standard temperature and pressure (stp) / standard condition room temperature and pressure (rtp) / room condition Past Paper Question 94 2A 1 a i 94 2A 1 a i 1a Gas container A and B each contain an ideal gas at low pressure and 298 K. The volume of container A is twice that of container B, but the number of moles of ideal gas contained in A is only half of that in B. i Calculate the ratio of the gas pressure in the two containers. 3 P A V A = n A RT P B V B = n B RT 1 mark P A P B = n A n B ( V B V A ) 1 mark = 1 2 ( 1 2 ) = 1 4 P A = 1 4 P B or P B = 4P A 1 mark II. Gaseous State Unit 2 Page 1 Topic II. Gaseous State Unit 2 Reference Reading 0.5.3.3 ILPAC 9 (2nd ed.), John Murray, pg 21–25 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., pg 158–160 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 143–144 Assignment Reading A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 168, 180–181 Syllabus Determination of molecular mass Notes C. Determination of molecular mass Although molecular mass of a substance is a microscopic properties which cannot be measured directly, ideal gas equation provides a link between molecular mass of a gas and some measurable quantities. PV = nRT where P is the pressure of the gas V is the volume of the gas n is the number of mole of gaseous particle R is the gas constant T is the temperature in Kelvin The quantity of no. of mole is related to the molar mass by the relationship, n = m M m , where m is the mass of the gas and M m is the molar mass of the gas. Since molar mass is numerically the same as molecular mass, molecular mass can be determined by measuring the mass, pressure, volume and temperature of a gas. PV = nRT PV = m M m (RT) where m is the mass of the gas M m is the molar mass of the gas M m = m PV (RT) or M m = ρ P (RT) where, ρ is the density of the gas = m V II. Gaseous State Unit 2 Page 2 1. Experimental determination of molar mass of a gas By determining the mass, pressure, volume and temperature of a gas (e.g. CO 2(g) ), the molar mass of the gas can be determined. In the following experiment, a volumetric flask is used to measure the volume and the mass of the gas. Measured values Mass of volumetric flask filled with air = 47.993 g Mass of volumetric flask filled with CO 2 = 47.998 g Mass of volumetric flask filled with water = 152.8 g Room temperature = 26 ºC Atmospheric pressure = 757 mmHg Given values Density of water at 26ºC = 1.00 gcm -3 Density of air at 26ºC and 757 mmHg = 0.00118 gcm -3 Gas constant (R) = 0.0821 atmdm 3 K -1 mol -1 Calculation Molar Mass of CO 2 = mass of CO 2 × gas constant × temp. in K pressure of CO 2 × volume of CO 2 = mRT PV Note ! Your should keep all the units in the intermediate steps. Mass of CO 2 Mass of water = Mass of volumetric flask filled with water - Mass of volumetric flask (filled with air) (the mass of air is neglected in this step) = 152.8 g - 47.933 g = 104.9 g Volume of the flask = mass of water density of water = 104.9 g 1.00 gcm -3 = 104.9 cm 3 Mass of air = Volume of flask × density of air = 104.9 cm 3 × 0.00118 gcm -3 = 0.124 g Mass of empty volumetric flask = Mass of volumetric flask filled with air - Mass of air = 47.933 g - 0.124 g = 47.809 g Mass of CO 2 = mass of volumetric flask filled with CO 2 - mass of empty volumetric flask = 47.998 g - 47.809 g = 0.189 g Pressure of CO 2 = atmospheric pressure Volume of CO 2 = volume of the flask Molar Mass of CO 2 = mass of CO 2 × gas constant × temp. in K pressure of CO 2 × volume of CO 2 = mRT PV = 0.189 g × 0.0821 atmdm 3 K -1 mol -1 × (273 + 26)K 757 760 atm × 105.0 1000 dm 3 = 44.4 gmol -1 II. Gaseous State Unit 2 Page 3 2. Experimental determination of molar mass of a volatile liquid The molar mass of a volatile liquid cannot be determined directly according to ideal gas equation, the liquid must be vaporized first. By injecting a known mass of volatile liquid into a heated gas syringe, the molar mass of the volatile liquid can be determined using the ideal gas equation. Molar Mass of the volatile liquid = mass of the liquid × gas constant × temp. in K pressure of the vapour × volume of the vapour = mRT PV The mass of the liquid is measured by a hypodermic syringe and a balance. It is determined by comparing the mass of the hypodermic syringe before and after injection. The hypodermic needle is equipped with a self-sealing rubber to ensure no loss of vapour from the gas syringe. The volume of the vapour is measured by the volume displaced by the vapour in the heated gas syringe. Precautions : 1. The needle of hypodermic syringe must be long enough to inject the liquid into the inner part of gas syringe where the temperature is higher. This ensures complete vaporization of liquid. 2. The temperature and volume readings must become steady before any record is taken. 3. To ensure accurate injection of liquid, the hypodermic syringe should be rinsed with the liquid first and any bubbles in the syringe should be expelled. Keep the syringe horizontal and avoid touching the piston or warming the barrel with you hand, either of which could result in loss of liquid. II. Gaseous State Unit 2 Page 4 Measured values Mass of hypodermic syringe and liquid before injection = 13.189 g Mass of hypodermic syringe and liquid after injection = 12.920 g Temperature of the vapour = 100 ºC Atmospheric pressure = 762 mmHg Volume of air in gas syringe before injection = 5.0 cm 3 Volume of air and vapour in gas syringe after injection = 66.0 cm 3 Given values Gas constant (R) = 0.0821 atmdm 3 K -1 mol -1 1 atm = 760 mmHg Calculation PV = nRT = m M m RT 762 760 atm × ( 66.0 - 5.0 1000 dm 3 ) = (13.189 -12.920 g) M m × 0.0821 atmdm 3 K -1 mol -1 × (100 + 273) K M m = 135 gmol -1 Mass of liquid used = mass of hypodermic syringe before injection - mass of hypodermic syringe after injection = 13.189 g - 12.920 g = 0.269 g Temperature of vapour = temperature of the steam jacket = 100 ºC = 373 K Pressure of vapour = atmospheric pressure = 762 760 atm = 1.00 atm Volume of vapour = Volume of air and vapour in gas syringe - Volume of air in gas syringe = 66.0 cm 3 - 5.0 cm 3 = 61.0 cm 3 = 0.0610 dm 3 Molar Mass of the volatile liquid = mass of the liquid × gas constant × temp. in K pressure of the vapour × volume of the vapour = mRT PV = 0.269 g × 0.0821 atmdm 3 K -1 mol -1 × 373 K 1.00 atm × 0.0610 dm 3 = 135 gmol -1 Glossary molar mass gas syringe hypodermic syringe self-sealing rubber steam jacket Past Paper Question 95 2A 3 a i 97 2A 1 b 99 1A 1 a i ii 95 2A 3 a i 3a i Derive an expression for the molar mass M of an ideal gas, in terms of its density d, pressure P, and absolute temperature T. 2 PV = nRT ½ mark n = m M where m is the mass of dry air ½ mark PV = m M (RT) M = m PV (RT) = d P (RT) 1 mark C i Generally well answered. However, some candidates did not express M in terms of d, P and T explicitly. 97 2A 1 b 1b At 360 K and 101 kPa pressure, the vapour produced by 0.226 g of a volatile liquid occupies 85.0 cm 3 . Assuming that the vapor behaves ideally, calculate the molar mass of the liquid. (1 kPa = 1 × 10 3 Nm -2 ) (Gas constant, R = 8.31 JK -1 mol -1 ) II. Gaseous State Unit 2 Page 5 99 1A 1 a i ii 1a At 433 K, 0.569 g of aluminium chloride produces a vapour having a volume of 96.0 cm 3 and a pressure of 80.0 kPa. i Assuming that the aluminium chloride vapour behaves as an ideal gas, calculate its molar mass. (1 kPa = 1 × 10 3 Nm -2 ) ii Deduce the molecular formula of aluminium chloride under the above conditions and suggest a structure consistent with this formula. II. Gaseous State Unit 3 Page 1 Topic II. Gaseous State Unit 3 Reference Reading 4.13.4 ILPAC 9 (2nd ed.), John Murray, pg 33–34 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., pg 156 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 140 Assignment Reading A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 169 Syllabus Dalton's law of partial pressure Relationship between partial pressure and mole fraction Notes D. Dalton’s law of partial pressure Dalton’s law of partial pressure – The total pressure (P Total ) exerted by a mixture of gases is equal to the sum of the partial pressure (p) of each constituent gas. P Total = p 1 + p 2 + p 3 ...p n The partial pressure is defined as the pressure the gas would exert if it was contained in the same volume as that occupied by the mixture. 1. Mole fraction Mole fraction is the percentage of a particular component in a mixture in term of number. e.g. For a mixture containing 0.2 mole of oxygen, 0.3 mole of hydrogen and 0.3 mole of carbon dioxide, the mole fraction oxygen in the mixture can be calculated by mole fraction of oxygen = no. of mole of oxygen molecules total no. of mole of molecules = 0.2 0.2 + 0.3 + 0.3 = 0.25 It is different from percentage by mass. In the above example, 0.2 mole of oxygen weighs 6.4 g, 0.3 mole of hydrogen weighs 0.6 g and 0.3 mole of carbon dioxide weighs 13.2 g. Percentage by mass of oxygen = mass of oxygen total mass = 6.4 g 6.4 g + 0.6 g + 13.2 g = 0.32 = 32% II. Gaseous State Unit 3 Page 2 2. Relationship between partial pressure and mole fraction According to the ideal gas equation, the pressure of a gas is directly proportional to the no. of gaseous particle or the no. of mole of particle present. P ∝ N or P ∝ n where N is the total no. and n is the no. of mole, they are related by the expression N = nL where L is Avogadro's constant. By combining with Dalton's law of partial pressure, the partial pressure exhibited by individual component in a gaseous mixture should also be proportional to the no. of mole of that kind of particle. 1 : P T ∝ n where P T is the total pressure and n is the total no. of mole of particles. 2 : p i ∝ n i p i is the partial pressure of a component and n i is the no. of mole of that component Combining 1 & 2, p i P T = n i n = mole fraction of individual component in the mixture The mole fraction of individual component can be calculated from the partial pressure p i and the total pressure P T . Glossary Dalton's law of partial pressure partial pressure mole fraction percentage by mass Past Paper Question 91 1A 2c 95 2A 3 a ii 96 2A 2 a ii 91 1A 2c 2c A container holds a gaseous mixture of nitrogen and propane. The pressure in the container at 200ºC is 4.5 atm. At -40ºC the propane completely condenses and the pressure drops to 1.5 atm. Calculate the mole fraction of propane in the original gaseous mixture. 3 n t RT 1 = P 1 V 1 (1) n t (1 - x)RT 2 = P 2 V 1 (2) where n t is total number of moles, x is mole fraction of propane (2) ÷ (1) (1 - x) T 2 T 1 = P 2 P 1 (1 - x) = 1.5 4.5 · 473.15 233.15 = 0.067646 x = 0.3235 1 mark for correct final answer (0.315 - 0.324) 2 marks for intermediate answers (partial pressure, partial volume, no. of mole), equations and method. 95 2A 3 a ii 3a ii At 98.6 kPa and 300 K, the density of a sample of dry air is 1.146 g dm -3 . Assuming that dry air contains only nitrogen and oxygen and behaves ideally, calculate the composition of the sample. (Gas constant, R = 8.31 J K -1 mol -1 ) 3 Molar mass of dry air = 1.146 gdm -3 98.6 kPa × 8.31 JK -1 mol -1 × 300 K = 1.146 × 10 3 gm -3 98.6 × 10 3 Nm -2 × 8.31 JK -1 mol -1 × 300 K = 28.98 gmol -1 1 mark Let mole fraction of N 2 be x 28.02 x + 32.0 (1 - x) = 28.98 1 mark x = 0.759 Composition of dry air 75.9% N 2 and 24.1% O 2 1 mark C ii A number of candidates made mistakes in the numerical calculation and in the units used. 96 2A 2 a ii 2a ii At 298 K 1.0 dm 3 of N 2 at 0.20 Pa pressure was mixed with 2.0 dm 3 of O 2 at 0.40 Pa in a 4.0 dm 3 container. Assuming that both N 2 and O 2 behave ideally, calculate the pressure of the gaseous mixture at 298 K. 2 In the mixture partial pressure of N 2 = ¼ × 0.20 Pa = 0.05 Pa ½ mark partial pressure of O 2 = ½ × 0.40 Pa = 0.20 Pa ½ mark pressure of mixture = (0.05 + 0.20) Pa = 0.25 Pa 1 mark Chemical Kinetics Page 1 Chemical Kinetics I. Rates of chemical reactions A. Definition of rate of reaction 1. Unit of rate of reaction B. Measuring the rate of reaction 1. Different approaches a) Constant amount approach b) Constant time approach 2. Interpretation of physical measurements made in following a reaction a) Volume of gas formed b) Colorimetric measurement II. Order of reaction A. Rate Law or rate equation 1. Differential form and integrated form of rate law 2. Graphical presentation of reaction rate 3. Order of reaction a) Experimental determination of order of reaction (1) By measuring the rates of reaction at different reactant concentration (2) By plotting graph of ln rate versus ln [X] B. Examples of different reaction 1. First order reaction a) Half life (t ½ ) b) Carbon-14 dating c) Examples of calculation 2. Second order reaction 3. Zeroth order reaction III. Collision theory and activation energy A. Collision theory 1. Distribution of molecular speed (Maxwell-Boltzman distribution) 2. Effect of temperature change on molecular speeds 3. Collision theory and activation energy a) Arrhenius equation B. Determination of activation energy 1. By measuring the rate of reaction at different temperature 2. By plotting graph of ln k versus 1 T 3. By plotting graph of ln rate versus 1 T Chemical Kinetics Page 2 IV. Energy profile of reaction A. Transition state 1. Order of reaction leads to interpretation of reaction mechanism at molecular level a) Multi-stages reaction (1) Alkaline hydrolysis of 2-chloro-2-methylpropane (2) Reaction between bromide and bromate(V) in acidic medium b) Rate determining step 2. Energetic stability and kinetic stability B. Effect of catalyst 1. Characteristics of catalyst 2. Theory of catalyst a) Homogeneous catalysis b) Heterogeneous catalysis 3. Application of catalyst a) Use of catalyst in Haber process and hydrogenation b) Catalytic converter c) Enzymes V. Factors influencing the rate of reaction A. Concentration B. Temperature C. Pressure D. Surface area E. Catalyst F. Light I. Rates of chemical reactions Page 1 Topic I. Rates of chemical reactions Reference Reading 5.1 A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 298–300 Assignment Reading Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 397–399, 401–403 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 352–354, 356–358 Syllabus Definition of rate of reaction Unit of rate of reaction Measuring of rate of reaction Notes I. Rates of chemical reactions A. Definition of rate of reaction In every reaction, reactant is converted to product. Reactant → Product Rate of reactions is reflected by either rate of appearing of product or rate of disappearing of reactant. Rate of reaction = change in amount of product time taken = - change in amount of reactant time taken The amount can be measured in term of mass, volume or any physical quantities related to amount. If the reaction is performed in solution, the amount can also be expressed in term of concentration. Rate of reaction = ∆ [product] ∆ t = - ∆ [reactant] ∆ t [ ] : concentration Note : negative sign is added for the ∆ [reactant] ∆ t because rate of reaction is always positive but the quantity ∆ [reactant] ∆ t is negative. Rate of a reaction depends on : i. concentration ii. temperature iii. catalyst iv. surface area v. pressure vi. light 1. Unit of rate of reaction If the rate of reaction is expressed as the change in concentration of reactant or product with respect to time, ∆ [product] ∆ t , the unit of rate of reaction would be conc n time -1 (e.g. mol dm -3 s , moldm -3 s -1 ) I. Rates of chemical reactions Page 2 B. Measuring the rate of reaction By definition, rate is the change in certain quantity with respect to time. For example, speed is the change in distance with respect to time. Rate = ∆ amount ∆ time Speed = ∆ distance ∆ time 1. Different approaches There are two approaches to measure the rate, i. constant amount approach ii. constant interval approach a) Constant amount approach Constant amount approach – Like Formula-1 car racing, a fixed no. of laps is set. The one who finishes all the laps first would be the fastest one and would be the winner. e.g. Rate of disproportionation of thiosulphate in acidic medium Sodium thiosulphate and hydrochloric acid solution is put into a beaker. In acidic medium, thiosulphate decomposes into yellow suspension of sulphur. Once the cross on the paper is masked by the yellow suspension, the time is noted. As constant amount of sulphur precipitate is needed to mask the cross and the time taken is noted, the rate of reaction can be estimated. Rate of reaction = Amount of sulphur precipitate formed time taken Rate ∝ 1 time taken Different concentration of thiosulphate and acid can be used to study the kinetic of the reaction. I. Rates of chemical reactions Page 3 b) Constant interval approach Constant interval approach – Like Daytona 24 hours car racing, a fixed interval e.g. 24 hours is set. The one who can finish more laps in 24 hours would be the winner. If the amount of a reactant or a product can be traced at constant intervals throughout the reaction, the rate of reaction can also be determined. For example, the reaction of disproportionation of thiosulphate, S 2 O 3 2- (aq) + 2H 3 O + (aq) → SO 2(g) + S (s) + 3H 2 O (l) , can also be followed by titrating a fixed amount (e.g. 25.00 cm 3 ) of reacting mixture at regular interval (e.g. 2 minutes) with standard I 2(aq) solution using starch indicator. I 2(aq) + 2S 2 O 3 2- (aq) → 2I - (aq) + S 4 O 6 2- (aq) Before titration, the disproportionation reaction must be quenched (stopped) first. This reaction is catalyzed by acid, therefore, quenching can be done by neutralizing the acid with excess alkali, e.g. sodium hydrogencarbonate (a weak alkali is preferred) or by sudden cooling of the reaction mixture. Rate = - d [S 2 O 3 2- ] d t The rate of reaction at any particular instant can be calculated from the slope. Rate = - slope It is found that the reaction is first order with respect to S 2 O 3 2- (aq) and zero order with respect to H 3 O + (aq) . Rate = - d [S 2 O 3 2- ] d t = k [S 2 O 3 2- ] In another example, for the catalytic decomposition of H 2 O 2(l) in the presence of MnO 2(s) catalyst, it can be quenched by an acidic medium where acid is the negative catalyst to the reaction. The concentration of the H 2 O 2(l) can be determined by titrating with standard MnO 4 - (aq) . 5H 2 O 2(aq) + 2MnO 4 - (aq) + 6H + (aq) → 2Mn 2+ (aq) + 8H 2 O (l) + 5O 2(g) I. Rates of chemical reactions Page 4 2. Interpretation of physical measurements made in following a reaction Besides titration, there are many other methods to follow a reaction. There are many quantities proportional to the amount of a chemical in the reaction mixture. e.g. volume of gas, colour, conductivity and optical activity a) Volume of gas formed In the reaction between magnesium and hydrochloric acid, hydrogen gas is evolved. Mg (s) + 2HCl (aq) → MgCl 2(aq) + H 2(g) Rate = - 1 2 d[HCl (aq) ] dt = d[MgCl 2(s) ] dt = d[H 2(g) ] dt The rate of reaction can be traced by monitoring the volume of the hydrogen gas evolved. N.B. All pure solids have a constant density or concentration, equals mass volume . Therefore, the rate of a reaction is usually depending on the size of the solid particle rather than the amount of the solid present. A graph of volume of hydrogen against time b) Colorimetric measurement If one of the reacting substances or products has a colour, the intensity of this colour will change during the reaction. This method relies on the measurement of intensity of colour, therefore, it is called colorimetry. An example will be the disappearance of the colour of bromine during the oxidation of methanoic acid by bromine: Br 2(aq) + HCO 2 H (aq) → 2Br - (aq) + 2H + (aq) + CO 2(g) The change in colour intensity could be followed by a photoelectric device in a colorimeter, as shown in the following figure. Since the colour intensity is directly proportional to the concentration of Br 2(aq) , therefore, Rate ∝ - d (colour intensity of the solution) dt Glossary rate of reaction constant amount approach constant time approach initial rate of reaction quenching colorimetry I. Rates of chemical reactions Page 5 Past Paper Question 93 1A 2 a iii 95 2A 2 b i 96 1B 4 d i ii iii 97 2A 3 a i ii iii 98 2A 3 a iii 99 1B 7 a 93 1A 2 a iii 2a Answer the following questions by choosing in each case one of the species listed below, putting it in the box and giving the relevant equation(s). Al (s) , AlCl 3(s) , AlO 2 - (aq) , Na (s) , CO 3 2- (aq) , Cu 2+ (aq) , P 4 O 10(s) , S (s) , S 2 O 3 2- (aq) , Zn 2+ (aq) iii Which species forms a pale yellow precipitate with dilute hydrochloric acid? 2 S 2 O 3 2- (aq) S 2 O 3 2- (aq) + 2H + (aq) → S (s) + SO 2(g) + H 2 O (l) 2 marks C Many candidates picked the wrong chemical species for the inorganic reactions in parts (i) to (iv). 95 2A 2 b i 2b In an aqueous solution, the decomposition of hydrogen peroxide in the presence of manganese(IV) oxide can be represented by the following equation: 2H 2 O 2(aq) MnO 2 → 2H 2 O (l) + O 2(g) i For a given amount of manganese(IV) oxide, outline how you would use a chemical method to determine the concentration of hydrogen peroxide, at different times, in the reaction mixture. 2 Withdraw a known volume of the reaction mixture ½ mark and add to excess dilute H 2 SO 4 , ½ mark titrate H 2 O 2 against standard MnO 4 - . 1 mark 5H 2 O 2 + 2MnO 4 - + 6H + → 2Mn 2+ + 8H 2 O + 5O 2 C This question was poorly answered. Candidates should brush up their skills in answering questions which involve the description of experimental procedures. 96 1B 4 d i ii iii 4d In an experiment to determine the concentration of H 2 O 2 contained in a rainwater sample, 5.0 cm 3 of the sample were mixed with an excess of a certain transition metal complex solution, giving a coloured mixture. The absorbance of the mixture was measured by a colorimeter and was found to be 0.0273. When 5.0 cm 3 of 1.46 × 10 -6 M H 2 O 2 (instead of the rainwater sample) were treated in the same way, an absorbance of 0.0398 was recorded. i Calculate the concentration of H 2 O 2 in the rainwater sample assuming that concentration is directly proportional to absorbance. 2 conc n . of H 2 O 2 in the sample = 1.46 × 10 -6 0.0398 × 0.0273 1 mark = 1.00 × 10 -6 M / 1.0 × 10 -6 M 1 mark (Deduct ½ mark for answers given in 1 × 10 -6 M or 10 -6 M) ii Why is the method of titration not suitable for the determination of the concentration of H 2 O 2 in the rainwater sample ? 1 The concentration of H 2 O 2 in rainwater is too low to be determined by titrimetric method. 1 mark iii Why is it not suitable to collect the rainwater sample for this experiment in an iron container ? 1 Fe can be oxidized by H 2 O 2 / transition metal ions can catalyse the decomposition of H 2 O 2 . 1 mark I. Rates of chemical reactions Page 6 97 2A 3 a i ii iii 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I 2(aq) + CH 3 COCH 3(aq) → CH 2 ICOCH 3(aq) + H + (aq) + I - (aq) 3a i The progress of the reaction can be monitored by a titrimetric method. Outline the experimental procedure. 6 ii State how the initial rate of the reaction can be determined from the titrimetric results. iii Suggest another method to monitor the progress of the reaction. 98 2A 3 a iii 3a The table below lists the rate constants, k, at different temperatures, T, for the first order decomposition of a dicarboxylic acid, CO(CH 2 CO 2 H) 2 , in aqueous solution : CO(CH 2 CO 2 H) 2(aq) → CH 3 COCH 3(aq) + 2CO 2(g) 9 T / K 273 293 313 333 353 k / s -1 2.46 × 10 -5 4.75 × 10 -4 5.76 × 10 -3 5.48 × 10 -2 ? iii Suggest a method to monitor the progress of the reaction. 99 1B 7 a 7a In a chemical kinetics experiment, samples of the reaction mixture are removed at regular time intervals for titrimetric analysis. Suggest TWO methods by which the reaction in the samples removed can be stopped or stowed down. II. Order of reaction Page 1 Topic II. Order of reaction Reference Reading 5.2 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 96–99, 103–106 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 197–198 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 403–410 Experiment – Determination of the order of reaction of disproportionation of thiosulphate in acidic medium Assignment Reading A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 302–308 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 358–365 Syllabus Order of reaction Rate Law Notes II. Order of reaction A. Rate Law or rate equation Rate Law or rate equation – A mathematical expression relates the concentration of a species with the rate of reaction or time. There are three different forms of rate law i. ordinary form ii. differential form iii. integrated form Examples of rate law (ordinary form) e.g. H 2(g) + Br 2(g) → 2HBr (g) Rate = k’[H 2 ][Br 2 ] ½ 1+k’’( [HBr] [Br 2 ] ) e.g. 2NO (g) + O 2(g) → 2NO 2(g) Rate = k[NO] 2 [O 2 ] e.g. 2H 2(g) + 2NO (g) → 2H 2 O (g) + N 2(g) Rate = k[NO] 2 [H 2 ] N.B. The rate equation has no direct relationship with the stochiometric coefficient. It can only be determined by experiment. According to collision theory, the rate of reaction always increases with concentration of reactant. concentration of reactant ↑ ⇒ rate of reaction ↑ In some reactions, the rate is directly proportional to the concentration. i.e. double the concentration of a reactant, only double the rate of reaction. But in some reactions, double the concentration of a reactant may cause 4 folds increase in reaction rate. The relationship needs not be linear. e.g. H 2(g) + I 2(g) → 2HI (g) For the above reaction, it is found experimentally that Rate ∝ [H 2 ] The reaction is first order with respect to H 2(g) Rate ∝ [I 2 ] The reaction is first order with respect to I 2(g) Rate ∝ [H 2 ][I 2 ] Rate = k [H 2 ][I 2 ] Rate law or rate equation (an overall second order reaction) k, the proportionality constant, is known as rate constant which depends on temperature only. II. Order of reaction Page 2 Rate has the unit (mole dm -3 s -1 ) and concentration has the unit (mole dm -3 ). Therefore, the unit of rate constant, k, depends on actually what the rate equation is. e.g. In the rate equation, Rate = k [H 2 ][I 2 ], k = Rate [H 2 ][I 2 ] unit of k = mole dm -3 s -1 mole dm -3 · mole dm -3 = mole -1 dm 3 s -1 e.g. For the reaction, 2H 2(g) + 2NO (g) → 2H 2 O (g) + N 2(g) , Rate = k[NO] 2 [H 2 ], k = Rate [NO] 2 [H 2 ] unit of k = mole dm -3 s -1 (mole dm -3 ) 2 · mole dm -3 = mole -2 dm 6 s -1 1. Differential form and Integrated form of rate law (The derivation of the integrated form of rate law is not required in A-Level) Rate of reaction = ∆ [product] ∆ t = - ∆ [reactant] ∆ t By using notation of calculus Rate of reaction = d [product] dt = - d [reactant] dt (differential form of rate law) For the reaction, H 2(g) + I 2(g) → 2HI (g) Rate = - d [H 2 ] dt = - d [I 2 ] dt = 1 2 d [HI] dt and Rate = k[H 2 ][I 2 ] ∴ Rate = - d [H 2 ] dt = k[H 2 ][I 2 ] Keep the [I 2 ] constant by using excess I 2 - d [H 2 ] dt = k’[H 2 ] where k’ = k [I 2 ], the reaction is first order with respect to H 2(g) - d[H 2 ] [H 2 ] = k’dt ⌡ ⌠ [H 2 ] 0 [H 2 ] - d[H 2 ] [H 2 ] = k' ⌡ ⌠ 0 f dt [H 2 ] is the concentration at time = t [H 2 ] 0 is the concentration at time = 0 - [ ln [H 2 ] ] [H 2 ] [H 2 ] 0 = k' t - (ln [H 2 ] - ln [H 2 ] 0 ) = k’t ln [H 2 ] 0 [H 2 ] = k’t or [H 2 ] 0 [H 2 ] = e k't (Integrated form of rate law) at t = 0 [H 2 ] 0 = [H 2 ] at t = ∞ [H 2 ] 0 [H 2 ] = ∞ ⇒ [H 2 ] = 0 II. Order of reaction Page 3 2. Graphical presentation of reaction rate For the reaction H 2 O 2(aq) + 2I - (aq) + 2H + (aq) → 2H 2 O (l) + I 2(aq) The completeness of the reaction can be monitored by following the concentration of one of the reactant or product. Rate of reaction = ∆ [product] ∆ t = - ∆ [reactant] ∆ t Using notation of calculus Rate = d [I 2 ] dt = - d [H 2 O 2 ] dt = 1 2 - d [I - ] dt = 1 2 - d [H + ] dt Note the rate of disappearing of I - (aq) is twice of the rate of disappearing of H 2 O 2(aq) . Although H 2 O (l) is one of the product, it is also the solvent. Because it is in large excess, its concentration would be rather constant. For the reaction of decomposition of NO 2(g) at 300 ºC 2NO 2(g) → 2NO + O 2(g) Rate = 1 2 - d [NO 2 ] dt = 1 2 d [NO] dt = d [O 2 ] dt Notation of calculus is preferable to ∆ [product] ∆ t or - ∆ [reactant] ∆ t since the former one reflects the rate of reaction at a particular instant instead of average rate over a period of time. II. Order of reaction Page 4 3. Order of reaction Order of individual reactant – The order of a reaction with respect to a given reactant is the power of that reactant’s concentration in the experimentally determined rate equation. Overall order of reaction – The (overall) order of the reaction is the sum of the powers of the concentration terms in the experimentally determined rate equation. e.g. 2H 2(g) + 2NO (g) → 2H 2 O (g) + N 2(g) Rate = k[NO] 2 [H 2 ] The reaction is second order with respect to NO (g) The reaction is first order with respect to H 2(g) The overall order of the reaction is 3 or the reaction is a third order reaction. A zeroth order means that the rate of reaction is independent of the concentration of that reactant. A first order means that the rate of reaction is directly proportional to the concentration of that reactant. A second order means that the rate of reaction is directly proportional to the square of the concentration of that reactant. In general, the higher the order of a reactant, the more dependent will be the rate of reaction on the concentration of that reactant. II. Order of reaction Page 5 a) Experimental determination of order of reaction (1) By measuring the rates of reaction at different reactant concentrations Suppose a reaction X + Y → Z has the rate law rate = k[X] a [Y] b The order of reaction with respect to X (i.e. a) can be determined by measuring the rate of reaction with different concentration of X, while the concentration of Y is kept constant. rate 1 = k[X] 1 a [Y] b (1) rate 2 = k[X] 2 a [Y] b (2) (1) ÷ (2) rate 1 rate 2 = k[X] 1 a [Y] b k[X] 2 a [Y] b = ( [X] 1 [X] 2 ) a ln rate 1 rate 2 = a ln ( [X] 1 [X] 2 ) a = ln rate 1 rate 2 ÷ ln ( [X] 1 [X] 2 ) (2) By plotting graph of ln rate versus ln [X[ The order of reaction with respect to X can also be determined by graphical method. Since more than 2 readings are used to determine the order, the order determined will be more accurately than the above method. For the same reaction X + Y → Z rate = k[X] a [Y] b rate = k'[X] a Keep [Y] constant by using excess Y, since b is also a constant. [Y] b will be a constant. ln rate = ln (k'[X] a ) ln rate = ln k' + a ln [X] According to the above equation, the order of a reaction (i.e. a) with respect to a reactant (i.e. X) can be determination by measuring the rates of reaction (i.e. rate) with different concentrations of the reactant (i.e. [X[). The order of reaction can be obtained from the slope of the graph with (ln rate) plotting against ln [X]. II. Order of reaction Page 6 B. Examples of different reaction 1. First order reaction First order reaction – the rate of reaction is directly proportional to the concentration of the reactant. Since the rate of decay is directly proportional to the amount of isotope remains. This form an exponential decay curve. After 5 half-lives, the fraction remaining will be = ½ × ½ × ½ × ½ × ½ = 0.03125 = 3.125% Radioactive decay is said to be a first order reaction, since the rate is proportional to the number of radioactive isotopes present. a) Half life (t ½ ) The rate of radioactive decay reaction is directly proportional to the abundance of the radioactive isotope. Radioactive decay is so special that the rate of reaction is even independent of temperature. This is the only reaction which is independent of temperature. Radiation ∝ No. of isotope ∝ Rate of decay Rate of decay ∝ I I : abundance of radioactive isotope ∝ no. of isotope Rate = k I = -d I dt By integration according to time. ln I 0 I = kt or I 0 I = e kt I 0 : abundance of radioactive isotope at time = 0 I : abundance of radioactive isotope at time = t By definition, half life (t ½ ), is the time taken by a given amount of radioactive isotope to decay to half of the original amount. ln I 0 I = kt when t = t ½ then I 0 = 2 I ln 2I I = kt ½ = ln 2 t ½ = ln 2 k = 0.693 k N.B.: It can be seen that the half life (t ½ ) of a first order reaction is related to the decay constant (k). II. Order of reaction Page 7 b) Carbon-14 dating In nature the, 7 14 N in the air, is transmutated to 6 14 C by the neutron from the cosmic ray. i.e. 7 14 0 1 N n + ÷ → ÷ + 6 14 1 1 C H 6 14 C isotope is radioactive. At the same time, it undergoes beta decay which is a first order reaction. 6 14 7 14 1 0 C N e ÷ → ÷ + − As a result of this simultaneous formation and decay of 6 14 C , the atmosphere contains a constant concentration of 6 14 C . The 6 14 C is present as 6 14 2 CO . This 6 14 2 CO enters plants via photosynthesis and animals via the food chain. Therefore, all living things have a constant proportion of their carbon in form of 6 14 C . However, when the animal or plant dies, replacement of 6 14 C ceases while decay of 6 14 C continues. It is assumed that the concentration of 6 14 2 CO in atmosphere remains unchanged throughout million of years. By comparing the content of 6 14 C in archaeological specimens with that in similar living materials living at the present time, it is possible to estimate the age of the specimens. c) Example of calculation Half life (t ½ ) of 6 14 C = 5730 years If 32 counts per minute per gram of carbon is emitted by living organisms, while a fossilized bone gives 8 counts per minute per gram of carbon, calculate the age of the fossil. The number of counts is proportional to the abundance of the radioactive isotope. ln I 0 I = kt and t ½ = ln 2 k = 0.693 k ln 32 8 = ln 2 t ½ · t ln 4 = ln 2 5730 · t t = ln 4 · 5730 ln 2 = 2 × 5730 = 11460 years II. Order of reaction Page 8 2. Second order reaction Second order reaction – the rate of reaction is directly proportional to the square of the concentration of the reactant. Consider the reaction A (l) + B (l) → C (l) + D (l) which is first order with respect to A, but second order with respect to B. Rate = k[A][B] 2 if we keep A in large excess, then [A] will remain a constant throughout the reaction. Rate = k’[B] 2 the reaction becomes second order with respect to B, where k’ = k [A] Rate = k’[B] 2 = - d[B] dt - k’dt = d[B] [B] 2 ⌡ ⌠ 0 t - k' dt = ⌡ ⌠ [B] 0 [B] d[B] [B] 2 - k' t = - 1 [B] [B] [B] 0 - k' t = \ | . | - 1 [B] - \ | . | - 1 [B] 0 - k' t = 1 [B] 0 - 1 [B] 1 [B] = k' t + 1 [B] 0 By plotting a graph 1 [B] versus t, the slope will equal to k’. Since k’ = k [A], the rate constant (k) can be determined. Half life (t ½ ) of second order reaction 1 [B] = k' t + 1 [B] 0 at t = t½, [B] = 1 2 [B] 0 1 ½[B] 0 = k' t ½ + 1 [B] 0 k' t ½ = 1 ½[B] 0 - 1 [B] 0 k' t ½ = 2 [B] 0 - 1 [B] 0 = 1 [B] 0 t ½ = 1 k ' [B] 0 N.B. In contrast to the constant half life of a first order reaction, half life of a second order reaction is depending on the initial concentration of the reactant. II. Order of reaction Page 9 3. Zeroth order reaction Zeroth order reaction – the rate of reaction is independent of the concentration of the reactant. Actually, there is no reaction with an overall order equals zero. But zero order reaction with respect to individual reactant is commonly found in catalytic reaction. In synthesis of ammonia by Haber process, N 2(g) + 3H 2(g) d 2NH 3(g) , it is found that the reaction is zero order with respect to both N 2(g) and H 2(g) . This is because Fe catalyst is used in the synthesis. The availability of the surface of Fe is the limiting factor of the reaction. Both N 2(g) and H 2(g) are in large excess comparing with Fe catalyst. For a general zero order reaction, A → product Rate = k[A] 0 = k(1) = k Rate = k = - d[A] dt - kdt = d[A] ⌡ ⌠ 0 t -k dt = ⌡ ⌠ [A] 0 [A] d[A] -kt = [ ] [A] [A] [A] 0 - kt = [A] - [A] 0 [A] = - kt + [A] 0 Half life of zero order reaction [A] = - kt + [A] 0 1 2 [A] 0 = - kt ½ + [A] 0 kt ½ = [A] 0 - 1 2 [A] 0 = 1 2 [A] 0 t ½ = [A] 0 2 k Glossary order of reaction rate law / rate equation rate constant differential form of rate law integrated form of rate law overall order of reaction half-life carbon-14 dating decay constant II. Order of reaction Page 10 Past Paper Question 90 1A 2 d i 91 2A 1 a iii 92 1A 2 a i ii 92 2A 2 a ii iii 93 2A 2 b i ii 94 1B 4 e i ii iii 95 1A 1 d i ii 95 2A 2 a ii 95 2A 2 b ii 96 1A 1 b i ii 96 2A 3 c i ii 97 2A 3 b i ii c 98 2A 2 b i ii 98 2A 3 a ii 99 2A 3 a i ii 90 1A 2 d i 2d Consider the reaction A (l) + B (l) → C (l) + D (l) which is first order with respect to A, but second order with respect to B. i Give a rate expression for this reaction 1 Rate = k[A][B] 2 1 mark 91 2A 1 a iii 1a The radioactive decay of 92 238 U may be represented by 92 238 U → 82 206 Pb + α + β. iii The age of a mineral containing 92 238 U and 82 206 Pb can be estimated from the mole ratio, 82 206 Pb : 92 238 U . It is assumed that all 82 206 Pb has arisen only by the decay of the uranium and that all the subsequent decays are small in comparison with that of 92 238 U which has a half-life of 4.51 × 10 9 years. Calculate the age of the mineral if the mole ratio of 82 206 Pb : 92 238 U is 0.231 : 1.000. Note : The integrated form of a first order reaction my be expressed as ln I I kt 0 = where I 0 is the initial concentration and I is the concentration at time t. 4 The decay constant k can be calculated from the half-life. k = ln 2 1 4.51 10 9 × = 1.5369 × 10 -10 year -1 1½ mark The ratio of Pb : U = 0.231 : 1.000 The amount of U at the beginning (I 0 ) = 1.000 + 0.231 = 1.231 ½ mark The amount of U at time t (I) = 1 ½ mark ∴ t = ln . . 1231 15369 10 10 × − = 1.35 × 10 9 years (no / wrong unit -½) (3 to 4 sig. fig. if not -½) 1½ mark C Many candidates were not able to establish the correct ratio of I 0 /I, and hence failed to obtain the correct answer. 92 1A 2 a i ii 2a For the first order reaction, A product k ÷ → ÷ , the integrated form of the rate equation is [A] [A] e 0 kt = − , where [A] and [A] 0 are the concentrations of A at time = t and time = 0 respectively. i Starting from this equation, derive the relationship between the half-life t ½ of the reaction and the rate constant k. 1 The half life of the reaction is defined by the concentration condition, [A] = [A] 0 /2. ½ mark Substituting into the integrated rate law give [A] 0 /2 = [A] 0 e -kt . This is followed by taking the natural logarithm to give ln(0.5) = -kt ½ . ½ mark Therefore, t ½ = 0.693/k. ii Without using the half-life method, show how you would determine the rate constant k from a set of experimental measurements of concentration at various times. 2 From the integrated rate equation, an experimental straight line plot can be obtained by taking the ln of the integrated rate equation, ln[A] = ln[A] 0 - kt. ½ mark Which can be rearranged to give an equation in the form of a straight line (y = mx +b) ln [A] = -kt + ln[A] 0 1 mark Therefore, k can be determined from the slope of the plot ln[A] vs t. ½ mark C A good example to demonstrate that most candidates lack the most basic mathematical application / manipulation skills. No knowledge of calculus was required in answering this question. The candidates should have the ability to take In and to recognize equations in the form of a straight line. Few candidates were able to score full marks in this question. II. Order of reaction Page 11 92 2A 2 a ii iii 2a In acid solution, chlorate(V) ions, ClO 3- , slowly oxidize chloride ions to chlorine. The following kinetic data are obtained at 25°C. [ClO 3 - ]/mol dm -3 [Cl - ]/mol dm -3 [H + ]/mol dm -3 Initial rate/mol dm -3 0.08 0.15 0.20 1.0 × 10 -5 0.08 0.15 0.40 4.0 × 10 -5 0.16 0.15 0.40 8.0 × 10 -5 0.08 0.30 0.20 2.0 × 10 -5 ii Determine the order of the reaction with respect to each reactant. 3 Since the rate of reaction Rate = k[ClO 3 - ] l [Cl - ] m [H + ] n 1 mark 10 10 4 0 10 0 08 015 0 2 0 08 015 0 4 1 4 2 4 4 1 2 2 5 5 . . ( . ) ( . ) ( . ) ( . ) ( . ) ( . ) log log × × = ⇒ = | \ | . | ⇒ − = ⇒ = − − l m n l m n n n n Similarly l = 1 and m = 1. with working max. 3 marks without working: all correct 2 marks or wrong 0 mark iii Determine the rate constant at this temperature. 2 Rate = k[ClO 3 - ][Cl - ][H + ] 2 1 mark k rate mol dm s mol dm mol dm s = = × = × − − − − − − − [ClO ][Cl ][H ] 3 - - + 2 10 10 0 08 015 0 2 2 08 10 5 2 3 1 3 4 2 3 9 1 . ( . )( . )( . ) ( ) . 1 mark no unit -½ mark C Many candidates gave wrong units for the rate constant. 93 2A 2 b i ii 2b Given the following data for the reaction at 298 K 2C + 3D + E → P + 2Q Experiment [C] / mol dm -3 [D] / mol dm -3 [E] / mol dm -3 initial rate / mol dm -3 s -1 1 0.10 0.10 0.10 3.0 ? 10 -3 2 0.20 0.10 0.10 2.4 ? 10 -2 3 0.10 0.20 0.10 3.0 ? 10 -3 4 0.10 0.10 0.30 2.7 ? 10 -2 i Deduce the rate law of the above reaction. 4 Rate = k[C] x [D] y [E] z Since rate 1 rate 2 = ( 0.10 0.20 ) x = 3.0 × 10 -3 2.4 × 10 -2 = 1 8 ∴ x = 3 1 mark rate 1 rate 3 = ( 0.10 0.20 ) y = 3.0 × 10 -3 3.0 × 10 -3 = 1 ∴ y = 0 1 mark rate 1 rate 4 = ( 0.10 0.30 ) z = 3.0 × 10 -3 2.7 × 10 -2 = 1 9 ∴ z = 2 1 mark Rate = k[C] 3 [D] 0 [E] 2 = k[C] 3 [E] 2 1 mark ii Calculate the rate constant. 2 Using results of experiment 1 3.0 × 10 -3 = k (0.10) 3 (0.10) 2 k = 300 (mol dm -3 ) -4 s -1 (1 mark for answer, 1 mark for correct unit) 2 marks C Some candidates could not do the calculation of the rate constant, due to their unfamiliarity in handling exponential values. II. Order of reaction Page 12 94 1B 4 e i ii iii 4e In an experiment to determine the rate constant at 298 K for the decomposition of sodium thiosulphate by a large excess of dilute hydrochloric acid, the time, t, taken for a certain amount of sulphur to appear was measured. Under these conditions: Rate = k’[S 2 O 3 2- ]. i Write a balanced equation for the reaction between hydrochloric acid and sodium thiosulphate. 1 S 2 O 3 2- (aq) + 2H + (aq) → H 2 O (l) + SO 2(g) + S (s) or Na 2 S 2 O 3 + 2HCl → 2NaCl + S + SO 2 + H 2 O 1 mark ii Describe how you would carry out this experiment, indicating the measurements you would take. 4 Make standard [S 2 O 3 2- ] solution of different molarity and place equal volume of these solutions in conical flasks. 1½ mark Measure temperature of solution / set thermostat to 298 K ½ mark Add a fixed volume of acid to Na 2 S 2 O 3 (using measuring cylinder) ½ mark Start stop clock ½ mark Note time, t, when a black “X” marked on a cardboard becomes invisible when looking through the flask. 1 mark iii What quantities would you plot on a graph for the determination of the rate constant k’ at 298 K? 1 1 t / s -1 and [S 2 O 3 2- (aq) ] / mol dm -3 1 or 0 mark C Many candidates have previous experience of this experiment, but nonetheless their answers were poor. The temperature control was omitted in most cases, and many candidates did not give a method for measuring the time taken for the formation of a certain amount of sulphur. Some utilized titration, while others used a colorimeter or filtering and weighing to quantitate sulphur, but none of these answers were acceptable. In (iii), candidates did not know that rate is inversely proportional to reaction time. 95 1A 1 d i ii 1d The iodination of propanone is catalysed by hydrogen ions. The overall equation is: CH 3 COCH 3(aq) + I 2(aq) → CH 3 COCH 2 I (aq) + HI (aq) Using four mixture B, C, D and E, the progress of the reaction was followed by colorimetric measurement. The results are tabulated below. Composition by volume of mixture /cm 3 Mixture propane water 1.00 M HCl 0.05 M I 2 in KI Initial rate / mol dm -3 s -1 B 10.0 60.0 10.0 20.0 4.96 × 10 -6 C 10.0 50.0 10.0 30.0 5.04 × 10 -6 D 5.0 65.0 10.0 20.0 2.45 × 10 -6 E 10.0 65.0 5.0 20.0 2.47 × 10 -6 i Determine the effects of the changes in concentration of each of the reactants (iodine and propanone) and the catalyst (hydrochloric acid) on the reaction rate. Write the rate expression for the reaction. 2 I 2 , use B and C : rate independent of [I 2 ] ½ mark propane, use B and D : rate ∝ [propanone] ½ mark H + , use B and E : rate ∝ [H + ] ½ mark rate = k [H + ] [CH 3 COCH 3 ] ½ mark ii For mixture B, calculate the rate constant for the reaction at the temperature of the experiment. (Density of CH 3 COCH 3 = 0.789 gcm -3 ) 2 rate = 4.96 × 10 -6 = k [0.1] [ 100 × 0.789 58.078 ] 1 mark k = 58.078 × 4.96 × 10 -6 0.1 × 100 × 0.789 = 3.65 × 10 -5 mol -1 dm 3 s -1 1 mark (½ mark for numerical answer; ½ mark for unit) C Many candidates were unable to deduce the quantitative relation between reaction rate and change in concentration of the reactants, merely stating that 'rate increases with increased concentration'. In some cases, the experimental nature of the rate results was not appreciated, so that the order with respect to iodine was calculated to be non-zero or fractional. Some candidates had difficulty in calculating the concentrations of the reactants from the data given. II. Order of reaction Page 13 95 2A 2 a ii 2a 88 228 Ra decays by the emission of β particles. The half-life for the decay is 6.67 years. ii A sample containing 0.50 g of 88 228 Ra is kept in a closed container. Calculate the mass of 88 228 Ra remaining after 5 years. (The decay of a radioisotope can be described by the equation, N t = N 0 e -kt , where N 0 and N t respectively represent the initial number and number at time t of the radioisotope, and k is the decay constant.) 3 k = ln 2 t½ = 0.104 year -1 1 mark ln 0.5 N t = 0.104 (5) 1 mark N t = 0.297 g 1 mark 95 2A 2 b i ii 2b In an aqueous solution, the decomposition of hydrogen peroxide in the presence of manganese(IV) oxide can be represented by the following equation: 2H 2 O 2(aq) MnO 2 ÷ → ÷÷ 2H 2 O (l) + O 2(g) i For a given amount of manganese(IV) oxide, outline how you would use a chemical method to determine the concentration of hydrogen peroxide, at different times, in the reaction mixture. 2 Withdraw a known volume of the reaction mixture ½ mark and add to excess dilute H 2 SO 4 , ½ mark titrate H 2 O 2 against standard MnO 4 - . 1 mark 5H 2 O 2 + 2MnO 4 - + 6H + → 2Mn 2+ + 8H 2 O + 5O 2 ii How would you use the results obtained in (i) to show graphically that the decomposition is first order with respect to H 2 O 2 and to find the rate constant of the decomposition ? 4 Plot [H 2 O 2 ] against t to obtain the following graph 2 marks for first order reaction t ½ = constant and k = ln 2 t ½ 2 marks OR Plot ln [H 2 O 2 ] against t to obtain a straight line 1 mark ln [H 2 O 2 ] [H 2 O 2 ] 0 = - kt ⇒ ln[H 2 O 2 ] - ln[H 2 O 2 ] 0 = -kt 2 marks for first order reaction, slope = -k. 1 mark C This question was poorly answered. Candidates should brush up their skills in answering questions which involve the description of experimental procedures. 96 1A 1 b i ii 1b Carbon-14, 14 C, is radioactive, emitting β-particles. i Write an equation for the decay of 14 C. 1 6 14 1 0 7 14 C e N → + − or 6 14 7 14 C N → + β 1 mark ii A charcoal sample from the ruins of an ancient settlement was found to have a 14 C/ 12 C ratio 0.60 times that found in living organisms. (1) Explain why the 14 C/ 12 C ratio in the charcoal sample is smaller than that in living organisms. (2) Given that the half-life for the decay of 14 C is 5730 years, calculate the age of the charcoal sample. (Note : The integrated form of the rate expression for radioactive decay can be represented by the following 3 II. Order of reaction Page 14 equation : log 10 ( N 0 N t ) = 0.301 t ½ t where N 0 is the initial number of radioactive nuclei; N t is the number of radioactive nuclei at time t; t ½ is the half-life for the decay) (1) In the piece of charcoal intake of C stops; ½ mark decay of 14 C continues while 12 C remains constant ∴ 14 C / 12 C ratio drops ½ mark (2) log 1.0 0.6 = 0.301 t 5730 ⇒ t = 4223 ± 3 year 1 mark + 1 mark (Deduct ½ marks for wrong /no unit) 96 2A 3 c i ii 3c The following equation represents the decomposition of N 2 O 5(g) : 2N 2 O 5(g) → 4NO 2(g) + O 2(g) The progress of the above decomposition can be followed by measuring the partial pressures of N 2 O 5(g) , P N O 2 5 , at different times, t. The table below lists the results of such an experiment : t / minute 0 100 200 300 400 550 800 1250 P N O 2 5 / kPa 13.3 10.9 8.9 7.2 5.9 4.4 2.7 1.1 i By plotting a suitable graph, find the order of the decomposition. 3 t 0 100 200 300 400 550 800 1250 ln(P N 2 O 5 ) 2.59 2.39 2.19 1.97 1.77 1.48 0.99 0.10 or log(P N 2 O 5 ) 1.12 1.04 0.95 0.86 0.77 0.4 0.43 0.04 Plotting the following graph 2 marks (1 mark for the straight line, 1 mark for correct labelling of the axes) the reaction is 1st order 1 mark ii Using your graph in (i), determine the rate constant of the decomposition. 2 From the plot of ln(P N 2 O 5 ) against t, - slope = k 1 mark = 1.99 × 10 -3 minute -1 (2.2 × 10 -3 – 1.8 × 10 -3 min -1 ) 1 mark (1 mark for numerical answer, ½ mark for unit) II. Order of reaction Page 15 97 2A 3 b i ii c 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I 2(aq) + CH 3 COCH 3(aq) → CH 2 ICOCH 3(aq) + H + (aq) + I - (aq) 3b The following initial rates and initial concentrations were obtained in an experiment at 298 K: 5 Initial concentration / mol dm -3 Initial rate / mol dm -3 s -1 [I 2(aq) ] [CH 3 COCH 3(aq) ] [H + (aq) ] 3.5 × 10 -5 2.5 × 10 -4 2.0 × 10 -1 5.0 × 10 -3 3.5 × 10 -5 1.5 × 10 -4 2.0 × 10 -1 5.0 × 10 -3 1.4 × 10 -4 2.5 × 10 -4 4.0 × 10 -1 1.0 × 10 -2 7.0 × 10 -5 2.5 × 10 -4 4.0 × 10 -1 5.0 × 10 -3 i Deduce the rate equation for the reaction. ii Calculate the rate constant for the reaction at 298 K. 3c Suppose that the reaction takes place in a buffer solution of pH 4. On the basis of your results in (b) deduce the half-life of the reaction at 298 K. 4 98 2A 2 b i ii 2b Potassium-40 is radioactive and decays to give the stable isotope, argon-40. The half-life of the decay is 1.27 × 10 9 years. 5 i In a rock sample, the ratio of 40 K to 40 Ar is 1 to 9. Estimate the age of the rock sample. ii The above method of estimation is based on several assumptions. One of the assumptions is that all 40 Ar present in the rock sample is derived from the decay of 40 K. Give one other assumption relating to 40 Ar. 98 2A 3 a ii 3a The table below lists the rate constants, k, at different temperatures, T, for the first order decomposition of a dicarboxylic acid, CO(CH 2 CO 2 H) 2 , in aqueous solution : CO(CH 2 CO 2 H) 2(aq) → CH 3 COCH 3(aq) + 2CO 2(g) 9 T / K 273 293 313 333 353 k / s -1 2.46 × 10 -5 4.75 × 10 -4 5.76 × 10 -3 5.48 × 10 -2 ? ii Estimate the rate constant of the reaction at 353 K and hence calculate the half-life of the reaction at the same temperature. 99 2A 3 a i ii 3a Consider the following data for the reaction A + B → products Initial concentration / mol dm -3 [A] [B] Initial rate / mol dm -3 s -1 4.0 × 10 -2 4.0 × 10 -2 6.4 × 10 -5 8.0 × 10 -2 4.0 × 10 -2 12.8 × 10 -5 4.0 × 10 -2 8.0 × 10 -2 6.4 × 10 -5 For this reaction, i deduce its rate equation, ii calculate the rate constant, and III. Collision theory and activation energy Unit 1 Page 1 Topic III. Collision theory and activation energy Unit 1 Reference Reading 5.3 Assignment Reading A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 172–173, 310–311 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 410–412 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 140–142, 266–367, 369 Syllabus Distribution of molecular speed Effect of temperature change on molecular speeds Collision theory and activation energy Arrhenius equation Notes III. Collision theory and activation energy A. Collision theory 1. Distribution of molecular speed (Maxwell-Boltzmann distribution) For a sample of gas at a specific temp, it is found that most particles possess speeds about average speed. Only very few have very high or very low speed (extreme speeds). This gives a Maxwell-Boltzmann distribution. Maxwell-Boltzmann distribution is different from normal distribution in the way that it is not symmetrical. Origin of Maxwell-Boltzmann distribution Gas particles are in constant random motion. During collisions, some particles gain energy from the others and some particles loss energy to the others. After a series of collisions, different particles will have different speed. Maxwell-Boltzmann distribution 2. Effect of temperature change on molecular speeds As temperature increases, collisions among the particles become more vigorous. This causes a wider spread in particle speed. The kinetic energy of the gas increases with increasing temperature, so the peak of most probable speed shifts to the right. The area under the curve represents the total no. of particle and should remain constant and independent of the temperature. This makes the height of the peak drops at higher temperature. III. Collision theory and activation energy Unit 1 Page 2 3. Collision theory and activation energy To simplify the discussion, a sample of ideal gas is considered first. Except in self decomposition, all reacting particles must get in contact with other for reaction to take place. The collision frequency (Z) is proportional to the velocity of the particle (c). Z ∝ c Conclusion 1 – Reaction takes place as a result of collision. The simple collision theory is contradictory to the experimental finding. For most reactions, every 10 K increase in temperature will double the reaction rate. Meanwhile, an increase in 10 K at room temperature will only cause 1.6 % increase in particles velocity. Reaction is something more than just collision. Experimental finding finds that reaction rate increases exponentially with increasing temperature. Refering to the Maxwell-Boltzmann distribution of molecular speed, by mathematical approximation, the fraction of the shaded area is represented by e E RT A − , where E A is called the activation energy. shaded area total area = no. of particles having energy greater than E A total no. of particles = e E RT A − The value e E RT A − increases exponentially with temperature. In order to explain the expontential increase of reaction rate with temperature, it is assumed that a reaction will only take place if the collision has an energy exceeds the activation energy E A . Different reaction has different value of E A . Rate ∝ Z and Rate ∝ e E RT A − therefore Rate ∝ Ze E RT A − Z : collision frequency R : (universal) gas constant = 8.314 JK -1 mol -1 E A : activation energy T : temperature in Kelvin Conclusion 2 – Reaction only occurs when the collision exceeds certain threshold energy, activation energy. III. Collision theory and activation energy Unit 1 Page 3 Since the rate of reaction is directly proportionally to the rate constant, k. Rate ∝ k k ∝ Ze E RT A − Z : collision frequency k = PZe E RT A − P : steric factor / orientation factor (a proportionality constant) k = Ae E RT A − where A = PZ A is called Arrhenius factor Through experiment, P is found to be a factor always smaller than 1. It is a factor reflects the steric factor and the fraction of collisions with effective orientation. Conclusion 3 – Reaction only occurs when the colliding particles are correctly oriented to one another. According to collision theory, a reaction will take place only if the following 3 conditions are satisfied and the collision is then called an effective collision. i. Collision ii. Exceed activation energy iii. Correct orientation of collision a) Arrhenius equation The equation k = Ae E RT A − is called Arrhenius equation. This equation relates the temperature and activation energy with rate constant, thus rate of reaction. k : rate constant (a constant depends on temperature only) A : Arrhenius constant E A : activation energy e E RT A − : activation state factor (fraction of particle has energy greater than E A at temperature T) Rate of reaction = k [Reactants] n = Ae E RT A − [Reactants] n From the above equation, it can be seen that the rate of a reaction is depending on 1. Temperature 2. Nature of the reaction (Activation energy, i.e. Steric factor & Orientation factor) 3. Concentration of reactants Glossary Maxwell-Boltzmann distribution collision theory activation energy collision frequency (Z) Gas constant (R) threshold Arrhenius constant (factor) III. Collision theory and activation energy Unit 1 Page 4 Past Paper Question 96 2A 1 b i ii 98 2A 3 b i ii 96 2A 1 b i ii 1b i Draw the Maxwell-Boltzmann curves for the distribution of molecular speeds at two different temperatures for an ideal gas. 2 Plot of Boltzmann distribution curve at two temperatures T 1 and T 2 , where T 2 > T 1 (½ marks for correct shape of the curves; ½ marks for indicating T 2 > T 1 ) ii Use your curves in (i) to explain why, for a fixed mass of an ideal gas at constant volume, the pressure increases as the temperature is raised. 2 As temperature increases, fraction of molecules with high average speed / kinetic energy increases. 1 mark Hence, force exerted by the collision of molecules on the container wall / the chance in momentum of molecules upon collision / frequency of collision increases. 1 mark 98 2A 3 b i ii 3b The exothermic reaction E(g) → E'(g) (1) is a single stage reaction. i Sketch curves to show the distribution of molecular kinetic energy of the reactant, E(g), at two different temperatures. ii With reference to your answer in (i), explain why the rate of reaction (1) increases with temperature. III. Collision theory and activation energy Unit 2 Page 1 Topic III. Collision theory and activation energy Unit 2 Reference Reading 5.4 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 199–200 Reading Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 312 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 413–414 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 100–101 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 365, 367–368 Syllabus Determination of activation energy Notes B. Determination of activation energy By rearranging the Arrhenius equation, the activation energy of a particular reaction can be determined experimentally. k = Ae E RT A − ln k = ln Ae E RT A − ln k = ln A + ln e E RT A − ln k = ln A - E A RT ln e ln k = ln A - E A RT where ln e = 1 1. By measuring the rate of reaction at different temperature Rate ∝ k Rate 1 Rate 2 = k 1 k 2 By measuring the rate of reaction at 2 different temperature, 2 set of data – Rate 1 , T 1 and Rate 2 , T 2 can be obtained. ln k 1 = ln A - E A RT 1 Eq. 1 ln k 2 = ln A - E A RT 2 Eq. 2 Eq. 1 - Eq. 2 ln k 1 - ln k 2 = (ln A - E A RT 1 ) - (ln A - E A RT 2 ) ln k 1 k 2 = - E A R ( 1 T 1 - 1 T 2 ) ln Rate 1 Rate 2 = - E A R ( 1 T 1 - 1 T 2 ) By putting the Rates and temperature determined, R = 8.314 J -1 K -1 mol -1 . The activation energy, E A , can be determined. III. Collision theory and activation energy Unit 2 Page 2 2. By plotting graph of ln k versus 1 T To be more accurate, A and E A can then be determined by plotting a graph with ln k versus 1 T k at different temperature can be determined by measuring the rates of a reaction at different temperatures according to the integral form of rate equation. (Refer to order of reaction and rate constant). 3. By plotting graph of ln rate versus 1 T Actually, it is very time consuming to use plotting of ln k versus 1 T to determine the activation energy. In order to determined a value of k at a certain temperature, a set of values of rate / concentration have to be determined first. (Refer to order of reaction and rate constant). To determine the E a , many sets of values of rate / concentration at different temperature have to be measured. Since Rate ∝ k and k = Ae E RT A − Rate ∝ Ae E RT A − Rate = rAe E RT A − r : a proportionality constant Rate = qe E RT A − q : another proportionality constant = rA ln rate = ln qe E RT A − ln rate = ln q + ln e E RT A − ln rate = ln q - E A RT ln e ln rate = ln q - E A RT where ln e = 1 ln rate = ln q - E A R 1 T By plotting a graph ln rate versus 1 T , the slope of the graph will equal to - E A R . This provides a more convenient way to determine the activation energy if the determination of the Arrhenius constant is not required. III. Collision theory and activation energy Unit 2 Page 3 Glossary Arrhenius equation Past Paper Question 91 2A 3 a i ii 92 1A 2 e 93 2A 2 a 94 2A 1 c iii 96 1A 1 b iii 97 2A 3 d 98 2A 3 a i 91 2A 3 a i ii 3a For the reaction 2XY (g) → X 2(g) + Y 2(g) , the rate constant is 3.91 × 10 -4 mol -1 dm 3 s -1 at 370ºC and 4.05 × 10 -2 mol - 1 dm 3 s -1 at 470ºC. Generally the rate constant of a reaction is related to the temperature by k = A exp(-E a /RT). Calculate i the activation energy, 4 ln ln k A E RT a = − ln k k E R ( 1 T 1 T ) T T a 1 2 2 1 = − 1 mark T 1 = 273 + 370 = 643 K T 2 = 273 + 470 = 743 K ½ mark ln . . . ( ) 4 05 10 391 10 8 31 1 643 1 743 2 4 × × = + − − E a ½ mark E a = 184.2 kJmol -1 (wrong unit -½) (not 3-4 sig.fig. -½) 2 marks ii the rate constant at 450ºC (Gas constant, R = 8.31 JK -1 mol -1 ) 2 450 ºC ≡ 723 K ln( . ) . . ( ) k 723 4 3 391 10 184 2 10 8 31 1 643 1 723 × = × − − k 723 = 1.77 × 10 -2 mol -1 dm 3 s -1 (wrong unit -½) 2 marks C i Some candidates were not familiar with calculations involving the use of natural logarithm. ii Some weaker candidates failed to realize that the absolute temperature should be used in the calculation. 92 1A 2 e 2e The diagram below gives the Maxwell-Boltzmann energy distribution of a system of molecules at two temperatures: What do the shaded areas of the curves represent and why are they different at different temperatures? 2 These curves plot the distribution of molecular kinetic energies at different temperatures. It is intended to reflect that the most sensitive parameter which affects the reaction rate is the activation energy, E a . ½ mark The shaded area under either one of these curves corresponds to E ≥ E a and is approximately equal to the fraction of molecules that collide with kinetic energy ≥ E a . ½ mark The increase of the reaction rate with temperature corresponds closely to the ratio of the corresponding areas where E a represents the minimum collision energy necessary for the reaction to occur. ½ mark Therefore, as the temperature changes, the area under the distribution curve for E ≥ E a changes. ½ mark C The poor performance to the first part of this question was disappointing. This is a standard question which has appeared almost on a yearly basis, yet candidates continued to fail to clearly explain the meaning of the various features of the Maxwell-Boltzmann distribution curve. III. Collision theory and activation energy Unit 2 Page 4 93 2A 2 a 2a Discuss, in terms of the Arrhenius equation, the effect of temperature on the rate of a reaction. 2 The rate of reaction is normally increased by raising the temperature. 1 mark Arrhenius Equation: ln k = - E a RT + C or k = Ae E RT a − Increase in T makes 1 T smaller and - E a RT larger, hence ln k (k) increases. 1 mark 94 2A 1 c iii 1c The table below lists the concentration of the reactant C as a function of time at 298 K for the following reaction. C → D Time / s 0 60 120 180 240 300 [C] /10 -2 mol dm -3 20.0 16.1 13.5 11.6 10.2 9.1 (Gas constant R = 8.31 JK -1 mol -1 ) iii The rate constant of the above reaction is found to be doubled when the temperature is raised from 298 K to 306 K. Determine the activation energy of the reaction. 2 Arrhenius Equation ln k = C - E a RT (1) ln k = C - E a R ( 1 298 ) (2) ln k + ln 2 = C - E a R ( 1 306 ) (2) - (1) ln 2 = E a R ( 1 298 - 1 306 ) 1 mark E a = 65.7 kJ (no unit -½ mark) 1 mark 96 1A 1 b iii 1b Carbon-14, 14 C, is radioactive, emitting β-particles. iii All radioactive decay has zero activation energy. Comment on the effect of temperature upon the rate of decay. 1 According to Arrhenius Equation, rate of decay is independent of temperature. 1 mark 97 2A 3 d 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I 2(aq) + CH 3 COCH 3(aq) → CH 2 ICOCH 3(aq) + H + (aq) + I - (aq) 3d For a given set of initial concentrations the initial rate doubles when temperature is increased from 298 K to 308 K. Calculate the activation energy of the reaction. 2 98 2A 3 a i 3a The table below lists the rate constants, k, at different temperatures, T, for the first order decomposition of a dicarboxylic acid, CO(CH 2 CO 2 H) 2 , in aqueous solution : CO(CH 2 CO 2 H) 2(aq) → CH 3 COCH 3(aq) + 2CO 2(g) 9 T / K 273 293 313 333 353 k / s -1 2.46 × 10 -5 4.75 × 10 -4 5.76 × 10 -3 5.48 × 10 -2 ? i Determine the activation energy for the reaction by plotting an appropriate graph. IV. Energy profile of reaction Unit 1 Page 1 Topic IV. Energy profile of reaction Unit 1 Reference Reading 5.5 Reading Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 313–315 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 417–420 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 373–375 Syllabus Energy profile Transition state Reaction mechanism Rate determining step Notes Consider the reaction between bromide and bromate(V) ion 5Br - (aq) + BrO 3 - (aq) + 6H + (aq) → 3Br 2(aq) + 3H 2 O (l) Kinetic studies show that this is an overall fourth order reaction. Rate = k[Br - ][BrO 3 - ][H + ] 2 The theory of collision and activation energy cannot account for the order of the reaction determined. IV. Energy profile of reaction During a reaction, bonds are in the process of making and breaking. The idea of transition state (activated complex) comes from an assumption that in any reaction, the reacting molecules/ions, if acquired sufficient energy to overcome the energy barrier, would first form a transition state. The reaction would then continue to proceed as the transition state, which is energetically unstable, immediately decompose to form the desired products. Consider a simple reaction of hydrolysis of 1-bromobutane in aqueous alkaline solution, as it proceeds via the following mechanism: A. Transition state Transition state is an activated complex at the top of the potential curve, which is so unstable and can never be isolated. In the hydrolysis of 1-bromobutane, the transition state is a pentavalent trigonal bipyramidal complex. The transition state is highly unstable having 10 valence electrons. Activation energy (E a ) – the energy difference between the reactant and the transition state (the threshold energy level leading to a reaction) The energy profile for the reaction can be depicted as follows. IV. Energy profile of reaction Unit 1 Page 2 1. Order of reaction leads to interpretation of reaction mechanism at molecular level Hydrolysis of 1-bromobutane is a second order reaction, the rate of reaction is directly proportional to the concentration of OH - (aq) ion and concentration of 1-bromobutane. Rate ∝ [OH - ][CH 3 CH 2 CH 2 CH 2 Br] Rate = k[OH - ][CH 3 CH 2 CH 2 CH 2 Br] It is proposed that the transition state is formed upon the collision between a OH - particle and a CH 3 CH 2 CH 2 CH 2 Br. Since the formation involves only 1 OH - particle and 1 CH 3 CH 2 CH 2 CH 2 Br molecule, the rate of formation of the transition state is directly proportional to [OH - ] and [CH 3 CH 2 CH 2 CH 2 Br]. Hydrolysis of 1-bromobutane is called a bimolecular nucleophilic substitution reaction (S N 2 reaction), since the formation of the transition state involves 2 molecules. The molecularity of the reaction is said to be 2. N.B. Molecularity is not the same as order the reaction. Molecularity is the number of the molecule involved in the formation of the transition state in a step of the reaction. While, order of reaction is the sum of the power of the concentration terms in the rate equation. They may or may not have the same value but their meaning are indeed different. a) Multisteps reaction (1) Alkaline hydrolysis of 2-chloro-2-methylpropane Alkaline hydrolysis of 2-chloro-2-methylpropane is found to be a first order reaction. C CH 3 CH 3 CH 3 Cl C CH 3 CH 3 CH 3 OH OH - (aq) Cl - (aq) + + Rate = k[(CH 3 ) 3 CCl] It is very clear that the reaction is not a S N 2 reaction and OH - ion is not involved in the formation of the transition state. Indeed, it is a multisteps reaction consists of 2 steps Step 1 (slow) Cl - C CH 3 CH 3 CH 3 + C CH 3 CH 3 CH 3 Cl + Step 2 (fast) C CH 3 CH 3 CH 3 + - OH C CH 3 CH 3 CH 3 OH As formation of the transition state for step 1 involves a higher activation energy E a ’, the step 1 would be slower and known as the rate determining step. The reaction is called unimolecular nucleophilic substitution (S N 1 reaction) because only 1 molecule is involved in the formation of the transition state of the slowest step. The reaction, therefore, is only first order with respect to (CH 3 ) 3 CCl. C CH 3 CH 3 CH 3 + is called a carbocation or carbonium ion. It is an intermediate of this reaction. Because it is at the bottom of a potential well, unlike the highly unstable transition state, it is possible to isolate the intermediate from the reaction mixture. IV. Energy profile of reaction Unit 1 Page 3 (2) Reaction between bromide and bromate(V) in acidic medium For the reaction, 5Br - (aq) + BrO 3 - (aq) + 6H + (aq) → 3Br 2(aq) + 3H 2 O (l) Rate = k[Br - ][BrO 3 - ][H + ] 2 If the transition state is formed by collision between Br - , BrO 3 - and 2 H + particles, the reaction rate would be extremely slow. The probability of having four particles collide at the same time is extremely low. b) Rate determining step For multisteps reaction, the rate of reaction is only controlled by the step with the slowest rate. The step with the slowest rate is called the rate determining step. Since the rate of the reaction is independent of the other steps. It is proposed that the reaction between bromide and bromate(V) in acidic medium involves 5 steps. Step 1 H + + Br - → HBr Fast Step 2 H + + BrO 3 - → HBrO 3 Fast Step 3 HBr + HBrO 3 → HBrO + HBrO 2 Slow Step 4 HBrO 2 + HBr → 2HBrO Fast Step 5 HBrO + HBr → H 2 O + Br 2 Fast Since Step 3 is the slowest step, it would be the rate determining step. Rate of reaction ∝ [HBr][HBrO 3 ] [HBr] ∝ rate of formation of HBr ∝ [H + ][Br - ] [HBrO 3 ] ∝ rate of formation of HBrO 3 ∝ [H + ][BrO 3 - ] Therefore rate of reaction ∝ [HBr][HBrO 3 ] ∝ ([H + ][Br - ])([H + ][BrO 3 - ]) ∝ [Br - ][BrO 3 - ][H + ] 2 Rate = k[Br - ][BrO 3 - ][H + ] 2 N.B. For the concentration of HBr, the rate of disappearing of HBr in Step 4 and 5 need not to be considered. Since Step 4 and 5 is following the rate determining step. The rate of Step 4 and 5 would only be the same as Step 3. IV. Energy profile of reaction Unit 1 Page 4 2. Energetic stability and kinetic stability The reaction N 2(g) + 3H 2(g) → 2NH 3(g) is exothermic. N 2(g) + 3H 2(g) → 2NH 3(g) ∆H = -92 kJmol -1 . However, N 2(g) does not react with H 2(g) at room temperature. This is because the reaction involves a very high activation energy, E A = 668 kJmol -1 . A mixture of N 2(g) and H 2(g) is said to be energetically unstable but kinetically stable. The high activation energy is assoicated with the high bond energies of N≡N and H–H bond. E(N≡N) 945.4 kJmol -1 E(H–H) 435.9 kJmol -1 Glossary energy profile transition state (activated complex) reaction mechanism bimolecular nucleophilic substitution (S N 2) molecularity unimolecular nucleophilic substitution (S N 1) carbocation / carbonium ion intermediate rate determining step energetic stability kinetic stability Past Paper Question 91 1A 2 b iii 93 1A 1 c i ii 94 2A 2 a iii 95 2B 4 c ii 99 2A 3 a i ii iii 91 1A 2 b iii 2b The energy profile of the reaction A (g) + B (g) d C (g) under two different catalysis X and Y are represented below. iii Compare the effect of increasing temperature on the rate of reaction in the system. 2 ↑ temp. will increase the rate of reaction of both systems, 1 mark the % increase in rate will be large for system X 1 mark C Many candidates discussed only one system when the question asked for the effect on each system. IV. Energy profile of reaction Unit 1 Page 5 93 1A 1 c i ii 1c Consider the energy diagram shown below for a certain reaction which takes place with three steps. i Which is the rate-determining step? Explain. 2 Step 3. It is the intermediate step with highest activation energy, ∴ it is the slowest step. 2 marks ii Is the reaction endothermic or exothermic? Explain. 2 Exothermic 1 mark since the potential energy of the products is lower than that of the reactant. i.e. ∆H is negative. 1 mark 94 2A 2 a iii 2a Given the following thermochemical data. Reaction ∆Ho 298 / kJ mol -1 C (graphite) + 2H 2(g) → CH 4(g) - 75.0 C (graphite) + O 2(g) → CO 2(g) - 393.5 H 2(g) + ½O 2(g) →H 2 O (l) - 285.9 iii The enthalpy change of combustion of diamond at 298K is -395.4 kJ mol -1 . Which allotrope of carbon, diamond or graphite, is energetically more stable? Explain why carbon does not convert from the less stable allotrope to the more stable one at room temperature. 2 ∆Ho comb. [C(diamond)] = - 395.4 kJmol -1 ∆Ho comb. [C(graphite)] = - 393.5 kJmol -1 ∴ Graphite is the relatively more stable allotrope. 1 mark Conversion from diamond to graphite involves the rearrangement of atoms in a giant covalent network would require a high activation energy. ∴ Conversion does not occur at room temperature. 1 mark C Many candidates erroneously said that diamond was the more stable allotrope of carbon. This indicated that their concept of 'energetic stability' was quite confused and/or that they could not relate the sign of enthalpy terms to the stability of compounds. 95 2B 4 c ii 4c Consider the data given below for the hydrogen halides and answer the questions that follow. Standard enthalpy change of formation / kJmol -1 Bond dissociation energy / kJmol -1 H–F -269.4 +562 H–Cl -92.8 +430 H–Br -36.8 +367 H–I +26.1 +298 ii At temperatures above 400 K, hydrogen iodide decomposes to produce violet fumes, but, hydrogen chloride and hydrogen bromide do not decompose. Briefly explain this difference and write a balanced equation for the decomposition of HI. 4 HI has the smallest bond dissociation energy, the activation energy for its decomposition is lowest. 2 marks And therefore it is the most easily decomposed HX. However, HBr and HCl do not decompose at temperature > 400 K. From the ∆H f values, HI is the most unstable hydrogen halide with respect to decomposition to its elements. 1 mark 2HI (g) → H 2(g) + I 2(g) 1 mark C ii Based on the given information candidates should have been able to point out that HI decomposes readily at moderately low temperatures because it has a small bond dissociation energy corresponding to a low activation energy for decomposition. The positive and therefore the largest enthalpy change of formation amongst the hydrogen halides means that HI is the least stable hydrogen halide with respect to decomposition into its elements. 99 2A 3 a i ii iii 3a Consider the following data for the reaction A + B → products Initial concentration / mol dm -3 IV. Energy profile of reaction Unit 1 Page 6 [A] [B] Initial rate / mol dm -3 s -1 4.0 × 10 -2 4.0 × 10 -2 6.4 × 10 -5 8.0 × 10 -2 4.0 × 10 -2 12.8 × 10 -5 4.0 × 10 -2 8.0 × 10 -2 6.4 × 10 -5 For this reaction, i deduce its rate equation, ii calculate the rate constant, and iii sketch a possible energy profile. (6 marks) IV. Energy profile of reaction Unit 2 Page 1 Topic IV. Energy profile of reaction Unit 2 Reference Reading 5.6 Reading Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 296, 315–317, 446 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 414–416 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 102 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 370–373 Syllabus Effect of catalyst Application of catalysts Notes B. Effect of catalyst According to collision theory, only the collision with an energy exceeding the activation energy may lead to a reaction. There are 2 ways to make more particles having an energy higher than activation energy, thus increase the reaction rate. 1. Increase the temperature of the particle ⇒ more particles possess an energy greater than E a . 2. Lower the activation energy ⇒ more particles at the original temperature posses an energy greater than the new E a '. This can be done by using of catalyst. 1. Characteristics of catalyst Catalyst – a regenerated reagent which modify the rate of reaction. In general, a catalyst is chemically involved in a reaction. It is consumed in one reaction step and regenerated in a subsequent step. It can thus be used repeatedly without undergoing any permanent changes. The following examples can be used to illustrate this: Using of catalyst provides an alternative multisteps reaction pathway. A lower activation energy is involved in the rate determining step of the catalyzed reaction. Thus, catalytic reaction has a faster rate of reaction. IV. Energy profile of reaction Unit 2 Page 2 Catalyst is usually highly specific, i.e. a particular catalyst usually can catalyze only one reaction. Since catalyst is involved in the formation of reaction intermediate, it may be physically changed after the reaction. e.g. from granule form to powder form. As ∆H remains unchanged, the equilibrium constant K of the reaction is not changed. The catalyst does not change the equilibrium position of a reaction. It speeds up both forward and reverse reaction for the same extent. Note : It can be proved that ln K = constant - ∆H RT Catalyst speeds up a reaction but has no effect on the yield of the reaction. 2. Theory of catalyst According to the physical state, catalysts can be classified into 2 categories. 1. Homogeneous catalyst – the catalyst has the same physical state as the reaction mixture 2. Heterogeneous catalyst – the catalyst has different physical state from the reaction mixture a) Homogeneous catalysis Example 1 Acid-catalysed esterification (will be further elaborated in organic chemistry) Esterification R' C O - O H O H R + R O H R' C O O H alcohol carboxylic acid charge separation involved r.d.s. R' C O - O H O R H + H 2 O + R' C O O R ester water In esterification, the nucleophilic oxygen atom on alcohol is added to the carbonyl carbon. It can be seen that an intermediate with charge separation in the structure is involved. This makes the reaction have a high activation energy, thus, a rather slow rate. Then, the proton possessed by the alcohol is shifted to the hydroxyl group to convert it to a better leaving group, water H 2 O. In contrast , esterification in the presence of any aqueous acid is found to be catalyzed. Acid Catalyzed Esterification C O R OH H + f C O R OH H O H R' + f C O R OH H + H O R' r.d.s. no charge separation involved C O R OR' H 2 O H + + + f + C O R O H O R' H H f In the acid catalyzed pathway, the carbonyl oxygen is first protonated. This activates the carbonyl group by increasing the polarity of the C=O bond and make the carbonyl carbon more positive and more vulnerable to the attack of the nucleophile, alcohol. Furthermore, no reaction intermediate with charge separation is involved in the acid catalyzed mechanism and a much lower activation energy would result. IV. Energy profile of reaction Unit 2 Page 3 Example 2 Reaction between aqueous solutions of iodide and persulphate ions: 2I - (aq) + S 2 O 8 2- (aq) → I 2(aq) + 2SO 4 2- (aq) The reaction rate of the above redox reaction is rather low. This is probably because the reaction involves collision among similarly charged particles. Addition of Fe 3+ (aq) or Fe 2+ (aq) into the reaction mixture speed up the reaction considerably. Oversimplified mechanism of the catalyzed reaction – Fe 3+ (aq) or Fe 2+ (aq) are served as a medium of e - transfer. 2Fe 3+ (aq) + 2I - (aq) → 2Fe 2+ (aq) + I 2(aq) Fe 3+ (aq) is consumed 2Fe 2+ (aq) + S 2 O 8 2- (aq) → 2Fe 3+ (aq) + 2SO 4 2- (aq) Fe 3+ (aq) is regenerated The catalyzed reaction has a lower activation energy since the alternative pathway involves only collision between oppositely charged particle. b) Heterogeneous catalysis Example 1 Decomposition of hydrogen peroxide in the presence of manganese(IV) oxide (MnO 2(s) ) catalyst Without the presence of catalyst, hydrogen peroxide will only decompose slowly into water and oxygen. 2H 2 O 2(aq) → 2H 2 O (l) + O 2(g) The rate of reaction can be traced by monitoring the volume of oxygen evolved. In the presence of catalyst, MnO 2(s) , the rate of reaction will be increased tremendously. This reaction is usually used in the laboratory in the preparation of oxygen. N.B. Please note that the presence of the catalyst only affect the rate of the reaction but it has no effect on the amount of oxygen obtained eventually. However, in the presence of acid, the reaction will be slowed down. The acid is said to be a negative catalyst. Beside the volume of the oxygen, the reaction can also be traced by quenching the reaction using excess acid. The amount of H 2 O 2(aq) can then be determined by back titrating with potassium iodide and sodium thiosulphate or directly with KMnO 4(aq) . H 2 O 2(aq) + 2I - (aq) + 2H + (aq) → 2H 2 O (l) + I 2(aq) or 5H 2 O 2(aq) + 2MnO 4 - (aq) + 6H + (aq) → 2Mn 2+ (aq) + 8H 2 O (l) + 5O 2(g) I 2(aq) + 2S 2 O 3 2- (aq) → 2I - (aq) + 2S 4 O 6 2- (aq) Although the MnO 2(s) granules used are chemically unchanged after the reaction, it will be physically changed into powder form. IV. Energy profile of reaction Unit 2 Page 4 3. Applications of catalysts a) Use of catalyst in Haber process and hydrogenation Haber process Iron is used as the catalyst in the Haber process to speed up the formation of ammonia. With the Fe catalyst, the new activation energy is much lower than in the uncatalyzed reaction. This makes the direct combination of N 2(g) and H 2(g) viable. N 2(g) + H 2(g) d 2NH 3(g) Hydrogenation of unsaturated oil All lipids are triesters of glycerol (propan-1,2,3-triol) and fatty acid (long chain carboxylic acid). The difference between oil and fat are only the degree of unsaturation of the fatty acids in the molecule. Oil contains more unsaturated fatty acid that fat does. Saturated carbon chain is larger than the unsaturated one with the same length. Therefore, the van der Waals' forces among the saturated carbon chain is stronger. That's why fat is a solid and oil is a liquid at room temperature. CH 2 CH CH 2 O O O O O O from propan-1,2,3-triol from different fatty acids Oil can be converted to fat by hydrogenation of the unsaturated carbon chain. This is done by bubbling H 2(g) through oil under heat and pressure with the presence of Ni (s) catalyst. The hydrogenated oil formed is called margarine. b) Catalytic converter All the cars imported into HK after 1 January 1993 are installed with catalytic converter. In the catalytic converter, some pollutants from car exhaust (e.g. carbon monoxide, nitrogen monoxide and unburnt hydrocarbons) are converted into relatively harmless substances (e.g. carbon dioxide, nitrogen and water) with the use of metal catalysts such as platinum (or palladium) and rhodium. 2CO (g) + 2NO (g) → 2CO 2(g) + N 2(g) The catalysts are coated on a honeycomb support to increase the surface area for better action. IV. Energy profile of reaction Unit 2 Page 5 c) Enzymes Enzymes are proteins that catalyse specific biochemical reactions. They are often called biological catalysts. Enzymes obtained from yeast have long been used in the production of alcohol by fermentation. The fermentation of glucose to form ethanol can be represented by the following equation : C 6 H 12 O 6 enzyme → 2C 2 H 5 OH + 2CO 2 Nowadays, enzymes are used in various aspects such as in the manufacture of biological washing powders. These washing powders contain enzymes which can break down stains caused by blood, egg, sweat as well as fats. They have the advantage of removing stains even at normal temperature. Glossary catalyst regenerated reagent homogeneous catalysis heterogeneous catalysis adsorption Haber process hydrogenation catalytic converter palladium rhodium enzyme Past Paper Question 91 1A 2 b iv 95 1A 1 e i 98 2A 3 b iii 91 1A 2 b iv 2b The energy profile of the reaction A (g) + B (g) d C (g) under two different catalysis X and Y are represented below. iv Why could the use of a different catalyst change the order of the reaction? 2 different reaction paths or different reaction mechanisms is involved with the catalyst added. 2 marks C In part (iv) some candidates related the order of reaction to activation energy. 95 1A 1 e i 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH 3 COOH (l) + C 2 H 5 OH (l) d CH 3 COOC 2 H 5(l) + H 2 O (l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. i Name a suitable catalyst for this reaction in the forward direction. 1 Concentrated sulphuric(VI) acid / hydrochloric acid / hydrogen chloride gas 1 mark IV. Energy profile of reaction Unit 2 Page 6 98 2A 3 b iii 3b The exothermic reaction E(g) → E'(g) (1) is a single stage reaction. iii In the presence of a catalyst, C, reaction (1) will proceed at a faster rate via the following mechanism : E(g) + C(g) → EC(g) EC(g) → C(g) + E'(g) (EC is the reaction intermediate.) Sketch labelled energy profiles for the conversion of E(g) to E'(g), with and without the catalyst. Explain why reaction (1) proceeds faster in the presence of the catalyst. V. Factors influencing the rate of reaction Page 1 Topic V. Factors influencing the rate of reaction Reference Reading 5.7 Assignment Reading A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 293–296 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 399–401 Laboratory experiments for general, organic & biochemistry (2nd ed.), Saunders College Publishing, 187–196, 522 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 354–356 Syllabus Factors influencing the rate of reaction Notes V. Factors influencing the rate of reaction A. Concentration The collision frequency among the particles increases with increasing concentration. B. Temperature Increase in temperature causes increase in energy of the particles. A larger fraction of particles will have energy greater than E A . Furthermore, the collision frequency also increase. C. Pressure For gaseous reactant, pressure is equivalent to concentration for solution. A gas under pressure means that there are more gaseous particle in a given volume. D. Surface area For solid reactant, a finer particles means a higher surface area to volume ratio. A larger surface area would lead to more collisions per unit mass of the solid reactant and higher reaction rate. E. Catalyst Use of catalyst lower the activation energy E A , this makes a larger fraction of particles having an energy greater than E A . F. Light Beside heat, light is also another source providing energy for the reaction. Glossary Past Paper Question 90 2A 1 c 90 2A 1 c 1c Give, with explanations, two factors that would increase the rate of a reaction. 4 Temperature: An increase in temperature increases the number of particles having sufficient energy to produce fruitful collision, and also that the average kinetic energy will be increased. This will result in higher velocities and hence more collision per unit time. 2 marks Concentration: An increase in concentration will give more particles per unit volume and hence more collision per unit time. 2 marks Catalyst: A catalyst that increases a reaction rate does so by providing a new mechanism. If the activation energy of the limiting step of the new mechanism is lower than the activation energy of the limiting step of the uncatalyzed reaction, then the rate is increased. 2 marks Maximium of 4 marks for any 2 correct answers. C Many candidates failed to point out explicitly that only in gas-phase reactions would an increase in pressure lead to an increase in reaction rate. Chemical Equilibria Page 1 Chemical Equilibria I. Nature of equilibrium A. Dynamic nature B. Examples of equilibrium 1. Bromine water in acidic and alkaline medium 2. Potassium dichromate in acidic and alkaline medium 3. Hydrolysis of bismuth chloride C. Equilibrium law 1. Introduction to different equilibrium constant (K) – K c , K p , K w , K a , K b , K In , K d 2. Equilibrium constant (K c and K p ) 3. Degree of dissociation (α) 4. Concentration of solid 5. Determination of equilibrium constant (K c ) a) Esterification b) Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) 6. Relationship between rate constant and equilibrium constant D. Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) 1. Concentration 2. Pressure 3. Temperature a) Equation : ln K = constant - ∆H T 4. Effect of catalyst on equilibria 5. Examples of calculation Chemical Equilibria Page 2 II. Acid-base Equilibria A. Acid-base Theory 1. Arrhenius definition 2. Bronsted-Lowry definition 3. Lewis definition B. Strength of acid 1. Leveling effect C. Dissociation of water 1. Ionic product of water D. pH and its measurement 1. Definition of pH 2. Temperature dependence of pH 3. Measurement of pH a) Use of pH meter (1) Calibration of pH meter b) Use of indicator (1) Colour of indicator E. Strong and weak acids/bases 1. Measuring of pH or conductivity of acid / base 2. Dissociation constant (K a and K b ) a) Dissociation of polybasic acid (1) Charge effect 3. Calculation involving pH, K a and K b a) Relationship between K a and K b (pK a and pK b ) b) Relationship between pH, pOH and pK w c) Some basic assumptions applied 4. Experimental determination of K a F. Buffer 1. Principle of buffer action a) pH of an acidic buffer b) pH of an alkaline buffer 2. Calculation of buffer solution G. Theory of Indicator H. Acid-base titration 1. Difference between equivalence point and end point 2. Titration using pH meter 3. Titration using indicator a) Choosing of indicator 4. Thermometric titration 5. Conductimetric titration a) Strong acid vs Strong base b) Weak acid vs Strong base c) Weak acid vs Weak base Chemical Equilibria Page 3 III. Redox Equilibria A. Redox reaction 1. Balancing of redox reaction a) By combining balanced half-ionic equation (1) Steps of writing balanced half-ionic equation (2) Combining half-ionic equations b) By the change in oxidation no. 2. Faraday and mole 3. Calculation of mass liberated in electrolysis B. Electrochemical cells 1. Measurement of e.m.f. a) Potentiometer b) High impedance voltmeter / Digital multimeter 2. Use of salt bridge a) Requirement of a salt bridge b) An electrochemical cell does not need salt bridge 3. Cell diagrams (IUPAC conventions) C. Electrode potentials 1. Standard hydrogen electrode 2. Relative electrode potential (Standard reduction potential) a) Meaning of sign and value 3. Electrochemical series 4. Use of electrode potential a) Comparing the oxidizing and reducing power b) Calculation of e.m.f. of a cell c) Prediction of the feasibility of redox reaction (1) Disproportionation reaction D. Secondary cell and fuel cell 1. Lead-acid accumulator 2. Hydrogen-oxygen fuel cell E. Corrosion of iron and its prevention 1. Electrochemical process of rusting 2. Prevention of rusting a) Coating b) Sacrificial protection c) Alloying I. Nature of equilibrium Part 1 Page 1 Topic I. Nature of Equilibrium Part 1 Reference Reading 6.0 – 6.1.0 Practical Chemistry (3rd ed.), Heinemann, 11–12 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 83–86, 228–229, 231 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 325–329 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 289–296 Syllabus Nature of equilibrium Dynamic nature Notes I. Nature of equilibrium A. Dynamic nature Equilibrium means a state of balance. There are two kinds of equilibrium : Static equilibrium and Dynamic equilibrium. Static equilibrium Dynamic equilibrium The positions of the two boys remain unchanged because they are not in motion. They are in a state of static equilibrium. A boy can also maintain a fixed position on a moving escalator. This can be done by running at the same speed of the escalator but at a different direction. The position of the boy remains unchanged but both the escalator and the boy are in motion. They are in a state of dynamic equilibrium. All equilibria studied in chemistry are dynamic in nature. In A-Level, we are going to study chemical equilibrium, acid-base equilibrium, redox equilibrium and phase equilibrium. Chemical equilibrium includes all equilibria involving any chemical reaction. Acid-equilibrium and redox equilibrium are 2 examples of chemical equilibrium. Phase equilibrium is the study of equilibrium involving interconversion of phases. For examples 1. Solid phase, liquid phase and gaseous phase. 2. Aqueous phase and organic phase. 3. Moving phase and stationary phase Equilibrium Static equilibrium Dynamic equilibrium Phase equilibrium Chemical equilibrium Acid-base equilibrium Redox equilibrium I. Nature of equilibrium Part 1 Page 2 All chemical equilibria are dynamic in nature. Consider the following reversible reaction, Reactant Product forward reaction backward reaction Rate of forward reaction = k 1 [Reactant] Rate of backward reaction = k -1 [Product] Time C o n c e n t r a t i o n / m o l d m - 3 Concentration of reactant Concentration of product 0 Concentrations of reactant and product At the beginning, there is only reactant but no product. Once the reaction proceeds, the concentration of the reactant drops and the concentration of the product rises. Since the rate of reaction increases with concentrations, the rate of the forward reaction will decrease and the rate of the backward reaction will increase. Time R a t e / m o l d m - 3 s - 1 Rate of forward reaction Rate of backward reaction Rate of reaction 0 At equilibrium, forward rate = backward rate When the increasing backward rate reaches the decreasing forward rate, there will be no net conversion between product and reactant. The amounts of reactant and product will remain constant and an equilibrium is established. However, the forward reaction and backward reaction do not stop after the equilibrium has established, there is always a constant conversion between the reactant and the product. Therefore, the equilibrium is said to be dynamic in nature. I. Nature of equilibrium Part 1 Page 3 Consider another example If H 2 O (l) is heated in a sealed pressure cooker at a constant temperature, H 2 O (l) vaporizes to steam H 2 O (g) . At the same time, steam also condenses back to water. Eventually, the amounts of water and steam will become constant. This system doesn’t involve any chemical change but only interconversion between liquid phase (water) and gaseous phase (steam). This is an example of phase equilibrium. All equilibria only exist in close systems. In the above example, no equilibrium will exit if the cover of the pressure cooker is removed. I. Nature of equilibrium Part 1 Page 4 B. Examples of chemical equilibrium 1. Bromine water in acidic and alkaline medium Br 2(aq) + H 2 O (l) d HBr (aq) + HBrO (aq) orange colourless colourless colourless Since HBr (aq) is a strong acid while HBrO (aq) is a weak acid, the equation may be rewritten as follow : Br 2(aq) + H 2 O (l) d H + (aq) + Br - (aq) + HBrO (aq) orange colourless colourless colourless colourless In this equilibrium, Br 2(aq) is orange in colour and all the other species are colourless. Once the equilibrium is established, the forward rate of reaction will be the same as the backward rate of reaction. And the concentrations of all species will remain constant. The equilibrium can be disturbed by either addition of acid or alkaline. Addition of acid Br 2(aq) + H 2 O (l) d H + (aq) + Br - (aq) + HBrO (aq) orange colourless colourless colourless colourless Upon addition of acid, the concentration of H + (aq) increases suddenly accompanying with a sudden increase in backward reaction rate. More product is converted back to reactant. The concentration of product drops and the concentration of the reactant rises. A new equilibrium position will be established when the concentrations of product and reactant reaches a level that the forward rate and the backward rate become the same again. At the new equilibrium position, the concentration of product will be lower that the original concentration while the concentration of reactant will become higher. The position of the equilibrium is said to be shifted to the left by the addition of acid. And the colour of the equilibrium mixture will become darker. I. Nature of equilibrium Part 1 Page 5 Addition of alkali Br 2(aq) + H 2 O (l) d H + (aq) + Br - (aq) + HBrO (aq) orange colourless colourless colourless colourless Conversely, the equilibrium position can also be shifted to the right by addition of alkaline. Upon addition of alkaline, the hydrogen ions are neutralized. H + (aq) + OH - (aq) → H 2 O (l) The decrease in concentration of H + (aq) slows down the backward reaction. Eventually, the equilibrium will adjust to a new position. The colour intensity of the equilibrium mixture will become lighter as the concentration of Br 2(aq) decreases. Questions : What if NaBr (s) is added ? 2. Potassium dichromate in acidic and alkaline medium 2CrO 4 2- (aq) + 2H + (aq) d Cr 2 O 7 2- (aq) + H 2 O (l) yellow colourless orange colourless Cr O - O - O O Cr O - O - O O H + H + O Cr Cr O O O O O - O - + + + + H 2 O chromate(VI) ion dichromate(VI) ion Upon addition of acid, the equilibrium position can be shifted to the right and becomes orange. While upon addition of alkali, the equilibrium position can be shifted to the left and becomes yellow. I. Nature of equilibrium Part 1 Page 6 3. Hydrolysis of bismuth chloride BiCl 3(aq) + H 2 O (l) d BiOCl (s) + 2HCl (aq) colourless colourless white ppt. colourless When solid bismuth chloride (BiCl 3 ) is mixed with water, a white suspension of bismuth oxychloride (BiOCl) is formed. Upon addition of drops of HCl (aq) , the white disappears and the white ppt. of bismuth oxychloride reappear upon addition of drops of water. Although not all reactions are reversible, it would be beneficial to consider them all reversible. And consider an irreversible reaction only an equilibrium lying completely onto one side. Glossary equilibrium / equilibria static equilibrium dynamic equilibrium chemical equilibrium acid-base equilibrium redox equilibrium phase equilibrium reversible reaction forward rate backward rate close system shifting in equilibrium position chromate(VI) ion dichromate(VI) ion bismuth chloride bismuth oxychloride Past Paper Question 92 2B 4 b i ii 98 2B 8 b i 92 2B 4 b i ii 4b i Draw the structures of the chromate(VI) and dichromate(VI) ions. 2 1 mark each ii Write an equation showing the action of aqueous acid on the chromate(VI) ion. 1 2CrO 4 2- + 2H + d Cr 2 O 7 2- + H 2 O 1 mark C i Candidates should take care in providing a 3-D diagram, which shows the correct bond orders and charge(s). Some candidates stated that the shape of the chromate(VI) ion is square-planar. ii the chromate(VI) - dichromate(VI) equilibrium is not a redox reaction, as stated by some candidates. 98 2B 8 b i 8b For each of the following, state the expected observation and write the relevant balanced equation(s). 6 i Dilute sulphuric(VI) acid is added dropwise to a solution of potassium chromate(VI). I. Nature of equilibrium Part 2 Page 1 Topic I. Nature of Equilibrium Part 2 Reference Reading 6.1.1 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 201–202 Chemistry – Experimental foundation, Prentice Hall Inc., 51–55 Assignment Reading Advanced Practical Chemistry, John Murray (Publisher) Ltd., 53–56 Calculations for A-Level Chemistry, Stanley Thornes (Publishers) Ltd, 230–238 A-Level Chemistry (3 rd ed.), Stanley Thornes (Publisher) Ltd., 86–87, 229–231, 232–233, 236–238 Chemistry in Context (4 th ed.), Thomas Nelson and Sons Ltd., 333–340 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 296–303 Syllabus Equilibrium law Equilibrium constant (K c and K p ) Notes C. Equilibrium law 1. Introduction to different equilibrium constant (K) – K c , K p , K w , K a , K b , K In , K d Considering the following reversible reaction a A + b B d c C + d D At equilibrium, it is found that the expression [C] c [D] d [A] a [B] b is a constant at a given temperature. The expression is called equilibrium constant, denoted by the symbol K or K eq . The unit of K is depending on the values of stoichiometric coefficients. K or K eq = [C] c [D] d [A] a [B] b This equation describes the relationship between the concentrations of reactant and product. The equation is also known as Law of equilibrium or Equilibrium law. a) Relationship between rate constant and equilibrium constant At equilibrium, the rate of forward reaction equals the rate of backward equation. Moreover, because the whole reaction is in equilibrium, all intermediate steps of the reaction are in equilibrium. Therefore, the rate of reaction is no longer depending on the rate determining step only. Instead, the rate of reaction is depending on the concentration and stoichiometric coefficients. Forward rate at equilibrium = k 1 [A] a [B] b where k 1 is the rate constant of forward reaction Backward rate at equilibrium = k -1 [C] c [D] d k -1 is the rate constant of backward reaction Forward rate at equilibrium = Backward rate at equilibrium k 1 [A] a [B] b = k -1 [C] c [D] d k 1 k -1 = [C] c [D] d [A] a [B] b = K ∴ K = k 1 k -1 N.B. This is applicable only if the reaction is at equilibrium. If the reaction is not at equilibrium, the rate equation must be determined experimentally. I. Nature of equilibrium Part 2 Page 2 There are many different kinds of expression of equilibrium constant depending on the type of equilibrium concerned. e.g. K c , K p , K w , K a , K b , K In , K d . Type Meaning Example Equilibrium Law Chemical Equilibrium constant in term of concentration, K c 2CrO 4 2- (aq) + 2H + (aq) d Cr 2 O 7 2- (aq) + H 2 O (l) K c = [Cr 2 O 7 2- (aq) ] [CrO 4 2- (aq) ] 2 [H + (aq) ] 2 Chemical Equilibrium constant in term of partial pressure, K p N 2(g) + 3H 2(g) d 2NH 3(g) K p = p NH 3 2 p N 2 · p H 2 3 Chemical Ionic product of water, K w H 2 O (l) d H + (aq) + OH - (aq) K w = [H + (aq) ][OH - (aq) ] Acid-base Acidity constant, K a CH 3 COOH (aq) + H 2 O (l) d H 3 O + (aq) + CH 3 COO - (aq) K a = [H 3 O + (aq) ][CH 3 COO - (aq) ] [CH 3 COOH (aq) ] Acid-base Bascity constant, K b NH 3(aq) + H 2 O (l) d NH 4 + (aq) + OH - (aq) K b = [NH 4 + (aq) ][OH - (aq) ] [NH 3(aq) ] Acid-base Indicator constant, K In HIn (aq) + H 2 O (l) d H 3 O + (aq) + In - (aq) K In = [H 3 O + (aq) ][In - (aq) ] [HIn (aq) ] Phase Partition coefficient / distribution coefficient, K d I 2(aq) d I 2(hexane) K d = [I 2(hexane) ] [I 2(aq) ] 2. Equilibrium constant (K c and K p ) The equilibrium constant K or K eq describes the relationship between the concentrations of reactants and products. However, there are many different ways to express concentration. In general, for a solution, concentration is expressed in term of molarity. for a gas, concentration is expressed in term of partial pressure (which is related to mole fraction). a) Concentration of pure solid and liquid Molarity is only one way of expressing concentration. Indeed, concentration is a very general concept concerning the relationship between the amount of a substance and the volume occupied by that substance. For a pure solid / liquid, the amount is usually measured in term of mass. Therefore, the concentration of a pure solid / liquid is the equivalent to its density which is a constant at a given temperature. Concentration of pure solid / liquid = amount of solid / liquid volume occupied = mass of solid / liquid volume occupied = density Example : Calculation of the concentration of water in term of molarity at 25ºC. Given : density of water at 25ºC is 1.00 gcm -3 = 1000.00 gdm -3 . Concentration of pure water in molarity = no. of mole of water volume occupied = mass / molar mass volume occupied = volume × density / molar mass volume occupied = density molar mass = 1000.00 gdm -3 18.0 gmol -1 = 55.6 moldm -3 I. Nature of equilibrium Part 2 Page 3 b) Equilibrium constant in term of concentration, K c Consider the following equilibrium 2CrO 4 2- (aq) + 2H + (aq) d Cr 2 O 7 2- (aq) + H 2 O (l) K = [Cr 2 O 7 2- (aq) ][H 2 O (l) ] [CrO 4 2- (aq) ] 2 [H + (aq) ] 2 Besides being a product, water is also the solvent in this equilibrium. Because it is in large excess comparing with the solutes, it can be considered as a pure liquid in here. Since the concentration of water is constant, it may be incorporated into the equilibrium constant and the expression will become K [H 2 O (l) ] = [Cr 2 O 7 2- (aq) ] [CrO 4 2- (aq) ] 2 [H + (aq) ] 2 = K c The expression excludes the concentration of pure solid and liquid is known as K c , equilibrium constant in term of concentration. In the writing of any equilibrium constant, the concentration of a pure solid or liquid in the equilibrium law should be excluded. The exclusion of concentration of pure solid / liquid in an equilibrium law is not only doctrinal. It has a significant physical meaning. Consider a saturated sugar solution, the concentration of the solution is not depending on the amount of sugar remains undissolved. sugar (s) + aq d sugar (aq) A saturated sugar solution A saturated solution with more undissolved sugar crystal Consider another equilibrium, Fe 2+ (aq) + Ag + (aq) d Fe 3+ (aq) + Ag (s) Equilibrium mixture with less Ag (s) suspension Equilibrium mixture with more Ag (s) suspension It is found that the addition of extra Ag (s) into the mixture has no effect on the concentrations of Fe 2+ (aq) , Ag + (aq) and Fe 3+ (aq) . Therefore, the equilibrium position is not depending on the amount of solid silver whose concentration is constant. This fact should be reflected in the equilibrium law. K c = [Fe 3+ (aq) ] [Fe 2+ (aq) ][Ag + (aq) ] I. Nature of equilibrium Part 2 Page 4 c) Equilibrium constant in term of partial pressure, K p Consider the thermal decomposition of CaCO 3(s) , CaCO 3(s) d CaO (s) + CO 2(g) K = [CaO (s) ][CO 2(g) ] [CaCO 3(s) ] [CaCO 3(s) ]K [CaO (s) ] = [CO 2(g) ] = K c Or if the concentration of CO 2(g) is expressed in term of partial pressure, the equilibrium constant is known as K p . K p = p CO 2(g) Consider the Haber process, N 2(g) + 3H 2(g) d 2NH 3(g) K c = [NH 3(g) ] 2 [N 2(g) ][H 2(g) ] 3 or K p = p NH 3(g) 2 p N 2(g) p H 2(g) 3 I. Nature of equilibrium Part 2 Page 5 3. Degree of dissociation (α) Sometimes, the equilibrium positions of dissociation reactions can be described by degree of dissociation. Degree of dissociation is the percentage of molecules that undergoes dissociation and represented by α. Consider the dissociation of carboxylic acid dimer in gaseous state, C O O H C H H H H H H C H O O C C O O H C H H H 2 (CH 3 COOH) 2(g) d 2CH 3 COOH (g) Consider 1 mole of dimer, the mole no. would be numerically the same as no. in percentage (i.e. mole fraction). Initial no. of mole 1 0 no. of mole at equilibrium 1 - α 2α Partial pressure at equilibrium p (CH 3 COOH) 2(g) p CH 3 COOH (g) The equilibrium law in term of partial pressure can be expressed as K p = p CH 3 COOH (g) 2 p (CH 3 COOH) 2(g) Let P be the total pressure, P = p (CH 3 COOH) 2(g) + p CH 3 COOH (g) Since mole fraction is related to partial pressure by the equation, mole fraction = n i n = p i P p i = n i n P = mole fraction × P p (CH 3 COOH) 2(g) = mole fraction of (CH 3 COOH) 2(g) × P = 1 - α (1 - α) + 2α P = 1 - α 1 +α P p CH 3 COOH (g) = mole fraction of CH 3 COOH (g) × P = 2α (1 - α) + 2α P = 2α 1 + α P K p = p CH 3 COOH (g) 2 p (CH 3 COOH) 2(g) = ( 2α 1 + α ) 2 P 2 1 - α 1 + α P = 4α 2 1 - α 2 P In this equilibrium, unlike equilibrium constant K p , dissociation constant (α) is also depending on the total pressure of the gaseous mixture on top of temperature. For example, T / K 298 303 308 313 K p / Nm -2 72.9 121.1 182.7 302 At 303 K and pressure of 1400 Nm -2 121.1 Nm -2 = 4α 2 1 - α 2 × 1400 Nm -2 α = 0.145 At 303 K and pressure of 200 Nm -2 121.1 Nm -2 = 4α 2 1 - α 2 × 200 Nm -2 α = 0.363 Thus, more carboxylic acid dimers will dissociate at a lower pressure. I. Nature of equilibrium Part 2 Page 6 4. Determination of equilibrium constant (K c ) The equilibrium constant K c can be determined by measuring the concentrations of different species in the equilibrium mixture. Indeed, sometimes it is not necessary to measure the concentrations of all the species. By knowing the initial concentration of the reactant and the equilibrium concentration of the reactant or product, the concentration of the other species can be determined according to the law of conservation of mass. Consider an equilibrium mixture with initial concentrations of 1 mol dm -3 Fe 3+ (aq) and 2 mol dm -3 of NCS - (aq) respectively. Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) iron(III) ion thiocyanate ion thiocyanato iron(III) ion pale yellow colour dark red complex e.g. Initially Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) 1 mol dm -3 2 mol dm -3 0 mol dm -3 At equilibrium Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) (1 - x) mol dm -3 (2 - x) mol dm -3 x mol dm -3 By knowing the equilibrium concentration of either Fe 3+ (aq) , NCS - (aq) or FeNCS 2+ (aq) , the equilibrium constant can be determined. K c = [FeNCS 2+ (aq) ] [Fe 3+ (aq) ][NCS - (aq) ] = x (1 - x) (2 - x) dm 3 mol -1 I. Nature of equilibrium Part 2 Page 7 a) Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) First of all, equilibrium constant is a value affected by temperature. It is better to keep the temperature constant by immersing the solution to be studied in a water bath. For the equilibrium, Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) , the initial concentrations of Fe 3+ (aq) and NCS - (aq) can be determined by prepared corresponding standard solution of Fe(NO 3 ) 3(aq) and KNCS (aq) . The problem is how to determine the equilibrium concentration of either Fe 3+ (aq) , NCS - (aq) or FeNCS 2+ (aq) . If titration is used to determine the concentration, the reaction must be quenched first. Otherwise, when the chemical is being titrated, the equilibrium position will also be shifted. Since FeNCS 2+ (aq) is deep red in colour, its concentration can be determined by colorimetric measurement without quenching. Firstly, a calibration curve which relates the concentration of FeNCS 2+ (aq) and the colour intensity must be constructed. This can either be done by using a calorimeter or visual comparison. In the construction of calibration curve, samples of standard FeNCS 2+ (aq) with various concentrations are prepared. These are prepared by adding excess NCS - (aq) into standard Fe 3+ (aq) . It is assumed that all Fe 3+ (aq) ions are converted to FeNCS 2+ (aq) . Therefore, the concentration of FeNCS 2+ (aq) can be determined from concentration of Fe 3+ (aq) . Equilibrium mixture with known initial Fe 3+ (aq) and NCS - (aq) concentrations are prepared. The equilibrium concentration of FeNCS 2+ (aq) in the mixture is determined by a colorimeter or by visual comparison of the intensity of the samples prepared before. Hence, the equilibrium constant K c = [FeNCS 2+ (aq) ] [Fe 3+ (aq) ][NCS - (aq) ] can be determined. Several trials with different concentrations should be done to obtain an average value. I. Nature of equilibrium Part 2 Page 8 b) Esterification CH 3 COOH (aq) + CH 3 CH 2 OH (aq) d CH 3 COOCH 2 CH 3(aq) + H 2 O (l) Both esterification and hydrolysis of ester are very slow, even in the presence of acid catalyst, it takes 48 hours for the system to reach a state of equilibrium. Hence, the equilibrium concentration of CH 3 COOH (aq) can be determined by titration even without quenching. In this example, water is not the solvent and is not in large excess. It cannot be considered as a pure substance and its concentration must be reflected in the equilibrium law. K c = [CH 3 COOCH 2 CH 3(aq) ][H 2 O (l) ] [CH 3 COOH (aq) ][CH 3 CH 2 OH (aq) ] Since aqueous acid is required to catalyze the reaction, it would be more convenient to approach the problem by treating it as a hydrolysis problem. This reduces the number of measurement of from 3 liquids (ethanoic acid, alkanol and aqueous acid) to 2 liquids at the preparatory stage. CH 3 COOCH 2 CH 3(aq) + H 2 O (l) d CH 3 COOH (aq) + CH 3 CH 2 OH (aq) 1 K c = K c ' = [CH 3 COOH (aq) ][CH 3 CH 2 OH (aq) ] [CH 3 COOCH 2 CH 3(aq) ][H 2 O (l) ] Procedure An accurately known volume of standard HCl (aq) is pipetted into a test tube containing known mass of CH 3 COOCH 2 CH 3(l) . The solution is shaken occasionally for 48 hours to ensure equilibrium is reached. The initial amount of water is determined by subtracting the mass of HCl from the mass of HCl (aq) . The initial amount of CH 3 COOCH 2 CH 3(l) is determined by weighing. The total no. of mole of HCl (aq) and CH 3 COOH (aq) at equilibrium are determined by titrating with standard NaOH (aq) . Experimental result Mass of the empty test tube = 12.01 g Mass of the empty test tube with CH 3 COOCH 2 CH 3(l) = 16.52 g Mass of the empty test tube with HCl (aq) and CH 3 COOCH 2 CH 3(l) = 5.20 g Volume of HCl (aq) added = 5.00 cm 3 Concentration of HCl (aq) = 1.953 moldm -3 Calculation I. Nature of equilibrium Part 2 Page 9 Glossary equilibrium law / law of equilibrium equilibrium constant (K/K eq ) equilibrium constant in term of concentration (K c ) equilibrium constant in term of partial pressure (K p ) degree of dissociation (α) dimer thiocyanate ion thiocyanato iron(III) ion calibration curve Past Paper Question 90 2A 1 a i ii 94 2A 1 b 95 1A 1 e ii 96 1A 1 f i ii 97 1A 2 a i 98 1A 2 c 99 2A 4 a i ii 90 2A 1 a i ii 1a Consider the following equilibrium at constant pressure: A 2(g) + 3B 2(g) d 2AB 3(g) A mixture of 4.0 mol of A 2(g) and 12.0 mol of B 2(g) was placed in a vessel of volume 20.0 dm 3 , and heated to 565K. When the system had reached equilibrium, it was found that 4.0 mol AB 3(g) was present. i Calculate the concentration of each species at equilibrium. 1½ A 2(g) + 3B 2(g) d 2AB 3(g) Since 4 moles of AB 3 will consume 2 moles of A 2 and 6 moles of B 2 ∴ at equilibrium: [B 2 ] = 12 - 6 20 = 6 20 moldm -3 ½ mark [A 2 ] = 4 - 2 20 = 2 20 moldm -3 ½ mark [AB 3 ] = 4 20 moldm -3 ½ mark ii Calculate the equilibrium constant of this reaction at 565K. 2½ K = [AB 3 ] 2 [A 2 ][B 2 ] 3 = ( 4 20 ) 2 ⋅ ( 20 2 ) ⋅ ( 20 6 ) 3 = 14.82 dm 6 mol -2 (wrong unit -½) 2½ mark C i Some candidates mistakenly used the initially given concentration rather than the equilibrium concentration in their calculation of equilibrium constant. ii Some weaker candidates failed to give the correct unit for the equilibrium constant. I. Nature of equilibrium Part 2 Page 10 94 2A 1 b 1b A colorimetric method can be used to provide data for the determination of the equilibrium constant of the following reaction. Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) Outline such a method for the determination of the equilibrium constant K c of the above equilibrium. 4 FeNCS 2+ (aq) ions absorb visible light in the electromagnetic spectrum / hence [FeNCS 2+ (aq) ] can be determined by measuring the absorbance of the solution. 1 mark Calibrate a colorimeter by measuring the absorbance of several solutions which contained known concentrations of FeNCS 2+ (aq) . (In these solutions, the NCS - ions are in excess, so that all Fe 3+ ions present can be considered as existing in the complex form, FeNCS 2+ ) 1 mark By mixing comparable volumes of standard Fe 3+ (aq) and NCS - (aq) solutions and measuring the [FeNCS 2+ ] formed with the colorimeter, [Fe 3+ (aq) ], [NCS - (aq) ] and [FeNCS 2+ (aq) ] at equilibrim can be found. 1 mark Hence, the equilibrim constant, K c = [FeNCS 2 + (aq) ] [Fe 3+ (aq) ][NCS - (aq) ] , can be determined. 1 mark C Poorly-answered. This indicated that most candidates were not familiar with the use of a colorimetric method in the determination of equilibrium constants. A large number of candidates did not mention the calibration of the colorimeter. 95 1A 1 e ii 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH 3 COOH (l) + C 2 H 5 OH (l) d CH 3 COOC 2 H 5(l) + H 2 O (l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. ii Calculate the equilibrium constant, K c , for the reaction at 298 K. 3 CH 3 COOH (l) + C 2 H 5 OH (l) d CH 3 COOC 2 H 5(l) + H 2 O (l) initial no. of moles 12.01 60.052 4.61 46.068 – – = 0.2 = 0.1 no. of mole at equilibrium 12.01 - 5.04 60.052 0.1 - 0.084 0.084 0.084 2 marks = 0.116 = 0.016 concentration at equilibrium 0.116 V 0.016 V 0.084 V 0.084 V K c = 0.084 V × 0.084 V 0.116 V × 0.016 V = 3.80 (no unit) 1 marks C A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. I. Nature of equilibrium Part 2 Page 11 96 1A 1 f i ii 1f SO 2(g) and O 2(g) were mixed in the mole ratio of 3 : 1 at 1000 K in the presence of a catalyst. When equilibrium was attained at 373 kPa pressure, one half of the SO 2(g) had been converted into SO 3(g) . i Write an expression for K p for the reaction of SO 2(g) and O 2(g) to give SO 3(g) . 1 K p = (P SO 3 ) 2 (P SO 2 ) 2 (P O 2 ) or (P SO 3 ) (P SO 2 )(P O 2 ) ½ 1 mark ii Calculate K p for the above reaction under the given conditions. 2½ 2SO 2 + O 2 d 2SO 3 initial 3 1 0 at equilibrium 1.5 0.25 1.5 total = 3.25 ½ mark P SO 3 = P SO 2 (= 1.5 3.25 × 373 kPa) ½ mark P O 2 = 0.25 3.25 × 373 kPa ½ mark K p = (P SO 3 ) 2 (P SO 2 ) 2 (P O 2 ) = 1 P O 2 = 3.25 0.25 × 373 kPa -1 = 0.035 kPa -1 1 mark or, K p = 1 P O 2 = 3.25 0.25 × 373 (kPa) -½ = 0.19 (kPa) -½ (1 mark) (½ mark for numerical answer, ½ mark for unit) (The answer and the expression (part (i)) for K p should be consistent with each other.) 97 1A 2 a i 2a Consider the following dissociation reaction: PCl 5(g) d PCl 3(g) + Cl 2(g) At 400 K and 101 kPa pressure, the percentage dissociation of PCl 5(g) is 86%. 4 i Calculate K p for the reaction at 400 K. 98 1A 2 c 2c At 4200 K, the equilibrium constant for the following reaction is 1.2 × 10 -2 . N 2(g) + O 2(g) d 2NO (g) 1.0 mol of O 2(g) and 2.0 mol of N 2(g) are allowed to react in a 2.0 dm 3 closed container. Calculate the concentration of N 2(g) , in moldm -3 . in the equilibrium mixture at 4200 K. 3 99 2A 4 a i ii 4a In the Haber process, ammonia is synthesized by the exothermic reaction of nitrogen and hydrogen at around 723K. N 2(g) + 3H 2(g) d 2NH 3(g) In a simulation of the process, a mixture of nitrogen and hydrogen was placed in a closed container. The initial concentrations of nitrogen and hydrogen were 0.50 mol dm -3 and 1.50 mol dm -3 respectively. When the equilibrium was attained at 723 K, 25.0% of the original nitrogen was consumed. i Calculate the respective concentrations of nitrogen, hydrogen and ammonia in the equilibrium mixture. ii Calculate K c for the reaction at 723 K. I. Nature of equilibrium Part 3 Page 1 Topic I. Nature of Equilibrium Part 3 Reference Reading 6.1.2.0–6.1.2.4 AS-Level Chemistry Syllabus for Secondary School (1991), Curriculum Development Council, 160 Assignment Reading A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 87, 231–232 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 343–354 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 305–309 Syllabus Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) Effect of catalyst on equilibria Le Chaterlier's principle Notes D. Effect of change in concentration, pressure and temperature on equilibria (Le Chaterlier's principle) 1. Concentration Br 2(aq) + H 2 O (l) d HBr (aq) + HBrO (aq) orange colourless colourless colourless Upon addition of extra Br 2(aq) into the equilibrium system, some Br 2(aq) added will be converted to HBr (aq) and HBrO (aq) . Therefore, the net change in concentration of Br 2(aq) is always less than the actual amount added. Because more products are formed upon addition of the reactant, the position of the equilibrium is said to be shifted to the right upon the addition of Br 2(aq) . K c = [HBr (aq) ][HBrO (aq) ] [Br 2(aq) ] An increase in [Br 2(aq) ] will make the value of the expression [HBr (aq) ][HBrO (aq) ] [Br 2(aq) ] smaller than the original K c . Therefore, the equilibrium will adjust in a way to restore the expression [HBr (aq) ][HBrO (aq) ] [Br 2(aq) ] back to the original K c . Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) iron(III) ion thiocyanate ion thiocyanato iron(III) ion pale yellow colour dark red complex K c = [FeNCS 2+ (aq) ] [Fe 3+ (aq) ][NCS 2+ (aq) ] Similarly, upon addition of Fe 3+ (aq) or NCS - (aq) , the solution will turn darker. It shift the equilibrium position to the right. However, upon addition of Na 3 PO 4(aq) or NaHPO 4(aq) , the colour will get paler. PO 4 3- (aq) ions form complex with free Fe 3+ (aq) and lower its concentration. This shifts the equilibrium position to the left. I. Nature of equilibrium Part 3 Page 2 2. Pressure Since only gas is compressible, pressure has only effect on the equilibrium involving gaseous species. 2NO 2(g) d N 2 O 4(g) brown colourless Once the plunger of the gas syringe is pressed, the colour of the gas gets darker because of an immediate increase of the concentration of brown NO 2(g) . However, the gas turns paler gradually. At equilibrium, forward rate = backward rate and forward rate = k 1 [NO 2(g) ] 2 backward rate = k -1 [N 2 O 4(g) ] The forward reaction is a second order reaction but the backward reaction is only a first order reaction. The increase in pressure increases the forward rate more than the backward rate. Hence a new equilibrium position is established eventually. I. Nature of equilibrium Part 3 Page 3 3. Temperature 2NO 2(g) d N 2 O 4(g) ∆H = -58.0 kJmol -1 brown colourless When the tube containing the mixture of NO 2(g) and N 2 O 4(g) is immersed in hot water, the colour of the gas gets darker. Conversely, when it is put inside icy water, the colour of the gas gets paler. This is an exothermic reaction. From the energy profile of the reaction, it can be seen that the forward reaction and backward reaction involve different activation energy. The forward reaction has a lower activation energy Ea 1 . i.e. Ea 1 < Ea -1 Once the temperature is increased, the rate of both forward reaction and backward reaction are increased. Moreover, from the diagram of distribution of molecular speed, the reaction with higher activation energy will have a higher percentage increase in rate. ∴ An increase in temperature will increase the values of k 1 and k -1 for different extent, thus the equilibrium constant and equilibrium position. The expression K = k 1 k -1 will become smaller. Therefore, a higher temperature will shift the equilibrium position of an exothermic reaction to the left. I. Nature of equilibrium Part 3 Page 4 a) Equation : ln K = constant - ∆H RT Or the effect of the temperature can be interpreted from another point of view. It can be proved that the equilibrium constant is related to the ∆H according to the equation ∆G = ∆H - T∆S = - RTln K where ∆G is Free energy change, which indicates the total free energy change of the system. T is the temperature in Kelvin. ∆H is enthalpy change of the system. It is a constant for a specific reaction. ∆S is entropy change of the system, which is a measure of the change in degree of disorderness. It is a constant for a specific reaction. N.B. Free energy change and entropy change are not required in A-Level. (1) Derivation of ln K = - ∆H RT + constant By rearranging the equation ∆G = ∆H - T∆S = - RTln K - RTln K = ∆H - T∆S ln K = - ∆H RT + T∆S RT ln K = - ∆H RT + ∆S R ln K = - ∆H RT + constant N.B. The actual derivation is not required in A-Level but you are expected to use the equation to explain the effect of temperature on an equilibrium. For an exothermic reaction, ∆H is negative, therefore, the expression - ∆H RT is positive. An increase in temperature will make the expression - ∆H RT less positive, hence K will become smaller. This shifts the equilibrium position to the reactant side. Conversely, the equilibrium position of an exothermic reaction will be shifted to the product side if the temperature is lowered. I. Nature of equilibrium Part 3 Page 5 4. Effect of catalyst on equilibria Addition of catalyst has no effect on concentrations nor equilibrium constant. Therefore, it has no effect on the position of an equilibrium. It only increases the forward rate and backward rate for the same extent and shortens the time required to attain equilibrium. 5. Le Chaterlier's principle The effect of a change in condition on the position of an equilibrium can be generalized as a principle, called Le Chaterlier's principle. Le Chaterlier's principle states that when a change is imposed on an equilibrium system, the system will response in a way to minimize the effect of the change. Effect of concentration Fe 3+ (aq) + NCS - (aq) d FeNCS 2+ (aq) pale yellow colourless dark red Upon addition of FeCl 3(s) , some Fe 3+ (aq) added with combine with NCS - (aq) to form FeNCS 2+ (aq) . As a result the net increase of concentration of Fe3+(aq) will be reduced and the equilibrium position is shifted to the right. If PO 4 3- (aq) is added to reduce the concentration of Fe 3+ (aq) , some FeNCS 2+ (aq) will dissociate to give more Fe 3+ (aq) . Consequently, the net decrease of concentration of Fe 3+ (aq) will be reduced and the equilibrium position is shifted to the left. Effect of pressure 2NO 2(g) d N 2 O 4(g) brown colourless Any increase in pressure can be reduced by converting 2 NO 2(g) molecules to 1 N 2 O 4(g) molecule Therefore, increase in pressure will shift the equilibrium to the right. If the pressure is decreased, the decrease in pressure can be reduced by converting 1 N 2 O 4(g) molecule to 2 NO 2(g) molecules. Effect of temperature 2NO 2(g) d N 2 O 4(g) ∆H = -58.0 kJmol -1 brown colourless The effect of decrease in temperature can be reduced by converting some NO 2(g) to N 2 O 4(g) because the reaction is exothermic. Therefore, a decrease in temperature will shift the equilibrium position of an exothermic reaction to the right. While a increase in temperature will shift the equilibrium position to the left. Le Chaterlier's principle is only a rule which helps to predict the outcome of the change. In order to explain the change, the changes in forward rate and backward rate must be considered. Glossary equilibrium position shifting of equilibrium position free energy change entropy change disorderness Le Chaterlier's principle I. Nature of equilibrium Part 3 Page 6 Past Paper Question 91 1A 2 b i ii 92 2C 7 c 94 2A 2 b ii 95 1A 1 e iv 95 2B 4 a iii 96 1A 1 f iii 97 1A 2 a ii 97 2B 8 a iii 97 2B 8 b i 91 1A 2 b i ii 2b The energy profile of the reaction A (g) + B (g) d C (g) under two different catalysis X and Y are represented below. i What is the effect of increasing temperature on the equilibrium of each system? 1½ ↑ temp. will shift the equilibrium to the left for system X and system Y. 1½ mark ii What is the effect of decreasing pressure on the equilibrium of each system? 1½ ↓ pressure will shift the equilibrium to the left for system X and system Y. 1½ mark C Many candidates discussed only one system when the question asked for the effect on each system. Many candidates did not distinguish kinetics from chemical equilibria. 92 2C 7 c 7 A carboxylic acid P, with a relative molecular mass less than 100, contains C, 55.8%; H, 7.0%; and O, 37.2% by mass . An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H 2 SO 4 gave the desired ester Q but with much of the starting material P unchanged. (Relative atomic masses : H 1.0; C 12.0; O 16.0) 7c Suggest, with explanations, two ways which would make the esterification go towards completion. 4 The reaction is reversible 1 mark RCOOH + CH 3 OH d RCOOCH 3 + H 2 O 1 mark Use large excess of CH 3 OH; mass action to shift equilibrium 1 mark Remove H 2 O as it is formed to shift equilibrium to the right 1 mark Use conc. H 2 SO 4 or other drying agent ½ mark C Some candidates misunderstood the question and gave alternative synthetic routes from acid to ester. A number of candidates failed to point out that esterification is reversible and the yield of product depends on the equilibrium position of the reaction. Some candidates confused equilibrium and rate of reaction thinking that by changing reaction conditions to increase the rate of reaction would make the reversible reaction go to completion. 94 2A 2 b ii 2b Account for each of the following: ii The melting point of ice decreases with an increase in pressure. 3 In ice, the intermolecular attraction is H-bond, each H 2 O molecule is tetrahedrally surrounded by 4 other H 2 O molecules. C.N. = 4, i.e. an open structure. 1 mark In liquid water, the molecules are packed closer together ∴ water has a higher density than water. 1 mark For the reaction, H 2 O (s) d H 2 O (l) , increase in pressure, shifts the equilibrium to the right ∴ m.p. of ice decreases with increase in pressure. 1 mark I. Nature of equilibrium Part 3 Page 7 95 1A 1 e i iv 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH 3 COOH (l) + C 2 H 5 OH (l) d CH 3 COOC 2 H 5(l) + H 2 O (l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. i Name a suitable catalyst for this reaction in the forward direction. 1 Concentrated sulphuric(VI) acid / hydrochloric acid / hydrogen chloride gas 1 mark iv Would the addition of more of the same catalyst affect the value of K c ? Explain. 1 No, catalyst can affect only the rate of (forward and backward reaction. 1 mark OR Catalyst does not affect position of an equilibrium, yield does not change. 1 mark C A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. 95 2B 4 a iii 4a Explain the following facts: iii Iodine is more soluble in aqueous potassium iodide solution than in water. 2 I 2 forms soluble complex I 3 - with KI in solution. Therefore, I 2 appears more soluble. 1 mark I 2(s) + KI (aq) d KI 3(aq) 1 mark or I 2(s) + I - (aq) d I 3 - (aq) 96 1A 1 f iii 1f SO 2(g) and O 2(g) were mixed in the mole ratio of 3 : 1 at 1000 K in the presence of a catalyst. When equilibrium was attained at 373 kPa pressure, one half of the SO 2(g) had been converted into SO 3(g) . iii If the above reaction takes place in the absence of the catalyst, but other conditions remain unchanged, will the value of K p increase, decrease or remain the same ? ½ remains the same ½ mark 97 1A 2 a ii 2a Consider the following dissociation reaction: PCl 5(g) d PCl 3(g) + Cl 2(g) At 400 K and 101 kPa pressure, the percentage dissociation of PCl 5(g) is 86%. 4 ii State the effect of an increase in pressure (I) on K p , and (II) on the percentage dissociation of PCl 5(g) . 97 2B 8 a iii 8a Suggest how the following nitrogen oxides can be prepared in the laboratory. In each case, state the reactant(s) used and the reaction conditions, and write balanced equation(s) for the reaction(s) involved. 6 iii dinitrogen tetraoxide, N 2 O 4 97 2B 8 b i 8b The synthesis of ammonia using the Haber Process involves the following: N 2(g) + 3H 2(g) d 2NH 3(g) ∆Ho = -92 kJ mol -1 7 i State the effect of a change in temperature on the reaction at equilibrium. I. Nature of equilibrium Part 4 Page 1 Topic I. Nature of equilibrium Part 4 Reference Reading 6.1.2.5 Assignment Reading A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 232–238 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 335–338 Syllabus 5. Examples of calculation Notes N.B. Since the equilibrium law describes the relationship of the concentrations of different species in an equilibrium system, all calculations must be based on the concentration ( n V ) instead of the absolute amount (n). 1. Write an equilibrium law(s) to describe the system. 2. Figure out the relationship among the equilibrium concentrations of different species, from initial concentrations, stoichiometric coefficients and law of conservation of mass. 3. Solve the equation(s). Example 1 H 2(g) + I 2(g) d 2HI (g) When 46.0 g of I 2(g) and 1.00 g of H 2(g) are heated to equilibrium at 470ºC, the equilibrium mixture contains 1.9 g I 2 . (a) How may moles of each gas are present in the equilibrium mixture? (b) Compute the equilibrium constant. (Given : Relative atomic masses: I, 126.9; H, 1.00) (a) 1. K c = [HI (g) ] 2 [H 2(g) ][I 2(g) ] = ( no. of mole of HI (g) V ) 2 no. of mole of H 2(g) V × no. of mole of I 2(g) V = no. of mole of HI (g) 2 no. of mole of H 2(g) × no. of mole of I 2(g) 2. initial no. of mole of I 2(g) = 46.0 g 126.9 gmol -1 × 2 = 0.181 mol initial no. of mole of H 2(g) = 1.00 g 1.00 gmol -1 × 2 = 0.500 mol no. of mole of I 2(g) at equilibrium = 1.90 g 126.9 gmol -1 × 2 = 0.0075 mol H 2(g) + I 2(g) d 2HI (g) initial produced used up at equilibrium I 2(g) 0.181 mol 0.181 - 0.0075 = 0.174 mol 0.0075 mol H 2(g) 0.500 mol 0.174 mol 0.500 - 0.174 = 0.326 mol HI (g) 0 0.174 × 2 = 0.348 mol 0 + 0.348 = 0.348 mol (b) 3 K c = no. of mole of HI (g) 2 no. of mole of H 2(g) × no. of mole of I 2(g) = (0.348 mol) 2 0.326 mol × 0.0075 mol = 50 I. Nature of equilibrium Part 4 Page 2 Example 2 H 2(g) + I 2(g) d 2HI (g) If 1.00 mole each of H 2(g) and I 2(g) are heated in a 30.0 dm 3 evacuated chamber to 470ºC. Using the value of K c from the Example 1, determine (a) how many mole of I 2(g) remain unreacted when equilibrium is established, (b) the total pressure in the chamber. (c) Now if one additional mole of H 2(g) is introduced into this equilibrium system, how many mole of the original iodine will remain unreacted? (Given : Universal gas constant (R) = 0.0821 atm mol -1 K -1 ) (a) 1. K c = no. of mole of HI (g) 2 no. of mole of H 2(g) × no. of mole of I 2(g) 2. Let x mole be the no. of mole of I 2(g) consumed. initial produced used up at equilibrium I 2(g) 1.00 mol x mol (1.00 - x) mol H 2(g) 1.00 mol x mol (1.00 - x) mol HI (g) 0 x mol × 2 2x mol 3. K c = (2x) 2 (1.00 - x)(1.00 - x) = 50 4x 2 1.00 - 2.00x + x 2 = 50 4x 2 = 50 - 100x + 50x 2 46x 2 - 100x + 50 = 0 x = 0.78 or 1.4 (rejected) no. of mole I 2(g) remaining = (1.00 - 0.78) mol = 0.22 mol (b) The number of moles of gas does not change as the reaction proceeds at 470ºC; hence 2.00 mol of gas remains at equilibrium. PV = nRT P = nRT V = 2.00 mol × 0.0821 atm mol -1 K -1 × 743 K 30.0 dm 3 = 4.07 atm (c) 1. K c = no. of mole of HI (g) 2 no. of mole of H 2(g) × no. of mole of I 2(g) 2. Since the final equilibrium position is not depending on when the additional H 2(g) is introduced, it can be assumed that it is introduced with the original H 2(g) . Let x mole be the no. of mole of I 2(g) consumed. initial produced used up at equilibrium I 2(g) 1.00 mol z mol (1.00 – z) mol H 2(g) (1.00 + 1.00) mol = 2.00 mol z mol (2.00 – z) mol HI (g) 0 z mol × 2 2z mol 3. K c = (2z) 2 (2.00 - z)(1.00 - z) = 50 4z 2 2.00 - 3.00z + z 2 = 50 4z 2 = 100 – 150z + 50z 2 46z 2 – 150z + 100 = 0 z = 0.93 or 2.3 (rejected) no. of mole I 2(g) remaining = (1.00 - 0.93) mol = 0.07 mol I. Nature of equilibrium Part 4 Page 3 Glossary equilibrium concentration stoichiometric coefficients law of conservation of mass Past Paper Question 90 2A 1 a i ii iii 95 1A 1 e ii iii 90 2A 1 a i ii iii 1a Consider the following equilibrium at constant pressure: A 2(g) + 3B 2(g) d 2AB 3(g) A mixture of 4.0 mol of A 2(g) and 12.0 mol of B 2(g) was placed in a vessel of volume 20.0 dm 3 , and heated to 565K. When the system had reached equilibrium, it was found that 4.0 mol AB 3(g) was present. i Calculate the concentration of each species at equilibrium. 1½ A 2(g) + 3B 2(g) d 2AB 3(g) Since 4 moles of AB 3 will consume 2 moles of A 2 and 6 moles of B 2 ∴ at equilibrium: [B 2 ] = 12 - 6 20 = 6 20 moldm -3 ½ mark [A 2 ] = 4 - 2 20 = 2 20 moldm -3 ½ mark [AB 3 ] = 4 20 moldm -3 ½ mark ii Calculate the equilibrium constant of this reaction at 565K. 2½ K = [AB 3 ] 2 [A 2 ][B 2 ] 3 = ( 4 20 ) 2 ⋅ ( 20 2 ) ⋅ ( 20 6 ) 3 = 14.82 dm 6 mol -2 (wrong unit -½) 2½ mark iii When the volume of the vessel was increased and the system allowed to come to a new equilibrium at the same temperature, 9.0 mol of B 2(g) was found to be present. Calculate the new volume. 3 At new equilibrium: 9 moles of B 2 is present. Let V be the new volume [B 2 ] = 9 V ½ mark [A 2 ] = 4-1 V = 3 V ½ mark [AB 3 ] = 2 V ½ mark K = 14.82 = ( 2 V ) 2 ⋅ ( V 3 ) ⋅ ( V 9 ) 3 ½ mark V = 14.82 × 9 3 × 3 4 = 90 dm 3 1 mark C i Some candidates mistakenly used the initially given concentration rather than the equilibrium concentration in their calculation of equilibrium constant. ii Some weaker candidates failed to give the correct unit for the equilibrium constant. 95 1A 1 e ii iii 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH 3 COOH (l) + C 2 H 5 OH (l) d CH 3 COOC 2 H 5(l) + H 2 O (l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. ii Calculate the equilibrium constant, K c , for the reaction at 298 K. 3 CH 3 COOH (l) + C 2 H 5 OH (l) d CH 3 COOC 2 H 5(l) + H 2 O (l) initial no. of moles 12.01 g 60.052 gmol -1 4.61 g 46.068 gmol -1 – – = 0.2000 mol = 0.100 mol no. of mole at equilibrium 12.01 - 5.04 60.052 0.100 - 0.084 0.084 0.084 2 marks = 0.116 = 0.016 concentration at equilibrium 0.116 V 0.016 V 0.084 V 0.084 V I. Nature of equilibrium Part 4 Page 4 K c = 0.084 V × 0.084 V 0.116 V × 0.016 V = 3.80 (no unit) 1 marks iii What additional mass of ethanol would be required in order to use up a further 0.60 g of ethanoic acid ? 2 0.60 g of ethanoic acid ≡ 0.010 mole CH 3 COOH (l) + C 2 H 5 OH (l) d CH 3 COOC 2 H 5(l) + H 2 O (l) no. of mole at equilibrium 0.0106 x 0.094 0.094 1 mark 3.80 = 0.094 2 0.106 x ∴ x = 0.0219 mole ½ mark Need to add 0.016 mole of ethanol = 0.74 g ½ mark C iii A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. Few candidates were able to complete part (iii). II. Acid-base Equilibria Part 1 Page 1 Topic II. Acid-base Equilibria Part 1 Reference Reading 6.2.1–6.2.2 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 255–258 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 219–222, 365–367, 369–370 Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 198–201, 326 Syllabus Acid-base Equilibria Acid-base Theory Dissociation of water Notes II. Acid-base Equilibria A. Acid-base Theory At the early stage of development of acid-base theory, acid was only described as a sour substance. And base is only a substance which neutralizes the sour taste of an acid. Nevertheless, scientists find that acids and bases bear some other properties. According to these properties, they developed a series of definitions to describe acid and base. They are called Arrhenius definition, Brønsted Lowry definition and Lewis definition. The coverage of the latter definition is boarder than the former one. Common Sense A r r h e n i u s d e f i n i t i o n B r φ n s t e d L owry d e f i n i t i o n L e w is defin it i o n 1. Arrhenius definition Arrhenius definition is based on ionization. In 1884, Swedish chemist Svante Arrhenius proposed that Acid – a hydrogen-containing compound that, when dissolved in water, produces hydrogen ions, H + (aq) . Base – a substance which reacts with an acid to give salt and water only. Examples of Arrhenius acid e.g. H 2 SO 4(l) + H 2 O (l) → H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) + H 2 O (l) → H 3 O + (aq) + SO 4 2- (aq) Examples of Arrhenius base e.g. NaOH (aq) + HCl (aq) → Na + Cl - (aq) + H 2 O (l) CuO (s) + HCl (aq) → Cu 2+ Cl - 2(aq) + H 2 O (l) This is the definition used in certificate level. Arrhenius theory is criticized that 1. Acid is restricted to hydrogen-containing species and base is restricted to oxide or hydroxide. 2. The theory is only applicable to aqueous medium where a lot of acid-base reaction takes place in the absence of water. II. Acid-base Equilibria Part 1 Page 2 2. Brønsted-Lowry definition In 1923, Danish chemist Johannes Brønsted and British chemist Thomas Lowry proposed a boarder definition, in which Acid – a proton donor Base – a proton acceptor e.g. NH 3(aq) + H 2 O (l) d NH 4 + (aq) + OH - (l) proton proton proton proton acceptor donor donor acceptor (base) (acid) (conjugate (conjugate acid of NH 3(aq) ) base of H 2 O (l) ) e.g. HCl (aq) + H 2 O (l) d H 3 O + (aq) + Cl - (aq) proton proton proton proton donor acceptor donor acceptor (acid) (base) (conjugate (conjugate acid of H 2 O (l) ) base of HCl (aq) ) When a Brønsted-Lowry acid donates a proton, it becomes a potential proton acceptor and is called conjugate base of the acid. Stronger a proton donor, weaker will be the conjugate base. Similarly, when a Brønsted-Lowry base accepts a proton, it becomes a potential proton donor and is called conjugate acid of the base. Stronger a proton acceptor, weaker will be the conjugate acid. Water is a very special example, it can behave as both proton acceptor and proton donor. Therefore, it is a Brønsted-Lowry acid and a Brønsted-Lowry base. e.g. H 2 O (l) + H 2 O (l) d H 3 O + (aq) + OH - (aq) proton proton proton proton acceptor donor donor acceptor (base) (acid) (conjugate (conjugate acid of H 2 O (l) ) base of H 2 O (l) ) The above reaction is also known as self-dissociation / self-ionization of water. All Arrhenius acids and bases can be classified into Brønsted-Lowry acid and base accordingly. 3. Lewis definition The American chemist Gilbert N. Lewis (1923) proposed an even broader definition for acid and base. He proposed that Acid – an electron acceptor Base – an electron donor This definition offers many advantages, including i. The acids are not limited to compounds containing hydrogen. ii. It works with solvents other than water. iii. It does not require formation of a salt or an acid-base conjugate pair. Since all chemical species are potential electron acceptor or electron donor, virtually, all chemical species are either Lewis acid or Lewis base. e.g. H 3 N: (g) + BF 3(g) → H 3 N → BF 3(s) electron electron donor acceptor (Lewis base) (Lewis acid) Most of the chemical reaction is caused by redistribution of electrons which leads to rearrangement of atoms and formation of a new substance. Therefore, all kinds of reaction can be considered as Lewis acid-base reaction. Concept of Lewis acid-base is very useful in describing reaction mechanisms. II. Acid-base Equilibria Part 1 Page 3 N.B. In the description of an acid-base equilibrium in A-Level, the Brønsted-Lowry definition is used. However, in the description of reaction mechanism, Lewis definition is used. B. Strength of acid Strength of an acid can be measured by an equilibrium constant called acidity constant or acid dissociation constant, K a . For an acid, HA (aq) in water HA (aq) + H 2 O (l) d H 3 O + (aq) + A - (aq) Equilibrium constant, K eq = [H 3 O + (aq) ][A - (aq) ] [HA (aq) ][H 2 O (l) ] Acidity constant, K a is defined as K a = [H 3 O + (aq) ][A - (aq) ] [HA (aq) ] = K eq [H 2 O (l) ] where [H 2 O (l) ] is a constant since water is the solvent in large excess. For a stronger acid, more HA (aq) molecules will dissociate into H 3 O + (aq) ion and A - (aq) ion and give a larger value for K a . Similar to pH, K a can also be expressed in a negative log scale. pH = - log [H + ] A low pH means a high concentration of H + ion. pK a = - log K a A low pK a means a larger value for K a and a stronger acid. Relative strength of selected acids and their conjugate bases Stronger an acid, weaker will be its conjugate base. Stronger a base, weaker will be its conjugate acid. N.B. Redox reaction can be considered as a competition for electron. oxidizing agent + e - d reducing agent In redox reaction, only strong oxidizing agent reacts with strong reducing agent. Similarly, acid base reaction can be considered as a competition for proton. conjugate base + H + d conjugate acid In acid base reaction, only strong acid react with strong base. The strength of an acid or a base is also an indicator of their stability. A strong acid or base is less stable than a weak acid or base. A strong acid tends to react with a strong base to form a weak conjugate base and weak conjugate acid. e.g. NH 2 - + HC≡CH → NH 3 + HC≡C - but not NH 3 + HC≡C - → NH 2 - + HC≡CH II. Acid-base Equilibria Part 1 Page 4 1. Leveling effect Acids stronger than hydroxonium ion H 3 O + (aq) , do not show any difference in acidity in water. The water will convert all those acid molecules into H 3 O + (aq) ions. HCl (aq) + H 2 O (aq) → H 3 O + (aq) + Cl - (aq) strong strong weak weak acid base acid base HBr (aq) + H 2 O (aq) → H 3 O + (aq) + Br - (aq) strong strong weak weak acid base acid base This is why we consider all HCl (aq) , H 2 SO 4(aq) and HNO 3(aq) are strong acid though they don't have the same strength. Similarly, a base stronger than hydroxide ion does not exist in aqueous medium. The water will convert all those base molecules into OH - (aq) ions. For example, NaNH 2(s) is a very strong base, a non-aqueous solvent (e.g. liquid NH 3(l) ) must be used. NaNH 2(s) + H 2 O (l) → Na + (aq) + OH - (aq) + NH 3(aq) This is known as the leveling effect of solvent. C. Dissociation of water Pure water is found to be a poor conductor of electricity but not an insulator. Therefore, it must contain a very small amount of ions. Indeed, the ions are from self-dissociation / self-ionization of water molecule. H 2 O (l) + H 2 O (l) d H 3 O + (aq) + OH - (aq) or simply H 2 O (l) d H + (aq) + OH - (aq) K = [H + (aq) ][OH - (aq) ] [H 2 O (l) ] 1. Ionic product of water H 2 O (l) d H + (aq) + OH - (aq) K = [H + (aq) ][OH - (aq) ] [H 2 O (l) ] In aqueous medium, H 2 O (l) is in large excess, thus [H 2 O (l) ] is a constant and can be incorporated in the equilibrium constant. K [H 2 O (l) ] = [H + (aq) ][OH - (aq) ] = K w K w is known as ionic product of water. Like other equilibrium constant, it is a quantity depending on temperature and has the value of 1.00 × 10 -14 mol 2 dm -6 at 25ºC. Pure water is neutral because the concentration of H + (aq) is the same as the concentration of OH - (aq) . i.e. [H + (aq) ] = [OH - (aq) ] Acidic Neutral Alkaline II. Acid-base Equilibria Part 1 Page 5 For pure water at 25ºC K w = [H + (aq) ][OH - (aq) ] = [H + (aq) ] 2 = 1.00 × 10 -14 mol 2 dm -6 [H + (aq) ] = 1.00 × 10 -7 moldm -3 Recall that pure water has the concentration of 55.6 moldm -3 . [H + (aq) ] [H 2 O (l) ] = 1.00 × 10 -7 moldm -3 55.6 moldm -3 = 1.80 × 10 -9 This means that only 1 out of every 556,000,000 ( 1 1.80 × 10 -9 ) water molecule will dissociate into a pair of H + (aq) and OH - (aq) . This explains the extremely low conductivity of pure water but it is not an insulator. Glossary acid-base equilibria Arrhenius acid-base Brønsted Lowry acid-base Lewis acid-base conjugate acid-base proton donor / acceptor electron acceptor / donor acidity constant / acid dissociation constant (K a ) pH pK a leveling effect self- dissociation / self-ionization of water ionic product of water (K w ) Past Paper Question 94 1A 1 f i 94 1A 2 c i ii 97 2A 4 c i 99 1A 4 a 94 1A 1 f i 1f i Write an equation to show that HPO 4 2- (aq) can act as a Brφnsted base in water. 1 HPO 4 2- (aq) + H 2 O (l) d H 2 PO 4 - (aq) + OH - (aq) 1 mark 94 1A 2 c i ii 2c i What is a “Brφnsted acid”? 1 “Brφnsted acid” is a proton donor 1 mark ii Write equations to show that nitric(V) acid is an acid in water, but a base in liquid hydrogen fluoride. 2 HNO 3 + H 2 O d H 3 O + NO 3 - 1 mark HNO 3 + HF d H 2 NO 3 + + F - 1 mark C i Many candidates thought that an aqueous medium was required for the proton donation. ii The equation for nitric(V) acid acting as an acid was given correctly, but few candidates gave the correct equation for it acting as a base. 97 2A 4 c i 4c A solution is formed by mixing equal volumes of 0.20 M CH 3 CO 2 H (aq) and 0.20 M CH 3 CO 2 Na (aq) . i Identify all Brφnsted acids and all Brφnsted bases in the solution. 99 1A 4 a 4a Write all the Bronsted acids present in aqueous ammonia. II. Acid-base Equilibria Part 2 Page 1 Topic II. Acid-base Equilibria Part 2 Reference Reading 6.2.3 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 258–259, 264, 268 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 367–369 Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 325–329 Syllabus pH and its measurement Temperature dependence of pH Notes D. pH and its measurement 1. Definition of pH pH is defined as the negative log of the concentration of H + (aq) in molarity. pH = - log [H + (aq) ] Because it is a log scale, every 1 unit increase in pH means ten folds decrease in the concentration of H + (aq) . pH [H + (aq) ] 0 1 mol dm -3 1 0.1 mol dm -3 2 0.01 mol dm -3 3 0.001 mol dm -3 4 0.0001 mol dm -3 For pure water at 25ºC, K w = [H + (aq) ][OH - (aq) ] = [H + (aq) ] 2 = 1.00 × 10 -14 mol 2 dm -6 [H + (aq) ] = 1.00 × 10 -7 moldm -3 pH = - log[H + (aq) ] = - log 1.00 × 10 -7 = 7.00 Therefore, pure water has pH 7 at 25ºC. Find the pH of a 0.1 M HCl (aq) solution. Find the pH of a 10 -8 M HCl (aq) solution. 2. Temperature dependence of pH Pure water has pH 7 only at 25ºC because K w is temperature dependent. The degree of dissociation of water, thus the K w , increases with increasing temperature. For example, at 50ºC, K w = 5.47 × 10 -14 mol 2 dm -6 K w = [H + (aq) ][OH - (aq) ] = [H + (aq) ] 2 = 5.47 × 10 -14 mol 2 dm -6 [H + (aq) ] = 5.47 × 10 -14 mol 2 dm -6 = 2.34 × 10 -7 moldm -3 pH = - log[H + (aq) ] = - log 2.34 × 10 -7 = 6.63 Therefore, water at 50 ºC has a pH 6.63. Temperature /ºC K w / mol 2 dm -6 0 0.11 × 10 -14 10 0.30 × 10 -14 20 0.68 × 10 -14 25 1.00 × 10 -14 50 5.47 × 10 -14 100 51.3 × 10 -14 N.B. A solution is said to be neutral because the concentration of H + (aq) is the same as the concentration of OH - (aq) , not because of the pH. A neutral solution has the pH 7 only at 25 ºC. Acidic Neutral Alkaline II. Acid-base Equilibria Part 2 Page 2 3. Measurement of pH pH meter, indicator and pH paper are commonly used to determine the pH of a solution. a) Use of pH meter pH meter is an electronic instrument which gives the pH value of a solution directly. However, pH meter needs to be calibrated before use. (1) Calibration of pH meter pH meter is calibrated by immersing the electrode into a buffer solution with known pH and adjusting the knob on the pH meter. Buffer solution is a solution whose pH is insensitive to the addition of small amount of acid or base. For example, blood is a buffer solution which has a constant pH of about 7.4 which is independent of the acidity of our diet. b) Use of indicator Indicator is a chemical which has different colour at different pH. Usually, it is a weak acid or a weak base whose conjugate acid and conjugate base have different colours. (1) Colour of indicator pH value Indicator 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 litmus | red | purple | blue | methyl orange | red |orange| yellow | phenolphthalein | colourless |pale pink| red | Universal indicator / pH indicator is a mixture of several indicators. As a result, the universal indicator gives a series of colour change at different pH instead of just 2. pH paper is a filter paper is soaked in universal indicator. Like other testing paper, it should be dipped into the solution to be tested and removed immediately. If the paper is immersed in the solution, the dye will diffuse into the solution and the colour observed would be paler than the standard. Glossary pH scale pH meter calibration of pH meter buffer solution indicator universal indicator. Past Paper Question 94 2A 3 b i 99 1A 4 d 94 2A 3 b i 3b Account for each of the following: i At 323K, the pH of pure water is less than 7.0. 2 The dissociation of water, 2H 2 O (l) d H 3 O + (aq) + OH - (aq) , is an endothermic process. 1 mark Increase in temperature, the equilibrium shifts to the right and hence [H 3 O + (aq) ] increases. In pure water, [H 3 O (aq) + ] at 323 K is higher than that at 298 K, ∴ pH of pure water at 323 K is less than 7. 1 mark C i Many candidates were not aware of the fact that the dissociation of water is an endothermic process. Some candidates erroneously pointed out that the low pH of water at 323 K was caused by the dissolution of some acidic gases. 99 1A 4 d 4d Constant boiling hydrochloric acid contains 20.2% by mass of HCl. Calculate the mass of constant boiling II. Acid-base Equilibria Part 2 Page 3 hydrochloric acid required to prepare 1.00 dm 3 of HCl (aq) of pH 2.0 at 298 K. II. Acid-base Equilibria Part 3 Page 1 Topic II. Acid-base Equilibria Part 3 Reference Reading 6.2.4 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 62–64 Modern Physical Chemistry, Bell and Hyman, 283 Assignment Reading A-Level Chemistry (3 rd ed.), Stanley Thornes (Publisher) Ltd., 258–262, 273 Chemistry in Context (4 th ed.), Thomas Nelson and Sons Ltd., 369 Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 329–330 Syllabus Strong and weak acids/bases Measuring of pH or conductivity of acid / base Dissociation constant (K a and K b ) Calculation involving pH, K a and K b Notes D. Strong and weak acids/bases An acid or a base is said to be strong if the degree of dissociation is high. Usually, this is measure in term of pH, the conductivity of the solution or the dissociation constant. 1. Measuring of pH or conductivity of acid / base Obviously, with the same concentration, a strong acid (HCl (aq) ) has lower pH than a weak acid (CH 3 COOH (aq) ). And a strong alkali (e.g. NaOH (aq) ) has higher pH than a weak base (e.g. NH 3(aq) ). Owing to the higher ionic concentration, the conductivity of a strong acid / base is also higher than that of a weak acid / base. Solution pH 1 M HCl (aq) 0 0.1 M HCl (aq) 1 0.1 M CH 3 COOH (aq) 3 pure water 7 0.1 M NH 3(aq) 11 0.1 M NaOH (aq) 13 1 M NaOH (aq) 14 2. Dissociation constant (K a and K b ) For the equilibrium, HA (aq) + H 2 O (l) d H 3 O + (aq) + A - (aq) K = [H 3 O + (aq) ][A - (aq) ] [HA (aq) ][H 2 O (l) ] Since H 2 O (l) is in excess, its concentration can be incorporated into the equilibrium constant and the new dissociation constant is called acidity constant K a . HA (aq) + H 2 O (l) d H 3 O + (aq) + A - (aq) K = [H 3 O + (aq) ][A - (aq) ] [HA (aq) ][H 2 O (l) ] K[H 2 O (l) ] = [H 3 O + (aq) ][A - (aq) ] [HA (aq) ] = K a e.g. CH 3 COOH (aq) + H 2 O (l) d CH 3 COO - (aq) + H 3 O + (aq) K a = [CH 3 COO - (aq) ][H 3 O + (aq) ] [CH 3 COOH (aq) ] = 1.7 × 10 -5 mol dm -3 II. Acid-base Equilibria Part 3 Page 2 Similarly, the dissociation constant of a base (basicity constant K b ) can also be defined in a similar way. B (aq) + H 2 O (aq) d HB + (aq) + OH - (aq) K = [HB + (aq) ][OH - (aq) ] [B (aq) ][H 2 O (l) ] K[H 2 O (l) ] = [HB + (aq) ][OH - (aq) ] [B (aq) ] = K b e.g. NH 3(aq) + H 2 O (l) d NH 4 + (aq) + OH - (aq) K b = [NH 4 + (aq) ][OH - (aq) ] [NH 3(aq) ] = 1.8 × 10 -5 mol dm -3 a) Dissociation of polybasic acid Consider the dissociation of tribasic acid H 3 PO 4(aq) in water, it dissociates in water through 3 consecutive steps. Acid K a (298K)/mol dm -3 Equilibrium H 3 PO 4(aq) 7.9 × 10 -3 H 3 PO 4(aq) d H + (aq) + H 2 PO 4 - (aq) K a 1 = [H + (aq) ][ H 2 PO 4 - (aq) ] [ H 3 PO 4(aq) ] H 2 PO 4 - (aq) 6.2 × 10 -8 H 2 PO 4 - (aq) d H + (aq) + HPO 4 2- (aq) K a 2 = [H + (aq) ][ HPO 4 2- (aq) ] [ H 2 PO 4 - (aq) ] HPO 4 2- (aq) 4.4 × 10 -13 HPO 4 2- (aq) d H + (aq) + PO 4 3- (aq) K a 3 = [H + (aq) ][ PO 4 3- (aq) ] [ HPO 4 2- (aq) ] Phosphoric acid is a tribasic acid which is not very strong. Only 25% of the molecules dissociate in water. The first dissociation constant K a 1 is low. Moreover, the second dissociation constant K a 2 is lower than the first one and the third dissociation constant K a 3 is even lower. Therefore, in a solution of H 3 PO 4(aq) , it consists of mainly water and H 3 PO 4(aq) molecules. Relative abundance of different species H 2 O (l) >> H 3 PO 4(aq) > H + (aq) > H 2 PO 4 - (aq) > HPO 4 2- (aq) > PO 4 3- > OH - (aq) Furthermore, the dissociation constants of a tribasic acid can be expressed as : H 3 PO 4(aq) d 3H + (aq) + PO 4 3- (aq) K a(overall) = K a 1 × K a 2 × K a 3 = [H + (aq) ][ H 2 PO 4 - (aq) ] [ H 3 PO 4(aq) ] × [H + (aq) ][ HPO 4 2- (aq) ] [ H 2 PO 4 - (aq) ] × [H + (aq) ][ PO 4 3- (aq) ] [ HPO 4 2- (aq) ] = [H + (aq) ] 3 [ PO 4 3- (aq) ] [ H 3 PO 4(aq) ] (1) Charge effect For a polybasic acid, the first dissociation constant K a 1 is always larger than the second dissociation constant K a 2 . This is because when the first proton is removed from the acid molecule, it is removed from an electrically neutral molecule. When the second proton is removed from the acid molecule, it is removed from a negatively charged species which is energetically less favorable. The effect of the charges on the dissociation of acid molecule is known as charge effect. II. Acid-base Equilibria Part 3 Page 3 3. Calculation involving pH, K a and K b a) Relationship between K a and K b (pK a and pK b ) Sometimes the value of K a and K b can also be expressed in negative log scale known as pK a and pK b . pK a = -log[K a ] pK b = - log[K b ] For the hydrolysis of a conjugate acid-base pair in water, HA (aq) + H 2 O (l) d H 3 O + (aq) + A - (aq) K a = [H 3 O + (aq) ][A - (aq) ] [HA (aq) ] A - (aq) + H 2 O (l) d HA (aq) + OH - (aq) K b = [HA (aq) ][OH - (aq) ] [A - (aq) ] K a × K b = [H 3 O + (aq) ][A - (aq) ] [HA (aq) ] × [HA (aq) ][OH - (aq) ] [A - (aq) ] = [H 3 O + (aq) ][OH - (aq) ] = K w K a × K b = K w = 1 × 10 -14 mol dm -3 OR -log (K a × K b ) = - log K w -log K a - log K b = - log K w pK a + pK b = p K w = 14 b) Relationship between pH, pOH and pK w K w = [H 3 O + (aq) ][OH - (aq) ] = 1 × 10 -14 mol dm -3 -log K w = -log ([H 3 O + (aq) ][OH - (aq) ]) -log K w = -log [H 3 O + (aq) ] - log [OH - (aq) ] pK w = pH + pOH = 14 II. Acid-base Equilibria Part 3 Page 4 c) Some basic assumptions Depending on whether the small quantity is omitted, there may be two possible answers for a single question. Calculate the pH of 0.100 M CH 3 COOH (aq) . K a of CH 3 COOH (aq) is 1.70 × 10 -5 mol dm -3 . 1. The equation is solved accurately CH 3 COOH (aq) d CH 3 COO - (aq) + H + (aq) concentration at equilibrium (M) 0.100 - x x x K a = [CH 3 COO - (aq) ][H 3 O + (aq) ] [CH 3 COOH (aq) ] = x · x 0.100 - x mol dm -3 = 1.70 × 10 -5 mol dm -3 x = 1.29 × 10 -3 or -1.31 × 10 -3 (rejected) pH = - log x = - log (1.29 × 10 -3 ) = 2.89 2. The equation is solved with the small quantity omitted. CH 3 COOH (aq) d CH 3 COO - (aq) + H + (aq) concentration at equilibrium (M) 0.100 - x x x K a = [CH 3 COO - (aq) ][H 3 O + (aq) ] [CH 3 COOH (aq) ] = x · x 0.100 - x mol dm -3 = 1.70 × 10 -5 mol dm -3 Since CH 3 COOH (aq) is a weak acid, x must be very small comparing with the original concentration (0.1M). i.e. 0.100 - x ≈ 0.100. x · x 0.100 = 1.70 × 10 -5 x = 1.30 × 10 -3 pH = - log x = - log (1.30 × 10 -3 ) = 2.89 It can be seen that no matter the small quantity is omitted or not, they give similar results. This is because the degree of dissociation of ethanoic acid is only about 1%. The dissociation constant of any weak acid with K a less than 1 × 10 -4 mol dm -3 can be neglected in the expression : (original concentration – x) M. This is applicable to most weak acids. II. Acid-base Equilibria Part 3 Page 5 3. Basic assumptions 1. Attention must be taken that the small quantity can be omitted only if it is added/subtracted to another number, but not multiplied/divided to another number. If x is very small, 0.1 - x ≈ 0.1 but 0.1 · x ≠ 0.1 2. All salts are strong electrolyte and completely ionize in water, only very small percentage is hydrolyzed. e.g. When NH 4 Cl (s) is dissolved in water NH 4 Cl (s) water → NH 4 + (aq) + Cl - (aq) NH 4 + (aq) + H 2 O (l) d NH 3(aq) + H 3 O + (aq) [NH 4 + (aq) ] >> [NH 3(aq) ] The amount of NH 4 + (aq) hydrolyzed is considered negligible comparing with the amount of NH 4 + (aq) from the ionization of salt (a strong electrolyte). 3. In alkaline medium, weak acid will be neutralized. CH 3 COOH (aq) + OH - (aq) → CH 3 COO - (aq) + H 2 O (l) And because the solution is alkaline, hydrolysis of CH 3 COO - (aq) is very minimal. CH 3 COO - (aq) + H 2 O (l) d CH 3 COOH (aq) + OH - (aq) [CH 3 COO - (aq) ] >> [CH 3 COOH (aq) ] Similarly, in acidic medium, weak alkali will also be neutralized. H 3 O + (aq) + NH 3(aq) → NH 4 + (aq) + H 2 O (l) Hydrolysis of the salt is very minimal in acidic medium NH 4 + (aq) + H 2 O (l) d NH 3(aq) + H 3 O + (aq) [NH 4 + (aq) ] >> [NH 3(aq) ] II. Acid-base Equilibria Part 3 Page 6 4. Example Some Useful Relationships K w = [H + (aq) ][OH - (aq) ] = K a × K b = 1 × 10 -14 mol 2 dm -6 p K w = pH + pOH = pK a + pK b = 14 Calculate the pH of 0.100 M CH 3 COO - Na + (aq) solution. K a of CH 3 COOH (aq) is 1.70 × 10 -5 mol dm -3 and K w of water is 1.00 × 10 -14 mol 2 dm -6 . Answer 1 CH 3 COO - Na + (aq) is a strong electrolyte which completely ionizes in water. CH 3 COO - Na + (aq) → CH 3 COO - (aq) + Na + (aq) A very small percentage of CH 3 COO - (aq) hydrolyzes in water and makes the solution alkaline. CH 3 COO - (aq) + H 2 O (l) d CH 3 COOH (aq) + OH - (aq) Inital concentration (mol dm -3 ) 0.100 0 0 Equilibrium concentration (mol dm -3 ) 0.100 - x x x K c = [CH 3 COOH (aq) ][OH - (aq) ] [CH 3 COO - (aq) ] = x · x 0.100 -x = [CH 3 COOH (aq) ][OH - (aq) ] [CH 3 COO - (aq) ] × [H + (aq) ] [H + (aq) ] = [CH 3 COOH (aq) ] [CH 3 COO - (aq) ][H + (aq) ] × [H + (aq) ][OH - (aq) ] = 1 K a × K w x · x 0.100 - x = K w K a Since x << 0.100, therefore 0.100 - x ≈ 0.100 x 2 0.100 = 1.00 × 10 -14 1.70 × 10 -5 x = 7.67 × 10 -6 pOH = - log [OH - (aq) ] = - log 7.67 × 10 -6 = 5.12 pH = 14 - pOH = 14 - 5.12 = 8.88 II. Acid-base Equilibria Part 3 Page 7 Anwser 2 Alternatively, the problem can be interpreted as hydrolysis of the base CH 3 COO - (aq) which is the conjugate base of CH 3 COOH (aq) . CH 3 COO - (aq) + H 2 O (l) d CH 3 COOH (aq) + OH - (aq) Inital concentration (mol dm -3 ) 0.100 0 0 Equilibrium concentration (mol dm -3 ) 0.100 - x x x Since the dissociation constant of the conjugate acid-base pair is related by the relationship K w = K a × K b . K b = K w K a = 1.00 × 10 -14 1.70 × 10 -5 = 5.88 × 10 -10 5.88 × 10 -10 = [CH 3 COOH (aq) ][OH - (aq) ] [CH 3 COO - (aq) ] = x · x 0.100 - x If the equation is solved exactly, x = 7.67 × 10 -6 or - 7.67 × 10 -6 (rejected) pOH = - log [OH - (aq) ] = - log 7.67 × 10 -6 = 5.12 pH = 14 - pOH = 14 - 5.12 = 8.88 4. Experimental determination of K a By half-neutralization By other methods Glossary conductivity dissociation constant acidity constant K a basicity constant K b polybasic acid charge effect pH pOH pK a pK b K w pK w hydrolysis of salt Past Paper Question 91 1A 3 b ii 92 2A 3 c ii 93 1A 1 e 94 1A 1 f ii 94 2A 3 c i 95 2A 2 c i ii iii 96 2A 1 c i 97 1A 2 b 91 1A 3 b ii 3b Write equation(s) to describe the reaction of ii NaHCO 3(s) with water. 1 NaHCO 3(s) → Na + (aq) + HCO 3 - (aq) HCO 3 - (aq) + H 2 O (l) d H 2 CO 3(aq) + OH - (aq) 1 mark C ii Most knew the hydrolysis of HCO 3 - (aq) to form H 2 CO 3(aq) and OH - (aq) . II. Acid-base Equilibria Part 3 Page 8 92 2A 3 c ii 3c ii 40.0 cm 3 of an aqueous solution of a weak acid, HA (aq) , was titrated with a strong base, MOH (aq) , at 298K. The initial pH, before the addition of base, was 2.70. At the equivalence point of the titration, the pH was 8.90. Calculate (I) the initial concentration of the acid. (II) the volume of the base added to reach the equivalence point. (III) the concentration of the base. (The dissociation constant of the weak acid and the ionic product of water at 298K are respectively K a = 1.8 × 10 -5 mol dm -3 K w = 1.0 × 10 -14 mol 2 dm -6 ) 5 (I) pH = 2.70 ⇒ -log[H + ] = 2.70 ⇒ [H + ] = 1.995 × 10 -3 Let a be the concentration of the weak acid HA d H + + A - a - x x x K H A HA x x a x a = = ⋅ − = × + − − [ ][ ] [ ] . 18 10 5 Since HA is a weak acid a - x ≈ a ∴ Concentration of weak acid = a = 0.2212M 2 marks (II) Let V be the volume (in cm 3 ) of MOH added. The total volume of the solution at the equivalence point = 40.0 + V (cm 3 ) Total no. of mole of HA present = 40 0 2212 1000 0 008847 × = . . mole At equivalence point, the amount of acid is the same as the amount of base. Thus, a pure salt solution is formed. It was because of hydrolysis, the solution is basic. A - + H 2 O d HA + OH - i.e. the [HA] = [OH - ] at equivalence point Since pH = 8.90 at equivalence point ⇒ [H + ] = 1.2589 × 10 -9 and [HA] = [OH - ] = 10 [H ] -14 + = 10 12589 10 14 9 − − × . = 7.943 × 10 -6 Also the amount of A - at equivalence point can be assumed to be equal to the amount of HA in the original solution because HA is a weak acid and the solution is alkaline. Therefore, HA d H + + A - at equivalence point, 7.943 × 10 -6 1.2589 × 10 -9 0.008847 (40.0 + V) / 1000 K a = [H ][A ] [HA] + - = 1.8 × 10 -5 = 1.2589 10 0.008847 (40.0 V) / 1000 7.943 10 9 6 × × + × − − 1.8 × 10 -5 = 12589 10 7943 10 9 6 . . × × − − × 0.008847 (40.0 + V) × 1000 (40.0 + V) = 77.89 cm 3 or V = 37.89 cm 3 2 marks (III) Concentration of MOH = 0.008847 37.89 × 1000 = 0.2335 M 1 mark (Comment : Wrong no. of significant figures in the calculation) C ii Few candidates obtained all three answers correctly, but marks were awarded for all intermediate steps. II. Acid-base Equilibria Part 3 Page 9 93 1A 1 e 1e A solution is prepared by dissolving potassium hydrogencarbonate in water at 298 K. Write chemical equations for four equilibrium reactions, each involving the hydrogencarbonate ions, that occur in the solution and calculate the value of the equilibrium constant for each reaction. Give that at 298 K: K 1 = 4.3 × 10 -7 moldm -3 ; K 2 = 4.8 × 10 -11 moldm -3 for H 2 CO 3 K w = 1.0 × 10 -14 mol 2 dm -6 for H 2 O 5 HCO 3 - (aq) d H + (aq) + CO 3 2- (aq) K 2 = 4.8 × 10 -11 moldm -3 HCO 3 - (aq) + H 2 O (l) d H 2 CO 3(aq) + OH - (aq) K = K w K 1 =2.33 × 10 -8 moldm -3 HCO 3 - (aq) + H + (aq) d H 2 CO 3(aq) 1 K 1 = 2.33 × 10 6 mol -1 dm 3 HCO 3 - (aq) + HCO 3 - (aq) d CO 3 2- (aq) + H 2 CO 3(aq) K = K 2 K 1 = 1.12 × 10 -4 HCO 3 - (aq) + OH - (aq) d CO 3 2- (aq) + H 2 O (l) K = K 2 K w = 4.8 × 10 3 mol -1 dm 3 4 marks(½ for equation, ½ for value) for any 4 of the above equilibria 1 mark for correct dimension of equilibrium constants C Most candidates could get the two equilibria involving HCO 3 - and H + , but failed to realize the two equilibria involving HCO 3 - and OH - indicating that the candidates were not familiar with the interplay between K a and K b of the respective conjugate acid and conjugate base. 94 1A 1 f ii 1f ii For the following equilibrium at 298 K, the equilibrium constant K c = 2.84 × 10 -22 (mol dm -3 ) 5 . Ag 3 PO 4(s) + H 2 O (l) d 3Ag + (aq) + HPO 4 2- (aq) + OH - (aq) Calculate the solubility, in mol dm -3 , of silver phosphate(V) in 0.10 M disodium hydrogenphosphate(V) solution at pH 10. Given : Ionic product of water at 298 K = 1.0 × 10 -14 (mol dm -3 ) 2 (Assume the extent of dissociation of HPO 4 2- ions at pH 10 is negligible. 3 At pH = 10 [OH - (aq) ] = 10 -4 moldm -3 [HPO 4 2- (aq) ] = 10 -1 moldm -3 K c = [Ag + (aq) ] 3 [HPO 4 2- (aq) ][OH - (aq) ] = 2.84 × 10 -22 ∴ [Ag + (aq) ] 3 = 2.84 × 10 -17 [Ag + (aq) ] = 3.05 × 10 -6 moldm -3 Q 1 mole of Ag 3 PO 4 produces 3 moles of Ag + (aq) ions. ∴ Solubility of Ag 3 PO 4 = 1.017 × 10 -6 moldm -3 3 marks 94 2A 3 c i 3c Given: K a for CH 3 (CH 2 ) 2 COOH = 1.5 × 10 -5 moldm -3 at 298K Calculate the pH of i an aqueous solution of 0.10M CH 3 (CH 2 ) 2 COOH; 2 K a = [H + ][PrCOO - ] [PrCOOH] Assumming [H + ] = [PrCOO - ], [PrCOO - ] = 0.1 1 mark 1.5 × 10 -5 = [H + ] 2 0.1 [H + ] = 1.225 × 10 -3 M pH = 2.91 1 mark II. Acid-base Equilibria Part 3 Page 10 95 2A 2 c i ii iii 2c The following reversible reaction occurs in an aqueous solution of ammonium ethanoate : CH 3 COO - (aq) + NH 4 + (aq) d CH 3 COOH (aq) + NH 3(aq) (I) i Write an expression for the dissociation constant, K a , of ammonium ion. 1 NH 4 + d NH 3 + H + K a = [NH 3 ][H + ] [NH 4 + ] 1 mark ii Calculate the equilibrium constant of reaction (I) at 298 K. (At 298 K, the dissociation constants of ethanoic acid and ammonium ion are 1.76 × 10 -5 mol dm -3 and 5.59 × 10 - 10 mol dm -3 respectively.) 2 NH 4 + + CH 3 COO - d NH 3 + CH 3 COOH K = [NH 3 ][CH 3 COOH] [NH 4 + ][CH 3 COO - ] 1 mark = K a (NH 4 + ) K a (CH 3 COOH) = 5.59 × 10 -10 1.76 × 10 -5 = 3.18 × 10 -5 1 mark iii For a 0.10 M solution of ammonium ethanoate at 298 K, calculate (1) the concentration of ammonia, and (2) the pH of the solution. 4 (1) Let x = [NH 3 ] = [CH 3 COOH] [NH 4 + ] = [CH 3 COO - ] = 0.100 - x 1 mark K = x 2 (0.100 - x) 2 = 3.18 × 10 -5 x = 5.60 × 10 -4 M 1 mark (2) Let y = [H + ] K a (CH 3 COOH) = y (100 - x) x y = 9.96 × 10 -8 M 1 mark pH = -log [H + ] = 7.00 1 mark OR K a (NH 4 + ) = y (x) (0.100 - 5.60 × 10 -4 ) 5.59 × 10 -10 = y (5.60 × 10 -4 ) (0.100 - 5.60 × 10 -4 ) y = 9.93 × 10 -8 M 1 mark pH = 7.00 1 mark C The performance of candidates in this question was poor, indicating that they were not able to apply what they had learnt to a slightly varied situation. 96 2A 1 c i 1c i At 298 K the pH of 0.050 M CH 3 CH 2 COOH is 3.10. Calculate the K a of CH 3 CH 2 COOH at 298 K. 2 CH 3 CH 2 COOH d CH 3 CH 2 COO - + H + K a = [CH 3 CH 2 COO - ][H + ] [CH 3 CH 2 COOH] ½ mark Assuming [CH 3 CH 2 COO - ] = [H + ] ½ mark K a = (10 -3.10 ) 2 (0.05 - 10 -3.10 ) = 1.26 × 10 -5 mol dm -3 (1.25 – 1.30 × 10 -5 mol dm -3 ) 1 mark (Deduct ½ mark for no / wrong unit) 97 1A 2 b 2b Account for the difference in K a values given in the following equilibrium reactions: H 2 SO 4(aq) + H 2 O (l) d H 3 O + (aq) + HSO 4 - (aq) K a = 7.94 × 10 2 mol dm -3 HSO 4 - (aq) + H 2 O (l) d H 3 O + (aq) + SO 4 2- (aq) K a = 0.10 mol dm -3 II. Acid-base Equilibria Part 4 Page 1 Topic II. Acid-base Equilibria Part 4 Reference Reading 6.2.5 Assignment Reading Advanced Practical Chemistry, John Murray (Publisher) Ltd., 65–67 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 271–272 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 374–377 Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 334–337 Syllabus Buffer Notes E. Buffer Buffer is a solution which resists to pH change upon addition of small amount of acid or alkali. Blood is an example of buffer solution, which has a constant pH of about 7.4. 1. Principle of buffer action Since buffer is capable to absorbs both H + (aq) and OH - (aq) , it must be a mixture of an acid and a base. However, strong acid and strong base cannot coexist, they neutralize each other. Buffer must be a mixture of a conjugate acid-base pair of a weak acid or weak base. e.g. A mixture of CH 3 COOH (aq) (acid) and CH 3 COO - (aq) (conjugate base), or A mixture of NH 3(aq) (base) and NH 4 + (aq) (conjugate acid) For the buffer consists of CH 3 COOH (aq) and CH 3 COO - Na + (aq). This involves the equilibrium of CH 3 COOH (aq) d CH 3 COO - (aq) + H + (aq) Upon addition of H + (aq) , H + (aq) will be removed by the base CH 3 COO - (aq) . CH 3 COO - (aq) + H + (aq) → CH 3 COOH (aq) Upon addition of OH - (aq) , OH - (aq) will be removed by the acid CH 3 COOH (aq) . CH 3 COOH (aq) + OH - (aq) → CH 3 COO - (aq) + H 2 O (l) Therefore, the pH of the solution can be maintained. II. Acid-base Equilibria Part 4 Page 2 a) pH of an acidic buffer For a buffer consists of an acid and its conjugate base (e.g. CH 3 COOH (aq) and CH 3 COO - Na + (aq) ), CH 3 COOH (aq) d CH 3 COO - (aq) + H + (aq) K a = [CH 3 COO - (aq) ][H + (aq) ] [CH 3 COOH (aq) ] [H + (aq) ] = K a [CH 3 COOH (aq) ] [CH 3 COO - (aq) ] pH = -log K a - log [CH 3 COOH (aq) ] [CH 3 COO - (aq) ] pH = pK a - log [acid] [salt] In a buffer solution, acid and its salt share the same volume ∴ pH = pK a - log ( no. of mole of acid / V no. of mole of salt / V ) = pK a - log no. of mole of acid no. of mole of salt The pH is depending on the nature of the acid i.e. pK a and the ratio of concentrations of the acid and its salt. Furthermore, the more concentrate the acid and its salt, the higher will be the buffering capacity of the solution because it will be able to absorb more base or acid. b) pH of an alkaline buffer For an buffer solution involving a weak base and its conjugate acid (e.g. NH 3(aq) and NH 4 + (aq) ) NH 3(aq) + H 2 O (l) d NH 4 + (aq) + OH - (aq) K b = [NH 4 + (aq) ][OH - (aq) ] [NH 3(aq) ] [OH - (aq) ] = K b [NH 3(aq) ] [NH 4 + (aq) ] pOH = pK b - log [NH 3(aq) ] [NH 4 + (aq) ] pOH = pK b - log [base] [salt] The base and salt share the same solution, therefore, pOH = pK b - log no. of mole of base no. of mole of salt Since pK w = pH + pOH = 14, pH = pK w - pOH pH = pK w - ( pK b - log [base] [salt] ) = 14 - pK b + log [base] [salt] ) II. Acid-base Equilibria Part 4 Page 3 2. Calculation of buffer solution Important relationships pH = pK a - log [acid] [salt] OR pH = pK a - log no. of mole of acid no. of mole of salt pOH = pK b - log [base] [salt] OR pOH = pK b - log no. of mole of base no. of mole of salt pK w = pH + pOH = 14 a) pH of buffer solution Calculate the pH of the buffer solution consists of 0.100 M NH 3(aq) and 0.0500 M NH 4 + (aq) respectively. K b of NH 3(aq) 1.58 × 10 -5 mol dm -3 . NH 3(aq) + H 2 O (l) d NH 4 + (aq) + OH - (aq) Initial concentration (mol dm -3 ) 0.100 0.0500 0 Equilibrium concentration (mol dm -3 ) 0.100 - x 0.0500 + x x Like the example mentioned in the last unit, the equation can be solved accurately or approximately. A. The equation is solved accurately. K b = [NH 4 + (aq) ][OH - (aq) ] [NH 3(aq) ] = (0.0500 + x) x 0.100 - x = 1.58 × 10 -5 x 2 + 0.0500 x - 1.58 × 10 -6 = 0 x = 3.16 × 10 -5 or - 0.0500 (rejected) pOH = - log x = - log (3.16 × 10 -5 ) = 4.50 pH = pK w - pOH = 14 - 4.50 = 9.50 B. The equation is solved with the approximation applied. It can be assumed that the 1. hydrolysis of ammonia is very minimal in the presence of NH 4 + . ∴ [NH 3(aq) ] = (0.100 - x) M ≈ 0.100 M 2. hydrolysis of ammonium ion is also very minimal in the presence of NH 3(aq) . [NH 4 + (aq) ] = (0.0500 + x) M ≈ 0.0500 M K b = [NH 4 + (aq) ][OH - (aq) ] [NH 3(aq) ] = 0.0500 · x 0.100 = 1.58 × 10 -5 x = 3.16 × 10 -5 pOH = - log x = - log (3.16 × 10 -5 ) = 4.50 pH = pK w - pOH = 14 - 4.50 = 9.50 II. Acid-base Equilibria Part 4 Page 4 C. Solving by applying the equation with approximations applied. pOH = pK b - log [NH 3(aq) ] [NH 4 + (aq) ] [NH 3(aq) ] = (0.100 - x) M ≈ 0.100 M [NH 4 + (aq) ] = (0.0500 + x) M ≈ 0.0500 M pOH = -log (1.58 × 10 -5 ) - log 0.100 0.0500 = 4.50 pH = pK w - pOH = 14 - 4.50 = 9.50 Practically, the equation pOH = pK b - log [NH 3(aq) ] [NH 4 + (aq) ] can only be used with the approximations applied. This is because a quadratic equation cannot be set up without considering the expression of K b . No matter which method is used, they all give the same result (pH = 9.50). This means that the approximations are valid. N.B. In the following examples, approximation is used whenever applicable. b) Effect of addition of acid / alkali into a non-buffered solution 1. Calculate the pH of a 0.200 M CH 3 COOH (aq) solution. K a of CH 3 COOH (aq) is 1.70 × 10 -5 mol dm -3 . CH 3 COOH (aq) d CH 3 COO - (aq) + H + (aq) At equilibrium 0.200 - x ≈ 0.200 x x x · x 0.200 = 1.70 × 10 -5 x = 1.84 × 10 -3 pH = - log [H + (aq) ] = - log (1.84 × 10 -3 ) = 2.74 2. Calculate the pH of the solution if 0.0500 mole of solid NaOH (s) is added into 600 cm 3 of 0.200 M CH 3 COOH (aq) . No. of mole of CH 3 COOH (aq) in 600 cm 3 of 0.200 M = 0.200 mol dm -3 × 0.600 dm 3 = 0.120 mol NaOH is a strong alkali which will neutralize 0.0500 mole of CH 3 COOH (aq) . In the equilibrium mixture, no. of mole of CH 3 COOH (aq) = (0.120 - 0.0500) mol = 0.070 mol no. of mole of CH 3 COO - (aq) = no. of mole of CH 3 COOH (aq) neutralized = 0.0500 mol pH = pK a - log no. of mole of acid no. of mole of salt = - log (1.70 × 10 -5 ) - log 0.070 0.050 = 4.6 3. Calculate the pH of the solution if 0.0500 mole of HCl (g) is dissolved in 600 cm 3 of 0.200 M CH 3 COOH (aq) . Since HCl (aq) is a strong acid, the solution will become very acidic. In acidic medium, the ionization of CH 3 COOH (aq) can be neglected comparing with the ionization from HCl (aq) . The effect of change in volume can be neglected when HCl (g) is being dissolved. [H + (aq) ] = [HCl (aq) ] = 0.0500 mol 0.600 dm -3 = 0.0830 mol dm -3 pH = - log 0.0830 = 1.08 II. Acid-base Equilibria Part 4 Page 5 c) Effect of addition of acid / alkali into a buffer solution 1. Calculate the pH of a buffer solution with 0.300 M of CH 2 ClCO 2 H (aq) and 0.200 M of CH 2 ClCO 2 - Na + (aq) respectively. K a of CH 2 ClCO 2 H (aq) is 1.21 × 10 -3 mol dm -3 . pH = pK a - log [acid] [salt] = - log (1.21 × 10 -3 ) - log 0.300 0.200 = 2.74 2. Calculate the pH of the solution if 0.0500 mole of solid NaOH (s) is added into 600 cm 3 of the buffer. no. of mole of CH 2 ClCOOH (aq) in 600 cm 3 of the buffer = 0.300 M × 0.600 dm 3 = 0.180 mol no. of mole of CH 2 COO - (aq) in 600 cm 3 of the buffer = 0.200 M × 0.600 dm 3 = 0.120 mol When 0.0500 mole of NaOH (s) is added, 0.0500 mole of CH 2 ClCO 2 H (aq) will be neutralized and 0.0500 mole of CH 2 ClCOO - (aq) will be produced. In the equilibrium mixture, no. of mole of CH 2 ClCOOH (aq) = (0.180 - 0.0500) mol = 0.130 mol no. of mole of CH 2 COO - (aq) = (0.120 + 0.0500) mol = 0.170 mol pH = pK a - log no. of mole of acid no. of mole of salt = - log (1.21 × 10 -3 ) - log 0.130 0.170 = 3.03 3. Calculate the pH of the solution if 0.0500 mole of HCl (g) is dissolved in 600 cm 3 of the buffer. no. of mole of CH 2 ClCOOH (aq) in 600 cm 3 of the buffer = 0.300 M × 0.600 dm 3 = 0.180 mol no. of mole of CH 2 ClCOO - (aq) in 600 cm 3 of the buffer = 0.200 M × 0.600 dm 3 = 0.120 mol CH 2 ClCOOH (aq) d CH 2 ClCOO - (aq) + H + (aq) When 0.0500 mole of HCl (g) is dissolved, 0.0500 mole of CH 2 ClCOOH (aq) will be produced and 0.0500 mole of CH 2 ClCOO - (aq) will be consumed. In the equilibrium mixture, no. of mole of CH 2 ClCOOH (aq) = (0.180 + 0.0500) mol = 0.230 mol no. of mole of CH 2 COO - (aq) = (0.120 - 0.0500) mol = 0.070 mol pH = pK a - log no. of mole of acid no. of mole of salt = - log (1.21 × 10 -3 ) - log 0.230 0.070 = 2.40 d) Comparison of the action of acid and alkali Original pH Alkali added Acid added Non-buffered Acid 2.74 4.6 1.08 Buffer solution 2.74 3.03 2.40 Comparatively, the pH of a buffer solution is more resistant to addition of alkali or acid. For a non-buffered acid, it contains no base to absorb the extra H + (aq) ions if acid is added. Furthermore, with the acid alone, addition of alkali will cause a big change in [acid] [salt] . This causes a relatively big change in pH. i.e. pH = pK a - log [acid] [salt] II. Acid-base Equilibria Part 4 Page 6 3. Preparation of buffer solution Usually, a buffer solution is prepared by mixing a weak acid / base with its salt. Or it can be prepared by half- way titration. For example, if 1dm 3 of 0.5 M CH 3 COOH (aq) requires 0.5 mole of NaOH (aq) for complete neutralization. 0.25 mole of NaOH (aq) can be added so that only half of the CH 3 COOH (aq) will be neutralized. CH 3 COOH (aq) + OH - (aq) d CH 3 COO - (aq) + H 2 O (l) Initial amount ≈ 0.5 mole ≈ 0.25 mole ≈ 0 mole Final amount ≈ 0.25 mole ≈ 0 mole ≈ 0.25 mole Since pH = pK a - log no. of mole of acid no. of mole of salt = pK a - log 1 = pK a This method is used to prepared a buffer solution with the pH equals pK a of the acid. For any pH not the same as pKa, the ratio of no. of mole of acid to no. of mole of salt can be adjusted to prepare a solution of desired pH. Glossary buffer hydrolysis half-way titration Past Paper Question 91 2A 2 c i ii 94 2A 3 c ii iii 96 2A 1 c i ii iii 97 2A 4 c ii 98 2A 1 a i ii 99 2A 4 c ii 91 2A 2 c i ii 2c A weak base MOH has an ionization constant K b = 2.0 × 10 -5 mol dm -3 , where K b = [M + ][OH - ]/[MOH]. Solution S is made up of 0.10 mol of MOH in 1.0 dm 3 , and a second solution T has 0.10 mol of MOH and 0.50 mol of MCl in 1.0 dm 3 . i Calculate the pH of solutions S and T. 4 MOH d M + + OH - Solution S: Since MOH d M + + OH - initial 0.1 0 0 at eqm. 0.1 - x x x ∴ K b = 2 × 10 -5 = x 2 0.1 - x , assuming (0.1 -x) ≈ 0.1 ∴ x = 2 × 10 -6 = 1.414 × 10 -3 M 1 mark pOH = -log(1.44 × 10 -3 ) = 2.85 pH = 14 - 2.85 = 11.15 1 mark Solution T: Since MOH d M + + OH - initial 0.1 0.5 0 at eqm. 0.1 - x 0.5 + x x x × (0.5 + x) 0.1 - x = 2 × 10 -5 with 0.5 + x ≈ 0.5 and 0.1 - x ≈ 0.1 x = 2 × 10 -5 × 0.1 0.5 = 4 × 10 -6 M 1 mark -log (4 × 10 -6 ) = 5.40 pH = 14 - 5.40 = 8.60 1 mark II. Acid-base Equilibria Part 4 Page 7 ii 0.01 mol of a strong acid HX is added separately to the solutions S and T. The new pH value for solution S is 10.26; calculate that for solution T. What conclusion can you draw from this result? (K w = 1.0 × 10 -14 mol 2 dm -6 ) 4 Solution T: Since MOH d M + + OH - Before addition of acid and hydrolysis 0.1 0.5 Upon addition of acid 0.1 - 0.01 0.5 + 0.01 0 Upon hydrolysis at eqm. 0.09 - x 0.51 + x x 1 mark x × (0.51 + x) 0.09 - x = 2 × 10 -5 with 0.51 + x ≈ 0.51 and 0.09 - x ≈ 0.09 x = 2 × 9 51 × 10 -5 = 3.53 × 10 -6 1 mark -log (3.53 × 10 -4 ) = 5.45 pH = 8.55 1 mark Solution T can resist the change in pH to a greater extent than solution S, it is generally known as a buffer. 1 mark C i Most candidates were not able to work out the final pH of solution T. 94 2A 3 c ii iii 3c Given: K a for CH 3 (CH 2 ) 2 COOH = 1.5 × 10 -5 moldm -3 at 298K Calculate the pH of ii an aqueous solution of 0.050M CH 3 (CH 2 ) 2 COONa and 0.050M CH 3 (CH 2 ) 2 COOH; and 2 Assuming [PrCOOH] = 0.050 M [PrCOO - ] = 0.050 M 1.5 × 10 -5 = [H + ] × 0.50 0.50 1 mark [H + ] = 1.5 × 10 -5 M pH = 4.82 1 mark iii 1.0 dm 3 of the solution in (ii) after the addition of 1.0 × 10 -3 mol of solid NaOH. 3 After addition of 1.0 × 10 -3 moles solid NaOH [PrCOOH] = 0.050 - 1.0 × 10 -3 = 0.049 M [PrCOO - ] = 0.050 + 1.0 × 10 -3 = 0.051 M 1 mark 1.5 × 10 -5 = [H + ] (0.051) (0.049) [H + ] = 1.44 × 10 -5 M 1 mark pH = 4.34 1 mark 96 2A 1 c i ii iii 1c i At 298 K the pH of 0.050 M CH 3 CH 2 COOH is 3.10. Calculate the K a of CH 3 CH 2 COOH at 298 K. 2 CH 3 CH 2 COOH d CH 3 CH 2 COO - + H + K a = [CH 3 CH 2 COO - ][H + ] [CH 3 CH 2 COOH] ½ mark Assuming [CH 3 CH 2 COO - ] = [H + ] ½ mark K a = (10 -3.10 ) 2 (0.05 - 10 -3.10 ) = 1.26 × 10 -5 mol dm -3 (1.25 – 1.30 × 10 -5 mol dm -3 ) 1 mark (Deduct ½ mark for no / wrong unit) ii Calculate the pH at 298 K of a solution which is 0.050 M with respect to CH 3 CH 2 COOH and to CH 3 CH 2 COONa. 2 In the solution, [CH 3 CH 2 COO - ] = [CH 3 CH 2 COOH] = 0.05 M K a = [H + ] ∴ pH = pK a 1 mark = 4.90 (4.88 – 4.90) 1 mark iii 5.0 × 10 -4 mol of solid NaOH were added separately to (1) 100 cm 3 of 0.050 M CH 3 CH 2 COOH and (2) 100 cm 3 of the solution in (ii) above. Assuming that the change in volume upon the addition of NaOH is negligible calculate the pH at 298 K of the solution in each case. Comment on the difference in the pH change of the two solutions. 5 (1) For 100 cm 3 of 0.05 M CH 3 CH 2 COOH [CH 3 CH 2 COO - ] = 5.0 × 10 -3 M ½ mark II. Acid-base Equilibria Part 4 Page 8 [CH 3 CH 2 COOH] = 4.5 × 10 -2 M ½ mark K a = [H + ] (5.0 × 10 -3 ) 4.5 × 10 -2 [H + ] = 1.13 × 10 -4 mol dm -3 (1.15 × 10 -4 mol dm -3 ) pH = 3.95 (3.93 – 3.96) 1 mark (2) For 100 cm 3 of the solution in (ii), [CH 3 CH 2 COO - ] = 5.5 × 10 -2 M ½ mark [CH 3 CH 2 COOH] = 4.5 × 10 -2 M ½ mark K a = [H + ] (5.5 × 10 -2 ) 4.5 × 10 -2 [H + ] = 1.03 × 10 -5 mol dm -3 (1.05 × 10 -5 mol dm -3 ) pH = 4.99 (4.97 – 5.00) 1 mark 0.05 M CH 3 CH 2 COOH is not a buffer ∴ pH changes significantly 1 mark or, the solution in (ii) is a buffer ∴ pH change is very small (1 mark) 97 2A 4 c ii 4c A solution is formed by mixing equal volumes of 0.20 M CH 3 CO 2 H (aq) and 0.20 M CH 3 CO 2 Na (aq) . 7 ii Calculate the concentration of each chemical species, excluding H 2 O, present in the solution at 298 K. (K a of CH 3 CO 2 H = 1.76 × 10 -5 mol dm -3 at 298 K) 98 2A 1 a i ii 1a Solution A is 0.15 M lactic acid, and solution B is a mixture of equal volumes of 0.30 M lactic acid and 0.10 M aqueous sodium hydroxide solution. Note : (1) Lactic acid is a monprotic acid and its K a at 298 K is 1.38 × 10 -4 moldm -3 . (2) For (ii) and (iii), you may assume that the volume changes are negligible. 9 i Calculate the pH of A and of B at 298 K. ii A few drops of dilute hydrochloric acid are added to 50.0 cm 3 of A and B. Compare the effect of such action on the pH of the two solutions. Explain your answer. 99 2A 4 c ii 4c ii Given the following materials and apparatus, describe a method to determine the dissociation constant, K a , of HCO 2 H. approximately 0.1 M aqueous HCO 2 H, approximately 0.1 M aqueous NaOH, phenolphthalein indicator, titration apparatus, and a calibrated pH meter II. Acid-base Equilibria Part 5 Page 1 Topic II. Acid-base Equilibria Part 5 Reference Reading 6.2.6–6.2.7 Assignment Reading Advanced Practical Chemistry, John Murray (Publisher) Ltd., 68–69 Experiment – Acid-base titration using method of double indicator A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 264–270 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 371–374 Chemistry in Context (5th ed.), Thomas Nelson and Sons Ltd., 331–334 Syllabus Indicators Acid-base titration Notes F. Theory of Indicator Acid-base indicator is a weak acid which has different colour at different pH. Actually, the conjugate acid- base pair of the acid have different colour. HIn (aq) + H 2 O (l) d H 3 O + (aq) + In - (aq) K in = [H 3 O (aq) + ][In - (aq) ] [HIn (aq) ] Colour I Colour II For example, methyl orange is a weak acid. The conjugate acid of methyl orange is red in colour while the conjugate base of methyl orange is yellow. Methyl orange (pK in of methyl orange = 3.7) N N S N H 3 C H 3 C O O O - + H 3 O + N N S N H 3 C H 3 C O O OH H 2 O + Conjugate acid (red) Conjugate base (yellow) In acidic medium, the concentration of H + (aq) is high and the equilibrium position is lying on the left. Therefore, it will be red in colour in acidic medium. For a similar reason, it is yellow in colour in alkaline medium. Phenolphthalein is also a weak acid. Its conjugate acid is colourless while the conjugate base is red. Because methyl orange and phenolphthalein have different K in , they change colour at different pH. Phenolphthalein (pK in of phenolphthalein = 9.3) C C O O OH HO C HO O - OH C O O - + 3H 2 O + 2H 3 O + Conjugate acid (colourless) Conjugate base (red) II. Acid-base Equilibria Part 5 Page 2 However, a naked eye cannot distinguish colours accurately. It cannot notice any difference unless the intensity of one colour is ten times of another. For methyl orange, pK in = 3.7 HMe (aq) + H 2 O (l) d H 3 O + (aq) + Me - (aq) K in = [H 3 O (aq) + ][Me - (aq) ] [HMe (aq) ] Therefore, it will become red at pH = , i.e. [acid] [salt] = 10 pH = pK in - log [acid] [salt] = pK in - log 10 = 3.7 - 1 = 2.7 The solution will become yellow at pH = , i.e. [acid] [salt] = 1 10 pH = pK a - log [acid] [salt] = pK in - log 1 10 = 3.7 + 1 = 4.7 The pH range from 2.7 to 4.7 is called the working range of methyl orange. N.B. Working range of a pH indicator = pK in ± 1 Phenolphthalein works in the same way as methyl orange. It has pK in 9.3, therefore, the working range of phenolphthalein is from pH 8.3 to 10.3. Because indicator is actually an acid, only minimal amount of indicator should be added during a titration. So that the pH of the solution will not be affected. G. Acid-base titration Titration is a process for determining the volume of one solution required to react quantitatively with a given volume of another. One solution is added to the other, a small amount at a time until just sufficient has been added to complete the reaction. The equivalence point is determined as an end point. Titrations may be carried out by hand from a burette or automatically. 1. Difference between equivalence point and end point Equivalence point – the stage in a titration when the reactants and products are present in equivalent amounts according to the stoichiometry of the reaction. End point – the stage in a titration when a change (e.g. colour, pH, temp., conductivity changes) is observed indicating the end of the titration. The accuracy of a titration is depending on how close the end point (experimental value) is to the equivalence point (theoretical value). II. Acid-base Equilibria Part 5 Page 3 2. Titration using pH meter About the equivalence point of most acid-base titration, the solution shows a great pH change upon addition of small amount of acid or base. By plotting a graph with the pH of the solution against the volume of titrant added, the end point can be determined. The graphy is usually called titration curve. For the titration between a weak acid and a weak base, plotting of a pH graph is not applicable because there is no abrupt pH change throughout the course. It is hard to decide where is the end point. II. Acid-base Equilibria Part 5 Page 4 3. Titration using indicator a) Choosing of indicator Choosing of indicator has a great impact on the accuracy of titration. A wrongly chosen indicator will cause a large error. The selection is depending on the shape of the pH curve of titration. The indicator is chosen so that the abrupt change on the pH curve will fall across the working range of the indicator. The value of pK in should be close to the value of the pH of the solution at the end-point, so that the colour change occurs as closely as possible to the equivalence point. In the titration of strong acid versus strong base, the equivalence point will be at pH 7 because the salt formed will not hydrolyse in water. Although neither methyl orange nor phenolphthalein changes colour at pH 7, the colour will change with a volume very close to the equivalence point. Therefore, both methyl orange and phenolphthalein can be used in the titration between a strong acid and a strong base. Conversely, in the titration of weak acid versus strong base, the equivalence point will be higher than pH 7, because of hydrolysis of the salt. A - (aq) + H 2 O (aq) d HA (aq) + OH - (aq) Furthermore, the vertical portion of the pH curve will also be shifted to the higher pH range. If phenolphthalein is used, the end point will be very close to the equivalence point. However, if methyl orange is used, the colour will change far before the equivalence point is reached. And a large error will be caused. Therefore, only phenolphthalein can be used in the titration between a weak acid and a strong base. II. Acid-base Equilibria Part 5 Page 5 Litmus is usually not used in titration because the colour change is not as obvious as methyl orange and phenolphthalein. It is only used in the form of litmus paper where a white background is available. In a titration of a strong acid and weak base, methyl orange should be used. In a titration of a weak acid and a strong base, phenolphthalein should be used. In a titration of a strong acid and a strong base, either methyl orange or phenolphthalein can be used. pH value Indicator 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 methyl orange | red |orange| yellow | phenolphthalein | colourless |pale pink| red | For the titration between a weak acid and a weak base, no indicator is appropriate because there is no abrupt pH change throughout the course. The end point of a titration between a weak acid and a weak base can neither be detected by any indicator nor a pH meter. The end points are usually determined by thermometric titration or conductimetric titration. 4. Thermometric titration Since neutralization is exothermic, the temperature of the solution will increase until the equivalence point is reached. However, the addition of any cold excess titrant into the warm solution will cause a drop in temperature. According to this principle, by plotting a graph with the temperature versus the volume of the titrant added, the end point can be determined. The graph is composed of 2 straight lines. One representing the exothermic reaction and another represents the cooling. The end point is where the two straight lines intersect each other. N.B. According to the theory behind thermometric titration, the points on the graph should not be joined together by a smooth curve. II. Acid-base Equilibria Part 5 Page 6 5. Conductimetric titration Depending on the nature of the acid, base and the salt formed, the conductivity / conductance of the solution will also change throughout the titration process. Conductivity of a solution depends on i) ionic concentration ii) mobility of the ions a) Strong acid vs Strong base If a strong acid is titrated against a strong base, the product will be water and the salt. For example, in the titration between NaOH (aq) and HCl (aq) , NaOH (aq) + HCl (aq) → H 2 O (l) + NaCl (aq) At the beginning, the solution contains only H + (aq) and Cl - (aq) . Upon neutralization, a pair of H + (aq) and OH - (aq) ions is converted to a H 2 O (l) molecule. Consequently H + (aq) in the acid is replaced by Na + (aq) which has lower conductivity. Therefore, the conductivity will drop before the end point. Beyond the end point, addition of a strong base (a strong electrolyte) increases the ionic concentration, thus the conductivity. Furthermore, the excess OH - (aq) ions are more conductive. The end point is determined by extrapolating the two sessions of the graph with 2 straight lines. b) Weak acid vs Strong base At the beginning, the acid solution contains mainly undissociated acid molecules and the conductance is minimal. The addition of a strong base will convert the molecular acid molecules to ionic salt. Thus, the conductivity will increase. Beyond the end point, the conductivity will increase even further upon the addition of excess strong base. c) Weak acid vs Weak base However, if the titration is between a weak acid and a weak base HA (aq) + BOH (aq) → H 2 O (l) + A - (aq) + B - (aq) Two covalent molecule reacts to form 1 water molecule and 2 ions. The increases in ionic concentration will cause an increase in conductivity, Beyond the end point, the addition of the weak acid will only dilute the solution and the conductivity will decreases. N.B. Sometimes, it is difficult to predict the actual shape of the graph but a sudden change in slope can be treated as the existence of end point. Glossary indicator methyl orange phenolphthalein K in naked eye working range titration equivalence point end point titration curve (pH curve) indicator thermometric titration conductimetric titration conductivity / conductance ionic concentration extrapolating II. Acid-base Equilibria Part 5 Page 7 Past Paper Question 90 1A 3 a 92 2A 3 c i 94 2A 3 b ii 95 1B 4 e i ii 98 2A 1 a iii 99 1A 4 b i ii 99 1A 4 c 90 1A 3 a 3a Explain why phenolphthalein turns pink in a solution of sodium carbonate, but remains colourless in a solution of sodium hydrogencarbonate. 2½ CO 3 2- and HCO 3 - hydrolyse to a different extent. ½ mark In a solution of Na 2 CO 3 , CO 3 2- + H 2 O d HCO 3 - + OH - ½ mark resulting in a pH sufficiently high ( and [H + ] sufficiently low) such that the equilibrium, In(red) + H + d InH + (colourless) 1 mark lies to the left and a pink colour is exhibited. In a solution of NaHCO 3 , whose pH is also above 7, ½ mark [H + ] is sufficiently high for the equilibrium to shift to the right. C Few candidates were able to answer that a solution of Na 2 CO 3 has a higher pH value than that of a solution of NaHCO 3 because the hydrolysis of CO 3 2- and HCO 3 - takes place to a different extent; and that at the higher pH, phenolphthalein exists in a form which has a resonance structure and the solution becomes pink. 92 2A 3 c i 3c i Explain the difference between the equivalence point and the end point of a titration. 2 Equivalence point is the point in the titration where moles of acid equals moles of base, corresponding to a salt solution. The end point is the point in a titration where a particular indicator changes colour which occurs at pH close to the value of pK a for the indicator. 2 marks 94 2A 3 b ii 3b Account for each of the following: ii At 298K, in a solution of pH 7.0, the indicator methyl orange shows its alkaline colour (yellow), while phenolphthalein shows its acidic colour (colourless). 3 Acid-base indicators are weak acids / bases. The dissociation of which can be represented by HIn (aq) + H 2 O (l) d H 3 O + (aq) + In - (aq) 1 mark The colour of an indicator depends on the relative concentrations of the acidic form, HIn and the alkaline form In - (aq) which are of different colours. ½ mark The dissociation constant K i of different acid-base indicators are different, thus they change colour over different pH range. The pH range of methyl orange is below 7, while that of phenolphthalein is above 7. ∴ at pH 7, methyl orange shows its alkaline colour, while phenolphthalein shows its acidic colour. 1½ mark C ii Very few candidates mentioned the fact that an acid-base indicator is a weak acid or weak base. They also failed to relate the colour of the indicator to its pH range. 95 1B 4 e i ii 4e For each of the volumetric analyses (i) to (iii), state whether an indicator is required. If an indicator is required, select the appropriate one from those given below: litmus; methyl orange; phenolphthalein; potassium dichromate(VI) solution; starch solution i ethanedioic acid titrated with sodium hydroxide 1 phenolphthalein 1 mark ii sulphuric(VI) acid titrated with aqueous ammonia 1 methyl orange 1 mark 98 2A 1 a iii 1a Solution A is 0.15 M lactic acid, and solution B is a mixture of equal volumes of 0.30 M lactic acid and 0.10 M aqueous sodium hydroxide solution. Note : (1) Lactic acid is a monprotic acid and its K a at 298 K is 1.38 × 10 -4 moldm -3 . (2) For (ii) and (iii), you may assume that the volume changes are negligible. iii 50.0 cm 3 of A and of B are titrated with a strong base. Compare the pH of the two solutions at the respective equivalence points of the titrations. Explain your answers. 99 1A 4 b i ii c 4b The graph below shows the variation of pH when 25.0 cm 3 of 0. 10 M HCl (aq) is titrated against 0. 10 M NaOH (aq) . II. Acid-base Equilibria Part 5 Page 8 i On the above graph, sketch a curve to represent the variation of pH when 25.0 cm 3 of 0. 10 M CH 3 CO 2 H (aq) is titrated against 0. 10 M NaOH (aq) . ii From the table below, choose an appropriate indicator for the titration in (i). Explain your choice. Indicator pH range of colour change bromocresol green 3.8 - 5.4 bromothymol blue 6.0 - 7.6 thymolphthalein 8.3 - 10.6 99 1A 4 c 4c Given an aqueous solution of Na 2 CO 3 and NaHCO 3 , suggest how to determine the concentrations of the two substances by titrimetric method using indicators. III. Redox Equilibria Part 1 Page 1 Topic III. Redox Equilibria Part 1 Reference Reading 6.3.1 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 188–189 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 74–80 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 21–29 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 186–187 Syllabus Redox Equilibria Faraday and mole Notes III. Redox Equilibria Redox equilibrium is similar to Acid-base equilibrium. An acid-base equilibrium can be considered as a competition of donation and acceptance of proton. Conjugate base + proton d Conjugate acid e.g. NH 3(aq) + H + (aq) d NH 4 + (aq) Upon donation of proton, a conjugated acid will be converted to a conjugate base. Redox equilibrium can also be considered as a competition of donation and acceptance of electron. Oxidizing agent + electron d Reducing agent e.g. Mg 2+ (aq) + 2e - (aq) d Mg (s) Upon donation of electron, a reducing agent will be converted to an oxidizing agent. A. Redox reaction A complete redox reaction consists of two systems of redox equilibria, thus two half ionic equation. One half ionic equation represents the reduction and another represents oxidation. Reduction (acceptance of electron) Ag + (aq) + e - (aq) d Ag (s) Oxidation (donation of electron) Cu (s) d Cu 2+ (aq) + 2e - Overall redox reaction 2Ag + (aq) + Cu (s) d 2Ag (s) + Cu 2+ (aq) 1. Balancing of redox reaction In a balanced equation, the no. of atoms and the no. of charges must be balanced at the two sides of the equation. Furthermore, in a redox reaction, the no. of electrons accepted by oxidizing agent must be the same as the no. of electrons donated by reducing agent. Therefore, a balanced redox reaction can be constructed in two ways a) By combining balanced half-ionic equation b) By the change in oxidation no. III. Redox Equilibria Part 1 Page 2 a) By combining balanced half-ionic equation (1) Steps of writing Balanced Half ionic equation 1. Write down the reactant and the product Cr 2 O 7 2- (aq) → Cr 3+ (aq) 2. Mass balance – by adding something already existed. Since the reaction is usually conducted in water medium, equation may be balanced by i. Adding H 2 O (l) for lack of O ii. Adding H + (aq) for lack of H (from dissociation of water, H 2 O d H + + OH - ) Cr 2 O 7 2- (aq) → 2Cr 3+ (aq) (Balance the no. of Cr) Cr 2 O 7 2- (aq) → 2Cr 3+ (aq) + 7H 2 O (l) (Add H 2 O (l) for lack of O) 14H + (aq) + Cr 2 O 7 2- (aq) → 2Cr 3+ (aq) + 7H 2 O (l) (Add H + (aq) for lack of H) 3. Charge balance – Add e - to balance the charge 14H + (aq) + Cr 2 O 7 2- (aq) → 2Cr 3+ (aq) + 7H 2 O (l) (14 × +1) + (-2) = +12 (2 × +3) + (0) = +6 14H + (aq) + Cr 2 O 7 2- (aq) + 6e - → 2Cr 3+ (aq) + 7H 2 O (l) (14 × +1) + (-2) + (6 × -1) = +6 (2 × +3) + (0) = +6 4. Check once again 14H + (aq) + Cr 2 O 7 2- (aq) + 6e - → 2Cr 3+ (aq) + 7H 2 O (l) Mass: 14H + 2Cr + 7O 2Cr + 14H + 7O Charge: (14 × +1) + (-2) + (6 × -1) = +6 (2 × +3) + (0) = +6 Note: Balancing of half-ionic equation can only be started with correct reactant and product, which can only be memorized. (2) Combining half-ionic equations In a redox reaction, an oxidizing agent must react with a reducing agent. i.e. Oxidizing agent does not react with another oxidizing agent. By combining an oxidation half equation and a reduction half equation, the ionic equation can be reconstructed. (First of all, the half ionic equations must be balanced.) For example, Reduction half ionic equation Ag + (aq) + e - → Ag (s) (Oxidizing agent: Ag + (aq) ) Oxidation half ionic equation Cu (s) → Cu 2+ (aq) + 2e - (Reducing agent: Cu (s) ) Since the no. of e - gained by Ag + (aq) must equal the no. of e - lost by Cu (s) , ∴ 2 Ag + are needed for each Cu. Reduction half ionic equation 2Ag + (aq) + 2e - → 2Ag (s) Oxidation half ionic equation Cu (s) → Cu 2+ (aq) + 2e - –––––––––––––––––––––––––––––––––––––––– 2Ag + (aq) + 2e - + Cu (s) → 2Ag (s) + Cu 2+ (aq) + 2e - (left and right must have the same no. of e - ) Overall ionic equation 2Ag + (aq) + Cu (s) → 2Ag (s) + Cu 2+ (aq) III. Redox Equilibria Part 1 Page 3 b) By the change in oxidation no. Since O.N. is the difference between the no. of electron and no. of proton associated with an atom, an increase in O.N. by 1 is equivalent to losing of 1 e-. Similarly, a decrease in O.N. by 1 is equivalent to gaining of 1 e-. Oxidizing agent MnO 4 - (aq) → Mn 2+ (aq) Each Mn gains 5 electrons +7 +2 Reducing agent Fe 2+ (aq) → Fe 3+ (aq) Each Fe loses 1 electron +2 +3 By the fact that the no. of electron gained by the oxidizing agent must be the same as the no. of electron lost by the reducing agent, each MnO 4 - (aq) must reacts with 5 Fe 2+ (aq) . Therefore, the equation will become MnO 4 - (aq) + 5Fe 2+ (aq) → Mn 2+ (aq) + 5Fe 3+ (aq) No. of O is not balanced, so add some H 2 O (l) . MnO 4 - (aq) + 5Fe 2+ (aq) → Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O (l) No. of H is not balanced, so add some H + (aq) . MnO 4 - (aq) + 5Fe 2+ (aq) + 8H + (aq) → Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O (l) Check the no. of atoms and charges again mass charge mass charge 1 Mn (-1) + 5×(+2) + 8×(+1) 1 Mn (+2) + 5×(+3) 4 O = +17 4 O = +17 5 Fe 5 Fe 8 H 8 H The equation, MnO 4 - (aq) + 5Fe 2+ (aq) + 8H + (aq) → Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O (l) , is balanced. This method is particularly useful in answering M.C. question where the completely balanced equation is usually not required. 2. Faraday and mole In counting the no. of atom, mole is used as the counting unit and method of counting by weighing is used. In counting the no. of electron, method of counting charges is used. The amount of charges carried by 1 mole of electrons is called 1 Faraday of charges (1 F). 1 Faraday of charges is found to be 96500 Coulombs (96500 C). 6.02 × 10 23 electron charges ≡ 1 F of charges ≡ 96500 C 1 electron charge ≡ 1 6.02 × 10 23 F ≡ 96500 6.02 × 10 23 C 1 electron charge ≡ 1.66 × 10 -24 F ≡ 1.60 × 10 -19 C By measuring the size of the current and the time taken, the amount of charges flowing through a circuit can be measured. Charges (Q) = Current (I) × Time (t) If the current is measured in ampere (A) and the time is measured in second (s), the unit of charge will be coulomb (C). Charges (Q) = Current (I) × Time (t) 1 C = 1 A × 1 s 1 C = 1 As III. Redox Equilibria Part 1 Page 4 3. Calculation of mass liberated in electrolysis In the electrolysis of CuSO 4(aq) using graphite electrodes, Cu (s) will deposit on the cathode and O 2(g) will evolve at the anode. By measuring the size of current and the time taken of electrolysis, the amount of Cu (s) and O 2(g) can be calculated. In the circuit, rheostat and ammeter must be installed to adjust and measure the current. For example, if a beaker of CuSO 4(aq) is electrolyzed at 200 mA for 20 minutes, what will be the mass of Cu(s) deposited on the cathode and the volume of oxygen evolved at the anode. All volume is measured at standard temperature and pressure. (Given : Relative atomic mass Cu : 63.55; O : 16.00 Molar volume of gas at s.t.p. is 22.4 dm 3 ) At cathode Cu 2+ (aq) + 2e - → Cu (s) At anode 4OH - (aq) → O 2(g) + 2H 2 O (l) + 4e - By the relationship Q = It, No. of coulombs flowing through the circuit = Current in ampere × time in second = 200 1000 A × (20 × 60) s = 240 C No. of mole of electron flowing through the current = charges in coulomb Faraday constant = 240 96500 C = 2.49 × 10 -3 mol Formation of each mole of Cu (s) requires 2 moles of electron, therefore, no. of mole of Cu (s) formed = no. of mole of electron 2 = 2.49 × 10 -3 mol 2 = 1.25 × 10 -3 mol Mass of Cu (s) deposited = no. of mole of Cu (s) × molar mass of Cu = 1.25 × 10 -3 mol × 63.55 g mol -1 = 0.0794 g Similarly, formation of each mole of O 2(g) requires losing of 4 mole of electron. no. of mole of O 2(g) formed = no. of mole of electron 4 = 2.49 × 10 -3 mol 4 = 6.23 × 10 -4 mol Volume of O 2(g) at s.t.p. = no. of mole of oxygen × molar volume of gas at s.t.p. = 6.23 × 10 -4 mol × 22.4 dm 3 mol -1 = 0.0140 dm 3 = 14.0 cm 3 Glossary redox equilibrium oxidizing agent reducing agent conservation of mass and charge faraday (F) coulomb (C) III. Redox Equilibria Part 1 Page 5 Past Paper Question 91 2A 2 b i ii 92 2A 2 a i 93 1A 1 d i ii 93 2A 3 a i ii 93 2B 6 b 94 1A 1 d e i ii 94 2B 4 e i ii 95 1A 2 b i 95 2B 4 b i ii 97 1A 1 b 99 2A 4 b i ii 91 2A 2 b i ii 2b i CrO 4 2- ions react with S 2 O 3 2- ions to form Cr(OH) 4 - and SO 4 2- ions in basic medium. Write balanced equations for each half reaction and for the overall reaction. 3 CrO 4 2- + 4H 2 O + 3e - → Cr(OH) 4 - + 4OH - 1 mark S 2 O 3 2- + 10OH - → 2SO 4 2- + 5H 2 O + 8e - 1 mark overall equation : 8CrO 4 2- + 3S 2 O 3 2- + 17H 2 O → 8Cr(OH) 4 - + 6SO 4 2- + 2OH - 1 mark ii What volume of 0.50M Na 2 CrO 4 solution is needed to react completely with 40.0 cm 3 of 0.20M Na 2 S 2 O 3 solution in basic medium? 2 No. of mole of S 2 O 3 2- = 0.2 × 40 × 10 -3 Since 3 moles of S 2 O 3 2- ≡ 8 moles of CrO 4 2- 1 mark ∴ no. of moles of CrO 4 2- = 0.2 × 40 × 8 3 × 10 -3 Volume of Na 2 CrO 4 = 0 2 40 8 3 10 050 3 . . × × × − = 0.04267 dm 3 = 42.67 cm 3 1 mark (no unit -½) (more than 2 decimal places -½) C i Some candidates failed to recognize that the given reaction takes place in basic medium. ii Most candidates were not able to write balanced equations for the half-reactions which take place in basic medium. 92 2A 2 a i 2a In acid solution, chlorate(V) ions, ClO 3 - , slowly oxidize chloride ions to chlorine. The following kinetic data are obtained at 25°C. [ClO 3 - ]/mol dm -3 [Cl - ]/mol dm -3 [H + ]/mol dm -3 Initial rate/mol dm -3 0.08 0.15 0.20 1.0 × 10 -5 0.08 0.15 0.40 4.0 × 10 -5 0.16 0.15 0.40 8.0 × 10 -5 0.08 0.30 0.20 2.0 × 10 -5 i Write the balanced equation for this reaction. 1 The balanced equation for this reaction ClO 3 - + 5Cl - + 6H + → 3Cl 2 + 3H 2 O 1 mark C i Weaker candidates were not able to give the balanced equation. 93 1A 1 d i ii 1d Write balanced equations for: i the reaction between MnO 2(s) and PbO 2(s) in acidic solution to give MnO 4 - (aq) and Pb 2+ (aq) . 1½ 2MnO 2(s) + 3PbO 2(s) + 4H + (aq) d 2MnO 4 - (aq) + 3Pb 2+ (aq) + 2H 2 O (l) 1½ mark ii the reaction between H 2 O 2(aq) and Cr(OH) 4 - (aq) in alkaline solution to give CrO 4 2- (aq) . 1½ 3H 2 O 2(aq) + 2Cr(OH) 4 - (aq) + 2OH - (aq) d 2CrO 4 2- (aq) + 8H 2 O (l) 1½ mark C A surprisingly large number of candidates could not do the balancing of equation for redox reactions which should be basic work for A-Level candidates. III. Redox Equilibria Part 1 Page 6 93 2A 3 a i ii 3a i Write balanced half equations for the redox reaction of Cr n+ and MnO 4 - in acidic solution. In terms of n, how many moles of electrons per mole of Cr n+ , are involved in the oxidation reaction? Write the balanced overall chemical equation for the above reaction. 5 The half reactions 7H 2 O + 2Cr n+ → Cr 2 O 7 2- + 14H + + (12 - 2n)e - 2 marks MnO 4 - + 8H + + 5e - → Mn 2+ + 4H 2 O 1 mark No. of moles of e - per Cr n+ is (6 - n) 1 mark Overall equation: 35H 2 O + 10Cr n+ + (12 - 2n)MnO 4 - + (12 - 2n)8H + → 5Cr 2 O 7 2- + 7OH - + (12 - 2n)Mn 2+ + (12 -2n)4H 2 O or 10Cr n+ + (12 - 2n)MnO 4 - + (26 - 16n)H + → 5Cr 2 O 7 2- + (12 - 2n)Mn 2+ + (13 - 8n)H 2 O 1 mark ii An acidified solution containing 1.50 × 10 -3 mol of Cr n+ is titrated with 0.0250 M KMnO 4 solution. The equivalence point of the reaction is reached after 48.00 cm 3 of KMnO 4 has been added. Calculate the value of n. 4 No. of moles of Cr n+ is 0.00150 No. of moles of MnO 4 - is 48.00 × 0.0250 1000 = 0.00120 1 mark 5 moles of Cr n+ ≡ 4 moles of MnO 4 - 1 mark 10 12 - 2n = 5 4 1 mark n = 2 1 mark C This part was unsatisfactorily answered. Many candidates started with wrong equations so that they were not able to obtain the correct final answer. The balancing of equations for the redox reactions should be given more attention. 93 2B 6 b 6b In aqueous solution, 1 mole of iodine reacts with 2 moles of thiosulphate ions, whereas 4 moles of bromine reacts with 1 mole of thiosulphate ions. Write balanced ionic equations for the two reactions. Predict the reaction between chlorine and thiosulphate ions in aqueous solution. Explain your prediction. 5 Equation for the reaction between I 2 and S 2 O 3 2- is I 2 + 2S 2 O 3 2- → 2I - + S 4 O 6 2- 1 mark for the reaction between 4Br 2 and S 2 O 3 2- , 4Br 2 + S 2 O 3 2- + 5H 2 O → 8Br - + 2SO 4 2- + 10H + 2 mark Chlorine will react similarly as bromine because Cl 2 is a stronger oxidising agent when compared with Br 2 and +6 (in SO 4 2- ) is the maximium oxidation state for S. 2 marks C Few candidates were able to construct the equation for the reaction of bromine and thiosulphate ions. Some candidates suggested that chlorine could oxidize thiosulphate ions to peroxodisulphate(VI); or that chlorine was a stronger reducing agent. 94 1A 1 d e i ii 1d An aqueous solution of titanium (Ti) salt was electrolysed by passing a current of 5.00 A for 2.50 hours. As a result, 5.60 g of metallic Ti were deposited at the cathode. Deduce the charge on the Ti ion in the solution. Give : 1 faraday = 96500 Cmol -1 3 Ti n+ + ne - → Ti i.e. 1 mole of Ti produced requires n moles of e - no. of moles of Ti liberated = 5.60 47.90 = 1.169 × 10 -1 1 mark no. of moles of e - passed = 5.00 × 2.50 × 60 × 60 96550 = 4.663 × 10 -1 1 mark ∴ n = 4.663 × 10 -1 1.169 × 10 -1 = 3.989 ≈ 4 1 mark 1e Write a balanced equation for each of the following reactions. i The reaction between BrO 3 - (aq) and Br - (aq) in acidic solutions to give Br 2(aq) . 1½ BrO 3 - (aq) + 5Br - (aq) + 6H + (aq) → 3Br 2(aq) + 3H 2 O (l) 1½ mark ½ mark for completing the equation 1 mark for balancing the equation ii The reaction between MnO 4 - (aq) and SO 2(g) in alkaline solution to give MnO 2(s) and SO 4 2- (aq) . 1½ 2MnO 4 - (aq) + 3SO 2(g) + 4OH - (aq) → 2MnO 2(s) + 3SO 4 2- (aq) + 2H 2 O (l) 1½ mark ½ mark for completing the equation 1 mark for balancing the equation C Generally well-answered, except that some candidates still put electrons in the overall equation. III. Redox Equilibria Part 1 Page 7 94 2B 4 e i ii 4e In concentrated HCl, 1 mole of hydrazine H 2 NNH 2 reacts with 1 mole of potassium iodate(V) to give 1 mole of iodine monochloride ICl with the evolution of a colourless gas. i Deduce, the oxidation state of nitrogen in the reaction product, and hence suggest what the colourless gas might be. 2 Q NH 2 NH 2 : IO 3 - = 1 : 1 In the reduction of IO 3 - , O.S. of I decreases from +5(IO 3 - ) to +1(ICl) ∴ In the oxidation of NH 2 NH 2 , 4 moles of e - should be released from each mole of NH 2 NH 2 . 1 mark In 1 mole of NH 2 NH 2 , there are 2 moles of N atoms. ∴ O.S. of N in the reaction product is 0. ½ mark The colourless gas in probably N 2 . ½ mark ii Write balanced half-equations and a balanced overall equation for the reaction. 3 IO 3 - + 5H + + HCl + 4e - → ICl + 3H 2 O 1 mark NH 2 NH 2 → N 2 + 4H + + 4e - 1 mark IO 3 - + H + + HCl + NH 2 NH 2 → ICl + N 2 + 3H 2 O 1 mark OR IO 3 - + 5H + + HCl + 4e - → ICl + 3H 2 O 1 mark NH 3 NH 3 2+ → N 2 + 6H + + 4e - 1 mark IO 3 - + HCl + NH 3 NH 3 2+ → ICl + N 2 + 3H 2 O + H + 1 mark C This question demanded application of knowledge to an unfamiliar situation and was poorly-answered. Only a few candidates worked out the identity of the colourless gas from the change in oxidation states of nitrogen. Some candidates gave I + , instead of ICl, as the product in the reduction of IO 3 - . 95 1A 2 b i 2b i What is the essential feature of a "redox reaction" ? 1 “Redox reaction” : a reaction involving the transfer of electron(s) from one reactant to another 1 mark OR A reaction involving a change in oxidation state / oxidation no. 1 mark C i Generally well answered, except that some candidates did not point out the essential feature of a redox reaction. They just wrote that a redox reaction involves reduction and oxidation. 95 2B 4 b i ii 4b Give the oxidation states of iodine in the two equations below. Balance the equation in each case. i S 2 O 3 2- (aq) + I 2(aq) → S 4 O 6 2- (aq) + I - (aq) 2 Oxidation state of I in I 2 = 0 1 mark Oxidation state of I in I - = -1 2S 2 O 3 2- (aq) + I 2(aq) → S 4 O 6 2- (aq) + 2I - (aq) 1 mark ii I - (aq) + IO 3 - (aq) + H + (aq) → I 2(aq) + H 2 O (l) 2 Oxidation state of I in I - = -1 1 mark Oxidation state of I in IO 3 - = +5 / V Oxidation state of I in I 2 = 0 5I - (aq) + IO 3 - (aq) + 6H + (aq) → 3I 2(aq) + 3H 2 O (l) 1 mark C ii Many candidates could not give the correct oxidation state of iodine in IO 3 - . 97 1A 1 b 1b In a nickel-plating experiment, after passing a current of 5.0 A through an electroplating bath containing a nickel compound for 5.5 minutes, 0.50 g of metallic nickel was deposited at the cathode. Assuming that the current efficiency is 100%, deduce the oxidation state of nickel in the compound. 3 99 2A 4 b i ii 4b In an electrolysis experiment, inert electrodes were used and a constant current of 1.00 A was passed for 1800 s through two electrolytic cells connected in series. i One of the cells contained an aqueous solution of silver(I) nitrate(V) and 2.02 g of silver was deposited on the cathode. Deduce the Faraday constant and the charge carried by an electron. ii The other cell contained an aqueous solution of a gold salt and 1.23 g of gold was deposited on the cathode. Deduce the oxidation state of gold in the gold salt. (Assume 100% current efficiency in this experiment.) III. Redox Equilibria Part 2 Page 1 Topic III. Redox Equilibria Part 2 Reference Reading 6.3.2 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 108–109 Physical Chemistry (3rd ed.), P.W. Atkins, 259–263 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 280–281 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 209 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 188, 278–280 Syllabus Electrochemical cells Measurement of e.m.f. Use of salt bridge Cell diagrams (IUPAC conventions) Notes B. Electrochemical cells If a reducing agent is mixed with an oxidizing agent, the electrons from the reducing agent flow to the oxidizing agent directly. The chemical energy of the two reagents will be converted to heat energy. e.g. Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s) strong strong weak weak reducing oxidizing oxidizing reducing agent agent agent agent However, if a reducing agent is connected to an oxidizing agent through an external circuit and completed with a salt bridge, the electrons from the reducing agent will flow through the external circuit. In this situation, the chemical energy is converted to electrical energy. And an electrochemical cell is constructed. III. Redox Equilibria Part 2 Page 2 1. Measurement of e.m.f. e.m.f. of a cell is defined as the potential difference across a cell when the circuit is open. That means the current is zero. If the current is not zero, owing to the internal resistance of a cell, the potential difference measured will be lower than the e.m.f. a) Potentiometer Potentiometer is the best device to measure the e.m.f. of a cell because the current flowing through the cell to be measured would be exactly zero. A standard accumulator is used as the reference in the potentiometer and the circuit is connected as the diagram shown. The sliding contacted is slided along the potentiometer wire until zero reading is attained on the galvanometer. The e.m.f. of the cell will be proportional to the potentiometer wire and given by the equation e.m.f. of the cell = e.m.f. of the accumulator × XB XY However, the setup and operation of a potentiometer is not very convenient. A voltmeter or multimeter may be used instead. b) High impedance voltmeter / Digital multimeter High impedance voltmeter or digital multimeter is not as good as potentiometer because the current flowing through the cell would not be zero. Although voltmeter and multimeter are not the ideal instruments for e.m.f. measurement, the operations of them are very simple. Moreover, both of them have very high resistance and the current flowing through them will be very minimal. The error will be within an acceptable level. 2. Use of salt bridge Salt bridge is made by a thread of cotton soaked in saturated KNO 3(aq) or NH 4 Cl (aq) solution. It is used to complete the circuit when the two half cells are connected together. a) Requirement of a salt bridge Not all kind of strong electrolytes can be used as a salt bridge. Criteria of a salt bridge 1. The electrolyte must have no reaction with the chemicals in the cell. For example, NH 4 Cl (aq) salt bridge should not be used with Ag (s) | Ag + (aq) system, otherwise precipitation will occur. Ag + (aq) + Cl - (aq) → AgCl (s) 2. The cation and anion in the salt bridge should have the same migration speeds. If they have different speeds, e.m.f. of the cell will be affected by the choice of salt bridge. For this reason, KNO 3(aq) and NH 4 Cl (aq) are the only two commonly used salt bridges. III. Redox Equilibria Part 2 Page 3 b) An electrochemical cell does not need salt bridge Salt bridge can be replaced by a porous partition if accurate supply of e.m.f. is not required. The tiny holes allows the ions to diffuse slowly but prevent the solutions from mixing. Although the setup is more convenient than the one with the salt bridge, it is not as good as salt bridge for two reasons. 1. The difference in diffusion speed of the ions in the two partitions affects the e.m.f. of the cell. 2. After a long period of time, the two electrolytes may mix together and an undesirable reaction may take place. Daniel cell invented in 1837 Moreover, some electrochemical cells even do not require the used of a porous partition. For the redox reaction 2AgCl (s) + H 2(g) d 2Ag (s) + 2HCl (aq) Reduction AgCl (s) + e - d Ag (s) + Cl - (aq) Oxidation H 2(g) d 2H + (aq) + 2e - Overall reaction 2AgCl (s) + H 2(g) d 2Ag (s) + 2Cl - (aq) + 2H + (aq) In this example the two half cells share the same electrolyte, HCl (aq) , therefore, neither salt bridge nor porous pot is required. 3. Cell diagrams (IUPAC conventions) Instead of drawing the actual diagram of an electrochemical cell, IUPAC has an agreed convention for writing an notation, called cell diagram, to represent a cell. In the notation, | phase boundary (boundary between solid, liquid or gaseous phase) || or MM salt bridge M porous partition , used to separate different species in the same phase [ ] used to enclose different species in the same chemical system + used between different species in the same chemical system N.B. The rule of writing of cell diagram is not very strict, it may be possible to construct more than 1 diagram for 1 electrochemical cell. III. Redox Equilibria Part 2 Page 4 Example 1 Example 2 Example 3 Zn (s) | Zn 2+ (aq) M Cu 2+ (aq) | Cu (s) Zn (s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu (s) Pt (s) | H 2(g) | HCl (aq) | AgCl (s) | Ag (s) Example 1 For the Daniel cell with the porous partition Zn (s) | Zn 2+ (aq) M Cu 2+ (aq) | Cu (s) e.m.f. of the cell = + 1.10 V By convention, the sign of e.m.f. of a cell is given by the polarity of the electrode at the right in a cell diagram. Therefore, if the cell diagram is written as Cu (s) | Cu 2+ (aq) M Zn 2+ (aq) | Zn (s) , the e.m.f of the cell will become - 1.10 V. Example 2 For the Daniel cell with salt bridge Zn (s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu (s) e.m.f. of the cell = + 1.10 V III. Redox Equilibria Part 2 Page 5 Example 3 Pt (s) | H 2(g) | HCl (aq) | AgCl (s) | Ag (s) e.m.f of the cell = + 0.22 V Sometimes " | " and " , " can be used interchangeably depending on how do you define a phase. AgCl (s) and Ag (s) can be considered as two different phases because they are different substances. Therefore it is written as above. Example 4 Electrochemical cell is not limited to metal | metal ion system. For those system doesn't involve metal, an inert electrode is required. Graphite and platinum are the two commonly used inert electrodes. For the system doesn't involve metal, the cell diagram is written in this way (Pt) | Red L , Ox L || Red R , Ox R | (Pt) Red : Reduced form Ox : Oxidized form The two half-ionic equations involved in the cell on the right are Reduction half equation MnO 4 - (aq) + 8H + (aq) + 5e - d Mn 2+ (aq) + 4H 2 O (l) Oxidation half equation Fe 2+ (aq) d Fe 3+ (aq) + e - The cell diagram is written as C (graphite) | Fe 2+ (aq) , Fe 3+ (aq) || [Mn 2+ (aq) + 4H 2 O (l) ] , [MnO 4 - (aq) + 8H + (aq) ] | C (graphite) Red L Ox L Red R Ox R Fe(II) Fe(III) Mn(II) Mn(VII) N.B. " , " is used to used to separate different species in the same phase " [ ] " is used to enclose used to enclose different species in the same chemical system Glossary electrochemical cell e.m.f potential difference potentiometer potentiometer wire accumulator galvanometer impedance multimeter salt bridge migration speed porous partition (pot) Daniel cell IUPAC cell diagram convention phase boundary inert electrode half-cell half-ionic equation Past Paper Question 90 2A 1 b i ii iv 95 2A 3 c i ii iii 96 2A 3 a ii 97 2A 4 a ii iii III. Redox Equilibria Part 2 Page 6 90 2A 1 b i ii iv 1b A cell based on the reaction: 6Fe 2+ (aq) + Cr 2 O 7 2- (aq) + 14H + (aq) → 6Fe 3+ (aq) + 2Cr 3+ (aq) + 7H 2 O (l) consists of one platinum electrode dipped into a beaker containing an acidified solution of K 2 Cr 2 O 7(aq) , and another platinum electrode dipped into a separate beaker containing FeSO 4(aq) , with the two solutions being connected by a salt bridge. With reference to this cell: i write balanced equations for the reactions occurring at the cathode and the anode; 2 Reduction occurs at the cathode Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - → 2Cr 3+ (aq) + 7H 2 O (l) 1 mark Oxidation occurs at the anode Fe 2+ (aq) → Fe 3+ (aq) + e - 1 mark ii state the direction in which electrons move when the electrodes are connected externally; 1 From anode to cathode 1 mark iv represent the cell using the I.U.P.A.C. convention 2 Pt|Fe 2+ (aq) ,Fe 3+ (aq) ||Cr 3+ (aq) ,[Cr 2 O 7 2- (aq) ,H + (aq) ]|Pt 2 mark C i Many candidates did not give state symbols in the half equation. 95 2A 3 c i ii iii 3c The electromotive force of a new zinc-carbon dry cell is 1.5 V. When it is producing an electric current, the following changes occur at the two electrodes: Anode : Zn (s) reacts to give Zn 2+ (aq) . Cathode: MnO 2(s) and NH 4 Cl (aq) react to give Mn 2 O 3(s) and NH 3(g) . i Write half equations for the reactions at the anode and at the cathode, and the equation for the overall reaction that occurs in the dry cell. 3 At anode Zn (s) → Zn 2+ (aq) + 2e - 1 mark At cathode 2MnO 2(s) + 2NH 4 + (aq) + 2e - → Mn 2 O 3(s) + 2NH 3(g) + H 2 O (l) 1 mark or 2MnO 2(s) + NH 4 + (aq) + 2e - → Mn 2 O 3(s) + OH - (aq) + NH 3(g) Overall equation Zn (s) + 2MnO 2(s) + 2NH 4 + (aq) → Zn 2+ (aq) + Mn 2 O 3(s) + 2NH 3(g) + H 2 O (l) 1 mark or Zn (s) + 2MnO 2(s) + NH 4 + (aq) → Zn 2+ (aq) + Mn 2 O 3(s) + NH 3(g) + OH - (aq) ii Write the cell diagram for the dry cell, using the IUPAC convention. 2 Zn (s) | Zn 2+ (aq) | MnO 2(s) | Mn 2 O 3(s) | C (s) 2 marks or Zn (s) | Zn 2+ (aq) M NH 4 + (aq) | MnO 2(s) | Mn 2 O 3(s) | C (s) or Zn (s) | Zn 2+ (aq) | [MnO 2(s) + NH 4 + (aq) ], [Mn 2 O 3(s) + OH - (aq) ] | C (s) or Zn (s) | Zn 2+ (aq) | [MnO 2(s) + 2NH 4 + (aq) ], [Mn 2 O 3(s) + 2NH 3(aq) ] | C (s) (1 mark for the correct species at cathode and anode ½ mark for using vertical solid lines to indicate phase difference, ½ mark for using vertical double line (dotted line) to represent salt bridge or using vertical single line to represent porous partition) iii Explain why the electromotive force of the dry cell drops, (1) after it has been used for some time; (2) after it has been stored for a long time without being used. 3 (1) If a current is drawn for some time, NH 3(g) will accumulate at the cathode, the equilibrium will shift to the left, leading to a drop in electrode potential. 1½ mark OR If a current is drawn for some time, Zn 2+ (aq) will accumulate at the anode, the equilibrium will shift to the left, making the electrode potential of Zn | Zn 2+ less negative. (2) If the cell is allowed to stand for some time, NH 4 + which is an acid will react with Zn ½ mark Zn (s) + 2NH 4 + (aq) → Zn 2+ (aq) + 2NH 3(aq) + H 2(g) ½ mark or Zn (s) + 4NH 4 + (aq) → Zn(NH 3 ) 4 2+ (aq) + 4H + (aq) + 2e - will decrease in [NH 4 + (aq) ], the electrode potential will also drop. ½ mark C As in previous years, this type of question on electrochemistry was poorly answered. The basic concept of the half cell was not clearly understood by most candidates, hence they were not able to write the correct half equations and the equation for the overall reaction. Many candidates could not write the IUPAC cell diagram for the zinc- carbon dry cell. 96 2A 3 a ii 3a ii Consider the following electrochemical cell : 6 III. Redox Equilibria Part 2 Page 7 (1) Identify, with explanation, the negative terminal of the electrochemical cell. (2) Comment on the change in concentration of Cu 2+ (aq) ions in beakers D and E as electrochemical reaction occurs. (3) Using the IUPAC convention, write the cell diagram for the electrochemical cell. (1) The left hand electrode / Cu electrode in beaker D is the negative terminal ½ mark [Cu 2+ (aq) ] on LHS is lower than that on RHS / 1.0 M 1 mark Oxidation occurs at left hand electrode / electrons flow from left hand electrode to right hand electrode ½ mark so that [Cu 2+ (aq) ] in the beaker E decreases / the difference in [Cu 2+ (aq) ] of the two beakers becomes smaller. (2) In beaker E, [Cu 2+ (aq) ] decreases because Cu 2+ is reduced to Cu ½ + 1 mark or Cu 2+ (aq) +2e - → Cu (s) In beaker D, [Cu 2+ (aq) ] remains almost constant ½ mark because Cu 2+ (aq) ions produced in the electrochemical reaction combine with OH - (aq) ions in the buffer to form Cu(OH) 2(s) 1 mark (3) Cu (s) | Cu(OH) 2(s) | OH - (aq) || Cu 2+ (aq) | Cu (s) 1 mark or Cu (s) | Cu(OH) 2(s) | OH - (10 -4 M) || Cu 2+ (aq) | Cu (s) (1 mark) or Cu (s) | Cu 2+ (aq) (2.0 × 10 -11 M) || Cu 2+ (aq) (1.0 M) | Cu (s) (1 mark) (Deduct ½ marks for each minor mistake) 97 2A 4 a ii iii 4a The diagram below shows an electrochemical cell connected to a digital voltmeter. An electromotive force of 0.83 V was recorded at 298K. 6 ii Write the overall equation for the electrochemical reaction. iii Write the cell diagram for the electrochemical cell, using the IUPAC convention. III. Redox Equilibria Part 3 Page 1 Topic III. Redox Equilibria Part 3 Reference Reading 6.3.3 Assignment Reading Advanced Practical Chemistry, John Murray (Publisher) Ltd., 110–112, 176–177 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 281–285 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 209–217 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 189–195 Syllabus Electrode potentials Standard hydrogen electrode Relative electrode potential (Standard reduction potential) Electrochemical series Use of electrode potential Notes C. Electrode potentials 1. Standard hydrogen electrode Since the actual e.m.f. of a cell is depending on both electrodes, the concentrations of electrolytes and temperature. It would be more convenient if a standard condition is agreed and a reference electrode is chosen for comparison. Standard condition is defined as 1 atm, 298 K with all solution having molar concentration. Hydrogen electrode is chosen by the scientist as the standard of reference. It consists of an inert platinum electrode coated with platinum black (platinum powder) which increases the area of contact. The electrode is enclosed in a hydrogen jacket at 1 atm and immersed in a 1M H + (aq) solution. Eventually, a redox equilibrium is set up and can be described by a half ionic equation. 2H + (aq) + 2e - d H 2(g) If electron is supplied to the electrode, H + (aq) will be converted to H 2(g) . Or, if electron is taken from the electrode, H 2(g) will be converted back to H + (aq) . Cell diagram of hydrogen electrode : Pt (s) H 2(g) | H + (aq) III. Redox Equilibria Part 3 Page 2 2. Relative electrode potential (Standard reduction potential) In order to compare the ease of losing or gaining electron of different chemical systems (different half cells), hydrogen electrode is chosen arbitrarily as the standard of comparison. Standard hydrogen electrode is assigned an (relative) electrode potential of 0 V. The values are also called reduction potentials because the half equations are always expressed in reduction form when listed in the electrochemical series. Reduction half equation 2H + (aq) + 2e - d H 2(g) Eo = 0 V oxidizing reducing agent agent At standard condition, the electrode potential measured is also called standard reduction potential. Determination of relative electrode potential (standard reduction potential) In the determination of the standard reduction potential of an unknown half cell, a cell with the hydrogen electrode as the left-hand electrode and the unknown electrode as the right-hand electrode is set up. The e.m.f. of the cell is then determined. For example when the half-cell Pt|Cl 2(g) |Cl -(aq) , is used as the right-hand cell and the hydrogen electrode is used as the left-hand cell, the e.m.f. is determined to be 1.36 V. However, a sign should be assigned to the e.m.f. value to denote the direction of flow of electrode (i.e. the relative affinity to electron). By definition, the polarity of the right-hand electrode is assigned to the e.m.f. value to do the job. Therefore, in this example, the e.m.f. of the cell is +1.36 V Cell diagram Pt | H 2(g) | H + (aq) || Cl - (aq) | Cl 2(g) | Pt Eo = + 1.36 V Furthermore, the relationship between the overall e.m.f. of a cell and the reduction potential of the 2 half cells is given by Eo cell = Eo right - Eo left When hydrogen electrode is chosen as the left-hand electrode, the value of Eo left will be zero. Eo cell = Eo right - Eo left = Eo right Therefore, by definition, electrode potential of an electrode is given by the e.m.f. of the cell if the LEFT electrode in the cell diagram is a hydrogen electrode. Cl 2(aq) + 2e - d 2Cl - (aq) Eo = + 1.36V III. Redox Equilibria Part 3 Page 3 Consider another example where the positions of the cell diagram and the physical setup are different. Cell diagram Pt | H 2(g) | H + (aq) || Sn 2+ (aq) | Sn (s) Eo = - 0.14 V Sn 2+ (aq) + 2e - d Sn (s) Eo = - 0.14 V N.B. When we are discussing the position of the electrode, we are referring to the cell diagram instead of the position of the actual set up. a) Meaning of sign and value A positive electrode potential means that the electrode accepts electrons more readily than a hydrogen electrode. As the sign and value of the electrode potential are depending on the readiness of a system to accept / donate electrons, electrode potential serves as a very good indicator of the oxidizing and reducing power of different species. When the reduction equations are listed according to the electrode potential, the list is called electrochemical series. III. Redox Equilibria Part 3 Page 4 3. Electrochemical series Reduction equation Eo / V Weak oxidizing agent Strong reducing agent K + (aq) + e - d K (s) - 2.92 PbSO 4(s) + 2e - d Pb (s) + SO 4 2- (aq) - 0.36 Sn 2+ (aq) + 2e - d Sn (s) - 0.14 2H + (aq) + 2e - d H 2(g) 0 4H + (aq) + SO 4 2- (aq) + 2e - d H 2 SO 3(aq) + H 2 O (l) + 0.17 AgCl (s) + e - d Ag (s) + Cl - (aq) + 0.17 PbO 2(s) + H 2 O (l) + 2e - d PbO (s) + 2OH - (aq) + 0.28 O 2(g) + 2H 2 O (l) + 2e - d 4OH - (aq) + 0.40 2H + (aq) + O 2(g) + 2e - d H 2 O 2(aq) + 0.68 Ag + (aq) + e - d Ag (s) + 0.80 2NO 3 - (aq) + 4H + (aq) + 2e - d N 2 O 4(g) + 2H 2 O (l) + 0.80 MnO 2(s) + 4H + (aq) + 2e - d Mn 2+ (aq) + 2H 2 O (l) + 1.23 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - d 2Cr 3+ (aq) + 7H 2 O (l) + 1.33 Cl 2(aq) + 2e - d 2Cl - (aq) + 1.36 PbO 2(s) + 4H + (aq) + 2e - d Pb 2+ (aq) + 4H 2 O (l) + 1.46 MnO 4 - (aq) + 8H + (aq) + 5e - d Mn 2+ (aq) + 4H 2 O (l) + 1.51 PbO 2(s) + SO 4 2- (aq) + 4H + (aq) + 2e - d PbSO 4(s) + 2H 2 O (l) + 1.69 MnO 4 - (aq) + 4H + (aq) + 3e - d MnO 2(s) + 2H 2 O (l) + 1.70 Strong oxidizing agent Weak reducing agent N.B. The values are only correct at standard conditions. The electrode potentials are measured at standard condition. If the condition is not standard, another value would be expected. For example, concentrated sulphuric acid has a concentration of about 18 M instead of 1 M. Reduction potential of 1M H 2 SO 4(aq) 4H + (aq) + SO 4 2- (aq) + 2e - d H 2 SO 3(aq) + H 2 O (l) Eo = +0.17 V Concentrated sulphuric acid has a much more positive electrode potential and can be considered as a strong oxidizing agent, while dilute sulphuric acid is not. 4. Use of electrode potential a) Comparing the oxidizing and reducing power On the electrochemical series, the chemical species are arranged according to their electrode potentials. Those oxidizing agents low on the list accept electron readily and are strong oxidizing agents. e.g. MnO 4 - /H + (aq) . And those reducing agents high on the list donate electron readily and are strong reducing agents. e.g. K (s) . III. Redox Equilibria Part 3 Page 5 b) Calculation of e.m.f. of a cell The e.m.f of a cell can be calculated from the values of the electrode potentials. Consider an overall cell reaction, Sn (s) + 2Ag + (aq) → Sn 2+ (aq) + 2Ag (s) It is comprised two half cells presented by 2 half equations Sn 2+ (aq) + 2e - d Sn (s) Eo = - 0.14 V Ag + (aq) + e - d Ag (s) Eo = + 0.80 V The electrode potential is reversed when the direction of reaction is reversed. Sn (s) d Sn 2+ (aq) + 2e - Eo = + 0.14 V Ag + (aq) + e - d Ag (s) Eo = + 0.80 V The overall electrode potential can then calculated by adding the 2 values together. Sn (s) + 2Ag + (aq) → Sn 2+ (aq) + 2Ag (s) Eo = (+ 0.14 V) + (+ 0.80 V) = + 0.94 V Or consider a cell diagram, Sn (s) | Sn 2+ (aq) || Ag + (aq) | Ag (s) The e.m.f. of the cell can be calculated by using the equation, Eo cell = Eo right - Eo left Sn 2+ (aq) + 2e - d Sn (s) Eo = - 0.14 V Ag + (aq) + e - d Ag (s) Eo = + 0.80 V Eo cell = Eo right - Eo left = (+ 0.80 V) - (- 0.14 V) = + 0.94 V Since the sign of the e.m.f is depending on the direction of reaction of a cell, when a cell diagram is converted to an equation or vice versa, the following convention should be observed. Red L + Ox R → Red R + Ox L Sn (s) + Ag + (aq) → 2Ag (s) + Sn 2+ (aq) Eo = + 0.94 V Furthermore, e.m.f. of a cell is depending on the nature of the two electrodes only. It is not depending on the current (or no. of electrons) flowing through the circuit. In the calculation of e.m.f. of cell, the electrode potentials are only simply added together and are not multiplied by the no. of electrons involved. In contrast, in the balancing of a redox reaction, the half equations are multiplied whenever necessary to make the no. of electrons conserve. In the above example Oxidation half ionic equation Sn (s) d Sn 2+ (aq) + 2e - Reduction half ionic equation Ag + (aq) + e - d Ag (s) × 2 ------------------------------------------- Overall ionic equation Sn (s) + 2Ag + (aq) → Sn 2+ (aq) + 2Ag (s) III. Redox Equilibria Part 3 Page 6 c) Prediction of the feasibility of redox reaction In general, oxidizing agent reacts with reducing agent. Indeed, only strong oxidizing agent reacts with strong reducing agent. For example, both Cu (s) and Ag (s) are reducing agents while Cu 2+ (aq) and Ag + (aq) are oxidizing agents. But only Cu (s) reacts with Ag + (aq) but Ag (s) doesn't reacts with Cu 2+ (aq) . Cu 2+ (aq) + 2e - d Cu (aq) Eo = + 0.34 V Ag + (aq) + e - d Ag (s) Eo = + 0.80 V Cu (s) + 2Ag + (s) → Cu 2+ (aq) + 2Ag (s) Eo cell = (- 0.34 V) + (+ 0.80 V) = + 0.46 V (energetically feasible) Cu 2+ (aq) + 2Ag (s) → Cu (s) + 2Ag + (s) Eo cell = (+ 0.34 V) + (- 0.80 V) = - 0.46 V (energetically infeasible) The reaction will only be energetically feasible if the sum of electron potentials is positive. N.B. Relationship between ∆G and Eo is not required in A-Level) When 1 e - jumps across a potential difference of 1V, the energy acquired by the electron is called 1 eV (electron volt). 1 eV = 1.60 × 10 -19 C × 1 V = 1.60 × 10 -19 J The e.m.f. of a cell is related to the ∆G by the expression, ∆Go = -nFEo cell where n is the no. of mole of e - involved, F is the Faraday constant Eo cell is the e.m.f. of the cell ∆Go is the total free energy change of the reaction. Its value is depending on the total no. of mole of electron passing through the cell. In contrast, Eo cell is a reflection of the electron pressure whose value is only depending on the nature of the two electrode. According to the expression ∆Go = -nFEo cell , a positive Eo cell means a negative ∆Go. This means an energetically feasible reaction. Note : ∆G = ∆H - T∆S = - RT ln K = - nFE III. Redox Equilibria Part 3 Page 7 (1) Disproportionation reaction Disproportionation – A process in which the same element undergoes oxidation and reduction simultaneously. e.g. 2Cu + (aq) → Cu 2+ (aq) + Cu (s) A chemical species will disproportionate if the reduction potential to the next lower oxidation state is more positive than that from the next higher state. i.e. E 2 > E 1 . Cu + (aq) + e - → Cu (s) Eo = + 0.52 V Cu + (aq) → Cu 2+ (aq) + e - Eo = - 0.15 V ---------------------------------------------------------------------------------------- 2Cu + (aq) → Cu (s) + Cu 2+ (aq) Eo = + 0.37 V (energetically feasible) Furthermore, it must be noticed that the overall reduction potential of Cu 2+ (aq) to Cu (s) is not the sum of the reduction potentials of the two steps. i.e. E 3 ≠ E 1 + E 2 . Two half-cell potentials cannot be added to give a reduction potential for a different half-cell. Glossary standard hydrogen electrode e.m.f. reference electrode standard condition platinum electrode relative electrode potential (standard reduction potential) electrochemical series feasibility disproportionation Past Paper Question 90 1A 2 a 90 2A 1 b iii 90 2B 6 a ii 92 2A 1 c i 95 2B 6 c iii 96 2B 6 d i ii e 97 1A 2 c i ii 97 2A 4 a i 99 2A 1 b i ii 90 1A 2 a 2a A convenient laboratory method for determining electrode potentials makes use of a standard hydrogen electrode. Draw a labelled diagram of a hydrogen electrode. What conditions should be maintained when a hydrogen electrode is used as a standard hydrogen electrode? State how the conditions can be met in the laboratory. 6 3 marks for correctly labelled diagram. Deduct 1 mark for each wrong/missing label Deduct ½ mark if ‘Platinized’ omitted Conditions: Maintaining the stated conditions: 298K/25ºC thermostat/water-bath 1 mark H 2 gas at 1 atm pressure pressure gauge 1 mark [H + ] = 1 M the use of standard acid solution 1 mark C Diagrams of a hydrogen electrode were generally not well drawn. Only about a third of the candidates mentioned the use of platinized platinum or platinum black, and even fewer considered the depth of its immersion in the solution containing hydrogen ions. III. Redox Equilibria Part 3 Page 8 90 2A 1 b iii 1b A cell based on the reaction: 6Fe 2+ (aq) + Cr 2 O 7 2- (aq) + 14H + (aq) → 6Fe 3+ (aq) + 2Cr 3+ (aq) + 7H 2 O (l) consists of one platinum electrode dipped into a beaker containing an acidified solution of K 2 Cr 2 O 7(aq) , and another platinum electrode dipped into a separate beaker containing FeSO 4(aq) , with the two solutions being connected by a salt bridge. With reference to this cell: iii calculate the e.m.f. of the cell, given that the standard reduction potentials of Fe 3+ /Fe 2+ and acidified Cr 2 O 7 2- /Cr 3+ are respectively +0.77V and +1.33V; and 1 e.m.f. = +1.33 - (+0.77) = 0.56V 1 mark 90 2B 6 a ii 6a The following species are either impossible to prepare or very unstable. Explain, in each case, why this is so. ii SI 6 1½ S(VI) has a high redox potential and would oxidize I - to I 2 . SI 6 is therefore very unstable with respect to disproportionation. 1½ mark 92 2A 1 c i 1c Given the following standard reduction potentials, Eo : (Cr 2 O 7 2- (aq) +14H + (aq) ),(2Cr 3+ (aq) +7H 2 O (l) )|Pt (s) 1.33V Cr 3+ (aq) ,Cr 2+ (aq) |Pt (s) -0.41V Zn 2+ (aq) |Zn (s) -0.76V i Predict the products of the reaction of zinc with dichromate(VI) solution. Explain your prediction. 3 In aqueous acid solution the overall reaction: Zn (s) → Zn 2+ (aq) + 2e - Cr 2 O 7 2- (aq) + 14H+ (aq) + 6e - → 2Cr 3+ (aq) + 7H 2 O (l) ------------------------------------------------------------------------------- Cr 2 O 7 2- (aq) + 14H+ (aq) + 3Zn (s) → 3Zn 2+ (aq) + 2Cr 3+ (aq) + 7H 2 O (l) Eo = -(-0.76) + 1.33 = 2.09 V is favourable because of the large positive Eo value. The overall reaction: Zn (s) → Zn 2+ (aq) + 2e - 2e - + 2Cr 3+ (aq) → 2Cr 2+ (aq) ------------------------------------------------- 2Cr 3+ (aq) + Zn(s) → Zn 2+ (aq) + 2Cr 2+ (aq) Eo = - (-0.76) + (-0.41) = 0.35 V is also favourable because of the positive Eo value. 2 marks Therefore, the products are Zn 2+ (aq) and Cr 2+ (aq) . 1 mark C i Candidates were not able to use all the given electrode potential data to predict the products : Zn 2+ and Cr 2+ . 95 2B 6 c iii 6c In aqueous solutions, TiO 2+ is colourless. It can be reduced to give a violet solution containing [Ti(H 2 O) 6 ] 3+ . iii The violet solution formed should be kept in a sealed vessel or handled in an inert atmosphere. Using the following data, predict, giving a balanced equation, what will happen to the violet solution if it is not kept in a sealed vessel or not handled in an inert atmosphere. Eo / V [Ti(H 2 O) 6 ] 3+ (aq) + e - → [Ti(H 2 O) 6 ] 2+ -0.37 TiO 2+ (aq) + 5H 2 O (l) + 2H + (aq) + e - → [Ti(H 2 O) 6 ] 3+ (aq) +0.11 O 2(g) + 2H 2 O (l) + 4e - → 4OH - (aq) +0.40 O 2(g) + 4H + (aq) + 4e - → 2H 2 O (l) +1.23 4 From the Eo values, O 2 is a stronger [O] agent than TiO 2+ . 1 mark At standard conditions, the Eo of the following reaction is +1.23 - (+ 0.11) V = +1.12 V 1 mark 4Ti 3+ + O 2 + 2H 2 O → 4TiO 2+ + 4H + 1 mark Therefore, in the presence of oxygen, [Ti(H 2 O) 6 ] 3+ will be oxidized to TiO 2+ (aq) / the violet solution will turn colourless. 1 mark C iii Candidates were expected to predict that if [Ti(H 2 O) 6 ] 3+ is not kept in a sealed container, the reaction 4[Ti(H 2 O) 6 ] 3+ + O 2 → 4TiO 2+ + 4H 3 O + + 18H 2 O takes place because the overall Eo for the reaction is +1.12 V. However, most of them failed to do so. 96 2B 6 d i ii e 6d You are provided with the following standard reduction potentials : Eo/V Cu 2+ (aq) + e - d Cu + (aq) + 0.15 Cu 2+ (aq) + 2e - d Cu (s) + 0.34 III. Redox Equilibria Part 3 Page 9 Cu + (aq) + e - d Cu (s) + 0.52 I 2(s) + 2e - d 2I - (aq) + 0.54 Cu 2+ (aq) + I - (aq) + e - d CuI (s) + 0.92 Using the above information i explain why copper(I) compounds are unstable in aqueous solutions; and 2 In aqueous solutions, Cu + disproportionates to give Cu (s) and Cu 2+ (aq) ½ + ½ mark 2Cu + (aq) → Cu (s) + Cu 2+ (aq) (1 mark) because the above reaction has a +ve Eº of + 0.52 - (+0.15) = + 0.37 V 1 mark ii predict what will be observed when a potassium iodide solution is added to a copper(II) sulphate(VI) solution. Explain your prediction and write a balanced equation for the reaction involved. 3 A brown solution and a (white) precipitate are formed because the reaction ½ + ½ mark 2Cu 2+ + 4I - → 2CuI + I 2 1 mark has a Eº value of (0.92 - 0.54) = + 0.38 V 1 mark 6e In acidic conditions, VO 2 + (aq) can be reduced by sulphur dioxide to give VO 2+ (aq) . Write a balanced equation for the reaction. 2 VO 2 + (aq) + 2H + (aq) + e - → VO 2+ (aq) + H 2 O (l) ½ mark +) SO 2(g) + 2H 2 O (l) → SO 4 2- (aq) + 4H + (aq) + 2e - ½ mark --------------------------------------------------------------------------------------------- 2VO 2 + (aq) +SO 2(g) → 2VO 2+ (aq) + SO 4 2- (aq) 1 mark (Award FULL marks if only the overall equation is given.) 97 1A 2 c i ii 2c Consider the standard reduction potentials listed below: [FeF 6 ] 3- (aq) + e - d [FeF 6 ] 4- (aq) Eº = + 0.40 V I 2(aq) + 2e - d 2I - (aq) Eº = + 0.54 V Fe 3+ (aq) + e - d Fe 2+ (aq) Eº = + 0.76 V Explain the following observations, giving a balanced equation in each case. 4 i When Fe(NO 3 ) 3(aq) is added to KI (aq) , a brown solution is formed. ii When concentrated KF (aq) is added to the resulting solution in (i) above, the brown colour fades. 97 2A 4 a i 4a The diagram below shows an electrochemical cell connected to a digital voltmeter. An electromotive force of 0.83 V was recorded at 298K. 6 i Write half-equations for the reaction at the anode and at the cathode, and give their corresponding standard electrode potentials. 99 2A 1 b i ii 1b i What is 'standard reduction potential' ? ii With reference to the standard reduction potentials listed below, explain why hydrochloric acid can be used to acidify potassium dichromate(VI) solution but cannot be used to acidify potassium manganate(VII) solution. Give a balanced equation for any reaction that takes place. Eo/ V Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - d 2Cr 3+ (aq) + 7H 2 O (l) +1.33 Cl 2(g) + 2e - d 2Cl - (aq) +1.36 MnO 4 - (aq) + 8H + (aq) + 5e - d Mn 2+ (aq) + 4H 2 O (l) +1.51 III. Redox Equilibria Part 4 Page 1 Topic III. Redox Equilibria Part 4 Reference Reading 6.3.4 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 288–290 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 217–219 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 196–198 Syllabus D. Secondary cell and fuel cell 1. Lead-acid accumulator Lead-acid accumulator is commonly used in vehicle to start the engine. Comparing with the other secondary cell, it produces a much larger current. Notes The two electrodes are lead alloy grids filled with spongy PbO 2(s) (cathode) and spongy Pb (s) (anode) immersed in the electrolyte of dilute H 2 SO 4(aq) . Upon discharge, thin layers of PbSO 4(s) will be coated on both the cathode and anode. However, the reaction is reversible. When the battery is charged up, PbSO 4(s) will be converted back to PbO 2(s) and Pb (s) . Question : What would happen if a Daniel cell is recharged ? PbO 2 cathode (Reduction) PbO 2(s) + 4H + (aq) + SO 4 2- (aq) + 2e - → PbSO 4(s) + 2H 2 O (l) Eo = + 1.69 Pb anode (Oxidation) Pb (s) + SO 4 2- (aq) → PbSO 4(s) + 2e - Eo = + 0.36 Overall reaction Pb (s) + PbO 2(s) + 2H 2 SO 4(aq) d 2PbSO 4(s) + 2H 2 O (l) Eo cell = + 2.05 V In a fully charged lead-acid accumulator, the 36% H 2 SO 4(aq) has the density 1.27 gcm -3 . Upon discharging, H 2 SO 4(aq) is consumed and H 2 O (l) is produced, the density of the electrolyte will decrease. The condition of charge in the cell can be determined by measuring the density of the acid with a hydrometer. Cell diagram Pb (s) | PbSO 4(s) | H 2 SO 4(aq) | PbSO 4(s) | PbO 2(s) | Pb (s) Red L Ox L Red R Ox R Pb(0) Pb(II) Pb(II) Pb(IV) No salt bridge or porous partition is required because the two electrodes share the same electrolyte, H 2 SO 4(aq) . III. Redox Equilibria Part 4 Page 2 2. Hydrogen-oxygen fuel cell Fuel cell is the battery used in spaceship to produce electricity. H 2 Anode (Oxidation) H 2(g) → 2H + (aq) + 2e - Eo = 0 V O 2 Cathode (Reduction) O 2(g) + 2H 2 O (l) + 4e - → 4OH - (aq) Eo = + 0.40 V Overall Reaction 2H 2(g) + O 2(g) d 2H 2 O (l) Eo = + 0.40 V On top of electrical energy, the fuel cell also provide the warmth and water for the astronaut in the spaceship. Glossary Secondary cell lead-acid accumulator alloy grid spongy hydrometer hydrogen-oxygen fuel cell Past Paper Question 91 1A 2 a i ii iii 93 2A 1 c i ii iii 98 2A 4 a i ii iii 91 1A 2 a i ii iii 2a An electrochemical cell containing an oxygen cathode and a hydrogen anode is shown below : The pistons above the gas chambers are frictionless. i Write balanced equations for the half reactions and for the overall reaction in the cell. 2 cathode: 2H + + ½O 2 + 2e - d H 2 O anode: H 2 d 2H + + 2e - overall: H 2 + ½O 2 → H 2 O 2 marks wrong coefficient -½ reverse labelling of cathode and anode reaction -½ wrong species in each reaction -½ wrong direction of reaction (e.g. if a single arrow) -½ III. Redox Equilibria Part 4 Page 3 ii How does the concentration of H 2 SO 4 affect the equilibria of the half reactions ? 1 ↑[H + ] will shift equilibrium of 2H + + ½O 2 + 2e - d H 2 O to right, and shift equilibrium of H 2 d 2H + + 2e - to left. ½ mark each iii If weights are added to the pistons of both chambers, how would the reading of the voltmeter change ? Explain your answer. 2 voltage reading increase 1 mark pressure in O 2 and H 2 chamber increase or [O 2 ] and [H 2 ] increase ½ mark equilibrium of H 2 + ½O 2 → H 2 O is shifted to right and favours the discharge of battery ½ mark C Surprisingly, many candidates could not apply knowledge of chemical equilibria to an electrochemical reaction. A fuel cell and an electrolytic cell could be the same problem if viewed from the equilibrium principles. Over half of the candidates gave OH - as the product in one or both of the half reactions. They also confused the anode and the cathode and the direction of overall reaction. 93 2A 1 c i ii iii 1c i Draw a labelled diagram of a cell that makes use of the reaction: 2H 2 SO 4(aq) + PbO 2(s) + Pb (s) → 2PbSO 4(s) + 2H 2 O (l) Indicate the direction of electron flow on your diagram. 2 Labelled diagram 1 mark Direction of electron flow 1 mark ii Write a balanced equation for the reaction at each electrode. 2 At PbO 2 (cathode) PbO 2(s) + HSO 4 - (aq) + 3H + (aq) + 2e - → PbSO 4(s) + 2H 2 O (l) or PbO 2(s) + 4H + (aq) + SO 4 2- (aq) + 2e - → PbSO 4(s) + 2H 2 O (l) 1 mark At Pb (anode) Pb (s) + HSO 4 - (aq) → PbSO 4(s) + 2e - + H + or Pb (s) + SO 4 2- (aq) → PbSO 4(s) + 2e - 1 mark iii Using the IUPAC convention, write the cell diagram for the above cell. 2 General format: (Pt)|Red L ,Ox L ||Red R ,Ox R |(Pt) Pb (s) |PbSO 4(s) |H 2 SO 4(aq) |PbSO 4(s) |PbO 2(s) |Pt (s) or Pb (s) |PbSO 4(s) |H 2 SO 4(aq) |PbO 2(s) |PbSO 4(s) |Pt or PbSO 4(s) |PbO 2(s) |H 2 SO 4(aq) |PbSO 4(s) |Pb (s) (without physical state -1) 2 marks C ii Many candidates had difficulties in writing balanced equations. iii Few candidates gave the right cell diagram. Many candidates were not aware that in writing cell diagrams, a vertical line " | ", instead of a comma ",", should be used to denote a phase boundary. 98 2A 4 a i ii iii 4a A lead-acid rechargeable cell is formed by dipping a lead plate coated with PbO 2(s) and another lead plate coated with PbSO 4(s) in H 2 SO 4(aq) . 8 i You are provided with the following standard reduction potentials at 298 K. Eo/V 2H 2 O (l) + 2e - d H 2(g) + 2OH - (aq) - 0.83 PbSO 4(s) + 2e - d Pb (s) + SO 4 2- (aq) - 0.36 Pb 2+ (aq) + 2e - d Pb (s) - 0.13 O 2(g) + 4H + (aq) + 4e - d 2H 2 O (l) + 1.23 PbO 2(s) + SO 4 2- (aq) + 4H + (aq) + 2e - d PbSO 4(s) + 2H 2 O (l) + 1.69 (I) Write half-equations for the reaction at the anode and at the cathode of the cell. (II) Write the overall equation for the electrochemical reaction and hence determine the standard electromotive force (e.m.f.) of the cell. III. Redox Equilibria Part 4 Page 4 ii Write the cell diagram in accordance with the IUPAC convention. iii Explain why (I) the voltage of the cell drops upon discharge: (II) the cell is rechargeable. III. Redox Equilibria Part 5 Page 1 Topic III. Redox Equilibria Part 5 Reference Reading 6.3.5 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 516–518 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 309–310 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 271–272 Syllabus Corrosion of iron and its prevention Notes E. Corrosion of iron and its prevention 1. Electrochemical process of rusting Formation of rust is a complicate process. Rusting only occurs when iron gets in contact with dissolved oxygen in water. Rusting can be considered as an electrochemical process. Initially, Fe (s) atom loses 2 electrons and becomes Fe 2+ (aq) ion. Anode (oxidation) Fe (s) → Fe 2+ (aq) + 2e - Eo = + 0.44 V The dissolved oxygen in water accepts the electrons and becomes hydroxide ion. Cathode (reduction) O 2(aq) + 2H 2 O (l) + 4e - → 4OH - (aq) Eo = + 0.40 V The overall e.m.f. of the cell = (+ 0.44 V) + (+ 0.40 V) = + 0.84 V (energetically feasible) Fe 2+ (aq) precipitates with OH - (aq) ion in water to form solid iron(II) hydroxide.. Fe 2+ (aq) + 2OH - (aq) → Fe(OH) 2(s) Iron(II) hydroxide is then further oxidized by the dissolved oxygen to iron(III) hydroxide. 4Fe(OH) 2(s) + O 2(aq) + 2H 2 O (l) → 4Fe(OH) 3(s) Upon standing, the iron(III) hydroxide changes to a reddish brown solid – rust, Fe 2 O 3 ·nH 2 O (s) . Since Fe 2 O 3 ·nH 2 O (s) has a very loose structure, it flakes off from the iron surface and exposes the inside of the iron block to further corrosion. Because rusting involves flowing of charges, rusting would be faster in the presence of an electrolyte. i.e. iron rusts much faster in sea water (sodium chloride solution) than in distilled water. Acidic condition also accelerates rusting by reacting with iron metal. III. Redox Equilibria Part 5 Page 2 Rust indicator - Potassium hexacyanoferrate(III) Since Fe 2+ (aq) ion is the one of the initial product in rusting, formation of rust can be monitored by detecting the presence of Fe 2+ (aq) ion. Potassium hexacyanoferrate(III), K 3 Fe(CN) 6(aq) , is a yellow chemical which turns deep blue in the presence of Fe 2+ (aq) ion. It serves as a very good indicator of rust. [Fe(CN) 6 ] 3- (aq) + Fe 2+ (aq) → [FeFe(CN) 6 ] - (aq) yellow deep blue When a drop of solution containing phenolphthalein and potassium hexacyanoferrate(III) is dropped onto an iron surface, the rim of the drop will turn pink and the center of the drop will turn blue gradually. This shows that the rim is the cathodic area where the concentration of dissolved oxygen is higher. And, the center is the anodic area. After a long time, most rust will form somewhere between the cathodic area and anodic area, where Fe 2+ and OH - ion meet.. 2. Prevention of rusting a) Coating Since rust is only formed in the presence of oxygen and water, iron can be protected from rusting by putting a coating on the surface. A. Paint or grease Painting is the most cost effective method for large iron object but if the painted surface is scratched, iron will be vulnerable. Grease or oil is used in the moving part of machinery where the paint will be scratched. B. Galvanising (zinc dipping) Iron is dipped into a tank of molten zinc or molten zinc is spray onto iron so that the surface of the iron is coated with a layer of zinc. The process is called galvanising or galvanization. The iron is called galvanised iron. Although zinc also reacted with oxygen in the air to form zinc oxide, zinc oxide has a very tight structure and protects the zinc underneath from further corrosion. The body of most automobiles are galvanized before the paint is applied. This make it more resistant to rusting. C. Tin-plating Since zinc is toxic, it is not used in protecting the food can from rusting. Instead, a layer of tin is coated on the surface of iron can to exclude the air and water. However, if the tin-plating is scratched and the iron underneath is exposed, the tin-plated can will rust even faster than an ordinary iron can. This is because iron is more reactive than tin and iron will lose electrons through tin quickly. III. Redox Equilibria Part 5 Page 3 D. Electroplating of other metal Chromium is another metal which is coated on iron to prevent rusting. Chromium is frequently used because of its silvery appearance. e.g. Most water taps are electroplated with chromium. b) Sacrificial Protection Iron is first changed to iron(II) ion in the rusting process. If the formation of iron(II) ion can be suppressed, iron can be protected from rusting. Fe (s) d Fe 2+ (aq) + 2e - A reactive metal loses electron more readily than a less reactive metal. By connecting iron to a more reactive metal, formation of Fe 2+ (aq) can be retarded. Once a Fe 2+ (aq) ion is formed, it will acquire the electrons from the more reactive metal and changes back to Fe. Consequently, iron is protected from corrosion but the more reactive metal is sacrificed instead. This method is preferable for the iron object which is hard to be repainted e.g. iron pipe under the ground, hull of ship. Sacrificial protection by galvanization Besides isolating iron from air and water, Zn (s) also protects iron by sacrificial protection because Zn (s) is more reactive than Fe (s) . For the same reason, the body of a car is always connected to the negative terminal of the car battery. This make the car body negatively charged and less vulnerable to rusting. c) Alloying Pure iron rust easily in moist air. But if appropriate amounts of carbon, chromium and nickel are added, iron will become stainless steel which is more resistant to rusting. Glossary rusting (corrosion of iron) rust (hydrated iron(III) oxide) anodic area cathodic area rust indicator (potassium hexacyanoferrate(III)) coating galvanising zinc dipping sacrificial protection tin-plating electroplating alloying Past Paper Question 97 2A 4 b i ii 99 2A 1 b iii 97 2A 4 b i ii 4b Discuss and write relevant equations for the electrochemical processes involved when broken surfaces of the following are exposed to moist air. i galvanized iron ii tin-plated iron 99 2A 1 b iii 1b iii Explain why cathodic protection can be used to prevent the corrosion of iron. (7 marks) Phase Equilibria I. One component system A. Water 1. Phase diagram of water 2. Isotherm of water 3. PVT surface of water B. Carbon dioxide 1. Phase diagram of carbon dioxide 2. Isotherm of carbon dioxide II. Two components system A. Ideal system 1. Raoult's law 2. Composition of the liquid mixture comparing with composition of the vapour 3. Boiling point-composition diagram a) Definition of boiling point b) Boiling point-composition diagram c) Fractional distillation B. Non-ideal solution 1. Deviation from Raoult's law a) Positive deviation b) Negative deviation c) Enthalpy change of mixing 2. Elevation of boiling point and depression of freezing point by an involatile solute 3. Boiling point of two immiscible liquid a) Steam distillation III. Three components system A. Partition of a solute between two phases 1. Solvent extraction a) Calculation 2. Paper chromatography a) R f value b) Separation of amino acids Phase Equilibria Unit 1 Page 1 Topic Phase Equilibria Unit 1 Reference Reading 7.1 Assignment Reading A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 239–240 Syllabus One component system PVT surface Phase diagram Isotherm Notes I. One component system One component system is a system consists of one substance. At different condition, the same substance may have different physical state or phase. There are 3 physical states : solid, liquid and gas. Moreover, the same substance with the same physical state may not be exactly the same. e.g. graphite and diamond are two allotropes of carbon, the two different phases of solid carbon. Diatomic oxygen and ozone is another pair of example. By measuring the physical properties of a substance at different pressure, volume and temperature, a PVT surface can be constructed. Different regions of the surface represents different phases of a substance. PVT surface is solely a summary of experimental findings and can only be obtained through experiment. Phase Equilibria Unit 1 Page 2 Depending on the properties, gaseous phase can be further divided into vapour and gaseous phases. Vapour is defined as a gas which can be liquefied by compression alone. Conversely, a gas cannot be compressed into liquid without cooling. However, it is rather difficult to study pressure, volume and temperature altogether. It would be easier to investigate a 2-dimensional projection of the 3-dimensional PVT surface along the temperature or volume axes. The projection along the volume axis is called phase diagram. The projection along the temperature axis is called isotherm. A. Water 1. Phase diagram of water A solid line on the diagram represents the condition at which two phases coexist. The temperature at which solid and liquid coexist is called melting point. The temperature at which liquid and gas coexist is called boiling point. The temperature at which solid and gas coexist is called sublimation point. The point where the 3 solid lines joined together is called triple point. It is the only condition at which solid, liquid and gaseous phases can coexist in equilibrium. It is an unique condition for a specific substance. For example, 6.03 × 10 -3 atm and 273.16 K for water 5.11 atm and 216.6 K for carbon dioxide. Phase diagram of water BD – melting/freezing curve AB – sublimation curve BC – vaporization/condensation curve EB – supercooling curve Water is a very special substance which expands upon freezing. Its freezing curve (BD) has a negative slope. Comparatively, carbon dioxide has a freezing curve with positive slope. The freezing curve (BD) represents the conditions at which water and ice are at equilibrium. i.e. melting point of ice at different pressure. From the negative slope of the freezing curve, it can be seen that the melting point decreases with increasing pressure. If pressure is exerted onto a cube of ice, it will melt because the freezing point of water is lowered. But, how is this related to the volume of ice ? This can be explained by Le Chatelier's principle. Water (liquid) d Ice (solid) higher density lower density (smaller volume) (larger volume) When pressure is exerted onto ice, the system will response in a way so that the effect of increased pressure can be reduced. This is done by converting ice to water which has lower volume. Once the pressure is removed, water will freeze again. At molecular level, when a high pressure is exerted onto ice, the open structure will be broken and ice will be converted to water. Like all kind of change, freezing process also involves activation energy, though it is very small comparing with other chemical process. If a beaker of water is cooled calmly, it may reach a temperature below 0ºC without freezing. The liquid water which has a temperature lower than 0ºC is called supercooled liquid. If the supercooled water is disturbed (i.e. activation energy is provided), the water will freeze immediately. The curve EB represents the supercooling curve of water. The existence of water extends into the ice region.. Supercooled water is energetically less stable than ice. Supercooled water is a metastable state, which means highly unstable. Phase Equilibria Unit 1 Page 3 2. Isotherm of water Isotherm is the curve representing the relationship between the pressure and volume of a substance at the same temperature. The isotherm has the shape of a hyperbola only at high temperature. This means that the pressure is inversely proportional to volume (i.e. P α 1 V which is stated by Boyles' Law). At high temperature, the gaseous molecules are moving at great speed. The intermolecular attractions among particles become insignificant comparing with thermal motions. This makes a gas behave as an ideal gas. At a temperature higher than the critical temperature, a gas cannot be liquefied by compression alone. This has important implication in the manufacturing of liquid air. If the temperature is too higher, the air can never be compressed into liquid no matter how high the pressure is. At a temperature lower than critical temperature, vapour can be compressed into liquid. The vapour pressure, at which the vapour and liquid coexist, is called saturated vapour pressure. Saturated vapour pressure is a quantity independent of volume. If the volume of the system is decreased, more vapour will condense into liquid and the saturated vapour pressure will remain unchanged. It is only depending on temperature. Water vapour d Liquid water 3. PVT surface of water The curve abcdef represents the change of water when it is warmed at 1 atm pressure. Along ab, ice expands a little bit. Along bc, ice starts to melt and contract. Ice and water coexist at melting point. Along cd, water is heated up and expands. Along de, water boils and turns to water vapour. Water and water vapour coexist at boiling point. Along ef, the temperature and volume of water vapour increase. At a temperature beyond critical temperature, it becomes a gas. The curve ghjk represents the compression of water vapour to liquid water at room temperature. The curve nopqrs represents the compression of water vapour at a temperature lower than triple point temperature. Along no, the vapour is compressed. Along op, the vapour sublimes directly into ice. Along pq, ice is compressed and the volume gets a little bit small. Along qr, ice melts because of the increasing pressure. Along rs, liquid water is compressed and the volume change is very minimal. PVT surface of water Phase Equilibria Unit 1 Page 4 B. Carbon dioxide Liquid carbon dioxide doesn't exist at atmospheric pressure. From the phase diagram of carbon dioxide, the triple point pressure (5.1 atm) is higher than atmospheric pressure. When dry ice is warmed, it sublimes directly to carbon dioxide gas. Carbon dioxide behaves like many other substances. It contracts upon freezing and has a freezing curve with a positive slope. 1. Phase diagram of carbon dioxide 2. Isotherm of carbon dioxide Glossary one component system sublimation phase allotrope PVT surface vapour gas projection phase diagram isotherm triple point freezing curve sublimation curve vaporization curve supercooled liquid supercooling curve metastable state hyperbola ideal gas critical point vapour pressure (saturated) vapour pressure Past Paper Question 90 1A 2 e i ii iii 92 1A 2 g i ii iii iv 98 1A 2 a i ii iii 90 1A 2 e i ii iii 2e Each of the graphs A to D in the above diagram shows the variation of pressure with volume for a particular substance at different temperatures T 1 , T 2 , T 3 and T 4 . i Name the point P and comment on its significance. 1½ Critical point ½ mark At points on isotherm above P, the gaseous state of the substance cannot be liquefied by the increased in pressure alone. 1 mark Phase Equilibria Unit 1 Page 5 ii What is the significance of the horizontal portion of graph D? 1 Along the horizontal portion, the gaseous and liquid states of the substance can co-exist. 1 mark Or The horizontal portion signifies the liquefaction (vaporization) of the gaseous (liquid) state of the substance. iii Which graph provides experimental evidence that PV = nRT? Explain your answer. 1½ Graph A ½ mark Q it has the shape of a hyperbola 1 mark or Q it represents the relationship that P is inversely proportional to 1/V. C iii Candidates were expected to know that Graph A provides experimental evidence that PV = nRT because the graph gives the relationship in the form of a hyperbola in which P is inversely proportional to V (at a given temperature). 92 1A 2 g i ii iii iv 2g For the phase diagram below: i Identify the point X and Y. 1 Point X - critical temperature (substance can no longer exist in the liquid state beyond this temperature) ½ mark Point Y - triple point (solid, liquid and vapour are all at equilibrium) ½ mark ii Which two points in the diagram define the sublimation curve? 1 Y, Z iii Mark on the diagram, using the symbol B, the normal boiling point. 1 Point B is the boiling point. 1 mark iv Suggest the meaning of the curve YW. 1 Metastable state or Supercooling occurs. 1 mark 98 1A 2 a i ii iii 2a The phase diagram for carbon dioxide is shown below. Phase Equilibria Unit 1 Page 6 i Identify points B and C. 1 ii What is the physical meaning of point C ? 1 iii Starting with a sample of CO 2(g) at room temperature and atmospheric pressure, suggest how the following can be obtained. (I) solid CO 2 (dry ice) (II) liquid CO 2 2 Phase Equilibria Unit 2 Page 1 Topic Phase Equilibria Unit 2 Reference Reading 7.2.1 Modern Physical Chemistry ELBS pg. 372–377 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 181–186 Syllabus Two components system Ideal system Raoult's law Notes II. Two components system Two components system is a system consists of 2 different substances. If a system is comprised of one more substance, the physical behavior will be more complicate. On top of pressure, volume and temperature, the physical properties of a mixture is also depending on the relative abundance (mole fraction) of individual component. A 4-dimensional space will be required to represent the experimental findings. It would be easier if the investigation is limited to a given temperature. Therefore, the effect of temperature can be excluded. The situation can further be simplified by considering the liquid and vapour phases only. The saturated vapour pressure of a liquid is also commonly known as the vapour pressure of the liquid. In the presence of a liquid, the saturated nature of the vapour is understood. At a given temperature in an one component system, the (saturated) vapour pressure is a constant independent of volume. If the volume of the mixture is decreased, some vapour will be compressed into liquid and the (saturated) vapour pressure will remain a constant. Liquid d Saturated vapour In general, the more volatile a liquid, the higher will be the (saturated) vapour pressure at a given temperature. However, if two volatile liquids are mixed, the (saturated) vapour pressure will no longer depend on temperature only. It will also be depending on the abundance of individual component. P A º = vapour pressure of pure A (more volatile) P B º = vapour pressure of pure B (less volatile) Phase Equilibria Unit 2 Page 2 By Dalton's law of partial pressure, Total vapour pressure = partial pressure of vapour A + partial pressure of vapour B P T = P A + P B A. Ideal system It is found that if the intermolecular forces among all molecules in a mixture have similar magnitude, i.e. the intermolecular forces ALA, ALB and BLB are all similar, the (saturated) vapour pressure will be jointly proportional to the mole fractions of the two components. The ideal behavior of the solution is further confirmed by the nearly zero enthaply change of mixing of the two components. 1. Rauolt's Law The system in which the total (saturated) vapour pressure is jointly proportional to the mole fractions of the two components is called an ideal system. The relationship P A = X A × P A º is called Raoult's law. Or in words, the partial vapour pressure of any volatile component of an ideal solution is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution. N.B. Meaning of superscript º : pure substance o : standard state Vapour pressure-composition diagram By Dalton's law of partial pressure, Total vapour pressure = partial pressure of vapour A + partial pressure of vapour B P T = P A + P B mole vapour mole vapour = fraction of × pressure of + fraction of × pressure of component A pure A component B pure B P T = X A × P A º + X B × P B º A vapour pressure-composition curve can also be considered as a kind of phase diagram. At an external pressure higher than the equilibrium vapour pressure, the vapour will tend to condense to form more liquid. And if the external pressure is lower than the equilibrium vapour pressure, the liquid will tend to vaporize to form more vapour. Phase Equilibria Unit 2 Page 3 2. Composition of the liquid mixture comparing with composition of the vapour Will the composition of the vapour be the same as the composition of the liquid ? Obviously, the more volatile component vaporizes more easily and will have a greater contribution in the total vapour pressure (which is related to concentration) at a certain temperature. Therefore, when an equal molar solution of 2 liquids evaporates, the vapour will be enriched by the vapour of the more volatile component. Quantitatively, the composition of the vapour of an ideal solution at a given temperature can be calculated according to Rauolt's law. This requires the liquid composition of the mixture and the (saturated) vapour pressure of individual component at the given temperature. Mole fraction of A in the liquid = X A Mole fraction of B in the liquid = X B = (1- X A ) Partial vapour pressure of A = P A = X A P A º Partial vapour pressure of B = P B = X B P B º Vapour pressure-composition diagram Vapour pressure of the liquid mixture (according to Rauolt's law) = P T = X A P A º + X B P B º = X A P A º + (1- X A ) P B º Mole fraction of A in the vapour (according to Dalton's law of vapour pressure) = X A ' = P A P A + P B = X A P A º X A P A º + X B P B º = X A P A º X A P A º + (1 - X A )P B º Data (Saturated) vapour pressure of A at 300 K = 1.0 × 10 4 Pa (Saturated) vapour pressure of B at 300 K = 3.0 × 10 3 Pa Calculation Determine the composition of the vapour of the equimolar mixture of A and B at 300 K. By repeating the calculation for the mixtures with different compositions, a vapour composition curve can be constructed on the same vapour pressure-composition diagram. Phase Equilibria Unit 2 Page 4 3. Boiling point composition diagram a) Definition of boiling point A liquid is said to be boiling when the (saturated) vapour pressure reaches the atmospheric pressure. A more volatile liquid has a higher (saturated) vapour pressure at a given temperature, thus it has a comparatively lower boiling point. b) Boiling point-composition diagram By measuring the boiling pointsof mixtures with different composition and the composition of the vapour, a boiling composition diagram can be constructed. A boiling point composition diagram also consists of 2 lines. The first one is the boiling point of the mixture with different composition. The second one is the composition of the vapour when the mixture boils. The diagram on the right is the boiling point- composition diagram of an ideal system consisting of propan-1-ol and 2-methylpropan-1-ol. Mole fraction of propan-1-ol in the liquid (prepared experimentally) Boiling point of the mixture / ºC (measured) Boiling point of the distillate / ºC (measured) Mole fraction of propan-1-ol in the vapour (read from the graph) A 0 109.0 109.0 0 B 0.10 105.2 99.1 0.24 C 0.20 100.9 94.2 0.41 D 0.30 97.4 91.3 0.53 E 0.40 94.3 88.6 0.65 F 0.50 92.0 86.8 0.74 G 0.60 89.5 85.3 0.82 H 0.70 87.3 84.4 0.90 I 0.80 85.6 84.0 0.94 J 0.90 84.2 83.7 0.98 K 1 83.3 83.3 1 Actually, it is quite difficult to determine the composition of the vapour chemically. It would be more convenient to determine the boiling point of the distillate (condensed from the vapour) and determine the composition of the vapour using the liquid composition curve as the calibration curve. (The boiling point composition curve cannot be constructed from the Rauolt's law, Rauolt's law does not provide any relationship between the vapour pressure and the temperature. It only provides the relationship between vapour pressure and mole fraction of the liquid.) Phase Equilibria Unit 2 Page 5 c) Fractional distillation The boiling point composition curve can be used to predict the purity of the distillate. For example, when a mixture of 0.2 mole propan-1-ol and 0.8 mole 2-methylpropan-1-ol is boiled, from the curve, the vapour will have 0.37 mole of propan-1-ol and 0.63 mole of 2-methylpropan-1-ol. The distillate will have the same composition as the vapour. It should be noted that boiling is a constant temperature process, in which the liquid mixture and the vapour will have the same temperature (Line CD). On the other hand, condensation of the vapour is accompanied with a decrease in temperature. (Line DE) If the 0.37 mole propan-1-ol and 0.63 mole 2-methylpropan-1-ol mixture is distilled again, from the curve, the distillate obtained will have 0.58 mole propan-1-ol and 0.42 mole 2-methylpropan-1-ol. If the distillation is continued, the final distillate will be 100% propan-1-ol. While more propan-1-ol is removed, the percentage of the less volatile 2-methylpropan-1-ol remain in the flask will increase. As a result, the boiling point of the liquid will increase gradually. Instead of separate distillations, successive distillations can be accomplished by using a fractionating column, thus fractional distillation. Upon heating, the liquid mixture vaporizes and condenses on the glass beads when it moves further from the heat source. The thin liquid film condensed on the glass bead will evaporate again while more hot vapour is coming up and the distillation will continue. In general, in the fractionating column, the composition will change when a liquid mixture is vaporized but the composition will not change when a vapour mixture is condensed. Glossary two components system 4-dimensional space (saturated) vapour pressure jointly proportional Rauolt's law ideal system/solution vapour pressure-composition diagram liquid composition curve vapour composition curve boiling point composition diagram boiling point fractional distillation fractionating column glass bead boiling process condensation process liquid film Past Paper Question 91 2A 3 b i ii 93 2A 1 a i ii 94 2A 2 c i 95 2A 1 c i 96 2A 2 c i 99 2A 2 a i ii 91 2A 3 b i ii Phase Equilibria Unit 2 Page 6 3b i A liquid is added as a solute to a solvent forming an ideal solution. Comment on the vapour pressure of the solution as compared with that of the pure solvent. 3 If a volatile solute is added to a solvent, the total vapour pressure will be the sum of the partial pressures of the solute and solvent. If the vapour pressure of the solute is higher than the vapour pressure of the solvent then the total vapour pressure is higher and vice versa. 3 marks ii The vapour pressures of pure methanol and ethanol at 25ºC are 89.2 mm Hg and 44.6 mm Hg respectively., A mixture containing 32.0 g of methanol and 184.0 g of ethanol behaves as an ideal solution. Calculate the partial vapour pressure of each component, the total vapour pressure of the mixture and the composition of the vapour. (Relative atomic masses : H, 1.0; C, 12.0; O, 16.0) 4 CH 3 OH mol. wt. = 32 C 2 H 5 OH mol. wt. = 46 No. of moles of CH 3 OH = 32 32 = 1 No. of moles of C 2 H 5 OH = 184 46 = 4 Mole fraction of CH 3 OH = 1 5 = 0.2 Mole fraction of C 2 H 5 OH = 4 5 = 0.8 1 mark By Raoult’s Law P A = x P A º 1 mark Partial pressure of CH 3 OH = 1 5 × 89.2 = 17.84 mmHg Partial pressure of C 2 H 5 OH = 4 5 × 44.6 = 35.68 mmHg 1 mark Total vapour pressure = 17.84 + 35.68 = 53.52 mmHg ½ mark The composition of CH 3 OH in the vapour = 17.84 53.53 = 1 3 The composition of C 2 H 5 OH in the vapour = 1 - 1 3 = 2 3 ½ mark i.e. ratio of CH 3 OH : C 2 H 5 OH is 1 : 2 C i Most candidates were not able to point out that the vapour pressure of the solution could be higher or lower than that of the pure solvent. ii Careless mistakes in the calculation of mole fractions were common. 93 2A 1 a i ii 1a Liquid A is miscible with Liquid B over the whole composition range, forming ideal solutions. The vapour pressures of pure A and pure B, measured at 350 K, are 24.0 kPa and 12.0 kPa respectively. i Calculate the vapour pressure of a liquid mixture consisting of 60 mole % A and 40 mole % B at 350 K. 3 P A = P A ºX A P B = P B ºX B 1 mark Partial pressure of A = 24 × 0.6 = 14.4 kPa Partial pressure of B = 12 × 0.4 = 4.8 kPa 1 mark Total vapour pressure P T = 14.4 + 4.8 = 19.2 kPa (wrong unit -½) 1 mark ii The mixture in (i) is distilled at 350 K and a small amount of the distillate is collected. If this distillate is distilled for a second time at 350 K, calculate the composition of the first droplet of the second distillate. 4 Composition of distillate (1st time) X A = P A P T = 14.4 19.2 = 0.75 X B = 1 - 0.75 = 0.25 1 mark Total vapour pressure of second distillation = 0.75 × 24 + 0.25 × 12 = 21 kPa 1 mark Composition of second distillate X A = 0.75 × 24 21 = 0.86 1 mark X B = 1 - 0.86 = 0.14 1 mark C ii Many candidates were not able to get the final answers correct because they did not realise that the ratio of partial pressures of A to B in the first distillate was the same as the mole ratio of A to B in the condensed phase. This indicated that the underlying principles of fractional distillation were not well understood by these candidates. Phase Equilibria Unit 2 Page 7 94 2A 2 c i 2c Two miscible liquids E and F form ideal solutions on mixing. At 298 K, the vapour pressure E is p 0 while that of pure F is 2p 0 . i Sketch a graph of vapour pressure versus mole fraction for the above solutions at 298K. For a mixture of E and F with mole ration E to F = 1 : 4, determine its vapour pressure, in terms of p 0 , at 298K. 3 Mole fraction of F = 4 4 + 1 = 4 5 or 0.8 From the graph, vapour pressure of the mixture = 1.8 p 0 OR BY CALCULATION P E = 0.2p 0 P F = 0.8 (2p 0 ) = 1.6p 0 ∴ vapour pressure of mixture = P E + P F = 0.2p 0 + 1.6p 0 = 1.8p 0 95 2A 1 c i 1c Two miscible liquids A and B form ideal solutions upon mixing. The boiling point of A is higher than that of B. i Draw the boiling point-composition graph at 1 atm for mixtures of A and B and label the graph in detail. 3 Boiling point-composition curve 1 mark for labelling the x- and y-axis1 mark for the liquid curve 1 mark for the vapour curve (deduct 1 mark if T B > T A ) C This question was well answered except that some candidates mistook the boiling point-composition diagram as the vapour pressure-composition diagram 96 2A 2 c i 2c i Two miscible liquids A and B form ideal solutions when mixed. At 298 K, the vapour pressures of pure A and pure B are 32 kPa and 16 kPa respectively. For a mixture of 1 mol of A and 3 mol of B at 298 K, calculate (1) the vapour pressure of the mixture; (2) the mole fraction of A in the vapour which is in equilibrium with the mixture. 4 (1) Vapour Pressure = 1 4 × 32 + 4 × 16 1 + 1 marks = 20 kPa 1 mark (Deduct ½ mark for wrong / no unit) (2) Mole fraction of A = 8 20 = 2 5 1 mark (Deduct ½ mark for including a unit in the answer.) 99 2A 2 a i ii Phase Equilibria Unit 2 Page 8 2a A liquid mixture consists of 70.0 mol % of methanol and 30.0 mol% of ethanol. Suppose that the mixture is an ideal solution. Calculate, at 298 K, (At 298 K, the vapour pressures of methanol and ethanol are 12.5 kPa and 9.9 kPa respectively.) i the vapour pressure of the mixture, and ii the concentration of methanol, in mol%, in the vapour phase at equilibrium. (3 marks) Phase Equilibria Unit 3 Page 1 Topic Phase Equilibria Unit 3 Reference Reading 7.2.2–7.2.3 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 203–204 Assignment Reading Advanced Practical Chemistry, John Murray (Publisher) Ltd., 78–80 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 186–187 Syllabus Non-ideal system Deviation from Raoult's law Enthalpy change of mixing Notes B. Non-ideal system However, most solutions do not behave ideally. Some of them will give a (saturated) vapour pressure higher than the expected and some of them will give a lower value. 1. Deviation from Raoult's law This can be interpreted as the change in the strength of intermolecular forces. Positive deviation If the intermolecular forces among the two kinds of particles (ALB) is weaker the original forces in pure components (ALA & BLB), it will be easier for the particles to escape from the liquid mixture and give a (saturated) vapour pressure higher than the expected at a given temperature. This is called positive deviation from Raoult's law. For an ideal solution, the liquid composition curve is linear in shape. If the solution is slightly positively deviated, the curve will curve up a little bit. If the solution is highly deviated, a curve with a maximium point will result. This means that the mixture with that particular composition is even more volatile than either component. Slightly deviated Highly deviated Negative deviation Similarly if the intermolecular forces among the two kinds of particles (ALB) is stronger, it will give a (saturated) vapour pressure lower than the expected at a given temperature. This is called negative deviation from Raoult's law. The minimium vapour pressure in the liquid composition curve represents a mixture which is less volatile than either component. Slightly deviated Highly deviated Phase Equilibria Unit 3 Page 2 a) Positive deviation A mixture of ethanol and benzene is a positive deviation from Raoult's law. It is a largely deviated system. The mixture has a minimium boiling point at 0.55 mole fraction of ethanol. At this point, both liquid and vapour have the same composition. This mixture is known as azeotropic mixture or azeotrope. Azeotropic mixture is a constant boiling mixture and cannot be separated by fractional distillation. If a 0.20 mole fraction of ethanol is distilled, the first distillate will contain 0.39 mole fraction of ethanol. If the distillate is distilled continuously, eventually, only a 0.55 mole fraction of ethanol distillate will be obtained. Boiling point-composition curve for ethanol / benzene Ethanol and water is another example of highly positive deviation. Fractional distillation can only purify the mixture up to 0.94 mole fraction ethanol or 97% by mass of ethanol. The remaining 3% water can only be removed by drying agent e.g. calcium oxide followed by distillation. This explains why absolute ethanol is much more expensive than 97% ethanol. Boiling point-composition curve for ethanol / water Phase Equilibria Unit 3 Page 3 b) Negative deviation Trichloromethane and propane is a system of highly negative deviation. The mixture has a maximium boiling point at 0.65 mole fraction of trichloromethane. If a 0.30 mole fraction of trichloromethane is distilled, the vapour will contain less trichloromethane (0.20 mole fraction) and more propanone. And the liquid remain in the flask will contain more trichloromethane. Eventually, a constant boiling mixture of 0.65 mole fraction of trichloromethane will be left in the flask and cannot be separated into trichloromethane and propanone. The distillate obtained will be pure propanone. Boiling point-composition curve for trichloromethane / propanone Similar if a 0.80 mole fraction of trichloromethane is distilled, the solution remain in the flask will be 0.65 mole fraction of trichloromethane. But the distillate obtained will be pure trichloromethane. Hydrochloric acid is also a highly negative deviation from Rauolt's law. Its azeotropic mixture has a composition of 0.11 mole fraction of HCl. Boiling point-composition curve for hydrogen chloride / water Phase Equilibria Unit 3 Page 4 c. Enthalpy change of mixing The change in strength of intermolecular forces can further be confirmed by the enthalpy change of mixing of components. When the components of a ideal system is mixed, the temperature change is almost zero. This means no special intermolecular attractions are formed or broken. When the components of a system showing positive deviation is mixed, the temperature decreases. This means that some intermolecular attractions are weakened. (i.e. bond breaking is an endothermic process.) When the components of a system showing negative deviation is mixed, the temperature increases. This means that some intermolecular attractions are strengthened. (i.e. bond formation is an exothermic process.) Glossary non-ideal system / solution positive deviation from Raoult's law negative deviation from Raoult's law azeotropic mixture / azeotrope / constant boiling mixture enthalpy change of mixing Past Paper Question 92 1A 2 f i ii iii 93 2A 1 b 94 2A 2 c i ii 95 2A 1 c i ii 96 2A 2 c ii 97 2A 1 c 99 2A 2 b i ii 92 1A 2 f i ii iii 2f The vapor pressure versus mole fraction diagram for a two component system (A and B) is shown below: i State the reason(s) which give(s) rise to the shapes of the curves in the diagram. 2 The attractive forces between A and B are weaker than those between molecules of the same kind. 1 mark As a result, there is an increased tendency to escape form solution and that the vapour pressure of each of the components is greater than that predicted by Raoult’s law. 1 mark ii What do points P and Q represent? 1 P, Q are the vapour pressure of pure A and pure B respectively. 1 mark iii What assumption(s) would have to be met for the curves to be linear? Give an equation to represent this behaviour. 1 The intermolecular attraction between molecule A and molecule B is similar to that between molecule A and molecule A and that between molecule B and molecule B. ½ mark Ideal behaviour (Obeys Raoult’s Law) P=X A P A º+X B P B º OR P A =X A P A º ½ mark C iii Some candidates confused ideal gas law with Raoult's Law for ideal solution. Phase Equilibria Unit 3 Page 5 93 2A 1 b 1b Draw vapour pressure versus composition graphs of two types of non-ideal solutions formed by mixing two miscible liquids. Interpret these graphs in terms of intermolecular attractions. 4 Vapour pressure – composition diagrams for negative and positive deviations from Rauolt’s Law are : (no dotted line -½) If the attraction between the molecules A & B is weaker (stronger) than the average intermolecular attraction in pure A & B, positive (negative) deviation results. Weaker (stronger) intermolecular attraction means molecules have greater (smaller) chance to escape, hence vapour pressure of the system is higher (lower) than that expected by Raoult’s Law. C Many ca`ndidates were not aware of the relation between intermolecular interactions and non-ideal behaviour of real solutions. 94 2A 2 c ii 2c Two miscible liquids E and F form ideal solutions on mixing. At 298 K, the vapour pressure E is p 0 while that of pure F is 2p 0 . i Sketch a graph of vapour pressure versus mole fraction for the above solutions at 298K. For a mixture of E and F with mole ration E to F = 1 : 4, determine its vapour pressure, in terms of p 0 , at 298K. 3 Mole fraction of F = 4 4 + 1 = 4 5 or 0.8 From the graph, vapour pressure of the mixture = 1.8 p 0 OR BY CALCULATION P E = 0.2p 0 P F = 0.8 (2p 0 ) = 1.6p 0 ∴ vapour pressure of mixture = P E + P F = 0.2p 0 + 1.6p 0 = 1.8p 0 ii If E and F do not form ideal solution on mixing, in what possible ways could the vapour pressure of the above mixture deviate from the value determined in (i)? Account for the deviation(s). 3 If deviation is negative, P < 1.8p 0 ½ mark If deviation is positive, P > 1.8p 0 ½ mark In case of negative (positive) deviation, the interaction betwee E and F is stronger (weaker) than the average of interaction between E and E and between F and F. The molecules has a smaller (greater) chance to escape, thus resulting in a lower (higher) vapour pressure. 2 marks [Max. 2 marks, if only one of the cases is discussed.] Phase Equilibria Unit 3 Page 6 95 2A 1 c i ii 1c Two miscible liquids A and B form ideal solutions upon mixing. The boiling point of A is higher than that of B. i Draw the boiling point-composition graph at 1 atm for mixtures of A and B and label the graph in detail. 3 Boiling point-composition curve 1 mark for labelling the x- and y-axis 1 mark for the liquid curve 1 mark for the vapour curve (deduct 1 mark if T B > T A ) ii Using your graph, explain how liquid B can be separated from a mixture containing A and B, by fractional distillation. 4 If a solution with composition X 1 is heated to T 1 , the gaseous phase in equilibrium with it has the composition X 2 which is richer in B. (1 marks for explanation; 1 mark for indication on the graph) This vapour is cooled to T 2 and condenses to liquid with a composition of X 2 . Vaporization of this liquid leads to a new vapour with a composition X 3 , which is richer in B than that with a composition X 2 and finally pure B is obtained. (1 mark for explanation; 1 mark for indication on the graph) C This question was well answered except that some candidates mistook the boiling point-composition diagram as the vapour pressure-composition diagram 96 2A 2 c ii 2c ii Predict whether the following pairs of liquids when mixed, would give solutions showing positive deviation or negative deviation from Raoult's Law. Explain your prediction. (1) CH 3 CH 2 OH and CH 3 (2) C 2 H 5 OC 2 H 5 and CHCl 3 4 (1) Mixtures of ethanol and methylbenzene will show positive deviation from Raoult's Law. ½ mark In pure EtOH, the intermolecular attraction is H-bond. ½ mark In the mixture EtOH molecules are surrounded by methylbenzene, the attraction between the molecules is mainly van der Waals' force. ½ mark Weakening of the intermolecular force causes an increase in vapour pressure / escaping tendency of molecule. ½ mark (2) Mixtures of C 2 H 5 OC 2 H 5 and CHCl 3 will show negative deviation from Raoult's Law. ½ mark In the pure, liquids the intermolecular attraction is dipole-dipole interaction. ½ mark In the mixture C 2 H 5 OC 2 H 5 forms H-bond with CHCl 3 Thus, the mixture will have a lower vapour pressure than the pure liquids. ½ mark 97 2A 1 c 1c Three types of boiling point - composition curves for two miscible liquids A and B at one atmosphere pressure are shown below. 7 Phase Equilibria Unit 3 Page 7 Account for the characteristics of these curves. 99 2A 2 b i ii 2b i At 1 atm pressure, the boiling points of pure nitric(V) acid and water are 359 K and 373 K respectively. At this pressure, nitric(V) acid and water form an azeotrope which boils at 394 K. The azeotrope consists of 35.3 mol %of HNO 3 , Sketch a labelled phase diagram for the nitric(V) acid-water system at 1 atm pressure showing (I) the boiling point - liquid composition relationship, and (II) the boiling point - vapour composition relationship. ii Describe all changes in the fractional distillation of an aqueous solution of 20 mol % of HNO 3 . (5 marks) Phase Equilibria Unit 4 Page 1 Topic Phase Equilibria Unit 4 Reference Reading 7.2.4 Chemistry (Structure and Dynamics), McGraw-Hill, 331–334 Assignment Reading A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd.,187–189 Syllabus Elevation of boiling point and depression of freezing point by an involatile solute Boiling point of two immiscible liquids Steam distillation Notes 2. Elevation of boiling point by an involatile solute A volatile solvent (A) has weak intermolecular forces among the solvent particles (ALA). A solute (B) is involatile only if the intermolecular forces among the particles (BLB) are strong. If an involatile solute (B) is soluble in the volatile solvent (A), the forces among the solute and solvent particles (ALB) must be stronger than the solvent-solvent interaction (ALA) and solute-solute interaction (BLB). Otherwise, the solute will not be soluble. Since the solute-solvent interaction is the strongest, when an involatile solute is dissolved in a volatile solvent, this must be a negative deviation from Rauolt's law. The vapour pressure of the mixture will be lower than the pure solvent at any temperature. Recall the phase diagram of pure water, pure water boils at 100 ºC and freezes at 0 ºC. If salt is dissolved in water, a higher temperature will be required to rise the vapour pressure of the mixture to atmospheric pressure. Therefore, the presence of involatile solute increases the boiling point of the solvent. For example, curve AB shows the vapour pressure of pure water at different temperature. Normally, its vapour pressure will reach atmospheric pressure at 100 ºC. Curve CD is the vapour pressure of the aqueous salt solution. At any particular temperature, the vapour pressure is always lower than that of pure water. And a higher temperature is required to make the solution boil. Phase Equilibria Unit 4 Page 2 If the lowering of vapour pressure is extended to the lower temperature range, the vaporization curve will meet the sublimation curve at a lower temperature. This will lead to a lower triple point temperature as well as a lower melting point. For this result, salt is sprayed onto the icy road to melt the ice in winter. The degree of elevation of boiling point and depression of freezing point depends on the nature of the solvent and the concentration of the solute. (Surprisingly, it is not depending on the nature of the solute.) 3. Boiling point of a mixture of two immiscible liquids For a two components system, if the intermolecular forces between the two components (ALB) are very small comparing with the original forces (ALA & BLB), the two liquids will be immiscible with each other. This is an extremely case of positive deviation from Rauolt's law. Under this circumstance, the two components will vaporize independently, the vapour pressure will be the sum of the vapour pressures of the two pure components and independent of the mole fraction of individual component. P T = P A º + P B º Comparing with an ideal solution : P T = X A P A º + X B P B º Since the total vapour pressure will be higher than the vapour pressure of either component at any temperature, the boiling point of the mixture will also be lower than either component. Nitrobenzene and water are two immiscible liquids. Nitrobenzene and water have boiling points 211 ºC and 100 ºC respectively. When they are mixed together, independent of the relative abundance, the mixture boils at 98 ºC. Phase Equilibria Unit 4 Page 3 a) Steam distillation According to the above principle, a heat sensitive organic compound can be separated by steam distillation. Sometimes, ordinary distillation cannot be used to purify heat sensitive organic compound. This is because some organic compounds decompose well before they boil. In order to separate them, the boiling point must be lowered. Most organic compounds are immiscible with water. Thus, water can be added to lower the boiling point of the compound. This is done by passing steam through the organic mixture to be separated or heat the mixture in the presence of large amount of water. Steam provides both the heat and water for the distillation. The method is called steam distillation. Steam distillation Since water and the organic compound are immiscible, water in the distillate can be separated by a separating funnel. Glossary elevation of boiling point vaporization curve sublimation curve triple point temperature depression of freezing point immiscible liquids steam distillation steam generator safety tube separating funnel Phase Equilibria Unit 4 Page 4 Past Paper Question 92 2A 2c ii 92 2A 2 c ii 2c ii The total vapour pressure exhibited by a mixture of immiscible liquids is equal to the sum of their individual saturated vapour pressures. (I) Explain why steam distillation enables some heat-sensitive organic compounds to be distilled at a lower temperature than their normal boiling points. (II) A mixture of two immiscible liquids F and G boils at 98°C. At this temperature, the vapour pressures of F and G are 733 mmHg and 27 mmHg respectively. Calculate the percentage of G by mass in the vapour when the mixture boils. (The relative molecular masses of F and G are 18.0 and 123.0 respectively.) 4 (I) A liquid boils when its vapour pressure reaches the value exerted by external pressure. Thus the total vapour pressure exerted by an immiscible mixture of liquids will reach atmospheric pressure at a temperature below the boiling-point of the heat sensitive / most volatile constituent. 1 mark (II) P F = vapour pressure of F = 733 torr P G = vapour pressure of G = 27 torr Since the no. of mole of F is proportional to P F . Similarly, the no. of mole of E is proportional to P G as PV = nRT. 1 mark Let m F and m G be the masses of F and G in the system P F P G = n F n G = m F M F m G M G or m f m G = P F M F P G M G = 733 × 18.0 27 × 123.0 = 3.97 ⇒ m f = 3.97 m G 1 mark Percentage by mass of G = m G m f + m G = m G 3.97 m G + m G = m G 4.97 m G = 0.201 or 20.1% 1 mark C ii To answer the questions, candidates had to realize that in the vapour phase, the number of moles of F and G are proportional to their vapour pressures and then the candidates needed to set up the relationship P G / P E = n G / n E = (m/M) G / (m/M) E Phase Equilibria Unit 5 Page 1 Topic Phase Equilibria Unit 5 Reference Reading 7.3 Assignment Reading A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 205–206 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd.,189–193 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 330–333 Syllabus Three components system Partition of a solute between two phase Solvent extraction Paper chromatography Notes III. Three components system The three components systems that we are going to study consist of 1 solute and 2 solvents. A. Partition of a solute between two phases When a solute is added into a mixture of two immiscible solvents, it dissolves in both of them. At equilibrium, the speed of diffusion from one solvent to another is the same as the reverse speed. Therefore, the concentrations of the solute in both solvents will remain constant. Solute in lower layer d Solute in upper layer And the equilibrium constant of the system is known as partition coefficient or distribution coefficient, K d . By convention, the concentration of solute in the less dense solvent (upper layer) will be used as the numerator in the equilibrium law. N.B. CCl 4 and CHCl 3 are the only two common organic solvents denser than water. K d = concentration in the upper layer concentration in the lower layer or [Solute (upper layer) ] [Solute (lower layer) ] K d can also be expressed as [Solute (upper layer) ] [Solute (lower layer) ] × molar mass of the solute molar mass of the solute = conc n in gram per dm 3 in the upper layer conc n in gram per dm 3 in the lower layer Since the numerator and denominator always have the same dimension, K d always has no unit. The concentration can be expressed in molarity, gcm -3 or gdm -3 , the choice of unit has no effect on the value of K d . Phase Equilibria Unit 5 Page 2 1. Solvent extraction Solvent extraction is an useful technique in organic chemistry. It allows partial removal of solute from one solvent into another immiscible one. This is done by shaking the sample to be purified with an immiscible solvent in a separating funnel. Ether is a slightly polar organic solvent lighter than water. It is commonly used in solvent extraction because it is immiscible with water and volatile. Once the solute is extracted, ether can be evaporated at reduced pressure to obtain the pure sample. N.B. Ether can never be evaporated by a naked flame because it is highly immflammable. High temperature is also undesirable because ether forms explosive peroxide with oxygen at high temperature. a) Calculation Example 1 At 291 K, the partition coefficient of butanoic acid, CH 3 CH 2 CH 2 COOH between ether and water is 3.5. Calculate the mass of butanoic acid extracted by shaking 100 cm 3 of water containing 10 g of butanoic acid with 100 cm 3 of ether. K d = [Solute (ethereal phase) ] [Solute (aqueous phase) ] = 3.5 Let x g be the mass of acid extracted by 100 cm 3 of ether. x / 100 (10 - x) / 100 = 3.5 x = 7.78 Therefore, 7.78 g of butanoic acid will be extracted. Example 2 For the same aqueous solution of butanoic acid, calculate the mass of butanoic acid extracted if two portions of 50 cm 3 ether are used. Let x 1 g and x 2 g be the masses of acid extracted in the two portions. In the first extraction, x 1 / 50 (10 - x 1 ) / 100 = 3.5 x 1 = 6.36 In the second extraction, The mass of acid remaining in the aqueous phase = original mass - mass extracted in the first extraction = (10 - 6.36) g = 3.64 g x 2 / 50 (3.64 - x 2 ) / 100 = 3.5 x 2 = 2.32 The total mass extracted by the two 50 cm 3 portions of ether = 6.36 g + 2.32 g = 8.68 g Although the same total amount of ether is used, the mass of acid extracted in two portions is larger. For this reason, solvent extraction is usually done with a small amount of solvent at a time. Phase Equilibria Unit 5 Page 3 2. Paper chromatography Partition of a solute between two phases is not limited to 2 immiscible liquids only. Paper chromatography is also a kind of distribution equilibrium. The solute partitions between the stationary phase and the mobile phase. K d = [Solute (moving phase) ] [Solute (stationary phase) ] In paper chromatography, stationary phase is the layer of water adsorbed on the natural cellulose fibres of filter paper. And, moving phase is the solvent used to carry the solute along the paper. In a mixture of dyes, different dyes have different partition coefficients between the moving phase and stationary phase. The dye with larger partition coefficient will move faster on the paper when the solvent is soaking up. Because of these differences, different dyes can be separated. a) R f value The colour pattern obtained from paper chromatography is called paper chromatogram. Each spot on the chromatogram is identified by a value called R f value. It is defined as : R f = distance moved by the coloured spot distance moved by the solvent front R f is for a particular compound depends upon the solvent used and the temperature. It always has a value smaller than 1. It is possible to characterize a particular compound separated from a mixture by its R f value. b) Separation of amino acids A mixture of unknown amino acids can be separated and identified by means of paper chromatography. However, all amino acid are colourless. They must be developed first. The positions of the amino acids in the chromatogram can be detected by spraying with ninhydrin (a developer), which reacts with amino acids to yield highly coloured products. Glossary three components system partition partition coefficient / distribution coefficient solvent extraction separating funnel explosive peroxide paper chromatography stationary phase mobile phase adsorbed R f value chromatogram characterize ninhydrin developer Phase Equilibria Unit 5 Page 4 Past Paper Question 92 2A 2 c i 96 1A 1 e i ii 98 2A 4 b i ii 99 1A 2 c i ii iii 92 2A 2 c i 2c i The distribution coefficient of a compound X between two immiscible liquids D and E is 12. X is more soluble in D. What mass of X can be extracted form 50cm 3 of a solution of E which initially contains 4g of X when shaken with 100cm 3 of D? 3 Let α g be the mass of X in 1 cm 3 of D and β g be the mass of X in 1 cm 3 of E Since distribution coefficient is 12. ∴ α = 12β and 100α + 50β = 4 100α + 50 12 α = 4 α = 48 1250 = 0.0384 The mass of X in liquid D = 100α = 3.84 g 3 marks C i Many candidates gave the mass of X remaining in E instead of the mass of X extracted from E. They probably did not read the question carefully. 96 1A 1 e i ii 1e At 298 K, 50 cm 3 of a solution of I 2 in CCl 4 were mixed and shaken with 200 cm 3 of distilled water in a separating funnel until equilibrium was attained. The two layers were then separated. 20.0 cm 3 of the CCl 4 layer required 12.15 cm 3 of 0.105 M S 2 O 3 2- (aq) for titration, whereas 50.0 cm 3 of the aqueous layer required 8.25 cm 3 of 0.0050 M S 2 O 3 2- (aq) . i Calculate the distribution coefficient of I 2 between water and CCl 4 at 298 K. 2 ratio of concentration of I 2 : water CCl 4 = 8.25 × 0.0050 50 12.15 × 0.105 20 1 mark = 0.012934 = 0.013 1 mark (Accept distribution coefficient expressed as [I 2 (CCl 4 )]/[I 2(aq) ], in which case the value calculated is 77.3 (77)) (Award ½ marks for a correct expression of K d ) ii If the 200 cm 3 of distilled water contain some dissolved KI, will this affect the value of the distribution coefficient ? Briefly explain. 1 No change ½ mark because distribution coefficient is a function of temperature only ½ mark (Also accept K will change because there is a change in ionic strength/activity (1/0)) 98 2A 4 b i ii 4b Even though the partition coefficient of ethanoic acid between water and 2-methylpropan-1-ol is 3.05 at 298 K, 2- methylpropan-1-ol is still used to extract ethanoic acid from aqueous solutions. Calculate the efficiency of ethanoic acid extraction (in terms of %) at 298 K by shaking 100 cm 3 of a 0.50 M aqueous solution of ethanoic acid with i 200 cm 3 of 2-methylpropan-1-ol; ii two successive portions of 100 cm 3 of 2-methylpropan-1-ol. Decide whether (i) or (ii) is the better extraction method. Explain your answer. 99 1A 2 c i iii 2c A mixture of amino acids, A, B and C, was separated by paper chromatography using an appropriate solvent. The chromatogram obtained is shown below: Phase Equilibria Unit 5 Page 5 (× is the starting point of the mixture) i Briefly describe the principle underlying the separation of A, B and C by paper chromatography. ii The amino acid spots are invisible to naked eyes. Suggest how to make them visible. iii Calculate the R f value for A. Energetics I. Enthalpy Change (∆H) A. B. C. Definition of a system Definition of enthalpy (H) and enthalpy change (∆H) 1. Constant pressure process a) Work done by a system Standard enthalpy change of formation (∆Hfo) 1. Definition of ∆Hfo a) Standard enthalpy change (∆Ho) b) Standard enthalpy change of formation (∆Hfo) c) Standard enthalpy change of formation of an element 2. Direct determination of ∆Hfo a) Use of simple calorimeter Other enthalpy change (∆H) 1. Examples of enthalpy change a) Enthalpy of combustion b) Enthalpy of atomization c) Enthalpy of neutralization d) Enthalpy of hydrogenation e) Enthalpy of solution f) Enthalpy of reaction Hess’s Law 1. Indirect determination of ∆Hfo a) Hess’s Law (1) (2) Use of energy level diagram / energy cycle Use of equation D. E. II. Bonding Energy A. B. C. Bond energy 1. Definition of bond energy 2. Determination of bond energy term (bond enthalpy) Strenght of covalent bond 1. Relationship between bond length and bond energy 2. Factors affecting the bond strength Estimation of enthalpy change by bond energy Electron affinity Lattice energy 1. Born-Haber cycle a) Determination of lattice energy by Born-Haber cycle 2. Factors affecting the value of lattice energy Stoichiometry of ionic compounds Limitation of enthalpy change (∆H) Entropy (S) and entropy change (∆S) Gibbs free energy change (∆G) III. Energetics of formation of ionic compounds A. B. C. IV. Gibbs free energy change (∆G) A. B. C. Chemical Bonding Page 1 Chemical Bonding I. Shape of molecule A. Lewis structure of molecule 1. Oxidation no. of individual atoms in a molecule 2. Formal charge of individual atoms in a molecule 3. Limitation of octet rule Valence Shell Electron Pair Repulsion Theory (VSEPRT) 1. Octet expansion of elements in Period 3 2. Position of lone pair 3. 3-dimensional representation of molecular shape Hybridization Theory 1. Overlapping of atomic orbital 2. Hybridization theory 3. Examples a) sp hybridizaion b) sp2 hybridizaion c) sp3 hybridizaion d) sp3d hybridizaion e) sp3d2 hybridizaion 4. Structure and shape of hydrocarbons a) Characteristic of Carbon-Carbon bond b) Shape of hydrocarbons (1) (2) (3) Saturated hydrocarbons Unsaturated hydrocarbons Aromatic hydrocarbons (a) Delocalization of π-electrons (b) Stability of benzene B. C. D. Molecular Orbital Theory II. Ionic bonding A. B. C. Strength of ionic bond - lattice energy X-ray diffraction 1. Electron density map of sodium chloride Periodicity of ionic radius 1. Definition of radius 2. Periodicity of ionic radius 3. Size of isoelectronic particles Chemical Bonding Page 2 III. Covalent bonding A. B. C. H2+ ion Electron diffraction Covalent radius 1. Definition of covalent radius 2. Addivity of covalent radius 3. Breaking down of addivity in covalent radius and bond energy a) Resonance (1) (2) (3) Resonance structure of benzene Resonance structure of nitrate ion Rules in writing resonance structures Delocalization energy b) D. Breaking down of addivity in bond enthalpies (1) Dative covalent bond 1. Examples of H3N→BF3 and Al2Cl6 a) H3N→BF3 b) Al2Cl6 IV. Bonding intermediate between ionic and covalent A. B. Differences between ionic bond and covalent bond Incomplete electron transfer in ionic compound 1. Electron density map of LiF comparing with those of NaCl and H2 2. Difference among lattice energies of Na, Ag and Zn compounds a) Lattice energies of sodium halide, silver halide and zinc sulphate b) Bonding intermediate between covalent and ionic 3. Polarization of ionic bond a) Fajans’ Rules in polarization of ionic bond Electronegativity 1. Definition of electronegativity 2. Pauling scale of electronegativity Polarity in covalent bond 1. Deflection of a liquid jet by an electric field 2. Dipole moment a) Vector quantity of dipole moment b) Polarity of molecule c) Factors affecting dipole moment (1) (2) (3) Inductive effect (I) Mesomeric effect / resonance effect (R) Presence of lone pair C. D. V. Metallic bonding A. B. C. D. Electron sea model of metal Strength of metallic bond Melting and boiling of metal Strength of ionic bond, covalent bond and metallic bond Intermolecular forces Intermolecular Forces I. Van der Waals’ forces A. B. Discovery of van der Waals' forces Origin of van der Waals' forces 1. Induced dipole-induced dipole attractions 2. Dipole-dipole interactions 3. Dipole-induced dipole attractions Relative strength of different origins of van der Waals' forces Nature Strength Solubility and hydrogen bond Intramolecular hydrogen bond Examples of hydrogen bonding 1. Ethanoic acid dimer 2. Simple hydrides - CH4, NH3, H2O and HF 3. Structure and physical properties of ice 4. Biochemical importance of hydrogen bond a) DNA b) Protein Strength of van der Waals' forces and hydrogen bond C. II. Hydrogen bond A. B. C. D. E. F. III. Relationship between structures and properties States of Matter I. Solid state A. Metallic structure 1. Hexagonal close packing 2. Cubic close packing - face centred cubic 3. Tetrahedral holes and octahedral holes 4. Body-centred cubic structure Giant ionic structure Covalent giant structure 1. Allotrope of carbon a) Diamond b) Graphite 2. Silicon(IV) oxide Molecular crystal 1. Iodine – face centred cubic 2. Carbon dioxide (Dry ice) – face centred cubic Gas laws 1. Charles' law 2. Boyle's law 3. Avogadro's law 4. PVT surface Ideal gas equation (Ideal gas law) 1. Molar volume Determination of molecular mass 1. Experimental determination of molar mass of a gas 2. Experimental determination of molar mass of a volatile liquid Dalton's law of partial pressure 1. Mole fraction 2. Relationship between partial pressure and mole fraction B. C. D. II. Gaseous state A. B. C. D. Rates of chemical reactions A. Collision theory and activation energy A. By plotting graph of ln rate versus T III. Second order reaction 3. Different approaches a) Constant amount approach b) Constant time approach 2. By measuring the rate of reaction at different temperature 1 2. Collision theory and activation energy a) Arrhenius equation Determination of activation energy 1. Definition of rate of reaction 1. Examples of different reaction 1. . Differential form and integrated form of rate law 2. B. Interpretation of physical measurements made in following a reaction a) Volume of gas formed b) Colorimetric measurement Rate Law or rate equation 1. Order of reaction A. First order reaction a) Half life (t½) b) Carbon-14 dating c) Examples of calculation 2. Zeroth order reaction Collision theory 1. B. Graphical presentation of reaction rate 3.Chemical Kinetics Page 1 Chemical Kinetics I. Order of reaction a) Experimental determination of order of reaction (1) (2) By measuring the rates of reaction at different reactant concentration By plotting graph of ln rate versus ln [X] II. Unit of rate of reaction Measuring the rate of reaction 1. Distribution of molecular speed (Maxwell-Boltzman distribution) 2. B. By plotting graph of ln k versus T 1 3. Effect of temperature change on molecular speeds 3. Energetic stability and kinetic stability Effect of catalyst 1. b) Rate determining step 2. Transition state 1. Order of reaction leads to interpretation of reaction mechanism at molecular level a) Multi-stages reaction (1) (2) Alkaline hydrolysis of 2-chloro-2-methylpropane Reaction between bromide and bromate(V) in acidic medium B. Application of catalyst a) Use of catalyst in Haber process and hydrogenation b) Catalytic converter c) Enzymes Concentration Temperature Pressure Surface area Catalyst Light V. Theory of catalyst a) Homogeneous catalysis b) Heterogeneous catalysis 3. E. . B. C. Characteristics of catalyst 2. Factors influencing the rate of reaction A.Chemical Kinetics Page 2 IV. Energy profile of reaction A. F. D. Bromine water in acidic and alkaline medium 2. Dynamic nature Examples of equilibrium 1. Ka. Concentration of solid 5. Hydrolysis of bismuth chloride Equilibrium law 1. . Determination of equilibrium constant (Kc) a) Esterification b) Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) 6. Kd 2. D. Introduction to different equilibrium constant (K) – Kc. Effect of catalyst on equilibria 5. B. Equilibrium constant (Kc and Kp) 3. Kp. Potassium dichromate in acidic and alkaline medium 3. Examples of calculation C. Nature of equilibrium A. Kw. Concentration 2. pressure and temperature on equilibria (Le Chaterlier's principle) 1. Pressure 3. Kb. Degree of dissociation (α) 4.Chemical Equilibria Page 1 Chemical Equilibria I. Relationship between rate constant and equilibrium constant Effect of change in concentration. KIn. Temperature ∆H a) Equation : ln K = constant T 4. Measurement of pH a) Use of pH meter (1) Calibration of pH meter Colour of indicator B. Leveling effect Dissociation of water 1.Chemical Equilibria Page 2 II. Ionic product of water pH and its measurement 1. Lewis definition Strength of acid 1. 4. Principle of buffer action a) pH of an acidic buffer b) pH of an alkaline buffer 2. Ka and Kb a) Relationship between Ka and Kb (pKa and pKb) b) Relationship between pH. H. D. pOH and pKw c) Some basic assumptions applied Experimental determination of Ka . Difference between equivalence point and end point 2. G. Buffer 1. Bronsted-Lowry definition 3. Definition of pH 2. C. Conductimetric titration a) Strong acid vs Strong base b) Weak acid vs Strong base c) Weak acid vs Weak base Calculation involving pH. Temperature dependence of pH 3. F. Thermometric titration 5. Arrhenius definition 2. Calculation of buffer solution Theory of Indicator Acid-base titration 1. Titration using indicator a) Choosing of indicator 4. Use of indicator (1) Strong and weak acids/bases 1. Measuring of pH or conductivity of acid / base 2. Acid-base Equilibria A. Dissociation constant (Ka and Kb) a) Dissociation of polybasic acid (1) Charge effect 3. Titration using pH meter 3. Acid-base Theory 1. b) E. Secondary cell and fuel cell 1.f. Standard hydrogen electrode 2.Chemical Equilibria Page 3 III.m. a) Potentiometer b) High impedance voltmeter / Digital multimeter 2. of a cell c) Prediction of the feasibility of redox reaction (1) Disproportionation reaction D. E. Hydrogen-oxygen fuel cell Corrosion of iron and its prevention 1. Cell diagrams (IUPAC conventions) Electrode potentials 1.m. Electrochemical process of rusting 2. Relative electrode potential (Standard reduction potential) a) Meaning of sign and value 3. Balancing of redox reaction a) By combining balanced half-ionic equation (1) (2) Steps of writing balanced half-ionic equation Combining half-ionic equations B. Use of salt bridge a) Requirement of a salt bridge b) An electrochemical cell does not need salt bridge 3. Electrochemical series 4. Faraday and mole 3. Redox reaction 1. Lead-acid accumulator 2. 2. Measurement of e. b) By the change in oxidation no. Use of electrode potential a) Comparing the oxidizing and reducing power b) Calculation of e. Prevention of rusting a) Coating b) Sacrificial protection c) Alloying . Redox Equilibria A. Calculation of mass liberated in electrolysis Electrochemical cells 1.f. C. Raoult's law 2. One component system A. Composition of the liquid mixture comparing with composition of the vapour 3. B. Deviation from Raoult's law a) Positive deviation b) Negative deviation c) Enthalpy change of mixing 2. Boiling point of two immiscible liquid a) Steam distillation Partition of a solute between two phases 1. Water 1. Elevation of boiling point and depression of freezing point by an involatile solute 3. Solvent extraction a) Calculation 2. Phase diagram of carbon dioxide 2. Three components system A.Phase Equilibria I. Isotherm of water 3. Phase diagram of water 2. PVT surface of water Carbon dioxide 1. Isotherm of carbon dioxide Ideal system 1. Paper chromatography a) Rf value b) Separation of amino acids B. III. II. Boiling point-composition diagram a) Definition of boiling point b) Boiling point-composition diagram c) Fractional distillation Non-ideal solution 1. . Two components system A. . Atomic Structure I. Determination of the ionization energy a) Through the convergence limit of Lyman series b) Through Rydberg constant Detail electronic configuration 1. Quantum numbers a) Energy of electrons in different orbitals 5. Atomic structure A. IV. Electronic configuration of first 36 elements a) Extra stability of half-filled and full-filled subshell 7. II. Electronic structure of atoms A. Successive ionization energy 3. B. Nature of radioactivity a) Penetrating power of radiation 2. p-. B. Building up of electrons in an atom a) Aufbau principle b) Pauli exclusion principle c) Hund's rule d) Building up of electrons e) Detail electronic configuration of some atoms 6. Radioactive decay a) Half life b) Use of radioactive isotope Emission spectrum of hydrogen 1. A. III. Origin of emission and absorption spectrum 2. Different model of atomic structure Isotopic mass and relative atomic mass 1. p and d orbitals Variation in atomic radius Variation in electronegativity C. Periodic Table .E. Atomic orbitals A. B. down a group 4. Mass spectrometer a) Determination of atomic mass b) Determination of molecular mass Radioactivity of nuclide 1.E. Pattern of energy levels of hydrogen 3. across periods b) First I. d. Definition of ionization energy 2. Position of s-. Periodicity of First ionization energy a) First I.and f-block elements Shape of s. The masses of other atoms are determined by comparing with this standard.B. Ramsden pg. 3–5.3 Chemistry in Context. 57–58 Physical Chemistry. 6–12 Chemistry in Context. 4th Edition. Model is just as good as what you want it to explain. 2. B.N.–4. Relative atomic mass – The relative atomic mass of an element is the weighted average of the isotopic masses of its natural isotopes on the Carbon-12 scale. Orbital model (1927) The exact position of the electron cannot be predicted or measured. Atomic Structure Unit 1 Syllabus Notes A. E. 5th Edition. Thomas Nelson and Sons Ltd. 2–4. an electron density cloud is used to describe the probability of finding an electron in certain region.1. . Different model of atomic structure Dalton’s model Thomson’s model Rutherford’s model Bohr’s model Wave model Orbital model Dalton’s model (1808) Atom is a solid. Bohr’s model (1913) The electron is moving around the nucleus in certain circular orbits. 3rd Edition. Wave model (1924) The electron is moving around the nucleus in a wave like motion. Atomic structure Unit 1 Page 1 Topic Reference Reading I. Isotopic mass and relative atomic mass Isotopic mass – The relative isotopic mass of a particular isotope of an element is the mass of one atom of that isotope on the Carbon-12 scale.I. 3rd Edition. 1. 10–17 A-Level Chemistry. Fillans pg. Thomson’s model (1899) Atom is made of electrons evenly spread throughout positive mass. 1. 54–62 Different model of atomic structure Mass spectrometer I. ELBS pg. (a mass has no unit). pg. Atomic Structure 1. Carbon-12 scale – A scale in which Carbon-12 isotope is chosen as the standard and 12 units is assigned to it by definition. N. indivisible sphere. Rutherford’s model (1911) Atom is made of electrons and nucleus where most of the space is empty. the magnetic field is adjusted so that ions of a particular mass are focused onto the ion-detector Detector – detect the signal and pass it on to a recorder. Ionization chamber – gas molecules are bombarded with fast speed electrons to form positive ions. Construction of Mass Spectrometer Heater – to vaporize the sample. the mass of the ion can be calculated.the various m/e signals and peak heights are processed by computer. .I. M(g) + e. Measurement of isotopic masses of an element.(fast) → M+(g) + 2e- Accelerating plate / Negatively charged plate – to accelerate the beam of positive ions across the electric field. in which a beam of ion is arranged in order of increasing mass to charge (m/e) ratio. So by using a mass spectrum of a sample of an element. Identification of compounds . Atomic structure 1. Mass Spectrum A spectrum is obtained with a mass spectrometer. Accelerating voltage(V) and the radius of the path(r). Vacuum pump – to evacuate the apparatus. This allow the positive ions to travel in straight lines without collision and prevents presence of other substances in the spectrometer that will ionize and affect the result Knowing the magnetic field strength(B).) Important uses of Mass Spectrometer 1. Mass Spectrometer Unit 1 Page 2 A device in which particles are ionized and the accelerated ions are separated according to their mass to charge (m/e) ratio. The detector current is directly proportional to the relative abundance of the ions reaching the detector. 2. either element or molecule. Measurement of molecular mass. 4. Magnet / Magnetic field – to deflect the ion beam. the relative atomic mass of the element can be determined. 3. Determination of ionization energies. (Calculation not required. It is caused by the existence of trace amount of C-13 isotope in the nature.81" Total height of the four peaks = 0. the peak on the mass spectrum with the highest (m/e) ratio corresponds to the original molecule. a very small peak (about 1% of that of 16) with m/e ratio 17 is found. It tends to break up into several fragments.07" +1.→ CH4+ + 2eCH4+ → CH3+ + H and CH4+ 16 12 CH3+ → CH2+ + H etc. 12 CH3+ 15 12 CH2+ 14 12 CH+ 13 C+ 12 Besides the peak with m/e ratio 16. it becomes energetic and unstable.66" The height of the peak at 208 = 2.81" + 1.35" Since.81" = 6. a molecule will be ionized upon bombardment with fast moving electrons. lead-206. For example. Once a molecule is ionized. the following is the spectrum of a sample of methane CH4.81" 1.35" 6. The height of the peak at 204 = 0.35" 6. Similar to an atom. Atomic structure Unit 1 Page 3 a) Determination of relative atomic mass from mass spectrum (Not to scale) The four isotopes of lead are lead-204.07" The height of the peak at 206 = 1. Since the fragment always has a smaller mass than the original molecule.07" 1.I.19) b) Determination of molecular mass Another important use of mass spectrometer is to determine the mass of a molecule. . Ionization Fragmentation Fragment m/e 12 CH4 + e. the height of the peak ∝ the relative abundance The relative atomic mass of lead = 204 × 0.66" 2.35" = 207(207.66" + 2. lead-207 and lead-208.81" The height of the peak at 207 = 1.35" 6.81" + 206 × + 207 × + 208 × 6. Glossary orbital – Loosely speaking. More accurately. C-12 was chosen as the standard for atomic mass and carbon-12 scale is established. the ions travel along until deflected by the magnetic field the magnetic field is adjusted so that ions of a particular mass are focused into the ion-detector knowing the magnetic field strength. in 1961. . accelerating voltage and the radius of the path. by definition. it is used to describe the geometrical figure which describes the most probable location of an electron. 6 C 2 marks the sample (an element) is bombarded by electrons to form positively charged ions. it would be more convenient if carbon is used as the standard of reference in mass spectrum.I. Carbon–12 scale (relative) isotopic mass (relative) atomic mass Past Paper Question 90 2A 3 a 96 1A 1 a i ii iii 98 2A 4 c 99 2A 1 a i 90 2A 3 a 3a Give an account of the use of a mass spectrometer for determining the relative masses of particles.the mass spectrometer responds to the mass-to-charge (m/e) of the ionized species 4 marks Many candidates did not point out that the mass spectrometer responds to the mass to charge ratio. Because all organic compound contains carbon. Indeed the experimental determination of the more precise value of Avogadro number is still the interest of many scientists. there is no wonder why Avogadro number is not a whole number. structure of numerous number of organic molecules have been studied. In carbon-12 scale. . Some candidates did not know the positively charged ions are accelerated by the electric field. 12 g of C-12 contains 1 mole of C-12 atoms. the atomic mass can be calculated. Since the definition is solely arbitrary. As a result. the ions are accelerated by the electric field between plates A and B. Atomic structure Carbon-12 scale Unit 1 Page 4 After the invention of mass spectrometer. The modern method of determining atomic masses uses the mass spectrometer. it is used to denote the allowed energy level for electrons. 003 1.).I.6605 × 10-27 × 6.m.m.m.u. 6e ½ mark ii The isotopic mass of 12C is 12.939 72.12 = 12. 6p.00 + 13. to the relative atomic mass of carbon. 13C . O2 21% and CO2 1%. Calculate the mass.12 Using the above data.m.85% (I) Suggest an experimental method to detect the isotopes of rubidium and state how the relative abundance of each isotope can be obtained. Avogadro constant.6605 × 10-27 kg.012) iii The following data were obtained from the mass spectrum of a carbon-containing compound : Ion Mass / a.0120 kg 1 mark (Accept answers which could round off to 0.000 × 100.000 100. as required.00 + 1.u. 6e ½ mark 13 C 7n.000 × 1. though some candidates did not give the unit. L = 6.12 1 mark relative atomic mass = (100.57 = 101.0221 × 1023 mol-1) Mass of 1 mole of 12C = 12. The table below lists the mass and relative abundance of each isotope.u. 12 C 6n. (1 a. Page 5 1 2 2 3 . 6p.) 99 2A 1 a i 1a i Rubidium occurs naturally in two isotopic forms. 12. of 1 mole of 12C atoms.12) 1214.003 × 1. Atomic structure Unit 1 96 1A 1 a i ii iii 1a i Write down the number of neutrons. Relative abundance 85 Rb 84. Many candidates wrongly assigned the unit a. calculate the relative atomic mass of carbon. Relative intensity 12 + C 12.u.01) (Deduct ½ mark if unit is given) C Well answered in general.00 13 + C 13. protons and electrons in one atom of carbon-12.0119996) = 0.15% 87 Rb 86.011 1 mark (Accept answers which could round off to 12. (You only need to consider the major isotope of each element. Isotope Mass / a.m.000 atomic mass unit (a. 12C. kg.937 27.u. and in one atom of carbon-13. (II) Calculate the relative atomic mass of rubidium. in kg.0221 × 1023 1 mark = (0. 98 2A 4 c 4c Sketch the expected mass spectrum for a gas sample having the composition : N2 78%. = 1. Emission α particle β particle γ radiation Nature He nuclei (2p + 2n) fast moving electrons electromagnetic wave Relative velocity 1/20 velocity of light 3 – 99% velocity of light 100% velocity of light Range in air a few cm a few m a few km Relative penetrating power 1 100 10000 1 1 Penetrating power is ∝ mass . all others exhibit some kind of radioactivity. Ramsden pg. pg. Only 238 nuclei are stable.I. neutron and electron or emit strong electromagnetic radiation. but owing to the presence of the isotope. Thomas Nelson and Sons Ltd.) 1 The deflection is ∝ charge and ∝ mass All of the three kinds of radioactivity have the ability to ionize the air.N. 3rd Edition ELBS pg. so a Geiger-counter can be used to measure the activity of the radioactive material. 76–81 Nature of radioactivity Radioactive decay C. ∝ velocity Emission α particle β particle γ radiation Relative mass 4 units 1 / 1840 unit – Relative charge +2 -1 0 Relative effect of electric and magnetic field small deflection large deflection no deflection Effect of magnetic field Effect of electric field (The magnetic field is going into the paper. Atomic structure Unit 2 Page 1 Topic Reference Reading I. a) Penetrating power of radiation There are 3 types of radioactivity. Nature of radioactivity There are totally 109 elements. ∝ charge . . E. Atomic Structure 1.2 Chemistry in Context. 83–89. Radioactivity of nuclide Unit 2 Syllabus Notes 1. 93–95 A-Level Chemistry. 13–20 Chemistry in Context. there are over 1600 different nuclei. 5th Edition. 3rd Edition. They eject proton. The process is called radioactive decay and the elements are said to be radioactive. 1. one element can be changed to another element by radioactive decay. The U-238 isotope can decay through multiple stages to a stable Pb-206 isotope . Beta Decay This is a decay common to most radioactive isotopes of elements with Z < 82.I. a large amount of energy is released in the process. It will only change from a high energy(excited) state to an low(stable) energy state. ii) the mass number and the atomic number in all nuclear reaction must be conserved. is usually omitted for simplification. 2. its mass no. In the reaction. atomic fission or atomic fusion.g. a) b) c) 238 92 226 88 224 88 U → 4 He + 234 Th 2 90 4 222 Ra → 2 He + 86 Rn 220 Ra → 4 He + 86 Rn 2 When an element undergoes alpha decay. an element emitting gamma radiation will not change to another element. unlike ordinary chemical reaction.g. e. will be reduced by 2. the energy is from the binding energy in the nucleus. will remain unchanged and the atomic no. Note: i) the charge of the particle. the charge on He nucleus. Atomic structure 2. nuclei are involved instead of electrons. a) 234 90 Th* → γ–ray + 234 90 Th Most of the heavy isotope will undergo multiple decays to form a series. The process in which an element changes to another element in nuclear reaction is known as transmutation. (high p: n ratio) e. a) b) c) 14 6 108 47 1 0 C → 0 e + 14 N −1 7 Ag → 0 e + 108 Cd −1 48 0 1 n → −1 e + 1 H When an element undergoes beta decay. its mass no. Gamma Radiation Unlike alpha and beta decay. (low p: n ratio) e. Alpha Decay This is a decay common to most of the radioactive isotopes of elements with Z > 82 (Pb). e.g.g. will increase by 1. will be reduced by 4 and the atomic no. Radioactive decay Unit 2 Page 2 In a nuclear reaction. 7 × 103 years 1. only depends on the amount of isotopes present. Atomic structure a) Half life of radioactive decay Unit 2 Page 3 Half life . e.03125 = 3.I.125% . the fraction remaining will be = ½×½×½×½×½ = 0.5 × 109 years 5.g.the time taken by a given amount of radioactive isotope to decay to half of the original amount. The changes in external factors such as temperature and pressure have no effect on it. The rate of radioactive decay. For example. Radioactive Isotopes Uranium-238 Carbon-14 Polonium-214 Half-life 4.5 x 109 years to decay to half of the original amount (50 Uranium-238 isotopes).5 × 10-4 seconds The stability of an isotope is directly proportional to its half life. it will take 4. This forms an exponential decay curve. After 5 half-lives. Rate of radioactive decay The rate of the decay decreases as the decay proceeds because the rate of decay is directly proportional to the amount of isotope remains. unlike rate of chemical reaction. if there are 100 Uranium-238 isotopes. atomic mass. In some cases.e. i. if the speeds of intruding neutrons are different. and energy state. 5. Use of radioactive isotope Treatment of cancer Study of metabolic pathways. Thickness gauges and empty packet detection. protons. Uranium-238 can be transmutated to Uranium-239 by bombarded with an neutron. penetrating power radioactive decay half-life . 6.I. Archaeological and geological dating. Atomic structure Artificial Transmutation of elements Unit 2 Page 4 Artificial transmutation reactions are usually achieved by bombarding an element with a stream of particles. Sterilizing surgical instruments. it will cause different modes of nuclear reaction. 2. i. such as carbon 14.e. a) b) U → 239 Np + 0 e 93 −1 239 Np → 94 Pu + 0 e −1 14 7 14 6 N in air. is transmutated to 14 1 14 1 i.g. alpha particles and other small nuclei.e. e. water pipes) Glossary nuclide . 7 N + 0 n → 6 C + 1 H In nature the. especially neutrons. 4. Detection of leakage. e.g.g. 238 92 U + 1 n → 239 U 0 92 239 92 239 93 The unstable Uranium-239 then decays in two stages by emitting β-particles. (e.A type of atom specified by its atomic number. 27 13 27 13 27 13 24 Al + 1 n → 11 Na + 4 He 0 2 27 Al + 1 n → 12 Mg + 1 p 0 1 28 Al + 1 n → 13 Al + γ-ray 0 C by the neutron from the cosmic ray (fast moving neutron). 3. a) b) c) b) 1. One of the isotopes 222 86 Rn decays to give 222 86 218 84 Po with a half-life of 3. Rn is considered hazardous to health. The penetrating power of β particles is 100 times stronger than α particles.67 years. 238 206 4 0 92 U → 82 Pb + 8 2 He + 6 −1 e Many candidates incorrectly stated that α particles are helium atoms. α particles are bare helium nucleus with two positive charge and four units of mass. . 1 mark 1 Ra → 228 89 Ac + 0 −1 e Many candidates did not make use of the Periodic Table given in the question paper to find the symbol (Ac) for the product of the decay. 99 1A 2 b i ii 2b Radioactive decay of radium (Ra) in rocks and soil produces radioisotopes of radon (Rn). The path of α and β particles are deflected/bent by a magnetic field but they are deflected in opposite direction. 222 86 i ii Write the nuclear equation for the decay of Explain why Rn .82 days.I. Compare the action of a magnetic field upon the paths of X and Y 229 225 i X is _______________ 90 Th → X + 88 Ra ii X is / helium nucleus / α particle 225 225 Y is _______________ 88 Ra → Y + 89 Ac Y is / fast moving electron / β particle. OR by a diagram 1 mark 1 mark 1 mark 1 mark 4 The magnetic field is going into the paper. which is negatively charged. Atomic structure Unit 2 Page 5 Past Paper Question 91 2A 1 a i ii 93 1A 1 a i ii 95 2A 2 a i 99 1A 2 b i ii 91 2A 1 a i ii 1a The radioactive decay of i 238 92 U may be represented by 238 92 U→ 206 82 Pb + α + β. β particles are fast electrons. 88 1 mark 1 mark i C Write a balanced equation to represent the decay of 228 88 228 88 Ra . and hence they arrived at incorrect answers. ii C State the nature of α and β particles and compare their penetrating power. Some candidates did not emphasize that the electrons in β particles are moving very fast. Bent motion Correct direction 95 2A 2 a i 228 2a Ra decays by the emission of β particles. and gives their charge and mass. Balance the above equation. 3 1 mark 1 mark 1 mark 2 2 marks 93 1A 1 a i ii 1a Name the particles X and Y in the following nuclear reactions. The half-life for the decay is 6. Origin of emission Emission spectrum i.1 Chemistry in Context. 1. for example. But some lights. like a rainbow. excited state. Electronic Structure of atoms 2. 3rd Edition ELBS pg. Thomas Nelson and Sons Ltd. Each transition involves a definite amount of energy. 64–68 Emission spectrum of hydrogen Determination of ionization energy II. 31–36 Chemistry in Context. 4th Edition. the light from the sodium lamp or green light from the flame test of copper. The atom is in the ground state because all electrons in the atoms are having the lowest possible energy. the spectrum is known as a line spectrum. ii. The greater the number of electrons making a particular transition. called a quantum. 69–71 Physical Chemistry. instead they are only mixtures of a limited number of wavelengths. 5th Edition. the more intense the spectral line is. E. The white light can be separated into different colour light by a prism and forms a continuous spectrum of visible light. pg. The relationship between the wavelength(λ) and frequency(υ) of any electromagnetic radiation is given by c where c is the velocity of light λ= υ vi. Fillans pg.II. Different transition has different energy and each transition corresponds to a single line in the emission spectrum. Electronic Structure of atoms Unit 1 Page 1 Topic Reference Reading II. This gives coloured emission lines on dark background. the electrons occupy the orbital of lowest energy. are not composed of a continuous range of wavelengths. The electrons takes up the energy and are promoted to a higher energy state. Emission spectrum of hydrogen Unit 1 Syllabus Notes Emission Spectroscopy White light from tungsten lamp consists of a continuous range of wavelengths of visible light. In this case. iv. Energy is given to the atoms by heating the element with Bunsen flame or hitting with electrons in discharge tube.N. Electronic Structure of atoms A. they tends to fall back to a lower energy state to dissipate their extra energy in form of radiation. nearly all the atoms in a given sample of substance are in the ground state. Since the electrons in excited state are not stable. Ramsden pg. v. 3rd Edition. 63–74 A-Level Chemistry. And. the relationship between energy of radiation is given by the equation ∆E = hυ where h is Planck’s constant and υ is frequency . iii. At room temperature. The emission line spectrum of hydrogen atom consists of many lines. corresponds to the transition of an electron from the third energy level (n=3) to the second energy level (n=2).II. iii. At n = ∞. where the spectral lines merge. . the electron has effectively left the atom. the line with longest wavelength in the visible region. Unit 1 Page 2 Pattern of energy levels of hydrogen Balmer series i. It gives out a quantum of light. ii. Electronic Structure of atoms 2. The visible part of the spectrum was first studied by the scientist Balmer. With increasing frequencies. That means the atom has been ionized. corresponds to the electron transition between energy levels with n = ∞ and n = 2. called photon. the lines get closer and closer until they merge. some of them fall in the visible region and some of them fall in the invisible region. Relationship between emission spectrum and energy levels iv. The red line. The point. this is because the difference between energy levels gets smaller as the frequency increases. This series of spectrum lines was named after him as Balmer series. v. II. . Electronic Structure of atoms Unit 1 Page 3 Complete Hydrogen Emission Spectrum i. n1 : the principal quantum number of the energy level to which the electron returns. ii. the only difference is the destination of the electrons. n2 : the principal quantum number of the energy level from which the electron leaves. with n2 greater than n1. the complete spectrum is composed of many series. 1 1 υ = cR H ( 2 − 2 ) where RH is Rydberg constant n1 n 2 c is velocity of light n1 and n2 are integers. The origin of other series is similar to that of Balmer series. called principal quantum number. Bohr obtained an expression for the frequencies of the lines in the spectrum. Besides Balmer series. E. υ2→1 = cRH ( 2 − υ∞→1 = cRH ( 2 − 1 1 1 1 1 ) 22 1 ) ∞2 ii. The frequencies of the first lines in Lyman series are measured. A graph of υ against ∆υ is plotted. it is quite difficult to determine the convergence limit accurately as the lines are getting close together when approaching the convergence limit and the spectrum becomes a bright continum beyond the limit. and the υ at y-intercept. Thereafter. Therefore.) = h υ∞→1 = hcRH where L is the Avogadro's constant The ionization energy per mole = LhcRH Glossary emission spectrum Paschen series absorption spectrum Rydberg equation Planck’s constant Balmer series Lyman series . ∆υ2 = (υ3 . The frequency corresponding to the ionization process is calculated by iii. The Ionization energy can be calculated by I.υ1). e. b) Through Rydberg constant For example..υ2) .iii. the limit is usually determined graphically. the Rydberg constant can be determined by measuring the frequency of the first line of Lyman series (υ2→1). The ionization energy(I. there is a convergence limit in the line spectrum. = hυ i. Electronic Structure of atoms 3. iv.). The Rydberg constant is calculated from the first line in the Lyman series. ii.II.E. The curve is extrapolated.E. i. the ionization energy of hydrogen can be calculated. by applying the Ryberg equation. It represents the point where the difference in energy levels diminished. ∆υ1 = (υ2 .g. v. the energy transition from the ground state and the continuum for one electron is Ionization energy(I. a) Through the convergence limit of Lyman series Ionization of hydrogen = ∆E = h υ∞→1 However. will correspond to the transition of electron from n = ∞ to n = 1. where ∆υ = 0. Unit 1 Page 4 Determination of the ionization energy The ionization energy of hydrogen can be determined by measuring the frequency the convergence limit (υ∞→1) of Lyman series or by determining the Rydberg constant through the frequency of a specific line. The differences in frequencies between two adjacent lines are calculated.. In Lyman series. 023 x 1023) 1 mark = 1.626 x 10-34 J s.n 2) 1 mark λ 2 where RH is the Rydberg constant n2 = 3. 97 2A 2 a 2a Describe and account for the characteristics of the emission spectrum of atomic hydrogen.E. L = 6.4. the convergence limit for the Lyman series occurs at 3. = hυ 1 mark = (6. 99 1A 2 a i ii 2a i Outline the origin of atomic emission.. Calculate the ionization energy of hydrogen. These spectral lines are the result of transitions of the electron from one energy level to another 2 marks with the lower energy level with quantum number n1 = 2.023 x 1023 mol-1) I.5 .II. Many candidates neglected to give the answer to (ii) in kJmol-1.26 × 10-34) × (3.275 x 1015 Hz. 1 mark ii In the atomic emission spectrum of hydrogen. 1 1 1 The wavelengths of this series satisfies the formula. ii Briefly explain why each series of atomic emission lines of hydrogen converges at the lower wavelength end. 95 1A 1 b i ii 1b i What can you deduce from the fact that the spectral lines in the atomic emission spectrum of hydrogen are not equally spaced ? The energy differences between electron shells in a hydrogen atom are not the same (converge). (3 marks) 4 1 2 7 .. It is a series of lines resulting from photons emitted or absorbed by the hydrogen atom with wavelength in the visible region. in kJ mol-1 (Planck constant. h = 6.307 × 106 Jmol-1 / 1307 kJmol-1 C Many candidates were unable to point out that the energy differences between two energy levels are not the same and hence the spectral lines in the emission spectrum are not equally spaced.275 × 1015 × 6. Avogadro constant. λ = wavelength 1 mark C Many candidates gave the correct formula for the relationship between wavelength and quantum numbers. = RH (22 . Electronic Structure of atoms Unit 1 Page 5 Past Paper Question 90 2A 2 c 95 1A 1 b i ii 97 2A 2 a 99 1A 2 a i ii 90 2A 2 c 2c Give a brief account for the Balmer series in the spectrum of the hydrogen atom. but only a few defined their symbols. 60 × 10-19 C × 1V = 1. 26–33. 37–42. 49–53 Chemistry in Context. 74–77. 64. iii. Normally measured in electron volts (eV) or kJmol-1. 3rd Edition ELBS pg.8. 8 in the second group.8. 39–41 Chemistry in Context.1. 68–71. It can be observed from the graph that the I. 73–74. pg.II. 173–175 Detail electronic configuration B. Electronic Structure of atoms Unit 2 Page 1 Topic Reference Reading II. 170. 5th Edition. 38–42. Fillans pg. 1 in the first group.N. is much higher than 1st I. ii. Ramsden pg. (1eV = charge of electron × 1V = 1.s of K atom can be divided into 4 groups.3 Modern Physical Chemistry ELBS pg. 8 in the third group and remaining 2 in the fourth.60 × 10-19 J) 2. Definition of first ionization energy Syllabus Notes X(g) → X+(g) + eFirst ionization energy – Energy required to remove an electron from a free atom in gaseous state. 71–78 Physical Chemistry. The 2nd I. Successive ionization energy (presence of electron shells) The information from successive ionization energy can be used to deduce the electronic configuration of an atom. Thomas Nelson and Sons Ltd. Detail electronic configuration 1.E.E.2–2. 4th Edition. 3rd Edition. E. 88–95 A-Level Chemistry.E. The ionization of Potassium (K) can take place in steps 1st ionization K(g) → K+(g) + e2nd ionization K+(g) → K2+(g) + e3rd ionization K2+(g) → K3+(g) + ei. This indicates that a K atom has four electron shells and electronic configuration of 2. ∆HI1 = 419 kJmol-1 ∆HI2 = 3051 kJmol-1 ∆HI3 = 4412 kJmol-1 The value of successive ionization energy increases as more electrons are removed from the atom. 80–84. . Electronic Structure of atoms Unit 2 2. This is because extra energy is needed to removed an electron from an ion of increasing positive charge. The attraction between the outermost electron and the nucleus decreases with increasing atomic size.E. in a 2-3-3 manner. in a 2-10-3-3 manner. The 1st I. Electronic Structure of atoms 3. d and f sub-shells) within an electron shell (n = 1. Besides the main trends. / kJmol-1 150 0 14 0 0 13 0 0 12 0 0 110 0 10 0 0 900 Cl Br I First I. / kJmol-1 Na 2000 18 0 0 16 0 0 14 0 0 12 0 0 10 0 0 K Rb Cs Ar Kr Xe 1st I. This suggest the presence of subshells. decreases as the number of electron shells or the size of the atom increases.II. the trend can be broken into 4 parts. of group 0 On down a group. from K to Kr. iv.E.) iii. This suggest the presence of another set of subshells. 1st I. ii. In period 2 and 3. b) First I. the 1st I.E.E..E. / kJmol-1 490 4 70 4 50 430 4 10 390 3 70 3 50 First I..E. This suggests the presence of subshells (s. some irregularities are found within a period.E. In period 4. 3 . p. . of group VI 1st I. the increasing trends can further be broken up into 3 parts. 2. increases roughly across a period due to the increase in nuclear charge.E. Unit 2 Page 2 Periodicity of First ionization energy (presence of electron sub-shell) a) First I.E.E. of group I i. across periods First ionisation Energy of the first 38 elements i. down a group 53 0 510 Li 170 0 16 0 0 F 2400 2200 He Ne First I. in the range of -l to +l Either +½ or -½ Electrons with different subsidiary quantum numbers can be denoted by the letters s. Since different orbitals have different energies. Quantum numbers To facilitate the discussion of the detail electronic configuration of an atom. A more energetic electron would be removed more readily and has a lower ionization energy. p orbital. iv. p. e. the energy of orbitals increases in following order. Allowed value Any natural number Depends on n. ii. The energy of orbitals is independent of magnetic quantum number (m) and spin quantum number (s). The orbitals are labelled according to the values of n. d orbital and f orbital. iii. s < p < d < f. The principal quantum no. p 2. specify the shape of the orbital. The orbitals with n = 2. Quantum numbers is the I. l and m.g. 0 and -1 are called 2px. s 1. 2 . Electronic Structure of atoms Unit 2 Page 3 A prevailing explanation for these variation in ionization energy is that different electrons occupy different orbitals in an atom. . in the range of 0. no of the electron in an atom. a set of four quantum numbers is assigned to each electron in an atom. or s orbital. specify the spin of the electron. With the same principle quantum no. d 3 f a) Energy of electrons in different orbitals i. (n-1) Depends on l. 1. the electron in different orbital shows difference in energy. d and f. specify the direction of the orbital. 4. (n) determines the principal energy level of the orbital. l = 0.D. (n). The four quantum numbers are Principal quantum number (n) Subsidiary quantum number (l) Magnetic quantum number (m) Spin quantum number (s) specify the principal energy level.. l = 1 and m = 1.II. 2py and 2pz orbitals. Of the available orbitals. the added electron will go into empty orbital. Each orbital may hold only two electrons. keeping electron spins the same. Unit 2 Page 4 Building up of electrons in an atom a) Aufbau principle In building up the electronic configuration of an atom in its ground state. (Hund's rule) According to the Aufbau principle. (Aufbau principle) ii. the following procedure of filling the orbitals shown be observed i. no two electrons can have all four quantum numbers the same. Hund’s rule (Rule of maximum multiplicity) Electrons in degenerate orbitals will occupy each orbital singly before electron pairing takes place. To generalize the three rules. When the atom becomes heavy. Where a number of orbitals of equal energy are available. and they must have opposite spin. (Pauli exclusion principle) iii. . before spin-pairing occurs. the energy of 4s orbital is lower than that of 3d orbital. For example. The order of energy levels need not to be exactly parallel to the principal quantum. the added electron will always occupy the one with the lowest energy. the electrons will fill the available orbitals with lowest energy first. the electrons are placed in the orbitals in order of increasing energy b) Pauli exclusion principle – c) In any atom. the high energy levels tend to overlap each other. Electronic Structure of atoms 5.II. The order of the energy levels mainly depends on the attraction between the orbital and the nucleus. 3p electrons also experience repulsion will Kshell. iii. This is known as shielding / screening effect of inner electrons. electrons in shells of lower principal quantum number are more effective shields than electrons in shells of higher quantum number and s-orbital electrons is more effective in shielding than p orbital electrons. Secondary shielding effect The shielding effect of the electrons/repulsion between electrons in the same orbital. the electrons in an atom experience attraction from the nucleus less than the actual nuclear charge.e. The effective nuclear charge is always less than actual nuclear charge. ii. (because of lower potential energy) This once again depends on the spatial distribution of the orbital. The stronger the attraction. Note : The completely filled shell can be denoted by K. N according to the principal quantum no Shielding effect and effective nuclear charge Owing to shielding effect. Penetration of 3s orbital Spatial distribution of 3s orbital has 2 small peaks close to the nucleus and make it experience more attraction. from the fact that 4s orbital is less energetic than 3p orbital. Distance from the nucleus 3s orbital is closer to the nucleus than 3p orbital. . The effect of penetration of penetration must be more significant than the effect of distance and shielding effect. Spatial distribution of 3s and 3p orbitals Moreover. This is much weaker than primary shielding effect. L. For example.II. i. Shielding / Screening effect Besides attraction with the nucleus. Electronic Structure of atoms Unit 2 Page 5 The order of energy of the orbitals can be predicted by the pattern shown on the left. the lower will be the energy of the orbital. M. L-shell and 3s electrons. 3s orbital has lower energy than 3p orbital because i. This is the shielding effect of the inner electrons on the outer electrons from the nucleus. thus the 3s orbital will experience more attraction. There is two kinds of shielding effect Primary shielding effect This is stronger than the secondary shielding effect. s = +½ m = 0. l = 0. The quantum no. n = 1. of the electron in hydrogen atom in ground state l = 0. 2p m = 0. n = 2. l = 0.II. ii. m = 0. 1s 2s l = 0. s = +½ n = 1. The quantum no. m = 0. n = 2. s = -½ m = 0. l = 1. 1s 1s22s22p2 Detail electronic configuration of some atoms The electronic configuration of argon atom in ground state 2s 2p 3s 3p 1s22s22p63s23p6 0 0 000 0 000 . l = 0. l = 0. n = 2. Unit 2 Page 6 The quantum no. l = 0. 1s 2s 1s22s1 l = 0. 1s 1 electron in box 1s1 Electron with spin quantum no. s = +½ m = 0. n = 2. of the six electrons in carbon atom in ground state 0 0 000 e) i. s = -½ m = 0. of the three electrons in lithium atom in ground state n = 1. s = +½ iv. s = +½ m = 0. 1s l = 0. Electronic Structure of atoms d) Building up of electrons i. s = +½ m = 0. n = 2. s = +½ m = 0. l = 0. s = -½ 2 electron in box 1s2 iii. s = +½ 0 0 n = 1. l = 1. of the two electrons in helium atom in ground state n = 1. s = -½ m = 1. +½ is represented by an upward arrow. n = 1. m = 0. The quantum no. n = 1. ii.E. Cr has the electronic configuration [Ar]3d54s1 Instead of [Ar]3d94s2.. completely/full-filled orbital. of Be and N in period 2.II. This is because i. a half-filled or full-filled d orbitals mean an even distributed electrons in an atom which make the electron repulsions weaker. 1 2 3 4 5 6 7 8 9 10 11 12 or [Ar]3d104s24p2 or K L M 4s24p2 Electronic configuration of first 36 elements H He Li Be B C N O F Ne Na Mg 1s1 1s2 1s22s1 1s22s2 1s22s22p1 1s22s22p2 1s22s22p3 1s22s22p4 1s22s22p5 1s22s22p6 1s22s22p63s1 1s22s22p63s2 13 14 15 16 17 18 19 20 21 22 23 24 Al Si P S Cl Ar K Ca Sc Ti V Cr 1s22s22p63s23p1 1s22s22p63s23p2 1s22s22p63s23p3 1s22s22p63s23p4 1s22s22p63s23p5 1s22s22p63s23p6 1s22s22p63s23p64s1 1s22s22p63s23p64s2 1s22s22p63s23p63d14s2 1s22s22p63s23p63d24s2 1s22s22p63s23p63d34s2 1s22s22p63s23p63d54s1 25 26 27 28 29 30 31 32 33 34 35 36 Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s22s22p63s23p63d54s2 1s22s22p63s23p63d64s2 1s22s22p63s23p63d74s2 1s22s22p63s23p63d84s2 1s22s22p63s23p63d104s1 1s22s22p63s23p63d104s2 1s22s22p63s23p63d104s24p1 1s22s22p63s23p63d104s24p2 1s22s22p63s23p63d104s24p3 1s22s22p63s23p63d104s24p4 1s22s22p63s23p63d104s24p5 1s22s22p63s23p63d104s24p6 a) Extra stability of half-filled and full-filled subshell It can be observed that Cr (Z=24) and Cu (Z=29) do not follow the general trend.4s2 is less favorable than 4s1 as pairing up of the 4s electrons creates greater repulsion between the 2 electrons. The electronic configuration of potassium atom in ground state 1s 2s 2p 3s 3p 4s 1s22s22p63s23p64s1 0 0 000 0 000 0 1s 2s 2p 3s 3p 3d iii. Be has a full-filled 2s orbital and N has a half-filled 2p orbital. The electronic configuration of germanium atom (Z=32) in ground state 1s 2s 2p 3s 3p 3d 4s 4p 0 0 000 0 000 00000 0 000 1s22s22p63s23p63d104s24p2 6. Instead of [Ar]3d44s2. . The electronic configuration of scandium atom (Z=21) in ground state 4s 0 0 000 0 000 00000 0 1s22s22p63s23p63d14s2 Note : or [Ar]3d14s2 or K L 3s23p63d14s2 Writing of electronic configuration follows the principal quantum no. not the order of placing electrons iv. Electronic Structure of atoms Unit 2 Page 7 ii. This also accounts for the anomaly of first I. half-filled orbital and ii. Cu has the electronic configuration [Ar]3d104s1 Extra stability is associated with i. Mn 1s22s22p63s23p63d54s2 f-block elements : Lanthanide and Actinide e. d.g. Na 1s22s22p63s1 p-block elements : Group IIIA to Group VIIA e.g. stability 1 mark Extra stability. 4a The first ionization energies of the second row elements are in the order Li < B < Be < C < O < N < F < Ne.II. p-. Be > B 1s22s2 half-filled electronic configurations.and f. Glossary First ionization energy electron shell/orbital electron sub-shell/sub-orbital quantum number Aufbau principle Pauli exclusion principle Hund's rule degenerate orbitals – orbitals with the same energy penetration shielding / screening effect effective nuclear charge electron in box half-filled orbital full-filled orbital s-block p-block f-block transition metal 90 2B 4 a 92 2A 1 a ii iii 93 1A 2 a v 95 1A 1 a i 96 1A 1 c i ii 97 2A 2 b i ii iii 98 1A 1 a d-block Past Paper Question 95 2B 5 a iv 98 2B 8 a 95 2B 6 a 90 2B 4 a 4 Account for the following observations. p-. the electron is added into the p. 4 . While moving across.block. N 1s22s22p3 d-block elements : Transition metal e. s-block elements : Group IA to Group IIA e.g. the shielding effect involved is only a secondary one 1 mark filled 2s orbital. Increase in effective nuclear charge with full explanation. the periodic table can be divided into different blocks. Pr 1s22s22p63s23p63d104s24p64d104f35s25p6 Across the s-block. no. i. of protons increase 1 mark From Li to Ne.e. Position of s-. half-filled stability 1 mark N > O 1s22s22p3 C Some candidates failed to mention or explain the abnormally high ionization potentials of Be and N.g. electrons are added in the same principal shell. electron is being added into the s sub-shell. Electronic Structure of atoms Unit 2 Page 8 7. d. d and f sub-shell.and f.block elements According to the electronic structure of elements. 95 2B 6 a 6a Write the electronic configuration of a titanium atom in its ground state. Electronic Structure of atoms Unit 2 Page 9 92 2A 1 a ii iii 1a ii Arrange the following elements in order of increasing first ionization energy : F. 1 To which group in the Periodic Table does A belong ? Group III / IIIA / 3 / 3A 1 mark 95 2B 5 a iv 5a The table below lists some properties of the alkali metals.060 520 -3. suggesting the 2nd electron is closer to the nucleus than the first.II.148 403 -2. 3 There is a gradual increase in the successive ionization energies in general because the nucleus becomes effectively more positively-charged as each electron is being lost. 2½ Na < F < Ne 1 mark Na has the least I. 1s22s22p63s23p63d24s2 3d 4s [Ar] 11000 2 (accept any correct form of electronic configuration) 1 mark 2 .and Lshell status and this requires a relatively high I. Standard electrode Melting point / ºC Element Atomic Ionic radius / nm First ionization potential / V radius / nm energy / kJmol-1 Li 0.93 39 Cs 0.216 0. It is for this reason that an electron in Ne requires the highest I.169 374 -2.95 29 iv Explain the trend in the first ionization energy from Li to Cs. 1½ mark iii Describe and explain the variation of the successive ionization energies of potassium. Na(s).shell status. 3rd 6900.157 0. Ne and Na. Cu2+(aq). Explain your order. Al(s). To lose an electron means that the F atom would move further away from the K. CO32-(aq). AlCl3(s).from the gaseous atom of the element.203 0.235 0. The increase is greatest when the 2nd electron are being ionized.133 418 -2. Zn2+(aq) v Which species has the following ionization energies: 1st 500. 1817.095 495 -2.and L. S(s).71 98 K 0.04 180 Na 0. P4O10(s). 4th 9500 kJmol-1? Na(s) 1 mark 1 95 1A 1 a i 1a i The first four successive ionization energies of an element A are 578.92 64 Rb 0. 3 marks 93 1A 2 a v 2a Answer the following questions by choosing in each case one of the species listed below. 2nd 4600.123 0. putting it in the box and giving the relevant equation(s). S2O32-(aq).’s occur when the 10th and 18th electrons are being ionized. because there is a single 3s-electron which is not strongly attracted towards the nucleus and losing it would accomplish the completed K. AlO2-(aq). C iv the effective attractive forces on the electron is greatest in the case of the smallest first atom.E. 2746 and 10813 kJ mol-1 respectively. Similar greater but less abrupt I.E.E.E. The atomic radius increases as the group is descended hence the outer-shell electrons become less strongly held 2 marks and it is easier to remove an e. In fact. This again showed that they did not grasp these concepts well but just produced their answers by citing some terms learnt in their lessons. 97 2A 2 b i ii iii 2b i What is an electron shell in an atom ? ii For each of the electron shells with principal quantum numbers 1. (DO NOT accept explanations in terms of effective nuclear charge. Some candidates incorrectly explained the difference in ionization energies between O and F in terms of electronegativity.shell / has a large atomic size than O. N and O 98 2B 8 a 8a Which of the following. is the ground state electronic configuration of chromium atom ? Explain your answer.of atom increases across a period. ½ mark ½ mark the attraction experienced by the outermost electron is smaller hence S has a smaller I. Explain your arrangement. But they explained their answers rather vaguely. ∴ outermost e. N+.of O is more easily removed than that of F. the effective nuclear charge experienced by the outermost shell electrons of sulphur was greater than that of oxygen. 2 and 3 list the subshells. Effective nuclear charge experienced by outermost e. Many candidates incorrectly attributed the difference in ionization energies to effective nuclear charge and/or screening effect. ½ mark Or. Such answers reflected the fact that many candidates did not fully understand the concepts associated with screening effect and effective nuclear charge. 1 mark C Many candidates pointed out that the outermost shell electron of fluorine experiences a greater effective nuclear charge. 1 F has 1 more proton in the nucleus. Electronic Structure of atoms Unit 2 Page 10 96 1A 1 c i ii 1c Explain why i the first ionization energy of oxygen is greater than that of sulphur. deduct ½ mark for any misconception) C The question was poorly answered.E. 1 S has 1 more e.E. iii Deduce the maximum number of electrons in each of the electron shells in (ii). ½ mark the effect of increase in nuclear charge has outweighed the screening effect of the additional electron hence F has a greater I. (I) or (II). 3d 4s (I) [Ar] (II) [Ar] 6 3 2 11111 1 11110 2 .II. 98 1A 1 a 1a Arrange the following chemical species in the order of increasing ionization enthalpy. ii the first ionization energy of oxygen is smaller than that of fluorine. dxy. 39–40 Chemistry in Context.4 Shape of atomic orbitals III. p and d orbitals s orbital s-orbital is spherical in shape. 78–80 A-Level Chemistry. The three porbitals have the same energy (degenerate). And then. For 1s orbital of hydrogen. Ramsden pg. Syllabus Notes p orbital p-orbital is dump bell in shape. 5th Edition. Shape of s. The three p orbitals are perpendicular to each other. Atomic orbitals In the orbital model of atom. d orbital There are 5 degenerate d-orbitals. The electron density at the nucleus is zero. Glossary probability density degenerate s orbital p orbital d orbital . an electron density diagram is therefore used to described the distribution of electrons. dxz and dyz orbitals lie on xy. dz2 orbital looks like pz orbital but with a ring at the centre. of finding an electron. 72 2. 3rd Edition. Atomic orbital is the region within which there is a high probability. Different orbital have different shape. dx2-y2 orbital lies on xy plate with the lobes are directing to the x and y axes. A.e 90%. size and energy. the electrons are moving in a wave like motion. Thomas Nelson and Sons Ltd.III. the electron density (probability density of finding a electron) increases as the distance from the nucleus increases and reaches a maximum at a distance equal to Bohr’s radius. 4th Edition. 41–42 Physical Chemistry. the electron density decreases as the distance from the nucleus increases. Because of its extremely small mass. pg. xz and yz plate respectively. i. Fillans pg. Atomic orbitals Page 1 Topic Reference Reading III. E. the position of individual electron cannot be located precisely by any phyical mean.N. Atomic orbitals Modern Physical Chemistry ELBS pg. . 1s orbital is spherical in shape. ½ mark Physical Significance ½ mark . 4 C Most candidates were able to give the correct shape of the 1s orbital. 96 1A 1d 1d Sketch the pictorial representation of a p orbital and indicate the location of the nucleus in your diagram. of the 1s-electron. 92 1A 2 c 2c Give the shape of a 1s-orbital and comment on its physical significance. Atomic orbitals Page 2 Past Paper Question 91 2A 3 c 92 1A 2 c 96 1A 1d 91 2A 3 c 3c Give a brief account of the electron density of the 1s orbital of a hydrogen atom. the most simple description of the surface of the sphere represents the probability of finding. the variation is exactly the same in any direction from the nucleus.The probability of finding an electron at various points in space is called an orbital. the region of highest probability of finding the electron. say 90% of the charge distribution is pictorially represented by a single surface along which the probability of finding the electron is constant. (The most probable distance of the electron from the nucleus of the hydrogen atom is equal to the Bohr’s radius. like clouds in reality. At a certain distance from nucleus.They should be thought of as shapes with fuzzy and indistinct edges. they failed to describe the variation of electron density as a function of distance from the nucleus. 2 1 nucleus (or along any axis) Many candidates did not do what the question asked for and drew three p-orbitals instead of one.III. like a cloud. 1 mark C .) The electron density is zero at the nucleus and at infinite distance from the nucleus. ½ mark C Only a handful of candidates were able to explain clearly the meaning of the spherical shape of the 1s-orbital and its physical significance. ½ mark . i.For a give 1s-orbital. 1s-orbital is spherical.e. there is a maximium probability of electron density. say 90 %. however. A majority of the candidates did not make the important point that orbitals should be thought of as shapes with fuzzy and indistinct edges. 4th Edition. 5th Edition. Thomas Nelson and Sons Ltd. resulting a contraction of the electron cloud of the ion. atomic radius will be referring to van der Waals’ radius. The electron in the outermost shell is effectively shielded from the nucleus by the inner electrons. 175–176 2. All electrons go into the same quantum shell and the increase in nuclear charge is not effectively shielded.N. a) Factor affect the size of atomic radius proton to electron ratio (p/e) – An increase in this ratio causes an increase in effective nuclear charge.5 Variation in atomic radii Variation in electronegativity IV. 85–88 A-Level Chemistry. Syllabus Notes If not specified. The ionic radius of a cation is. Effective nuclear charge increases across a period and has a greater attraction on the outermost electrons drawing them closer to the nucleus. of electron shells increases. of course. . However. Periodic Table Page 1 Topic Reference Reading IV. atomic radius is normally referring to covalent radius. Ramsden pg. 37-41 Physical Chemistry. 111–112. for those atoms where formation of chemical bond is impossible. Chemistry in Context. Across a period – Atomic size decreases as atomic number increases. 334–335 Chemistry in Context. Variation in atomic radius Covalent radius – Van der Waals’ radius – Half of the distance between two nuclei in an element where the atoms are chemically combined together. smaller than the atom itself. Half of the distance between two nuclei in an element where the atoms are not chemically combined together. Periodic Table Modern Physical Chemistry ELBS pg. 170–171. pg. E. Fillans pg. 3rd Edition.IV. Down a group – Atomic size increases as no. Periodic Table A. 3rd Edition ELBS pg. 8 Li 1.5 Mg 1. Variation in electronegativity Electronegativity is a measure of tendency of an atom in a stable molecule to attract electrons within a bonds.IV. The scientist Pauling computed a scale of electronegativity according to bond dissociation energies.5 S 2. ionization energy and electron affinity values of an atom.8 Sn 1.0 Sr 1. it is closely related to the effective nuclear charge experienced by the bonding electrons.7 Be 1.8 Ge 1. electronegativity values increase from left to right across each period and decrease down each group. Glossary atomic radius covalent radius van der Waals radius proton to electron ratio electronegativity .8 I 2.8 Rb 0.0 Cl 3.5 Se 2.1 As 2. In Pauling scale.0 Al 1. fluorine.5 In general. 4 is assigned to the most electronegative atom.9 O 3. Periodic Table Page 2 Variation of atomic radii B.5 _ _ N 3. Noble gases do not has any value of electronegativity because they do not form any stable molecule at all.0 Br 2.0 P 2.8 Cs 0.0 Ba 0.0 Sb 1. Therefore.0 Na 0.1 C 2.9 B 2.4 Te 2.2 Ca 1.5 Si 1. Pauling electronegativity values of some elements (The shaded ones are more electronegative than C) H 2.1 F 4.9 K 0. therefore larger the atomic radius.93 39 Cs 0.increase in electron affinity / electronegativity from C to F . 1 .123 0.92 64 Rb 0. Periodic Table Page 3 Past Paper Question 89 2B 4 d ii 92 2A 3 b i ii 95 2B 5 a I 89 2B 4 d ii 4d Comment on the relative oxidizing properties of ii the non-metal C.71 98 K 0.133 418 -2.IV.235 0.157 0.increase in nuclear charge . ½ mark 95 2B 5 a i 5a The table below lists some properties of the alkali metals. O and F.95 29 i Explain why the atomic radii increase from Li to Cs.2p electrons are less effectively shielded .095 495 -2. the 'next' atom has one more shell of electrons around the nucleus.060 520 -3.148 403 -2.decrease in atomic size . 1 mark C as the group is descended. the next atom has another shell of electrons around the nucleus.04 180 Na 0. of electron increases.216 0.F is particularly reactive because of the weak F–F bond due to lone pair-lone pair repulsion 3 1 mark 2 marks 92 2A 3 b i ii 3b i Define the covalent radius of an atom. 1 mark ii State and explain the trends in the covalent radius on going down any group and going across a short period of the 2 periodic table. 1 The covalent radius is defined as on-half the distance between two atoms of the same kind held together by covalent bond. Going down any group of the periodic table the covalent radius increases ½ mark because the elements become larger as the no.203 0. N. Standard electrode Melting point / ºC Element Atomic Ionic radius / nm First ionization potential / V radius / nm energy / kJmol-1 Li 0.169 374 -2. As the group is descended. C<N<O<F . ½ mark Going across any period the covalent radius decreases ½ mark because the nuclear charge increases. Bonding Energy A. B. B. Constant pressure process a) Work done by a system Standard enthalpy change of formation (∆Hfo) 1. C. Indirect determination of ∆Hfo a) Hess’s Law (1) (2) Use of energy level diagram / energy cycle Use of equation D. IV. II. Gibbs free energy change (∆G) A. C. Definition of bond energy 2. Determination of bond energy term (bond enthalpy) Strenght of covalent bond 1. B. Energetics of formation of ionic compounds A. E. Born-Haber cycle a) Determination of lattice energy by Born-Haber cycle 2. Direct determination of ∆Hfo a) Use of simple calorimeter Other enthalpy change (∆H) 1. Relationship between bond length and bond energy 2. C.Energetics I. . Enthalpy Change (∆H) A. Examples of enthalpy change a) Enthalpy of combustion b) Enthalpy of atomization c) Enthalpy of neutralization d) Enthalpy of hydrogenation e) Enthalpy of solution f) Enthalpy of reaction Hess’s Law 1. B. Factors affecting the bond strength Estimation of enthalpy change by bond energy Electron affinity Lattice energy 1. Factors affecting the value of lattice energy Stoichiometry of ionic compounds Limitation of enthalpy change (∆H) Entropy (S) and entropy change (∆S) Gibbs free energy change (∆G) III. Definition of ∆Hfo a) Standard enthalpy change (∆Ho) b) Standard enthalpy change of formation (∆Hfo) c) Standard enthalpy change of formation of an element 2. Definition of a system Definition of enthalpy (H) and enthalpy change (∆H) 1. Bond energy 1. C. 204 – 206 Chemistry in Context. Fillans pg. B. 146–147 3. The absolute enthalpy (H) of a substance depends on i. 99 – 101 A-Level Chemistry. 4th Edition. the flask.I. 3rd Edition. gas syringe and everything inside them is defined as the system.0–3. The energy change associated with a reaction provides another useful tools to investigate the chemistry involved. Definition of enthalpy (H) and enthalpy change (∆H) The enthalpy of a substance. The electronic structure is only the result of deduction from some macroscopic observations (ionization energies and successive ionization energies). Enthalpy change (∆H) can be arbitrarily defined as the heat change (q) at constant pressure. we can measure the enthalpy change (∆H) of a system. the region or particular quantity of matter of interest is defined as the system. Nobody have seen the atom directly. ii. The atmosphere and everything outside the flask and syringe is defined as the surroundings. . Instead. is an indication of the total energy content of the substance. Ramsden pg. 3rd Edition ELBS pg. kinetic energy possessed by the atoms and sub-atomic particles (kinetic energy) The absolute enthalpy (H) of a substance is not measurable since it is still not possible to measure all the interaction between the sub-atomic particles in an atom. 5th Edition. E. Thomas Nelson and Sons Ltd. Enthalpy Change (∆H) Unit 1 Page 1 Topic Reference Reading I. Enthalpy Change (∆H) Chemistry in Context.N. A.1 Definition of a system Enthalpy and enthalpy change I. 165–166 Physical Chemistry. Enthalpy Change (∆H) Unit 1 Syllabus Notes The discovery of detailed electronic structure of atom serves as a very good example in how does a chemist work. The surroundings are defined as everything other than the system. sometimes called its heat content. For example. Definition of a system In the study of energetics. potential energy inherent in the electrical and nuclear interactions of the constituent particles (chemical energy). pg. B.Heat capacity of the system × temperature change of the system (∆T) N. Then the amount of the heat evolved can be calculated from the heat capacity of the system. . Since the quantity of the heat change (q) is difficult to be measured. in order to determine the enthalpy change. so a negative sign is added in the expression. Strictly speaking. it would be more convenient if we assume that all the heat energy evolved in the above example is absorbed by the system.H1 = q Since H1 is larger than H3 in this exothermic reaction. we allow the magnesium and hydrochloric acid to react. the temperature has increased due to heat energy released in the reaction. We measure all the heat transferred between system and surroundings in order to return the system to 25ºC. Constant pressure process Unit 1 Page 2 Consider the reaction between magnesium and hydrochloric acid : Mg ( s ) + 2 HCl( aq ) → MgCl2 ( aq ) + H 2 ( g ) Initial stage 1. ∆H = heat change (q) = . Final stage 3. ∆T is positive but q and ∆H are negative. When the system is cooled down to 25 ºC from 50 ºC. By definition. the amount of heat exchanged (q) between the system and the surrounding have to be determined. some energy is lost to the surrounding in form of heat energy(q). Starting from room temperature (25ºC). this is the enthalpy change for the reaction. Enthalpy Change (∆H) 1. according to the definition. H1 and H2 must be equal.Hinitial = H3 .I. Immediate after reaction 2. ∆H = Hfinal . In an exothermic reaction. it can be seen that ∆H of an exothermic reaction is always negative. We let the system cool back to room temperature (25ºC). Immediately after the reaction. By law of conservation of energy. Vinitial = = = is positive if the volume of the system is increasing. the pressure will no longer be a constant. More heat will evolve in this reaction since the work done against atmospheric pressure is now exhibited in form of heat. the heat change (q) measured would not be the same as the value of ∆H. hydrogen gas evolves.I. Enthalpy Change (∆H) a) Work done by a system Unit 1 Page 3 Upon mixing of magnesium and dilute hydrochloric acid in the flask. The volume of the system increase and the system has done a work against the atmospheric pressure. according to the definition of ∆H. and is negative if the volume of the system is decreasing. Work done is defined as Force(F) × Distance(d) [Pressure(P) × Area(A)] × [Volume change(∆V) ÷ Area(A)] Pressure(P) × Volume change(∆V) P∆V ∆V = which Vfinal . Therefore. Glossary system surrounding enthalpy change (∆H) heat change / transfer work done . If the volume is kept constant in this case. if the pressure of the system is not kept constant. the standard enthalpy change at 298K should be written as ∆Ho(298K). Definition of ∆Hfo a) Standard enthalpy change (∆Ho) The enthalpy change (∆H) for a reaction which occurs under standard conditions is called the standard enthalpy change. it is assumed to be 298K. e. in writing of equation. 3rd Edition. Ramsden pg. by convention. ii. at a specified temperature. in the pure state if they are crystalline substances or liquids ii. a standard state has been arbitrarily defined. 5th Edition.2 Modern Physical Chemistry ELBS pg. Note : The superscript o does not contain the information of the temperature. 2H2(g) + O2(g) → 2H2O(l) ∆H = -484 kJmol-1 ∆H = -572 kJmol-1 It can be seen that the value of enthalpy change depends on the physical state of the reactant and product. All the reactants and products i. Enthalpy Change (∆H) Unit 2 Page 1 Topic Reference Reading I. Thomas Nelson and Sons Ltd. . activated complex infinite dilution (with respect to all solutes present) * t ∞ 1. 4th Edition. i. 2H2(g) + O2(g) → 2H2O(g) ii. Enthalpy Change (∆H) 3. If the temperature is not specified. 215–222 Chemistry in Context. Standard state Unit 2 Syllabus Notes For the basis of comparison and tabulation of data. usually. It is represented by the symbol ∆Ho. the physical state of individual species must be specified. Fillans pg.g. 3rd Edition ELBS pg. E.I. pg. 208 – 210 Chemistry in Context. 165 – 169 Physical Chemistry. To be specific. at 1 atmosphere pressure if they are gases iii. 102 – 104 A-Level Chemistry. 148–150 Standard enthalpy change of formation (∆Hfo) Determination of enthalpy change Other enthalpy change C. but not necessarily. 298K (25ºC) The condition of 1 atm and a specified temperature is called standard condition. Therefore. Standard enthalpy change of formation (∆Hfo) i. Meaning of superscripts o º = = = = = standard pure substance excited electronic state transition state. at a concentration of 1M if they are dissolved in a solvent iv.N. Standard enthalpy of formation of carbon dioxide ∆Hfo(298K) = -393. under standard conditions.g.g. If the heat capacity of the system is known.g. Enthalpy Change (∆H) Unit 2 Page 2 b) Standard enthalpy change of formation (∆Hfo) Standard enthalpy change of formation (∆Hfo)– The enthalpy change which occurs when 1 mole of a compound is formed.3 kJmol-1 2C(graphite) + 2H2(g) → C2H4(g) Standard enthalpy of formation give us a rough ideal on the energetic stability of the compound respect to its constituent elements. there is no change. O2(g) → O2(g). 2. oxygen. e.5 kJmol-1 C(graphite) + O2(g) → CO2(g) ii. For an endothermic reaction. the temperature will fall. Direct determination of ∆Hfo Some of the enthalpy of formation can be determined directly by measuring the heat change in direct synthesis of the compound. ∆Hfo[SO2(g)].I. is the enthalpy change associated with the following change. Actually. the temperature will raise. A more negative standard enthalpy formation means more energetically stable. from its elements in their standard state. ∆Hfo[CO2(g)]. The heat energy evolved or absorbed can be calculated from the temperature change. a) Use of simple calorimeter A simple calorimeter can be used to determine the amount of heat absorbed or released in a reaction. standard enthalpy change of formation of an element. . e. c) Standard enthalpy change of formation of an element According to the definition. the standard enthalpy change of formation of an element is always zero. i. For an exothermic reaction. e. Standard enthalpy of formation of ethane ∆Hfo(298K) = -52. Since the polystyrene cup is an insulator.40ºC. Determination of enthalpy of reaction of Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) An excess of zinc powder was added to 50.10ºC and it rose to 25. ∆H + mcp∆T = 0 where m is mass of the solution cp is the specific heat capacity of the solution under constant pressure ∆T is the temperature change in K v.30 K = -0.18 kJkg-1K-1 × 4. due to reaction (at constant T) + change in heat energy of solution = 0 iv. a graph is plotted and extrapolated to determined the change in temperature.00500 mole = -360 kJ vii. Usually. the temperature was 21. The value -0. . Enthalpy Change (∆H) Unit 2 Page 3 e. but the values of ∆H and ∆Ho would be very close. As there is no lost of energy to the surroundings. iii. the total energy change in the system is zero.899 kJ vi. Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) ∆H = -360kJmol-1 viii. In order to obtain the value for 2 mole of silver ions as specified by the equation. ∆H.899 kJ corresponds to 0.1000 kg × 4. i.18 kJkg-1K-1. 50. Strictly speaking. Initially. All the chemical energy released in the reactions transformed into heat energy which raises the temperature of the solution. Calculate the enthalpy change for the reaction : Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) Assume that the density of the solution is 1. ∆H cannot be written as ∆Ho in this question because the conditions of the experiment were not standard.00500 mole of silver ions. 2 mole The enthalpy change using 2 mole of Ag+ ions = -0.899 kJ × 0. its heat capacity is assumed to be zero and there is no energy exchanged between system and surroundings. ii. ∆T.I. ix.0 ∴ ∆H = -mcp∆T = .g.0 cm3 of 0.00 gcm-3 and its specific heat capacity is 4. Ignore the heat capacity of the metals.100M AgNO3(aq) in a polystyrene cup. ∆Hohydro = -238.g. Enthalpy Change (∆H) Unit 2 Page 4 D. d) Standard enthalpy of hydrogenation ∆Hohydro It is the heat change when 1 mole of an unsaturated compound is converted to saturated compound by reaction with gaseous hydrogen at 1 atm.2 kJmol-1 Neutralization of a weak acid is found to be less exothermic than that of a strong acid because energy is required to ionize the molecules of weak acid first. the value would be the enthalpy change involved in atomization of 1 mole of the compound being concerned. CH2CH=CHCH2(g) + 2H2(g) → CH3CH2CH2CH3(g) e) Standard enthalpy of solution ∆Hosol’n It is the heat change when 1 mole of a substance is dissolved at 1 atm in a stated amount of solvent or infinite amount of solvent.5 kJmol-1 b) Standard enthalpy of atomization ∆Hoat It refers to the formation of 1 mole of gaseous atoms from the element in the defined physical state under standard conditions. ½H2(g) → H(g) ∆Hoat[½H2(g)]= +218 kJmol-1 For a compound. 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) Glossary standard enthalpy of formation standard state calorimeter enthalpy of atomization enthalpy of neutralization enthalpy of combustion . Other standard enthalpy change (∆Ho) 1.6 kJmol-1 e. C(graphite) + O2(g) → CO2(g) ∆Hco[C(graphite)] = -393.g. ∆Hoat[C(graphite)]= +715 kJmol-1 i.1 kJmol-1 e. e.100H O) ∆Hosol’n = -35 kJmol-1 i. LiCl (s) + 100H 2 O (l) → LiCl (aq.g.g.8 kJmol-1 e. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) ∆Honeutralization = -55. e.g. 2 ii. e.I.g. ∆Horxn = -845. CO2(g) → C(g) + 2O(g) c) Standard enthalpy of neutralization ∆Honeutralization It is the heat change when an acid and a base react to form 1 mole of water under standard condition. ∆Honeutralization = -57. C(graphite) → C(g) ii. Examples of enthalpy change a) Standard enthalpy of combustion ∆Hco It refers to the complete combustion of 1 mole of the substance under standard conditions. ∆Hoat[CO2(g)] = +1606 kJmol-1 e. LiCl( s ) → LiCl( aq ) H2O( l ) ∆Hosol’n = -37.2 kJmol-1 f) Standard enthalpy of reaction ∆Horxn It is the heat change in a reaction at 1 atm between the number of moles of reactants shown in the equation for the reaction.g. 2 1 . State THREE sources of error in the result obtained in such an experiment. The enthalpy of neutralization of a strong acid (HCl) by a strong base (NaOH) is -57.2 kJ mol-1 while that of hydrochloric acid is -57. the temperature rise was recorded by a thermometer which also served as a stirrer. 92 2A 1 b i 1b i Define the standard enthalpy of formation of a compound. using CH3OH(l) as an illustration. When a solution of an acid was poured into a solution of an alkali in the calorimeter. a polystyrene foam cup was used as a calorimeter.3 kJ mol-1. e. 99 1B 7 b 7b In an experiment to determine the enthalpy change of neutralization.3 kJ mol-1. The enthalpy of neutralization of ethanoic acid with NaOH is -55. Enthalpy Change (∆H) Unit 2 Page 5 Past Paper Question 91 2A 1 b ii 92 2A 1 b i 99 1B 7 b 91 2A 1 b ii 1b ii The enthalpy of neutralization of ethanoic acid with aqueous sodium hydroxide is -55. Account for the difference in these two values. C(s) + 2H2(g) + ½O2(g) → CH3OH(l) 1 mark C i Many candidates did not include temperature and pressure in their definition of standard enthalpy of formation.2 kJ mol-1 because ethanoic acid is a weak acid and is only 2 marks slightly ionized. and the difference is due to the enthalpy of ionization.I. The standard enthalpy of formation of a compound is the standard enthalpy change that occurs when one mole of the compound is made from its constituent elements under standard conditions (298 K and 1 atmospheric pressure).g. inevitably some C(s) will remain and some CO2(g) will be present. Not all compounds can be synthesized directly from its constituent elements at standard state. e.3 Modern Physical Chemistry ELBS pg.N. The standard enthalpy change of the reaction. Hess’s Law 1.g. 2Na(s) + O2(g) → Na2O2(s) For the ∆Hfo of the compound which cannot be determined directly. 62–75 A-Level Chemistry.I. Presence of side products / side reactions. pg. 5th Edition. Indirect determination of ∆Hfo Unit 3 Syllabus Notes Not ∆Hfo of all compounds can be determined directly. But. Thomas Nelson and Sons Ltd. The enthalpy change for a reaction is constant and is the sum of enthalpy changes for the intermediate steps. law of conservation of energy (Hess’s Law) can be used to determine its value. 150–157 Hess’s Law Applications of Hess’s Law E. 4Na(s) + O2(g) → 2Na2O(s). 222–226 Chemistry in Context. e. CH4(g) ii. Ramsden pg. Fillans pg. This is because i. a) Hess’s Law Hess's Law The change in enthalpy accompanying a chemical reaction is independent of the pathway between the initial and final states. the following 2 data can be determined experimentally by calorimetry. ∆Ho = -394 kJmol-1 C(s) + O2(g) → CO2(g) CO(g) + ½O2(g) → CO2(g) ∆Ho = -283 kJmol-1 . 170–174 Physical Chemistry. NH3(g) iii. cannot be determined directly since the extent of the reaction is difficult to control.g. 210 – 213 Chemistry in Context. C(s) + ½O2(g) → CO(g). E. the extent of the reaction is difficult to control. In some reactions. 3rd Edition. Enthalpy Change (∆H) Unit 3 Page 1 Topic Reference Reading I. 3rd Edition ELBS pg. Enthalpy Change (∆H) 3. (-283 kJmol-1) = -111 kJmol-1 ii. Write down the common reference to which both reactant and product are related. constituent elements in their standard states iii. 4. an energy level diagram may be used. . Steps of drawing an energy level diagram 1.I. 3. 3. 1 mole of CO2(g).∆H3 ∴ ∆Hfo[CO(g)] = ∆H2 = -394 kJmol-1 . there are three popular common reference used in construction of energy cycles. Write down an equation according to the definition of the problem. the heat change accompanying Route 1 should equal to that of Route 2. Energy Level Diagram To represent the relative enthalpy of the substance involved. The combustion of CO(g) gives the same product. constituent atoms N.B. Enthalpy Change (∆H) Unit 3 Page 2 (1) Use of energy level diagram / energy cycle i. The combustion of 1 mole of graphite produces CO2(g) and releases 394 kJ. combustion products ii. The gap represents the heat of formation of CO(g). The energy released is 283 kJ. 2. ∴ ∆H1 = ∆H2 + ∆H3 ∆H2 = ∆H1 . it is CO2(g). Complete the cycle / diagram. Draw a line to represent the enthalpy of 1 mole of graphite and 1 mole of oxygen at standard condition. They are i. 2. Steps of constructing energy cycle / energy level diagram 1. In general. Energy Cycle By Hess's Law. In the above example. The generalized statement of this approach in the calculation of ∆Horxn is ∆Horxn = Σ∆Hfo(products) . Reverse the second equation and the sign of ∆Hfo.g. Enthalpy Change (∆H) (2) Use of equation Unit 3 Page 3 The standard heat of reaction can also be determined if the standard heat of formation of the reactants and products are given. ∆Hfo = -394 kJmol-1 C(s) + O2(g) → CO2(g) CO2(g) → CO(g) + ½O2(g) ∆Hfo = +283 kJmol-1 --------------------------------------------------------------------------------------------C(s) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½O2(g) --------------------------------------------------------------------------------------------∆Hfo = -111 kJmol-1 C(s) + ½O2(g) → CO(g) Glossary Hess’s Law energy level diagram energy cycle . by doing some manipulations on the equations. For the reaction. ∆Hfo of any element. C(s) + ½O2(g) → CO(g). O2. C(s) + O2(g) → CO2(g) CO(g) + ½O2(g) → CO2(g) ∆Hfo = -394 kJmol-1 ∆Hfo = -283 kJmol-1 If you want to determine the heat of formation of CO(g). CO(g) + ½O2(g) → CO2(g).{∆Hfo[CO(g)] + ∆Hfo[½O2(g)]} = (-394 kJmol-1) . is zero.I. ii.(-283 kJmol-1) = -111 kJmol-1 By definition.Σ∆Hfo(reactants) e. e.(∆Hfo[CO(g)] + 0 kJmol-1) = (-394 kJmol-1) . following method may be used. C(s) + O2(g) → CO2(g) CO(g) + ½O2(g) → CO2(g) ∆Hfo = -394 kJmol-1 ∆Hfo = -283 kJmol-1 i.Σ∆Hfo(reactants) = {∆Hfo[CO2(g)]} . Add the two equations together. Or simply.g. ∆Horxn = -283 kJmol-1 ∆Horxn -283 kJmol-1 -283 kJmol-1 ∆Hfo[CO(g)] Note : = Σ∆Hfo(products) . i Give equations of suitable reactions for this determination. Enthalpy Change (∆H) Unit 3 91 2A 1 b i Page 4 Past Paper Question 90 1A 2 b 91 1B 4 b i ii iii 92 2A 1 b ii 93 2A 3 b i ii 94 2A 2 a i ii 95 2A 1 b 97 2A 2 c 98 2A 2 c i ii 90 1A 2 b 2b Show how ∆Hof for CuSO4·5H2O(s) can be determined using the following data: ∆Hosoln CuSO4·5H2O(s) = + 8 kJ mol-1 . inappropriate acid e. Many errors arose as a result of a poorly constructed Born-Haber cycle. Many incorrect explanations like "heat loss to surroundings" and "large heat of formation" were given for (iii).773 kJ mol-1 -1 ∆Hosoln CuSO4(s) = .286 kJ mol-1 aq + Cu(s) + S(s) + 9/2 O2(g) + 5H2(g) ∆Hf CuSO4(s) + 5 ? ∆Hf H2O(l) aq + CuSO4(s) + 5H2O(l) ∆Hsoln CuSO4(s) ∆Hf CuSO4?H 2O(s) CuSO4?H 2O(s) + aq ∆Hsoln CuSO4?H 2O(s) CuSO4(aq) + 5H2O(l) 3 C . stirrer. many failed to give balanced coefficients and states of the species. 2 2 2 . 3 Ca(s) + C(s) + O2(g) → CaCO3(s) 2 Side reactions: 2Ca(s) + O2(g) → 2CaO(s). numeric and units) Deduct ½ mark for wrong units. insulation. temperature bath.directly combustion of Ca can be violent . calcium carbonate. the enthalpies of formation of carbon dioxide(g) and water(l) are available. Calorimeter (beaker or plastic cup. and water.side reactions .66 kJ mol . CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) 2 marks (missing or incorrect physical state -½. 91 1B 4 b i ii iii 4b You are required to determine indirectly the enthalpy of formation of calcium carbonate(s). Very few candidates used the more analytical approach of setting up equations. C(s) + O2(g) → CO2(g) 2 marks for any one explanation: . You are give the reagents: calcium.286 kJ aq + CuSO4(s) → CuSO4(aq) . ∆Hof H2O(l) = . Only about half of the candidates did this numerical problem correctly.extent of reaction cannot be controlled .773 kJ Cu(s) + S(s) + 2O2(g) → CuSO4(s) 5H2(g) + 2½O2(g) → 5H2O(l) 5 × . Of those who gave the equations with correct species. H2SO4 and HNO3 -½) ii Name the major pieces of apparatus needed for this determination. burette or pipette ½ mark balance ½ mark iii Suggest one experimental difficulty in the direct determination of the enthalpy of formation of calcium carbonate from its elements.g. Most candidates scored marks by pointing out a very common piece of equipment and giving a partial explanation of the experimental difficulty. ∆Hof CuSO4(s) = .8 = -2277 kJ mol-1 2 marks for working (cycle/equations and expressions) 1 mark for correct answer (sign.cannot react or cannot react to form CaCO3 (1 mark only) C This section showed that the candidates were weak in applying general chemical knowledge to experiments.66 kJ CuSO4(aq) + 5H2O(l) → CuSO4·5H2O(s) + aq -8 kJ ------------------------------------------------------------------------------------------Cu(s) + S(s) + 5H2(g) + 4½O2(g) → CuSO4·5H2O(s) ∴∆Hfo for CuSO4·5H2O(s) = (-773) + 5(-286) + (-66) . wrong coefficient -½. electric heater) ½ mark thermometer or thermocouple ½ mark measuring cylinder. a strong acid.I. 6 kJ mol-1 Na(s) + ½O2(g) + ½H2(g) + aq → NaOH(aq) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ∆Ho = -57.3 H2(g) + ½O2(g) → H2O(l) -285.9 kJ mol-1 1 mark ∆Ho = (-425.(2) ∆(g) + H2(g) → CH3CH2CH3(g) 1 mark ∆H = -2091 + (-285. hence propene is more stable because in the structure of cyclopropane there is squeezing of bond angles / ring is highly strained / poor overlapping of atomic orbital.5 kJ mol-1 missing sign(-1).9) + (-164.9) = -164.(-2220) = -156.(∆Hfo(CH3OH) + 0) ∴ ∆Hfo(CH3OH) = -238.8)) . Comment on the relative stabilities of cyclopropane and propene.3 + (-71.6 NaCl(s) + aq → NaCl(aq) +3.9 Na(s) + ½O2(g) + ½H2(g) + aq → NaOH(aq) -425. ∆Ho = -726. Some weaker candidates did not even give state symbols in the chemical equations. resulting in an exothermic reaction.6 kJmol-1 CH3OH(l) + 3 2 O2(g) → CO2(g) + 2H2O(l) ∆Ho = Σnp(∆Hfo)p .3 kJ mol-1 H2O(l) → H2(g) + ½O2(g) ∆Ho = +285. 1 mark Weaker candidates produced wrong answers because they used the wrong signs and some could not correlate the fact that the cyclopropane molecule was highly strained.(-2058) = -33 kJmol 1 mark 1 mark The conversion from cyclopropane to propene is exothermic.8 Calculate the enthalpy change involved in the hydrogenation of cyclopropane to propane.6 = (-393.3 HCl(g) + aq → HCl(aq) -71. and that it would release the excess energy upon conversion to propene.Σnr(∆Hfo)r ∆Ho = (∆Hfo(CO2) + 2∆Hfo(H2O) .8) .8 kJmol-1 Standard enthalpy of formation of H2O(l) Calculate the standard enthalpy of formation of CH3OH(l) at 298K.8 kJmol-1 1 mark (1) + (3) .9 ½H2(g) + ½Cl2(g) → HCl(g) -92.9 ∆Ho = -425.(∆Hfo(CH3OH) + 3 2 ∆Hfo(O2)) Page 5 4 3 Since ∆Hfo(O2) is zero.5 + 2 × (-285. 9 (1) ∆(g) + 2 O2(g) → 3H2O(l) + 3CO2(g) ∆H = -2091 kJmol-1 (2) CH3CH2CH3(g) + 5O2(g) → 4H2O(l) + 3CO2(g) ∆H = -2220 kJmol-1 (3) H2(g) + ½O2(g) → H2O(l) ∆H = -285.2) + (-3.(4) ∆(g) → H2C=CH–CH3(g) 1 mark -1 ∆H = -2091 . Enthalpy Change (∆H) Unit 3 91 2A 1 b ii 1b i Calculate the enthalpy of formation of NaCl(s) from the following data : Reaction ∆Ho/kJmol-1 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) -57.8 kJmol-1 2 marks Calculate the enthalpy change involved in the conversion of cyclopropane to propene. 4 4 .1 kJ mol-1 (no unit -½) 3 marks C Some candidates were not able to give a correct energy cycle and hence failed to obtain the correct answer.3) + (+285. 9 (4) H2C=CH–CH3(g) + 2 O2(g) → 3CO2(g) + 3H2O(l) ∆H = -2058 kJmol-1 (1) .9) = -365.2 kJ mol-1 NaCl(aq) → NaCl(s) + aq ∆Ho = -3. ∴ -726.6 kJmol-1 -393.6) + (-57. missing unit(-½) 93 2A 3 b i ii 3b Given the following thermochemical data at 298 K: -1 -1 Compound ∆Hocombustion /kJ mol ∆Hoformation /kJ mol correct steps 2 marks correct answer 1 mark i ii C cyclopropane (g) -2091 — propene (g) -2058 — propane (g) -2220 — water (l) — -285.5 kJmol-1 Standard enthalpy of formation of CO2(g) -285. 92 2A 1 b ii 1b ii Given the following thermochemical data at 298K : Standard enthalpy of combustion of CH3OH(l) -726.9 kJ mol-1 ½H2(g) + ½Cl2(g) → HCl(aq) ∆Ho = -92.I. 56) .9 kJ mol-1 calculate the standard enthalpy change of formation of CH3CH2OH(l) at 298 K.(+105.3 kJ mol 1 mark C Many candidates gave +88.5 H2(g) + ½O2(g) →H2O(l) .890.3 kJ mol-1 –––––––––––––––––––––––––––––––––––––––––––––––––––––– ∆H = + 88.7 kJ mol-1 CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) .5 + 2(-285. Reaction ∆Ho98 / kJ mol-1 2 C (graphite) + 2H2(g) → CH4(g) .I.07 .890. ∆H = .) C Some candidates could not construct the energy cycle correctly.75.(-75. Mistakes in the numerical calculation were common. 1 mark The reaction AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) is equivalent to Ag+(aq) + Cl-(aq) → AgCl(s) ∆H = Σ∆Hf(products) .6 kJ mol-1 1 mark 2H2O(l) → 2H2O(g) -1 ∆Hovaporization(H2O) = + 44.801. 97 2A 2 c 2c Given the following thermochemical data at 298 K: = -393.9) .Σ∆Hf(reactants) 1 mark = -127.15) 2 mark = -65.7 kJ mol-1 for the following reaction.6 kJ.15 2 Cl2(g) + aq + e → Cl (aq) 1 Ag(s) + 2 Cl2(g) → AgCl(s) -127.3 kJ mol-1 for the enthalpy change of vaporization of water.48 kJmol-1 (Accept any correct method using an energy cycle.8 kJ mol-1 Standard enthalpy change of formation of H2O(l) Standard enthalpy change of combustion of CH3CH2OH(l) = -1336.(-167.285.393. 2 2 marks ∆Ho98 = -393.801.5 kJ mol-1 Standard enthalpy change of formation of CO2(g) = -285.56 1 3 -207. 2 CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) Calculate the enthalpy change of vaporization of water at 298K. instead of + 44.36 2 N2(g) + 2 O2(g) + aq + e → NO3 (aq) 1 -167.3 kJmol-1 2 ii The enthalpy change ∆Ho98 is .0 C (graphite) + O2(g) → CO2(g) .07 Calculate the standard enthalpy change for the reaction AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq). Page 6 2 2 4 4 .CH4(g) + 2O2(g) → 2H2O(l) + CO2(g) ∆H = .9 i Calculate the enthalpy change ∆Ho98 for the reaction CH4(g) + 2O2(g) → 2H2O(l) + CO2(g). Enthalpy Change (∆H) Unit 3 94 2A 2 a i ii 2a Given the following thermochemical data.0) = . 95 2A 1 b 1b You are provided with the following thermochemical data: Reaction ∆Ho298/kJmol-1 + Ag(s) + aq → Ag (aq) + e +105. I. Enthalpy Change (∆H) Unit 3 98 2A 2 c i ii 2c Both H2(g) and CH3OH(l) are possible fuels for powering rockets. Page 7 i ii . Deduce which of the above two fuels is more effective in powering rockets. H2(g) + ½O2(g) → H2O(g) CH3OH(l) + 1½O2(g) → CO2(g) + 2H2O(g) Compound molar mass / g ∆Hfo / kJmol-1 44 -394 CO2(g) 18 -242 H2O(g) 32 -239 CH3OH(l) For each of the above reactions. Their combustion reactions are shown below. calculate the enthalpy change at 298 K per kg of the fuel-oxygen mixture in the mole ratio as indicated in the stoichiometric equation The effectiveness of a fuel can be estimated by dividing the enthalpy change per kg of the fuel-oxygen mixture in its combustion reaction by the average molar mass of the product(s) in g. bond energy term does not refer to one particular bond but it is the average bond energy of a particular type of bond. e. it is important to specify the process involved. Not all C–H bonds are equivalent. it is half of D.II. 5th Edition. D: The energy required to break one mole of a particular covalent bonds in molecules.g. Thomas Nelson and Sons Ltd. Usually. exothermic and bond breaking process absorbs energy. Definition of bond energy In discussion of bond energy. the bond formation process evolves energy. For example. E(X–Y): 2 H C H H H H 1 C OH By convention. E. pg. Bond Energy Terms. endothermic. 157–159 Definition of bond enthalpies Determination of bond enthalpies II.e. Cl–Cl bond in Cl2 molecule In contract. 3rd Edition. for a diatomic molecule. ii. two definitions have to be made clear. 214 – 215 Chemistry in Context. Otherwise. Unit 1 Syllabus Notes Energy change always involves with bond formation and bond breaking. The standard enthalpy change for the reaction. Bond energy 1. 174–180 Physical Chemistry. in CH4(g) = +1646 kJmol-1 ÷ 4 = 413 kJmol-1 In contrast with the bond dissociation energy. e. is an average value of bond dissociation energies for a given type of bond. . 3rd Edition ELBS pg. Bond Dissociation Energy.N. Enthalpy changes for four successive dissociation reactions : Reaction CH4(g) → CH3(g) + H(g) CH3(g) → CH2(g) + H(g) CH2(g) → CH(g) + H(g) CH(g) → C(g) + H(g) ∆Ho(298K) kJmol-1 +425 +470 +416 +335 Since dissociation energy refers to a particular bond. i. Bonding Energy Unit 1 Page 1 Topic Reference Reading II. C–H bond at position 1 is stronger than the bond at position 2. E(C–H). The presence of the electronegative oxygen makes the C–H bonds become more polar and stronger. i. the bond will be weak or will not form at all.1 Modern Physical Chemistry ELBS pg. CH4(g) → C(g) + 4H(g). Ramsden pg.g. even in the same molecule. A. if not specified. ∆Hoat[CH4(g)] = (+425) + (+470) + (416) + (+335) = +1646 kJmol-1 The bond energy terms.e. Also known as average bond energy or bond enthalpy. enthalpy of atomization is the energy required to produce one mole of gaseous atom. 76–83 A-Level Chemistry.4. Bonding is just a balance between attractions and repulsions. If the attractions between atoms are stronger that the repulsions. Bonding Energy This is the principles of energetics and bonding energy again. Fillans pg. the bond will form. 229–233 Chemistry in Context. i. bond energy is referring to bond energy term (average bond energy). E(C–H) is the average value of bond dissociation energies of C-H bond in all kinds of hydrocarbon. Bonding Energy 3. thermochemical methods (by Hess’s Law) ii. Stage 1 Atomization of graphite ∆Hoat[C(graphite)] = +715 kJmol-1 C(graphite) → C(g) Stage 2 Atomization of oxygen ∆Hoat[O2(g)] = +498 kJmol-1 O2(g) → 2O(g) Stage 3 Enthalpy of formation of carbon dioxide ∆Hfo[CO2(g)] = -393 kJmol-1 C(graphite) + O2(g) → CO2(g) Stage 4 Enthalpy of atomization of carbon dioxide ∆Hoat[CO2(g)] CO2(g) → C(g) + 2O(g) By Hess’s Law ∆Hoat[CO2(g)] = . ∆Hoat[CO2(g)] The enthalpy changes of the first three stages can be determined experimentally. spectroscopic methods (not required) iii.(-393 kJmol-1) + 715 kJmol-1 + 498 kJmol-1 = +1606 kJmol-1 E(C=O) = ∆Hoat[CO2(g)] ÷ 2 = +1606 kJmol-1 ÷ 2 = +803 kJmol-1 Glossary bond energy bond energy terms (average bond energy) enthalpy of atomization bond dissociation energy Past Paper Question . Bonding Energy Unit 1 Bond C–H C–C C–Cl E(X-Y) kJmol-1 413 347 346 Page 2 After the collection of a number of bond dissociation energies for a particular type of bond. Determination of bond energy term (bond enthalpy) Bond energy term can be determined by i. the bond energy term can be determined.∆Hfo[CO2(g)] + ∆Hoat[C(graphite)] + ∆Hoatm[O2(g)] = . 2.II. electron impact methods (not required) Determination of E(C=O) in CO2 by using an energy level diagram E(C=O) should be half of the enthalpy of reaction of CO2(g) → C(g) + 2O(g). 3rd Edition ELBS pg.141 0. Therefore.161 Bond energy +568 kJmol-1 +432 kJmol-1 +366 kJmol-1 +298 kJmol-1 ii. the bond strength is inversely proportional to the bond length. the bond would be very strong. the closer will be the nuclei drawn together. bond polarity i.154 0. number of bonding electrons / bond order Bond Bond energy / kJmol-1 C–C +346 C=C +610 C≡C +837 N–N +163 N=N +410 N≡N +945 . number of bonding electrons / bond order iii.092 0. Ramsden pg. Therefore. atomic size ii. 229–233 Chemistry in Context. Strength of covalent bond Factors affecting the bond strength B. this is because of the repulsion between lone pairs on the two atoms which weakens the bond.228 I–I 0. the bond energy increases with increasing no. the more energy will be required to break the bond.4. owing to the small size of F. Bonding Energy Unit 2 Page 1 Topic Reference Reading II.II. Bond C–C C=C 2.199 Br–Br 0. C–C bond.127 0. 76–83 A-Level Chemistry. For identical atoms. Bonding Energy 3. However if F is bonded to an atom with no lone pair. Bond F–H Cl–H Br–H I–H Bond length/nm 0. 3rd Edition. e. Bond length / nm 0.142 Cl–Cl 0. atomic size Bond Bond length/nm F–F 0. Fluorine shows anomaly in group VII. Fillans pg.N. Relationship between bond length and bond energy Unit 2 Syllabus Notes The bond strength of a bond can be measured by the bond energy.3 Modern Physical Chemistry ELBS pg. 174–180 Physical Chemistry.4.g. the strength of the bond decreases with increasing atomic size. of bonding electron. The stronger the bond. The stronger the bond. For the same kind of bond. This explains why F four a larger variety of compounds that the other elements because the compound formed is energetically more stable than the others. the bonding electrons are further from the nucleus and the shielding effect of the electron is getting more significant.2–3. Strength of covalent bond 1. bond strength can be served as a indicator of bond strength.134 Bond energy +347 kJmol-1 +612 kJmol-1 Factors affecting the bond strength i.266 Bond energy +158 kJmol-1 +242 kJmol-1 +193 kJmol-1 +151 kJmol-1 As the size of the atoms gets bigger. E. 95 2B 4 a ii 4a Explain the following facts: ii The bond dissociation energy of F2 is less than that of Cl2.8 +430 H–Br -36. 1 mark Explanation : The H–X bond lengths increase as the atomic radii of the halogens increase. Some candidates thought that the Cl–Cl bond is stronger than the Br–Br bond because 'chlorine is more electronegative' or because 'the effective nuclear charge of Br is smaller'. N2O4 d 2NO2 H2O2 → 2HO· or homolytic bond fission/state the products weak N–N and O–O bond strengths due to lone pair-lone pair repulsions in O2 and long N–N bond in N2 3 1 mark 1 mark 1 mark 91 2B 4 c i 4c i Why is the Cl-Cl bond stronger than the F-F or Br-Br bonds? 2 We expect that the X-X bond energy decreases on descending the group because the distance between the nuclear protons and the bond-pair electron increases. 98 1A 1 b ii 1b ii Explain why the carbon-oxygen bond lengths in CO and CO2 are different. 2 3 2 . or to 'electron repulsion' and 'lone-pair repulsion'.N.1 +298 i Explain briefly the trend in the bond dissociation energy of the hydrogen halides. 2 marks C Many candidates did not know that the small bond dissociation energy of F2 is due to the strong repulsion of lone electron pairs of the fluorine atoms in the F2 molecule. bond polarity Glossary Past Paper Question bond length 90 2B 4 c 91 2B 4 c i 95 2B 4 a ii 98 1A 1 b ii bond order molecular orbital theory 95 2B 4 c I 90 2B 4 c 4 Account for the following observations. 2 marks C Some candidates did not know that the H–X bond lengths increase as the atomic radii of the halogens increase from F to I and that the longer the bond. Unusually short F–F distance leads to high repulsion due to lone pair electrons of fluorine atoms. The gives the bond certain ionic attraction on top of covalent bond. so the attractive force decreases. 4c Dinitrogen tetraoxide and hydrogen peroxide are both unstable to heat.8 +367 H–I +26. This is so for Cl2 and Br2. the weaker it is and the bond dissociation energy is smaller. the weaker it is and the smaller is the bond dissociation energy. 95 2B 4 c i 4c Consider the data given below for the hydrogen halides and answer the questions that follow.9 Unit 2 Page 2 Bond energy +388 +463 +562 Polarization of the bond gives the bond certain ionic character. N–H 0. 1 mark C i The abnormally weak F–F bond strength was attributed to 'internuclear repulsion'. bond polarity Bond Difference in E.4 +562 H–Cl -92.9 O–H 1. Two points were required for the second part of the answer : a mention of the change in distance between nuclear protons and shared (bond-pair) electrons.II. Therefore F–F bond is weaker than expected. to 'the small size of the fluorine atom'. Bond dissociation energy / kJmol-1 Standard enthalpy change of formation / kJmol-1 H–F -269. and the consequent change in attractive force. and few answers cited non-bonding electron repulsion between the two fluorine atoms.4 F–H 1. 1 mark The F-F bond is weaker than that of Cl2 due to non-bonding electron repulsion between lone pairs. Trend : Bond dissociation energy decreases with increasing molecular mass of hydrogen halides. The longer the bond. on the F nuclei. Bonding Energy iii. Estimation of Enthalpy of formation of methane.8 kJmol-1 -890. Since energy is required in bond breaking and energy is released in bond formation. Bonding Energy Unit 3 Page 1 Topic Reference Reading II.5 kJmol-1 -285.3 kJmol-1 ∆Hfo[CH4(g)] = (∆Hco[C(graphite)] + 2 × ∆Hco[H2(g)]) . 157–158 Estimation of bond enthalpies and bond length C.8)) . Fillans pg. 3rd Edition ELBS pg. Bonding Energy 3. This is because bond energy terms is only the average bond energy and not the actual value of the bond being concerned. Therefore. 76–83 Chemistry in Context.8 kJmol-1 It can be seen that the value estimated by bond energy terms differs from the value determined by Hess’s Law. 5th Edition. pg. 229–233 Chemistry in Context.(-890.∆Hco[CH4(g)] = (-393. Estimation of enthalpy change by bond energy Unit 3 Syllabus Notes With the atomization energies and Hess's Law. . 174–180 Physical Chemistry.4 Modern Physical Chemistry ELBS pg.II. the value from Hess’s Law is always preferred. Thomas Nelson and Sons Ltd. bond energy terms can be used for estimation of enthalpy changes. the overall enthalpy can be estimated by the equation.(4 × 413) = -65 kJmol-1 Value determination by Hess’s Law ∆Hco[C(graphite)] ∆Hco[H2(g)] ∆Hco[CH4(g)] -393. CH4(g) E(C–H) ∆Hoat[C(graphite)] ∆Hoat[½H2(g)] +413 kJmol-1 +715 kJmol-1 +218 kJmol-1 Value estimation by bond energy terms ∆Hfo[CH4(g)] = energy required .3) = -74.energy released = ΣE(bond broken) . i.4.5 + 2 × (-285.energy released = (∆Hoat[C(graphite)] + 4 × ∆Hoat[½H2(g)]) .ΣE(bond formed).(4 × E(C–H)) = (715 + 4 × 218) . ∆H = energy required . 0 kJmol-1 ∆H = energy required .4) .7 kJmol-1 (experimental value 184.4(413) + 2(611) + 2(436) (1 mark) 1 mark = -250 kJmol-1 (0 marks for omitting the eve sign. H2C=CH–CH=CH2(g) + 2H2(g) → CH3CH2CH2CH3(g) the experimental molar enthalpy change is -239 kJmol-1.0) = -184. by bond energy terms E(C–H) E(C–C) E(C–Cl) ∆Hoat[C(graphite)] ∆Hoat[½H2(g)] ∆Hoat[½Cl2(g)] +413 kJmol-1 +347 kJmol-1 +346 kJmol-1 +715 kJmol-1 +218 kJmol-1 +122 kJmol-1 ∆Hfo[CH3CH2Cl(g)] = energy required .energy released = ΣE(bond broken) .energy released = (2 × ∆Hoat[C(graphite)] + 5 × ∆Hoat[½H2(g)] + ∆Hoat[½Cl2(g)]) .(2 × E(H–Cl)) = (435. i Estimate the molar enthalpy change for the above hydrogenation using the bond energy terms below : 2 Bond Bond energy term / kJmol-1 H–H 436 C–H 413 C–C 346 C=C 611 1 mark ∆H = -2EC–C . deduct ½ mark for no units) .(5 × E(C–H) + E(C–C) + E(C–Cl)) = (2 × 715 + 5 × 218 + 122) .3-diene. Bonding Energy Unit 3 Page 2 ii.II. H–H(g) + Cl–Cl(g) → 2H–Cl(g) E(H–H) E(Cl–Cl) E(H–Cl) +435.4 kJmol-1 +432.4EC–H + 2EC–C + 2EH–H or = -2(346) .(5 × 413 + 347 + 346) = -116 kJmol-1 iii.9 + 243. Estimation of Enthalpy of formation of 1-chloroethane.9 kJmol-1 +243.(2 × 432.6 kJmol-1) Glossary Past Paper Question 96 2A 1 a i 96 2A 1 a i ii 1a For the hydrogenation of buta-1. CH3CH2Cl(g).ΣE(bond formed) = (E(H–H) + E(Cl–Cl)) . Use of bond energy term to predict the heat of reaction For the reaction. i. 69–70.5. 85–86.will expand and the incoming electron will occupy a position further from the nucleus which weakens the attraction with the nucleus. The value of lattice energy is directly proportional to the strength of the ionic bond.e. Furthermore. ∆Holat = -788. the electron affinity of O..1–3. it must have a negative value. Fillans pg. 5th Edition. 74.N. Electron affinity Electron affinity. Energetics of formation of ionic compounds A. A(g) + e. i. 226–228 Chemistry in Context.e. 1. Energetics of formation of ionic compounds Page 1 Topic Reference Reading III. . Born-Haber cycle Born-Harber cycle – A thermodynamic cycle derived by application of Hess’s Law. ∆Holat – the enthalpy change which occurs when one mole of the ionic compound is formed.III. pg. Energetics of formation of ionic compounds 3. 216 – 218 Chemistry in Context. 160–162 Lattice energy Electron affinity and magnitude of electron affinity Determination of lattice energy by Born-Haber cycle III. 180–182 Physical Chemistry. Na+(g) + Cl-(g) → NaCl(s) Since lattice energy always associates with bond formation. B. we construct an energy diagram called BornHaber cycle to determine the lattice energy. i.→ O-(g) ∆Heo = -142 kJmol-1 i.(or second electron affinity of O) O-(g) + e.5 kJmol-1 e. O(g) + e. ∆Heo or EA – the enthalpy change which occurs when one mole of electrons are added to one mole of neutral atom in the gaseous state to form one mole of gaseous ions under standard conditions.e. Thomas Nelson and Sons Ltd. The second EA of O is endothermic because in the presence of the extra electron. E. 3rd Edition. the electron affinity of O. the extra electron charge will offer a stronger repulsion with the incoming electron. as a crystal lattice.e. exothermic. Lattice energy Lattice energy. Ramsden pg.g. the electron cloud of O.2 Modern Physical Chemistry ELBS pg. 84–90 A-Level Chemistry. Commonly used to calculate lattice energies of ionic solids and average bond energies of covalent compound. Syllabus Notes The value of the electron affinity is depending on the attraction between the incoming electron and the nucleus and the shielding effect offered by the existing electrons.→ O2-(g) ∆Heo = +791 kJmol-1 The first EA of O is exothermic because the attraction between the incoming electron and the nucleus is atronger than the repulsion between the incoming electron and the existing electrons.→ A-(g).5. from its constituent gaseous ions. The lattice energy cannot be determined directly by experiment. 3rd Edition ELBS pg. : radii of cation and anion . r . The charge density of the constituent ions – higher the charge density of the ion. Factors affecting the value of lattice energy Note: ∆Hio – ionization energy. ∆Heo – electron affinity. Energetics of formation of ionic compounds a) Determination of lattice energy by Born-Haber cycle Generalized Born-Haber cycle for an ionic compound Detail Born Haber cycle of sodium chloride Page 2 Transformation of elements in their standard states to gaseous ions.A. ∆H1 ∆Hoat[Na(s)] = +108 kJmol-1 ∆Hio[Na(g)] = +500 kJmol-1 ∆Hoat[½Cl2(g)] = +121 kJmol-1 ∆Heo[Cl(g)] = -364 kJmol-1 ∆H1 = ∆Hoat[Na(s)] + ∆Hio[Na(g)] + ∆Hoat[½Cl2(g)] + ∆Heo[Cl(g)] = 108 + 500 + 121 + (-364) = 365 kJmol-1 Standard heat of formation of the ionic crystal lattice. ∆Holat = ∆Holat[NaCl(s)] Finally. The magnitude of the lattice energy depends on i. ∆H2 ∆H2 = ∆Hof[NaCl(s)] = -411 kJmol-1 Lattice energy. The packing efficiency of the crystal lattice – small and similarity in size of ions result in better packing and more negative lattice energy ii. stronger will be the electrostatic attraction and the lattice energy will be more negative.III. I. by Hess’s Law ∆H1 + ∆Holat[NaCl(s)] = ∆H2 ∆Holat[NaCl(s)] = ∆H2 − ∆H1 = (-411) .E. 1 Lattice energy ∝ r + r + r +.(+365) = -776 kJmol-1 3. E. Born-Haber cycle for Ca+Cl-. By summing up the terms. Stoichiometry of ionic compounds Page 3 The calculation of the lattice energies and use of Born-Haber cycle can also be used to explain why group I. II and III metal always form M+. e. The octet rule only helps to predict the formation of Ca2+ ion but it doesn't offer explanation. the enthalpy of formation of the 3 hypothetical ionic crystal can be calculated. Ca2+(Cl-)2 and Ca3+(Cl-)3 The lattice energies are calculated based on the mathematical model. M2+ and M3+ ion respectively. ∆Hfo = -155 kJmol-1 Ca(s) + ½Cl2(g) → Ca+Cl-(s) 2+ Ca(s) + Cl2(g) → Ca (Cl )2(s) ∆Hfo = -763 kJmol-1 Ca(s) + 2 Cl2(g) → Ca3+(Cl-)3(s) 3 ∆Hfo = +1356 kJmol-1 The enthalpy of formation of CaCl2(s) is most exothermic among the three. The rationale behind the formation of Ca2+ ion is solely from an energetic point of view. Glossary lattice energy electron affinity Born-Haber cycle intermediate type of bond . Energetics of formation of ionic compounds C.III.g. This explain why calcium chloride has the empirical formula CaCl2 because CaCl2(s) is the most stable one among the three. 349 Electron affinity of Cl(g) . The lattice energy decreases in magnitude on descending the group.outweighs the 1 mark repulsion between the e. Cl has a smaller size / a smaller no.makes the second electron affinity an endothermic process. ii Explain why the first electron affinity of chlorine is more negative than that of bromine. 1 1 mark O-(g) + e. Addition of an e. C Most candidates did not give a precise definition for 'electron affinity'. NaCl 771. 3 94 1A 1 a i ii iii 1a i Write an equation to represent the change related to the second electron affinity of oxygen. After addition of an e-. the size of O expands. Explain why they have 2 opposite signs. and the experimental lattice enthalpy of AgCl(s) are given in the table below. They did not point out that the electron affinity of an element X is the molar enthalpy change when atoms of X in the gaseous state take up electrons to form X-(g) ions under standard conditions.→ O2-(g) ii The first and second affinities of oxygen are -142 kJmol-1 and +791 kJmol-1 respectively. hence the attraction for another electron is weakened. CsCl 645 kJ mol-1 1 mark 2 marks This is due to increasing size of cation results in smaller attraction between cation and anion. i What is the meaning of 'electron affinity' ? Electron affinity is the molar enthalpy change ½ mark ½ mark for the following process at standard state/conditions X(g) + e.in O atom and the incoming e-. Energetics of formation of ionic compounds Page 4 Past Paper Question 92 1A 3 d 94 1A 1 a i ii iii 96 2B 5 a i ii 98 1A 1 c i 99 2A 1 a ii 92 1A 3 d 3d Briefly describe and explain the change in the lattice energy of Group I chlorides on descending the group.411 Enthalpy change of formation of NaCl(s) 2 2 3 .→ X-(g) Or ½ mark electron affinity of an element X is the enthalpy change 1 mark when 1 mole of X-(g) is formed from gaseous atoms of X ½ mark under standard conditions.to the O(g) is exothermic Q attraction between the nucleus and the incoming e. 1 mark iii Explain why all the inert gases have positive first electron affinities.A.905 i Calculate the experimental lattice enthalpy of NaCl(s) using the following thermochemical data. 96 2B 5 a i ii 5a The first electron affinities of chlorine and bromine are -364 kJmol-1 and -295 kJmol-1 respectively. and the repulsion between the O-(g) ion and the e. 1 Screening effect of the stable p6 configuration of inert gases (s2 in the case of He) makes the addition of an extra ean unfavourable process ∴ inert gases have +ve first E. ∆Ho/kJ mol-1 108 Atomization enthalpy of Na(s) 495 First ionization enthalpy of Na(g) 239 Bond dissociation enthalpy of Cl2(g) . Compound Theoretical lattice enthalpy Experimental enthalpy NaCl(s) . of electron shells 1 mark ∴ Cl can exert a stronger attraction on an extra electron 1 mark 98 1A 1 c i 1c The theoretical lattice enthalpies of NaCl(s) and AgCl(s).833 . 1 mark C It was surprising to find that many candidates could not distinguish between electron affinity and electronegativity and hence they performed badly in this part.770 ? AgCl(s) .III. III. Energetics of formation of ionic compounds 99 2A 1 a ii 1a ii Using the thermochemical data given below. ∆Ho298 / kJ mol-1 Rb(s) → Rb(g) 82 Rb(g) → Rb+(g) + e403 Cl2(g) → 2Cl(g) 242 Cl(g) + e → Cl (g) -348 2Rb(s) + Cl2(g) → 2RbCl(s) -861 Page 5 . calculate the lattice enthalpy of rubidium chloride. Therefore. It can be generalized that the state with a high degree of disorderness is always favourable in nature. Chemistry in Context. .0 Modern Physical Chemistry ELBS pg. The scientist called the disorderness in arrangement of a system – entropy S and the change in the disorderness – entropy change ∆S. Although the heat content of water is higher than ice. E. Some processes take place spontaneously even though they are endothermic. the products of an endothermic change is less stable than the reactants. is not a very good description of the relatively stability of systems. 3rd Edition. melting is an endothermic change. why does it change to water ? Therefore. A. Gibbs free energy change (∆G) 3. However. In the above example. Ramsden pg. bond energy and lattice energy. Entropy (S) and entropy change (∆S) Consider another example. Gibbs free energy change (∆G) (not required in A-Level) So far. we have applied the concept of energetic to different chemical systems such as ionization energy. Gibbs free energy change (∆G) Page 1 Topic Reference Reading IV. Assignment Reading Syllabus Notes Gibbs free energy change IV. enthalpy change ∆H.e. in certain instances. the free expansion of a balloon in vacuum.e. ∆H > 0. ∆S > 0. ∆H = 0. does that mean ice is more stable than water ? If ice is more stable than water. i. ∆H Water melting process Ice B. Since no work is done in the expansion. free exp ansion in vacuum→ The expansion is spontaneous which means that the bigger balloon is more stable than the smaller balloon though the ∆H is 0. the arrangement and the movement of the molecules are getting more random. 5th Edition ELBS pg. the arrangement of the molecules becomes more random and the entropy of the system increases. Limitation of enthalpy change (∆H) For example. the temperature will remain constant.IV. changes in enthalpy are only part of the driving force for the change. 339–350 A-Level Chemistry.e. Ice melts readily at room temperature. The size of the balloon increases and more spaces are available for the gaseous molecules to travel.N. The observation seems to be contradictory. i. i. i.e. Moreover. the value of ∆G is always smaller than zero.e.IV. ice tends to melt. free energy change ∆G only indicates the energetic stability of a system. Gibbs free energy change is defined as ∆G = ∆H . All spontaneous changes take place in the direction of decrease in free energy. whether ice will melt or freeze is depending on temperature. ∆G < 0.15K) 0ºC (273.0 kJmol-1 Temperature -10ºC (263. ice has a more ordered packing that water. Whether the change will take place is also depending on the kinetic stability of the system which will be discussed in the chapter of rate of reaction.15K) 10ºC (283. melting is a process associated with an increase in disorderness i. Gibbs free energy change (∆G) Page 2 In order to address the limitation of enthalpy change ∆H in describing the relative stability of different systems.T∆S where ∆H is enthalpy change T is temperature in Kelvin ∆S is entropy change For example. At any temperature above 0ºC. Thus. ∆S > 0. However.15K) ∆H of melting 6010 kJmol-1 6010 kJmol-1 6010 kJmol-1 -T∆S of melting -5790 kJmol-1 -6010 kJmol-1 -6230 kJmol-1 ∆G of melting 220 kJmol-1 0 kJmol-1 -220 kJmol-1 Obviously. At any temperature below 0ºC. Gibbs free energy change (∆G) C. Willard Gibbs defined another function – Gibbs free energy change ∆G. Enthalpy change (∆H) of melting = 6010 kJmol-1 Entropy change (∆S) of melting = 22. Glossary Past Paper Question spontaneous reacton disorderness Gibbs free energy change (∆G) entropy (S) entropy change (∆S) . water tends to freeze. The scientist J. Hybridization theory 3. Structure and shape of hydrocarbons a) Characteristic of Carbon-Carbon bond b) Shape of hydrocarbons (1) (2) (3) Saturated hydrocarbons Unsaturated hydrocarbons Aromatic hydrocarbons (a) Delocalization of π-electrons (b) Stability of benzene B. Lewis structure of molecule 1. Overlapping of atomic orbital 2. Formal charge of individual atoms in a molecule 3. C. Molecular Orbital Theory II. Size of isoelectronic particles . Strength of ionic bond . Limitation of octet rule Valence Shell Electron Pair Repulsion Theory (VSEPRT) 1. Electron density map of sodium chloride Periodicity of ionic radius 1.Chemical Bonding Page 1 Chemical Bonding I. Shape of molecule A. 3-dimensional representation of molecular shape Hybridization Theory 1. Octet expansion of elements in Period 3 2. Definition of radius 2. B. D. Oxidation no. C.lattice energy X-ray diffraction 1. Position of lone pair 3. Examples a) sp hybridizaion b) sp2 hybridizaion c) sp3 hybridizaion d) sp3d hybridizaion e) sp3d2 hybridizaion 4. Periodicity of ionic radius 3. Ionic bonding A. of individual atoms in a molecule 2. H2+ ion Electron diffraction Covalent radius 1. C. Pauling scale of electronegativity Polarity in covalent bond 1. Difference among lattice energies of Na. silver halide and zinc sulphate b) Bonding intermediate between covalent and ionic 3. Breaking down of addivity in bond enthalpies (1) Dative covalent bond 1. Definition of covalent radius 2. Ag and Zn compounds a) Lattice energies of sodium halide. Covalent bonding A. Dipole moment a) Vector quantity of dipole moment b) Polarity of molecule c) Factors affecting dipole moment (1) (2) (3) Inductive effect (I) Mesomeric effect / resonance effect (R) Presence of lone pair C. B. D. B. Electron sea model of metal Strength of metallic bond Melting and boiling of metal Strength of ionic bond. Metallic bonding A. Definition of electronegativity 2. V. D. C. Polarization of ionic bond a) Fajans’ Rules in polarization of ionic bond Electronegativity 1. Deflection of a liquid jet by an electric field 2. Breaking down of addivity in covalent radius and bond energy a) Resonance (1) (2) (3) Resonance structure of benzene Resonance structure of nitrate ion Rules in writing resonance structures Delocalization energy b) D.Chemical Bonding Page 2 III. Electron density map of LiF comparing with those of NaCl and H2 2. B. Differences between ionic bond and covalent bond Incomplete electron transfer in ionic compound 1. covalent bond and metallic bond . Examples of H3N→BF3 and Al2Cl6 a) H3N→BF3 b) Al2Cl6 IV. Bonding intermediate between ionic and covalent A. Addivity of covalent radius 3. of bond pair = ((2 × 3 + 8) . Steps of drawing Lewis structure i. 8–14 Lewis structure Chemical Bonding I. Shape of molecule Unit 1 Page 1 Topic Reference Reading Syllabus Notes I. NO3iii. 2 for H. in H2O. of bond pairs can be calculated by the following method. If the molecule is negatively charged. of electron available) ÷ 2 no. Put the rest of the electrons into the diagram until octet is fulfilled and include the charge if any. Determine the no. Drawing the skeleton of the molecule Within a covalent molecule. 8 for O or N. Lewis structure of molecule Before the shape of a molecule can be derived. of electrons accordingly. of the bonds have to be formed. Valence Shell Electron Pair Repulsion Theory – repulsions among electron centres (bond pair and lone pair electrons) have to be minimized. of bonds The total no. add extra no. Solomons. Determination of central atom Not all atoms can be the central atom in a molecule. Shape of molecule Unit 1 The shape of a molecule can be predicted and explained by several different theory i. ii. subtract the no.0–4.g.1 Modern Physical Chemistry ELBS pg.I. Count the electron available Count the number of valence electrons provided by each atom.g. of bond pair = ((8 × 4) .24) ÷ 2 = 4 e. Determination the no.to fill its valence shell e. Hybridization Theory – overlapping of hybridized atomic orbitals iii. 6th Edition pg. there should be at least 1 bond pair between 2 atoms. of electrons accordingly. . Assuming each atom in the molecule needs the max. of electron required .g. e. NH3 no. it would be easier to draw the Lewis structure (cross-dot diagram) of the molecule first.g. of bond pair = (no. 84–86 Organic Chemistry. Shape of molecule 4. H can only be the peripheral atom since it can form only 1 bond. If the molecule is positively charged.8) ÷ 2 = 3 e.g. e. there should be 8 electrons for N and 2 electrons for each H atom. ii. Molecular Orbital Theory – overlapping of atomic orbitals to form molecular orbital A. of electrons required if the outermost shells of all the atoms have to be filled. in NH3. iv. Count the electron required Count the total no. of e. no.no. Formation of each bond can save the use of 2 electrons. no. I. of bond = (14 . iv. H N H H Total 3 N–H Remaining 8 6 2 H N H H 3 N–H 1 lone pair Total 6 2 8 no.no.g. of e. Central atom – N ii. Oxidation no. is related to the difference in electronegativity of the atoms in the molecule. – The difference between the number of electrons associated with an atom in a compound as compared with an atom of the element. of electron in neutral O atom . Central atom – N ii.N.N.g. 8 8 + no. of the doubly bonded oxygen = no.e. of e.available N 4H e. of e. results in negative oxidation no. the 4 electrons in the double bond is attracted towards the oxygen. It is assumed that the bonding electrons will be attracted to the more electronegative atom from the less electronegative.no. of e.no. e. N 3O -1 e.8 = -2 O. Oxidation no. of electrons on C in CO2 = 4 .no.available N 3H e.no.in neutral O atom .N.in neutral N atom . Formal charge of individual atoms in a molecule . Oxidation no. of carbon = no.8 = -2 O. Shape of molecule Examples NH3 i.required 5 6×3 1 24 8 8×3 32 O N O O Total 3 N–O Remaining 24 6 18 O N O O 1 N=O 2 N–O 8 lone pairs Total 4 4 16 24 no. of oxygen = no.8 = -2 Oxidation no.8) ÷ 2 = 4 NO3i.on O in NO3= 6 .of N in NO3= 5 . the more electronegative atom will has more electrons than if neutral. N 4H +1 (charge) e. of e. of bond = (32-24)÷2 = 4 1. N 3H e.available N 3O e. Central atom – N ii. of electron in neutral C atom . of e.required Unit 1 Page 2 5 1×3 8 8 2×3 14 iii. CO2 Oxygen is more electronegative than carbon.in neutral O atom .0 = +4 Oxygen is more electronegative than nitrogen O C O e.0 = +5 2. i. 8 8 0 H H N H H 4 N–H Total iv. of individual atoms in a molecule Oxidation no. NO3- - O N O O O. of nitrogen = no.8) ÷ 2 = 3 NH4+ i.required 5 1×4 -1 8 8 2×4 16 iii. of electrons on O in CO2 = 6 . of the singly bonded oxygen = no. H H N H H Total 4 N–H Remaining iii. of bond = (16 . iv.on O in NO3= 6 . some formal charges of some atoms can be memorized. the sum of formal charges equals the overall charge carried by the molecule. of e. Formal charge of an atom is calculated by assuming that the bond pair electrons are shared equally between the two atoms bonded together. formal charge is independent of difference in electronegativity.B. of e.in neutral N atom . of e.in neutral O atom . of e. Similar to oxidation no. CO2 Formal charge of oxygen = no.no. and formal charges.associated with O in NO3=6-6=0 Formal charge of the singly bonded oxygen = no. Shape of molecule Unit 1 Page 3 Unlike oxidation no. of e. of e.in neutral O atom .no. To minimize the fuzziness of calculation. of e..7 = -1 Formal charge of nitrogen = no. NO3- + O N O O N.I.associated with N in NO3= 5 .in neutral C atom .in neutral O atom .no.no. O C O e. The positive sign and negative sign must be accompanied for all oxidation no.associated with C in CO2 =4-4=0 Formal charge of the doubly bonded oxygen = no..associated with O in NO3= 6 . e.no. of e. of e.g. .associated with O in CO2 =6-6=0 Formal charge of carbon = no. of e.4 = +1 Non-zero formal charge is usually indicated on the Lewis structure.g. S has 6 valence electrons and each O is capable to form 1 double bond. P has 5 valence electrons and each Cl is capable to form 1 single bond. However. This is why P can form both PCl3 and PCl5 but N can only form NCl3. there are 3d orbitals available which is capable to accommodate 10 more electrons. For example. NH3 ½ mark each 2 92 2B 6 Ba i 6Ba i Write the formula in each case of ONE compound or ion in which sulphur has the oxidation state -2. NO3-. This is because. NO2-.I. SO2. Xe has 8 outermost electrons. -1 and -3. Shape of molecule 3. the no. for some structures do not obey octet rule. another method have to be used to draw the Lewis structure. +2. in SO2.g. However. Though the largest no. PCl5. e.ion. +3. O N O there are totally ( 5 + 6 + 6 ) = 15 valence electrons which is an odd number. Therefore. e. SO42½ mark each 2 . sometimes memorization is required. there are only 2s. This phenomenon is called expansion of octet (or expansion of coordination sphere) and will be explained in the chapter of hybridization theory. S has 6 valence electrons . the Lewis structure will become O S O . e. the Lewis structure will become Cl Cl Cl P Cl + Cl . Therefore. in PCl5. Each orbital can accommodate 2 electrons and makes up to a total of 8. 4 electrons about Xe will remains as the lone pairs. As a result. NH2OH.g. S2-. can be less than 8. S2O32-. Therefore. the Lewis structure will F F become F Xe F .g. 2 O atoms with O O -1 formal charge each forming 1 bond. Glossary Past Paper Question 6Aa Lewis structure 92 2B 6 Aa i Oxidation no. +4 and +6.g. for principal quantum number 2. for period 3 elements. SO32-. each F is capable to form 1 bond. in SO42. 2py and 2z orbitals available. 2px. 2 O atoms with 0 formal charge each forming 2 bond . For those which does not obey octet rule. the Lewis structure will become O- S O- . It is assumed that all valence electrons are capable to form bond with peripheral atoms. Limitation of octet rule Unit 1 Page 4 The forth-mentioned method of drawing Lewis structure is only limited to the elements in Period 2. there are 8 electrons about each O atom and only 7 eletrons about N. 92 2B 6 Ba i formal charge 92 2B 6 Aa i i Write the formula in each case of ONE compound or ion in which nitrogen has the oxidation state +5. of valence electron would be 8. e. Actually. It is not possible for all atoms to have an octet structure. And the peripheral will form the required number of bond to achieve octet. XeF4 is a noble gas compound not obeying octet rule. Therefore. e.g. in NO2. Thomas Nelson and Sons Ltd. 126–143 Organic Chemistry. double bond or triple bond. 90–92 4. 9–14 A-Level Chemistry. single bond. 2. The repulsion between lone-pair/lone-pair is greater than that between lone-pair/bond-pair which is. Fillans pg. Unit 2 Syllabus Notes Electron-pairs. Ramsden pg. greater than that between bond-pair/bond-pair. Valence Shell Electron Pair Repulsion Theory (VSEPRT) The shape of a molecule was related to the number of electrons in the outer shell of the central atom. 3. Where there are more than four pairs of electrons to consider. 6th Edition pg. pg. arrange themselves in space so as to minimize their mutual repulsions. Shape of molecule Modern Physical Chemistry ELBS pg.N.I. 1. . Shape of molecule Unit 2 Page 1 Topic Reference Reading I. 5th Edition. of electron centre in the outer shell of the central atom 2 3 Shape of the framework linear trigonal planar 120º BF3 Bond angle Example 180º BeCl2 4 tetrahedral 109º28' CH4 NH3 H2O 5 trigonal bipyramidal 120º 90º PF5 SF4 ClF3 6 octahedral 90º SF6 IF5 7 pentagonal bipyramidal 72º 90º IF7 Note: An electron centre can either be a lone pair. the interactions where the electron-pairs make an angle greater than 90° at the central atom are ignored. Solomons.2 Shape of molecule Valence Shell Electron Pair Repulsion Theory B. whether in bonding orbitals or lone-pair orbitals. E. No. 84–86 Physical Chemistry. 3rd Edition. 125–128. 131–132 Chemistry in Context. Count the number of electron centre around the central atom and choose the corresponding framework. there would be only 2 90º lone pair / bond pair repulsions. The 120º repulsion is neglected as it is much weaker than 90º repulsion. Unit 2 Octet expansion of elements in Period 3 Page 2 Elements in period 3 and onwards are capable to use the low lying energy d-orbitals available to form more bonds. Position of lone pair Since lone pair electrons. 2. there would be 3 90º lone pair / bond pair repulsions. For the molecule with 5 electron centres. Put the electron centres (with the atom if applicable) on the vertex of the framework so that the repulsion among them are minimized. 3. In the determination of the shape of a molecule. in SF4 If the lone pair occupies the equatorial position. Therefore. 4. . 2. If the lone pair occupies the axial position. only repulsions with the lone pair need to be considered. 3. comparing with bonding pair. Steps of using VSEPR theory to determine the shape of a molecule.g. Determine the bond angle. 1. 3-dimensional representation of molecular shape a bond coming out of the paper a bond going into the paper a bond on the paper a lone pair Usually two bonds are used to defined the plane of the paper and then the other bonds are added with respect to this plane. The position of the lone pair is not counted.I. the lone pair always occupies the equatorial position to minimize the repulsion. Shape of molecule 1. e. the element in period 3 can exceed the octet by expansion of its coordination sphere. is closer to the central atom and has greatest repulsion with other electron centres. only the geometry of the atoms are considered. This would be further explained in hybridization theory. Draw the Lewis structure. 109½º. NH3 and H2O has 4 electron centres. Shape of molecule Example BF3 F Unit 2 Lewis structure F B F Page 3 Shape of molecule trigonal planar F F B F no. 109½º is the angle for a perfect tetrahedral. they possess slightly different bond angle.I. the repulsion with lone pair is greater than the repulsion with bond pair. 107º and 105º respectively. However. The presence of a lone pair on N squeezes the bond angle to 107º. the two lone pairs on O squeezes the bond angle even further to 105º Example PCl5 Lewis structure Cl Cl Cl P Cl no. of electron centre 5 single bonds Total : 5 Framework chosen Shape of molecule trigonal bipyramidal Cl Cl P Cl Cl Cl Bond angle ∠ Cl–P–Cl : 120º × 3 ∠ Cl–P–Cl : 90º × 6 Cl SF6 F F S F F F F 6 single bonds Total : 6 4 single bonds octahedral F F F S F F F ∠ F–S–F : 90º NH4+ H H N H H + tetrahedral H + N H H H ∠ H–N–H : 109½º Total : 4 2 single bonds 2 lone pairs Total : 4 NH2- H N H angular / bent / V-shaped H N H ∠ H–N–H : ≈ 105º CO2 O C O 2 double bonds Total : 2 linear O C O ∠ O=C=O : 180º SO2 O S O 2 double bonds 1 lone pair Total : 3 angular / bent / V-shaped O S O ∠ O=S=O : ≈ 120º . of electron centre 3 single bonds Total : 3 4 single bonds Framework chosen Bond angle ∠ F–B–F : 120 º CH4 H H C H H tetrahedral H H C H H ∠ H–C–H : 109½º Total : 4 3 single bonds 1 lone pair Total : 4 NH3 H N H H trigonal pyramidal H N H H ∠ H–N–H : ≈ 107º H2O H O H 2 single bonds 2 lone pair Total : 4 angular / bent / V-shaped H O H ∠ H–O–H : ≈ 105º Although all CH4. I. Shape of molecule Unit 2 Oxidation no. 91 2B 5 e 93 1A 3 a ii formal charge Valence shell electron pair repulsion theory Page 4 Glossary Past Paper Question Lewis structure octet expansion 91 1A 3 d i ii 92 1A 3 g i ii 93 1A 1 b i ii 94 1A 1 c i ii 95 1A 2 e i ii 96 1A 2 b i ii 97 1A 3 b i ii iii 98 1A 3 a i ii iii 99 2A 3 c i 98 1A 4 a i 98 2A 1 c i 91 1A 3 d i ii 3d Draw the molecular shapes of i PCl5(g) Cl Cl P Cl Cl Cl 1 1 mark 1 ii SF4(g) F S F F F 1 mark 91 2B 5 e 5e Why is the bond angle in NF3 smaller than that in NH3? (Valence Shell Electron Pair Repulsion approach) Bond pair electrons nearer to F than H since F is more electronegative ∴ bond-pair/bond pair repulsion is smaller in NF3 than in NH3. H N H H N F F F 2 ½ mark ½ mark 1 mark or (Hybridization approach) The primary scheme of hybridization of N is sp3 the p orbital is more directional than s orbital and have been attracted strongly towards the electronegative F as a result the N in NF3 has higher s-character than N in NH3 Bond angle ∝ s-character e.g. Ideal bond energy of following hybridization: sp = 180º, sp2 = 120º, sp3 = 109½º 92 1A 3 g i ii 3g Draw diagrams showing the shapes of the following molecules. Indicate the lone pairs (if any) on each central atom. i ICl2 1½ Cl I Cl Linear shape 1 mark Lone pairs ½ mark XeOF4 1 e- only ii 1½ O note: Xe is [Kr]4d105s25p6 Square-pyramidal1 mark Lone pair ½ mark F Xe F F F I. Shape of molecule Unit 2 Page 5 93 1A 1 b i ii 1b For each of the following molecules, draw a three-dimensional structure and state the molecular geometry. i SiF4 F F Si F F 2 tetrahedral F 2 marks 2 2 marks ii OF2 O F bent / V-shaped 93 1A 3 a ii 3a Consider the following compound, X: ii Suggest expected values for the bond angles α, β and γ. α: 108º-110º β: 178º-182º γ: 118º-122º 1½ ½ mark each 94 1A 1 c i ii 1c For each of the following molecules, draw a three-dimensional structure showing the positions of the bond electron pairs and lone electron pairs (if any). In each case, state the molecular geometry and whether the molecule possesses a non-zero dipole moment. i BF3 2 F F B F 1 mark ½ mark ½ mark 2 ii trigonal / triangular planar no dipole moment ClF3 F F Cl F T-shaped Possesses a net dipole [0 mark if shape of ClF3 is described as triangular planar] 1 mark ½ mark ½ mark 95 1A 2 e i ii 2e For each of the following species, draw a three-dimensional structure showing the bond electron pairs and lone electron pairs of the central atom. State the shape of the species in each case. i ICl4Cl Cl I Cl Cl 1½ ii Square planar SCl2 S Cl Cl (1 mark, no charge -½ mark, no lone pair -½ mark) ½ mark 1½ 1 mark ½ mark V-shaped / bent / angular 96 1A 2 b i ii I. Shape of molecule Unit 2 2b For each of the following chemical species, draw a three-dimensional structure showing the bond electron pairs and lone electron pair(s) of the central atom underlined. State the shape of the species in each case. i ClO3- Page 6 1½ O O Cl O 1½ mark 1½ ii pyramidal (Deduct ½ mark for omitting the negative charge) NOF N F O 1½ mark angular or V-shaped (Deduct ½ mark for missing out the lone pair) (1 mark for 3-D structure; ½ mark for the shape) 97 1A 3 b i ii iii 3b For each of the following sulphur-containing chemical species, state its shape and the oxidation state of sulphur. i H2S ii SO2 iii SO4298 1A 3 a i ii iii 3a For each of the nitrogen-containing chemical species below, state its shape and the oxidation state of nitrogen. i NO2+ ii NH3 iii NO398 1A 4 a i 4a Alcohol E has the structure CH3CH(OH)C2H5. i Draw a three-dimensional representation of E. 98 2A 1 c i 1c i Draw the three-dimensional structure of BF3. 99 2A 3 c i 3c Consider the hydrides of three Period 3 elements : SiH4, PH3 and H2S i For each hydride, draw a three-dimensional structure showing the bond electron pairs and lone electron pair(s), if any, of the central atom. 3 1 1 1 1 4 I. Shape of molecule Unit 3 Page 1 Topic Reference Reading I. Shape of molecule Modern Physical Chemistry ELBS pg. 84–86 Physical Chemistry, Fillans pg. 126–143 Organic Chemistry, Solomons, 6th Edition pg. 26–37 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 133–140 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 419 4.3 Hydridization Theory C. Hybridization Theory Actually, the electrons in the molecule are not stationary, they undergo wave like motion. The valence shell electron pair repulsion theory is not good enough to explain this. Therefore, the scientist developed another theory, hybridization theory. Unit 3 Syllabus Notes Questions that cannot be answered by VSEPRT 1. electrons are not stationary 2. the strength of the 2 bonds in a double bond are not the same. 3. No. of unpaired e- available for bond formation. 4. shape of some molecules 5. strength of some bonds 1. Overlapping of atomic orbital Instead of fixed electron centre, covalent bond is now viewed as an overlapping of atomic orbitals. In hydrogen, two hydrogen 1s atomic orbitals are overlapped to form a covalent bond. The two 1s orbitals interfere with each other constructively. This causes a higher electron density along the internuclear axis and create the attraction to bring the two atoms together. There are two kinds of overlaps i. head-on overlap ii. sideway overlap Sideway overlap (Laterally overlap) of the orbital give π (pi) bond. Head-on overlap (End to End overlap) of the orbital give σ (sigma) bond. Since the bond strength of a covalent bond is directly proportional to the area of overlap, in most of the case, the σ bond will be stronger than the π bond. Also, the π electrons, unlike σ electrons, are not concentrated along the internuclear axis. In large atoms, π bonds are not favorable because, being removed from the line between the centres of the atoms, the π bond rapidly weakens as the size of the atom increases. e.g. C=C double exists but Si=Si double bond doesn't exist. I. Shape of molecule Unit 3 Page 2 σ bond and π bond can be distinguished by the shape of their cross section. σ bond has a cylindrical cross section about the bond axis but π bond does not has a cylindrical cross section. Orbital representation of one σ bond and two π bonds ` 2. Hybridization theory Cross section Hybridization means mixing. Hybridization theory addresses many questions that cannot be explained by Valence Shell Electron Pair Repulsion Theory. i. electrons are not stationery, they undergo wave like motion. ii. The strength of C=C (612 kJmol-1) is not twice of the strength of C–C (347 kJmol-1) bond. i.e. the two bonds in double bond are not equvalent. iii. There are not enough unpaired electrons in a C atom in ground state (1s22s22p2) for the formation of 4 single bonds. iv. The four single bonds in CH4 are equivalent and the shape is tetrahedral. Consider CH4 molecule, experimental findings tell us that the 4 C–H bonds are equivalent and the molecule is tetrahedral A C atom in ground state has only 2 unpaired electron in 2 different 2p orbitals which can only form 2 single bond. If one electron is promoted from 2s orbital to 2p orbital, you will get 4 unpaired electrons for bond formation. However, if the 4 unpaired electrons overlap with 1s electron of the 4 H atoms independently, the bond formed will not have equivalent strength. This is because the 4 unpaired electrons are not equivalent. One is a 2s electron and 3 are 2p electrons. Furthermore, the bond angle will not be 109½º any more. The angle between different p orbitals are only 90º. Therefore, scientist proposed that the 4 unpaired electrons interfere with each other first (i.e. mixed with each other) to form 4 equivalent orbitals before any bond is formed. Because the new orbitals are formed from one s orbital and three p orbital. They are called sp3 hybridized atomic orbitals or sp3 hybrid orbitals. 1s hybridized C* sp3 excited state C * 1s ground state C 2s 2p 2 2 110 1s 2s 2p 2 1 111 2 1111 It can be seen that the main purpose of developing hybridization theory, or any new theory, is to explain the experiemental findings. I. Shape of molecule Unit 3 Page 3 The process of combining the orbitals is called hybridization. Note : Hybridized atomic orbitals = ∑ atomic orbitals 4 - r t l oba s i no. of hybridized atomic orbitals = no. of atomic orbitals used hybridization excited state hybridized atomic orbital overlapping of orbitals (bond formation) In some of the hybridization processes, electron is promoted to the higher orbital first. e.g. from 2s to 2p orbital in C. This requires energy. Since bond formed will be stronger with hybridization, this will be paid off by the energy released during bond formation. Since hybridization is just a kind of interference, only the orbitals with similar energy can be hybridized. i.e. 2s orbital cannot hybridize with 3s orbital promotion ground state ∆H (more stable product) Hybridized orbital sp sp2 sp3 3 3 sp d or dsp (not required) sp3d2 or d2sp3 (not required) * Table needed to be remembered. 3. Examples No. of electron pairs (bond & lone pairs) 2 3 4 5 6 Geometrical arrangement Linear Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral Typical examples BeCl2, C2H2, CO2 BeF3, C2H4 CH4, NH3, H2O PCl5 Fe(CN)63+,SF6 a) sp hybridization e.g. C2H2 (linear) At first, one 2s electron is promoted to 2p orbital. One 2s orbital and one 2p orbital undergo hybridization to form two hybridized orbitals. sp hybridized orbitals consists of two lobes with linear geometry. One sp hybridized orbital overlaps with the 1s orbital of H to form a C–H σ bond. Another sp hybridized orbital overlaps with the sp hybridized orbital of another C to form a C–C σ bond. The two p orbitals on one C overlap with the two p orbitals of another C laterally to form 2 C–C π bond. It can be seen that not all the three C–C bond are equivalent. Because of the cylindrical symmetry of the σ bond, the C–H bond is free to rotate. Rotation of the C≡C bond will break the two π bonds, therefore C≡C bond is not free to rotate. ground state C 1s 2s 2p 2 2 110 1s 2s 2p * promotion→ excited state C 2 1 111 1s sp 2p hybridization→ hybridized C * 2 11 11 Orbital representation of ethyne (H–C≡C–H) g. there is still an empty 2p orbital and it does not fulfill octet. Since the π bond is not free to rotate. the unhybridized p orbital of the C at right angle to the plane has an unpaired electron. BF3 (trigonal planar) 1s ground state B 2s 2p Unit 3 Page 4 1s 2s sp2 2p 2p 2 2 100 promotion→ hybridization excited state B* 1s 2 1 110 2 111 0 → hybridized B* The three sp2 hybridized orbitals overlap with 2p orbitals of the three fluorine to form 3 σ bonds. . e.I. The p orbitals from the two C overlap laterally to form a π bond. sp2 hybridized orbitals consists of three lobes with trigonal planar geometry. ethene has a planar structure.g. Shape of molecule b) sp2 hybridization e. C2H4 (trigonal planar) 1s ground state C 2s 2p one 2s and two 2p orbitals sp2 hybridized orbitals 1s 2s sp2 2p 2p 2 2 110 promotion→ excited state C* * 1s 2 1 111 2 111 1 hybridization→ hybridized C Unlike BF3. BF3 is an electron deficiency species and ready to accept electron from other molecule. On B. This is the lone pair electrons of the molecule.g. NH4 e. being symmetrical. Shape of molecule c) sp3 hybridization Unit 3 Page 5 e. In H2O molecule. the larger will be the bond angle. the bond angle is decreased by 2. CH4 e. sp – 180º (50% s character).I. CH4 (tetrahedral) 1s ground state C 2s 2p 1s 2s sp3 2p 2 2 110 promotion→ excited state C* 1s 2 1 111 2 1111 * hybridization→ hybridized C sp3 hybridized orbitals consists of four lobes with tetrahedral geometry directed to the four vertices of a tetrahedron. sp3 – 109½º (25% s character) The higher the s character. In NH3 molecule. NH3 (trigonal pyramidal) 1s ground state N 2s 2p 1s sp3 2 2 111 hybridization→ hybridized N 2 1112 One of the sp3 hybridized orbital is occupied by an electron pair. . sp2 – 120º (33% s character). H2O It can be observed that the bond angle is related to the percentage of s character in the hybridized orbital. H2O (angular) 1s ground state O 2s 2p 1s sp3 2 2 211 hybridization→ hybridized O 2 1122 In CH4 molecule.g.g. adopts the bond angle of regular tetrahedron 109º29'. The tend can be predicted by the valence shell electron pair repulsion theory which state that the lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion and a lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion. the bond angle is decreased by 5.0º (104º31') from the tetrahedral value.2º (106º45'). g. As the principle quantum number increases. This is impossible for elements in period 2. .e.g. this is known as expansion of the octet. the energy gap between the orbitals gets smaller. Shape of molecule Unit 3 Page 6 d) sp3d hybridization (not required in A-Level) e. P. XeF4 (square planar) [Kr] 4d105s25p6 5s Xe [Kr] 4d 10 5p 5d 5s 5p sp3d2 5d 5d 2 222 00000 → promotion hybridization Xe [Kr] 4d 10 2 211 11000 221111 000 F F Xe F F → Xe [Kr] 4d 10 Not all noble gases can form compound. PF5 (trigonal bipyramidal) 3s 3p P [Ne] 3d 3s 3p sp3d 3d 3d 2 111 00000 → promotion P * [Ne] 1 111 10000 0000 * hybridization→ P [Ne] 11111 The hybridization process involves the promotion of an electron from 3s orbital to 3d orbital. as there is no low lying energy d orbital available. electron from lower energy level may be promoted to higher energy level to yield unpaired electrons for bonding. does not obey octet rule. They overlap with the p orbitals of the six fluorine atoms to form a perfect octahedral. Energy required to promote 2p electrons to 3s orbital will be large that cannot be paid off in the bond formation process. i. e) sp3d2 hybridization (not required in A-Level) e. * e. For heavy noble gas atom. SF6 (octahedral) [Ne] 3s23p4 3s S [Ne] 3p 3d 3s 3p sp3d2 3d 3d 2 211 00000 promotion→ S * [Ne] [Ne] 1 111 11000 111111 000 hybridization→ S The six sp3d2 hybrid orbitals are identical. The central atom.g.I. formal charge Valence shell electron pair repulsion theory Hybridization theory σ(sigma) bond π(pi) bond ground state .I. Shape of molecule Unit 3 Page 7 Glossary Lewis structure octet expansion excited state Oxidation no. while nitrogen can form only NCl3. 1½ mark 2v [PO4] P cannot have oxidation state higher than +5.I. of I = +7 1 mark 1 mark Example : IF7. 4d Oxygen usually forms compounds with oxidation numbers -2. why this is so. a. ½ mark 98 2B 5 a i 5a Consider the following compound F. Hightest O. ∴ It can form only 3 covalent bonds. in each case.in its valence shell and cannot expand its octet because N has no low lying energy orbitals for bonding. promotion of e. Sulphur has empty low lying energy 3d orbital . 1 mark 90 2B 6 a i iii v 6a The following species are either impossible to prepare or very unstable. IO495 1A 2 a 2a Explain why phosphorus can form PCl3 and PCl5. P has low-lying empty 3d orbital / energy level. Explain. gives 5 unpaired electrons. 1½ mark In N.S.is not possible. It is capable to form 5 covalent bond.to expand its coordination sphere. 1 mark Oxygen cannot expand its octet/has no low lying energy orbital for bonding. thus capable of forming compounds with a wide range of oxidation number. i NCl4 Nitrogen cannot have 9e. 1½ mark iii ArCl2 Ar does not form compounds because of the stability of close-shell electronic configuration. whereas sulphur can form compounds with many more oxidation numbers ranging from -2 to +6. -1 and 0. 1½ mark 3 1½ 1½ 1½ 94 2B 4 c 4c What is the highest oxidation state of iodine? Give an example of an iodine-containing compound in which iodine 2 is in this oxidation state. there is no low-lying d orbitals. Shape of molecule Unit 3 90 2B 6 a i iii v Page 8 Past Paper Question 90 2B 4 d 94 2B 4 c 95 1A 2 a 98 2B 5 a i 90 2B 4 d 4 Account for the following observations. CH3 a b c F CH CH C CH d 2 5 i Give the hybridization states of the carbon atoms. excitation of e. c and d. .to 3d. 1 mark Oxygen has a much higher electronegativity than S and therefore it usually acts as an oxidant. b. b) Shape of hydrocarbons (1) Saturated hydrocarbons In saturated hydrocarbons. 141–143 The unique nature of carbon Structure and Shape of hydrocarbons 4. Bond C–C N–N F–F Cl–Cl Br–Br I–I Bond energy +346 kJmol-1 +163 kJmol-1 +158 kJmol-1 +242 kJmol-1 +193 kJmol-1 +151 kJmol-1 ii. Structure and shape of hydrocarbons Unit 4 Syllabus Notes a) Characteristic of Carbon–Carbon bond i. since σ bond has cylindrical (circular) cross section and the area doesn't change when the bond rotates. Furthermore. 122–129 Organic Chemistry. Ability to form multiple bond Furthermore. E. 3rd Edition. This characteristic is called catenation of C atom.4 Principles of Organic Chemistry. Therefore. C–C bond is the one with the highest bond strength. 2nd Edition pg.I. Shape of molecule 4.S. Bond strength Comparing with other single bond. 110–111. 52–59. 33–34. Shape of molecule Unit 4 Page 1 Topic Reference Reading I. Solomons. There is over millions of different organic compounds identified and may be billions not identified. C is capable to form relative stable double bond and triple bond. 14–23. Peter R. Bond Bond energy / kJmol-1 C–C +346 C=C +610 C≡C +837 This make a lot of different kinds of connection possible in the organic molecule. Murray. 61–65 A-Level Chemistry. The arrangement can be viewed as the repulsion of the four bonding electron pairs or the result of sp3 hybridization. 6th Edition pg. it is capable to form a stable long chain molecule.N. Ramsden pg. all σ bond are free to rotate. all C atoms are tetrahedrally bonded. Shape of methane Shape of ethane . . This gives benzene a planar structure which guarantees maximium p orbital overlapping. The p orbitals overlap above and below the plane laterally to form an extensive π orbital. Another lobe overlap with 1s orbital of H to form C–H σ bond. the actual benzene is more stable that the Kekulé structure.134 0.154 0. This is due to the delocalization of the π electrons / resonance of π electrons (a) Delocalization of π-electrons 2s C ground state [He] 2p 2 110 2s 2p → promotion C excited state [He] 1 111 sp2 p hybridization → C sp2 hybrid atomic orbital [He] 111 1 The six carbon atoms undergo sp2 hybridization. Shape of ethene (H2C=CH2) (3) Aromatic hydrocarbons By X-ray diffraction. therefore C=C and C≡C bond are not free to rotate.I.139 0. Shape of ethyne (H–C≡C–H) Type of bond Single C–C Benzene C–C Double C=C Triple C≡C Bond length / nm 0. it was found that the six C–C bonds in benzene are equivalent and with the bond length intermediate between single C–C bond and double C=C bond. One lobe of each C sp2 hybrid orbital overlaps with lobe of another C to form C–C σ bond.120 Kekulé structure According to energetic studies. Shape of molecule (2) Unsaturated hydrocarbons Rigidity of double and triple bond – Unit 4 Page 2 Rotation of the C=C and C≡C bonds will break the one π bond and two π bonds respectively. (Oxidation) Purple manganate(VII) turns to black ppt. of manganese(VI) oxide. No reaction. No reaction. immediately. (Oxidation) Purple manganate(VII) turns to green manganate(VI). No reaction. (Radical substitution) Cyclohexene Cyclohexene Decolorize immediately. (Radical substitution) No reaction. cyclohexane and cyclohexene H H H H Benzene Reaction bromine in CCl4 (in dark) Hexane Benzene No reaction. H H H H H H H H H H H H H H H H H H H H H H H H Cyclohexane Cyclohexane Similar to hexane. Upon heating. This will be explained with more details in the section of reaction mechanism. For example. (Oxidation) The chemical properties of benzene is so different from ordinary alkene e. (Electrophilic substitution) No reaction. No reaction. This is due to the extra stability from delocalization of π electrons. No reaction. its reaction with bromine is a substitution reaction (electrophilic substitution) instead of addition reaction (electrophilic addition) in other typical alkenes. Hexane No reaction. Glossary catenation electrophilic substitution electrophilic addition reaction mechanism . MnO2(s). Shape of molecule (b) Stability of benzene Unit 4 Page 3 Chemical properties of benzene. No reaction.g.I. Purple manganate(VII) decolorize immediately. (Electrophilic addition) acidified potassium manganate(VII) KMnO4(aq)/H+(aq) neutral potassium manganate(VII) KMnO4(aq) alkaline potassium manganate(VII) KMnO4(aq)/OH-(aq) No reaction. Br2 decolorizes slowly with evolution of HBr. Br2 decolorizes slowly only in the presence of Fe catalyst with evolution of HBr.(cyclohexene). hexane. MnO42(aq). It resembles the properties of alkane instead. 1 mark H H C C H H σ bond π bond Since for sp2 hybridization. The C’s in ethene are sp2 hybdrized 1 mark The C–H bond is formed by overlapping of a sp2 hybrid orbital with an atomic orbital of H. PVC is more rigid and durable than PE.I. The C=C bond consisted of a s bond and a p bond. b : 180º. b and c. a: sp3. a : 108-110º. 99 2A 3 b 3b Briefly describe the bonding in benzene and give TWO pieces of supportive evidence. There are 4 C–H bonds and 1 C=C bond in an ethene molecule. b sp. These bonds are formed by the overlapping of two sp2 hybride orbitals and of the two unhybridized p orbitals on C’s respectively. c: sp2 Suggest expected values for the bond angles α. a: sp3 b: sp c: sp2 1½ ½ mark each 94 2C 9 b i 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. the hybrid orbitals are arranged on a triangular plane. ∴ Ethene molecule is planar and the C–C–H bond angles are 120º 96 1A 3 c i ii 3c Consider the following structure : 1 mark i ii Give the hybridization states of the carbon atoms a. 3 . β and γ. but incineration of PVC causes a more serious pollution problem. g : 119-121º (Deduct ½ mark for missing out the degree notation) 1½ 3 × ½ mark 1½ 3 × ½ mark 97 1A 4 c i 4c i Give the hybridization state of the carbon atoms and the bond angles in ethene. X: i Give the hybridization state of the carbon atoms indicated by a. i Describe the bonding and shape of the ethene molecule in terms of the type and spatial arrangement of the orbitals 3 involved. b and c. Shape of molecule Unit 4 Page 4 Past Paper Question 93 1A 3 a i 94 2C 9 b i 96 1A 3 c i ii 97 1A 4 c I 99 2A 3 b 93 1A 3 a i 3a Consider the following compound. The unhybridized 2p orbitals of the overlaps laterally to form a π bond. Shape of molecule Unit 5 Page 1 Topic Reference Reading Syllabus Notes I. 84–86 Organic Chemistry. Scientist developed another theory called molecular orbital theory to explain this. Similar to Aufbau principle in building up of electrons. in H2 molecule. Solomons. In anti-bonding molecular orbital. Each molecular orbital can accommodate 2 electrons. 6th Edition pg. the molecular orbital with lower energy will be filled first. For example. N. 1s ground state O 2s 2p 1s sp2 2p Unit 5 2 2 211 → hybridized O hybridization 2 221 1 O undergoes sp2 hybridization. bonding orbital has lower energy than antibonding orbital. of atomic orbital . 23–26 4. the two 1s orbitals can interfere constructively to create a bonding molecular orbital or destructively to create a anti-bonding molecular orbital. As a result. This tends to break the two hydrogen atoms apart. Molecular Orbital Theory (Not required in A-Level) According to hybridization theory. ∑ atomic orbital = molecular orbital = no. of molecular orbital no. the electron density is concentrated on the two sides of the molecule. In bonding molecular orbital. a phenomenon arisen from unpaired electron.5 Molecular Orbital Theory D. also there is a node(an area with zero electron density) between the two hydrogen atoms. all electrons in O2 should be paired up. thus tends to bring the two hydrogen atoms together. Shape of molecule Modern Physical Chemistry ELBS pg. This approach fails to explain why liquid O2 is paramagnetic. Electron pair in bonding orbital will have opposite spin and electron pair in anti-bonding orbital will have similar spin.B. the electron density is concentrated along the internuclear axis. The hybridized sp2 orbital of the two O overlaps with each other to form a σ bond.I. no.in bonding orbital . The overlappings of 2s. No. of e. 2 unpaired electron can be found in the 2 degenerate anti-bonding π* 2p molecular orbital. This explain why oxygen is paramagnetic. Shape of molecule Molecular orbitals of oxygen Unit 5 Page 2 When the two oxygen atoms is approaching each other. 2py and 2pz give rise to 6 different energy levels. the atomic orbitals start to overlap.6) ÷ 2 =2 Glossary Past Paper Question Molecular orbital theory paramagnetism . The 12 electrons fill up the molecular orbitals according to the energy. of electrons in bonding orbital = 10 No. of e. 1s orbital is not involved in the overlapping since it is inner orbital. of electrons in anti-bonding orbital = 6 Bond order = (no.I. 2px.in anti-bonding orbital) ÷ 2 = (10 . II. Ionic bonding Page 1 Topic Reference Reading II. Ionic bonding 4.7 Modern Physical Chemistry ELBS pg. 64–66 Chemistry in Context, 3rd Edition ELBS pg. 133, 192–195 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 99–104, 145–146, 149–151 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 83–86, 115–118, 125–127, 171–173 Strength of ionic bond X-ray diffraction Atomic and ionic size II. Ionic bonding Like all other kinds of bonding, ionic bond is a result of balance between attraction and repulsion. Syllabus Notes There are electrostatic attractions among oppositely charged particle and repulsion among similarly charged particle. The ions will come to equilibrium when the attractive and repulsive forces balance one another an the ions will stay in positions in the crystal lattice. The structure of ionic compound is called giant ionic structure or giant ionic lattice. II. Ionic bonding Different ionic lattices Page 2 CsCl - Simple cubic r+ r- = 0.933 NaCl - Face-centered cubic (Octahedral) r+ r- = 0.524 Two factors govern the packing of ions to form giant lattice: Each ion tends to have the highest no. of neighbours of opposite charge, as this would increase the lattice stability. The number of nearest neighbours is called the coordination umber of the central ion. ii.. Relative sizes of the ions. The arrangement of the ions in an ionic crystal is determined by the ionic radii ratio (r+/r-): Coordination no. 8 6 4 Shape Simple cubic Face centered cubic (Octahedral) Face-centered cubic (Tetrahedral) r+/r0.73 and above 0.41 - 0.73 0.22 - 0.41 Example CsCl NaCl ZnS i. Ionic bonding is non-directional in nature. An ion is surrounded by many oppositely charged ions in all directions, i.e. no particular orientation is favoured. Characteristics of ionic compounds 1. 2. 3. 4. Good electrolytes. Hard solids because the inter-ionic forces within an ionic crystal are usually strong. High melting point. Soluble in water and insoluble in organic solvents. A. Strength of ionic bond - lattice energy By definition, mMn+(g) + nXm-(g) → MmXn(s) ∆Holat Lattice energy can be served as a good estimate for the strength of the ionic bond. It has a typical value range from 600–1000 kJmol-1. II. Ionic bonding B. X-ray diffraction The structure of an ionic crystal can be determined by a technique called X-ray diffraction. When X-rays strike a crystal, they are diffracted by the electrons in the atoms or ions. The larger the atom, the more electrons it possesses and the brighter the spot will be on the diffraction pattern. From the pattern of spots on the X-ray film, the distribution of electron in the crystal can be calculated. Points of equal density in the crystal are joined by contours giving an electron density map. Page 3 1. Electron density map of sodium chloride From the electron density map, it can be seen that electrons are concentrated near the nuclei in ionic compounds. In NaCl, ions are not perfectly spherical but tend to expand to fill up any empty space present, until their charge clouds experience repulsion from that of the neighbouring ions. The electron density between the ions fall almost to zero. Electron density map of sodium chloride C. Periodicity of ionic radius 1. Definition of ionic radius Ionic radius – the effective radius of ions in crystal lattices. It is the distance measured from the centre of the ion to the region where the electron density is zero. 2. Periodicity of ionic radius Similar to atomic radius, the size of an ion is mainly governed by the strength of the effective nuclear charge. On top of this, it is also affected by the presence of other ions. If an ion is present in a surrounding with few ions around, its electron cloud would tend to expand to fill up the space, until it is repelled by the electron clouds of other ions Size of cation It is always smaller than the size of the corresponding atom as electron is removed. The proton to electron ratio increases. The nuclear charge will have a greater effect on the fewer remaining electrons, which makes the electron cloud contracts. Size of anion It is always bigger than the size of the corresponding atom as electron is added. The additional electron enters the highest energy level without an increase in nuclear charge. This causes greater repulsion among the electrons. The above discussion applies to simple ion only. Certainly, a polyatomic ion is much larger than a simple ion because it contains a larger no. of atoms. II. Ionic bonding 3. Size of isoelectronic particles Page 4 O2- ion, F- ion, Na+ ion, Mg2+ ion are said to be isoelectronic because they contain the same no. of electron and have the same electronic configuration, – 2,8. Glossary Past Paper Question Therefore, their sizes show the order of O2- > F- > Na+ > Mg2+. electrovalent bond coordination no. X-ray diffraction isoelectronic 92 1A 3 b 95 2B 5 a ii iii 97 1A 1 c electron density map ionic radius 92 1A 3 b 3b The size of three anions is in the order: Br- < H- < I-. Briefly explain why H- is smaller than I-, but larger than Br-. I- has more e- shells ([Kr]5s25p6) than H- (1s2). 1 mark 1 mark H- is large due to electron repulsion between 2 electrons in 1s orbital. 2 95 2B 5 a ii iii 5a The table below lists some properties of the alkali metals. Standard electrode Melting point / ºC Element Atomic Ionic radius / nm First ionization potential / V radius / nm energy / kJmol-1 Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 ii Explain why the atomic radius is larger than the ionic radius in each element and why the ratio, atomic radius : 3 ionic radius is greatest in the case of Li. The atom has one more electron shell than the cation. 1 mark In M+, the nuclear charge outweighs the screening effect of the electrons and the electron cloud of M+ experiences a stronger attraction. Therefore, the ionic radii are smaller than the atomic radii. 1 marks Atomic radius : ionic radius ratio is highest in the case of Li, because it has the smallest size. In Li+, the eexperience the strongest attraction ∴ contraction in size is greatest. 1 mark iii Explain why the ionic radius of K+ is larger than that of Ca2+, although they have the same electronic 2 configuration. Electronic configuration of K+ and Ca2+ are both 1s22s22p63s23p6. 1 mark Both number and arrangement of electrons are the same in two cations. Ca2+ is doubly charged, while K+ is singly charged, the outermost e- in Ca2+ experiences a stronger attraction (effective nuclear charge). Hence the radius of 1 mark Ca2+ is smaller. 97 1A 1 c 1c Arrange, with explanation, the following chemical species in the order of decreasing size. F, O and O- 3 III. Covalent bonding Page 1 Topic Reference Reading III. Covalent bonding 4.6.0–4.6.2 Modern Physical Chemistry ELBS pg. 78–82, 112–113 Chemistry in Context, 3rd Edition ELBS pg. 101–105 Organic Chemistry, Solomons, 5th Edition pg. 15–18, 506–509 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 114–115 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 86–90, 92–94 H2+ ion Additivity of covalent radius Breaking down of additivity due to resonance Dative covalent bond III. Covalent bonding A. H2+ ion For simplicity, H2+ ion can be used as the simplest model of covalent bond. In H2+ ion, only 1 electrons and 2 protons need to be considered. Syllabus Notes There are only 3 forces involved. 1 repulsion term between the 2 protons and 2 attraction terms between the electron and 2 protons. The bond length (equilibrium distance between two nuclei) depends on the balance of the repulsion and the attraction. Electron density map for the H 2 ion + The two positive nuclei are bond together by sharing the negative electron cloud. This arrangement leads to a lower potential energy than if the electron cloud were not shared. And a lower P.E. results in greater stability. Electron density is not uniform on the electron density map, but is greater along the internuclear axis. The high concentration of electron between two positive nuclei serves to bind the nuclei together - covalent bond. The attraction is even greater with the neutral H2 molecule, where two electrons are available for bonding. This is revealed in the decrease in the internuclear distance and increase in bond dissociation energy. + H2 H2 Internuclear distance (nm) 0.104 0.074 Bond Dissociation energy (kJmol-1) 257 435 III. Covalent bonding B. Electron diffraction Page 2 The electron density map of hydrogen cannot be obtained by X-ray diffraction / X-ray crystallography because hydrogen is volatile and cannot be crystallized easily. And also, the electron density of the hydrogen atom is too low to diffract the X-ray. Another similar method called electron diffraction is used to determine the structure of a gaseous molecule. When a beam of electrons is passed through a gas or vapour at low pressure. The nuclei of atoms in gas molecule scatter (diffract) the electrons. From the diffraction pattern, the distance between the nuclei and the structure of the molecule can be determined. C. Covalent radius 1 Definition of covalent radius Covalent radius – half the distance between two atoms of the same kind held together by covalent bond. 2. Additivity of covalent radius If there is a molecule A–B, the interatomic distance A–B is approximately equal to the arithmetic mean of the A–A and B–B distance. A− B = A− A + B−B 2 and because of this simple relationship, it is possible to obtain interatomic distance of certain compound by simple addition of covalent radii. e.g. Average bond length of C–C bond : 0.154 nm : 0.074 nm Bond length of H–H bond in H2 0.154 nm + 0.074 nm By calculation, bond length of C–H bond = = 0.114 nm 2 Average bond length of C–H bond determined by experiment : 0.108 nm The two values rather agree with each other. 3. Breaking down of additivity in covalent radius and bond energy This simple additive relationship of covalent radii was found to fail for certain molecules. e.g. benzene. The reason is the molecules are actually resonance hybrid of two or more possible structures. There are compounds whose bond length differ from calculated values. e.g. benzene and nitrate ion. Compound benzene C=C C–C N–O N=O Calculated 0.134 nm 0.154 nm 0.136 nm 0.115 nm Bond lengths (nm) Experimental 0.139 nm 0.121 nm NO3− All bond lengths are found to be equivalent. This is due to the delocalization of electrons / reasonance of electrons. The actual molecule or ion will be better represented by a hybrid of these structures. ii. N.III. the bond order of the C C bond is said to be 1½ . only electrons are allowed to move. the electrons are not shifting forth and back in the structure. Resonance structures are only representations on paper. This make all the six C–C bonds equivalent. six carbon atoms are joined together in a ring by overlap of carbon sp2 hybrid orbitals. Two Lewis structures are used to represent the real structure. sp2 O sp2 O N O sp2 sp2 (3) Rules in writing resonance structures i. Furthermore.B. they are evenly distributed in the structure. The real structure of the molecule cannot be represented by any individual resonance structure. They do not exist. The energy of the actual molecule is lower than the energy that might be estimated for any contributing structure. iii. Delocalization of electron is also called resonance of electrons. Since one π bond is shared among three N–O bonds. forming C–C σ bonds. Orbital representation is always the better way to represent the actual structure. The electrons are delocalized in a system of π orbitals above and below the ring. . Instead. Page 3 or Resonance structures of benzene Representation of hybrid Since one π bond is shared between two C–C bonds. C–H σ bonds are formed by further overlap with hydrogen 1s orbitals. iv. Covalent bonding a) Resonance / Delocalization of electrons (1) Resonance structure of benzene In benzene molecule. Resonance structures exist only on paper. In writing resonance structures. (2) Resonance structure of nitrate ion Usually. formal charge is included in the resonance structure and the hybrid structure to see the distribution of the electron. the bond order of the N O bond is said to be 12 . the p-orbitals of each carbon atoms overlap with each other to formal a new molecular orbital. (3 × E(C–C) + 6 × E(C–H)) = (3 × 612 + 3 × 436) .(3 × 347 + 6 × 413) = -375 kJmol-1 (experimental value -210 kJmol-1) The calculated value is so differed from the experimental value.(-375) = 165 kJmol-1 . To verify the validity of the structure.g hydrogenation Hydrogenation of benzene H H H H Benzene H + 3H2(g) H (l) H H H H H H H H H H H H Cyclohexane Estimation of enthalpy of hydrogenation by bond energy terms Bond broken 3 C=C 3 H–H Bond formed 3 C–C 6 C–H E(C–H) E(H–H) E(C–C) E(C=C) +413 kJmol-1 +436 kJmol-1 +347 kJmol-1 +612 kJmol-1 ∆H = energy required . The difference between these values indicates that the actual structure of benzene is more stable than Kekulé structure by (-210) . some reactions of benzene are studied.energy released = ΣE(bond broken) . Covalent bonding b) Breaking down of additivity in bond enthalpies The presence of delocalization of π electrons is further confirmed by the breaking down of addivitiy in bond enthalpies.ΣE(bond formed) = (3 × E(C=C) + 3 × E(H–H)) .III. e. Page 4 H H H C C C C H C C H H or H H H H Kekulé structure of benzene H H or German scientist Kekulé proposed that benzene is a ring compound with alternating C–C single bond and C=C double bond joining the carbon atoms together. less heat is released than the calculated value. instead. they delocalized around the ring and make the 6 C–C bonds equivalent. . 165 kJmol-1 (delocalization energy) -375 kJmol-1 H H H H Resonance structure of benzene -210 kJmol-1 H H or The breakdown of the additivity rule demonstrates the importance of bond energy terms in understanding the chemical bonding in benzene. Page 5 H The double bond electrons are not localized at fixed positions.III. Covalent bonding (1) Delocalization energy This extra stability is attributed to the delocalization of the bonding electrons over all 6 carbon atoms. This distribution minimizes the repulsion among the electrons and give a more stable structure. The dative covalent bond is also known as coordinate bond. 1 Examples of H3N→BF3 and Al2Cl6 a) H3N→BF3 Boron trifluoride does not possess an octet of electrons. BF3 gas reacts with NH3 gas to give a white solid with the composition NH3BF3. this is a suitable compound with which to introduce dative covalency. two AlCl3(g) molecule dimerizes to form Al2Cl6(g) molecule. The formation of a dative bond requires an unbonded(lone) pair of electrons in one atom (donor) and an empty orbital in another atom (acceptor). both electrons being provided by one of the atoms. BF3 has a planar structure and a vacant 2p orbital.III. The vacant 3p orbital accepts lone pair electrons on chlorine atom to form dative bond. Dative covalent bond Dative covalent bond – Page 6 The linkage of two atoms by a pair of electrons. The 2p orbital accepts the lone pair from NH3 and the hybridization of B changes from sp2 to sp3. 3s 3p 3s 3p Al [Ne] 2 100 promotion→ hybridization Al* [Ne] 1 110 * sp2 3p → hybridized Al [Ne] 111 AlCl3(g) has a planar structure similar to BF3(g). 0 Cl Al Cl Cl Cl Al Cl Cl Glossary Octet rule electron diffraction additivity of covalent radius resonance delocalization delocalization energy dative covalent bond / coordinate bond hybrid structure . b) Al2Cl6 In gaseous state. 1s ground state B 2s 2p 1s 2s sp2 2p 2p 2 2 100 → promotion hybridization excited state B* 1s 2 1 110 2 111 0 → hybridized B* B undergoes sp2 hybridization and uses the three sp2 hybridized orbitals to overlap with 2p orbitals of fluorine. Covalent bonding D. It is the shortest. but the students should be able to deduce these by "book-keeping" of the lone pair and bond pair electrons. 1 mark C–C bond length of CH3–CH3 = 2 × 0. Many candidates could not give the correct structure for CO. .154 nm C–C bond in ethane is a single bond which is longer than the carbon-carbon bond in benzene. CO2 and CO32. The measured carbon-carbon bond length in benzene is 0. 92 2B 5 a iii 5a iii Draw resonance electronic structures of CO and CO2. ½ mark This question was not well answered.116 and 0. O C O O C O 95 1A 1 c 1c Account for the fact that the carbon-oxygen bond lengths in CO. CO32. The answer required descriptions or diagrams of the electronic structures of the three compounds and the deduction of the bond orders.139 nm.III. 0. Estimate the carbon-carbon bond length in ethane. The charges on atoms in these structures are neglected in many textbooks. it is 1 + 1 mark because of delocalization electrons in the ring. It is more meaningful to show lone pairs of electrons. showing lone pair electrons and charges where appropriate. Explain any difference in the carbon-carbon bond lengths in these two molecules. and few described the resonance of the CO32. Covalent bonding Page 7 92 2B 5 a iii Past Paper Question 92 2A 3 b iii 95 1A 1 c 96 2A 1 a i ii 98 2A 1 c ii 92 2A 3 b iii 3b iii The covalent radius of carbon is 0.129 nm respectively. and formal charges in these electronic structure diagrams.+ 3 ½ mark ½ mark ½ mark ½ mark Bond order = 3 / the bond formed between C & O is triple bond. 3 2 C O O C O C O C O Any 2 structures C Any 2 structures ½ mark each Very few candidates were able to draw resonance structures for CO and CO2.It is a resonance hybrid of the following : O -O C O-O -O C O O OC O- C ½ mark 1 Bond order = 13 (intermediate between C–C and C=C bond). In benzene. CO2 O C O Bond order = 2 / the bond formed between C & O is double bond. It is the longest.077 = 0.077 nm. CO C O .structures. The average bond order is larger than 1.are 0.113. Account for the formation of the adduct and draw its threedimensional structure. deduct ½ mark for no units) ii Explain why the estimated value differs from the experimental value.4EC–H + 2EC–C + 2EH–H or = -2(346) . i Estimate the molar enthalpy change for the above hydrogenation using the bond energy terms below : 2 Bond Bond energy term / kJmol-1 H–H 436 C–H 413 C–C 346 C=C 611 1 mark ∆H = -2EC–C . ∴ energy released in hydrogenation is less than that in non-conjugated system.4(413) + 2(611) + 2(436) (1 mark) 1 mark = -250 kJmol-1 (0 marks for omitting the eve sign. or Page 8 H H C H or H2C CH CH CH2 H C C H (1 mark) + H2C CH CH CH2 etc.3-diene. BF3·NH3.3-diene. Covalent bonding 96 2A 1 a i ii 1a For the hydrogenation of buta-1. . C 98 2A 1 c ii 1c ii BF3 reacts with NH3 to form an adduct.3-diene is a conjugated system / delocalization of π electrons / resonance / 1 mark mesomerism occurs in buta-1. C H (1 mark) The molecule is thus stabilized. 2 Buta-1. H2C=CH–CH=CH2(g) + 2H2(g) → CH3CH2CH2CH3(g) the experimental molar enthalpy change is -239 kJmol-1. 1 mark (Award 1 mark for the explanation: the bond energy given is only the average bond energy / bond energy of a covalent bond depends on its bonding environment) Very few candidates related the difference in ∆H values to the delocalization of π electrons in the conjugated system.III. 38–41 A-Level Chemistry. B. Bonding intermediate between ionic and covalent 4.N. Bonding intermediate between ionic anc covalent Page 1 Topic Reference Reading IV. the electron density is highest between the bonding atoms and decreases outwards. Electron density map of LiF comparing with those of NaCl and H2 NaCl (nearly purely ionic) LiF (intermediate between ionic and covalent) H2 (purely covalent) All the bonding arise from the interaction between the electrons and the nuclei. In the LiF. Incomplete electron transfer in ionic compound 1. . From the electron density map. Thomas Nelson and Sons Ltd. Moreover. its electron density map shows that it is somewhere between a typical ionic compound and covalent compound. 96–99. oppositely charged ions attract electrostatically to one another. Ionic compounds are formed by complete transfer of electron. i. both of them are electrostatic in nature. 5th Edition. Ramsden pg.IV. 6th Edition pg. In ionic compounds. In case of covalent molecule. 111–113 Organic Chemistry. 162 Intermediate type of bond Incomplete electron transfer in ionic compound Electronegativity Polarization of ionic bond Polarity in covalent bond IV. The electron density is concentrated along a line between 2 nuclei. Syllabus Notes ii. Distribution of electrons In case of ionic solid. In covalent compounds. pg. Different arrangements of the electrons and atoms will give different kind of bonding. Bonding intermediate between ionic and covalent A. iii.8 Chemistry in Context. Covalent molecules are formed by sharing of electron among the bonding nuclei. 3rd Edition. the electron density between the ions fall almost to zero. E. in certain region. the positive charged nuclei also attract electrostatically the electrons shared by the atoms involved. Solomons. 3rd Edition ELBS pg. Differences between ionic bond and covalent bond The similarity in nature of ionic and covalent bonds is reflected by the same order of magnitude of lattice energies (600-1000 kJmol-1) and covalent bond energies (300-1000 kJmol-1). one can see that electrons are concentrated near the nuclei in ionic compounds while there are substantial electron density between the nuclei of hydrogen atoms. 111–114 Chemistry in Context. 9% 15.IV. The geometrical arrangement of the ions within the crystal and the interionic distances are known. The experimental values also suggest that the bonds are stronger than the expected. Bonding intermediate between ionic anc covalent 2. These can be determined by a method called X-ray diffraction. This is also an evidence of the presence of bonding intermediate between covalent and ionic.e. the theoretical values calculated from the mathematical model close to the values determined by the Born-Haber cycle. This agrees with the basic assumption of the model. ii.5 % 1.8% 5. This is because the difference in electronegativity between sodium and halogen is very large. thus the electron transfer is almost complete and shows no evidence of electron sharing. the bond possesses some covalent characters. Assumptions of the mathematical model for calculation: i. The great discrepancy of between the theoretical values and the experimental values of silver halides and zinc sulphide implies that our model for calculation is not realistic. but it is worthy to note the basic assumptions and the limitation of the model. The bonding within the crystal is completely ionic and all ions are spherical in shape. 71–73 of Modern Physical Chemistry ELBS. silver halide and Zinc sulphide Compound NaCl NaBr NaI AgCl AgBr AgI ZnS Theoretical lattice energy -777 kJmol-1 -731 kJmol-1 -686 kJmol-1 -768 kJmol-1 -759 kJmol-1 -736 kJmol-1 -3427 kJmol-1 Experimental lattice energy from Born-Haber cycle -781 kJmol-1 -742 kJmol-1 -699 kJmol-1 -890 kJmol-1 -877 kJmol-1 -867 kJmol-1 -3615 kJmol-1 Percentage error 0. Ag and Zn compounds Page 2 Besides the Born-Haber cycle. Difference among lattice energies of Na.5% 17.9 % 15. The mathematical treatment of the calculation is beyond the scope of the syllabus. i.5% b) Bonding intermediate between covalent and ionic In sodium halide.5 % 1. Owing to the small size of the silver ion which polarizes the large halide ion easily. . no sharing of electron. the lattice energies can also be calculated in terms of electrostatic interactions among ions. Further reading about the calculation of lattice energy can be found on pg. a) Lattice energies of sodium halide. iii. This causes a displacement of the outer electrons of Y. Polarization of an ion representing a transition from ionic bonding to covalent. This is caused by a higher concentration of electron density between the nuclei than in the crystal. Thus Na+ is less polarizing than Cu+. 3. i.239 This implies a stronger bond than in the crystal. The polarizing power of a cation is increased as (a) the size is reduced and (b) the charge is increased. the X+ ion would tend to attract the outer electrons in the charge cloud of Y-. Ions of transition metal are usually much smaller than the ions of main group metals) .e.300 Internuclear separation (nm) Vapour 0. a) Fajans’ Rules in polarization of ionic bond i. From spectroscopic studies of the vapours of alkali metal halides.back to X+.IV. Polarization of ionic bond When an ionic bond is formed between the cation X+ and anion Y-. Page 3 Partial covalent properties is due to incomplete electron transfer in ionic compound and this is called polarization of ions. This arises from distortion of the electron cloud of one ion or both from a spherical distribution. Covalent character is stronger for cations which do not have a noble gas electron arrangement (d electrons have poor shielding properties).275 0. so that the outer electron of X is not completely transferred to Y. owing to its charge. Halide LiBr LiI Crystal 0. Bonding intermediate between ionic anc covalent The discrepancies implies the presence of an appreciable proportion of covalent bonding. ii. The polarizability of anion increases as (a) the size increases and (b) the charge is increased.217 0. The ionic bond formed is said to have some covalent character. the internuclear distance in these molecule is less than in the corresponding ionic bond. 1 C 2.1 As 2. Definition of electronegativity Page 4 Electronegativity – A measure of tendency of an atom in a stable molecule to attract electrons within a bond.5 Mg 1. Pauling scale of electronegativity There are several method to measure the electronegativity of an element.8 Li 1.9 O 3.5 S 2. Bonding intermediate between ionic anc covalent C.4 Te 2.8 Sn 1. An ebonite rod rubbed with fur develops a negative charge A number of liquids are tested Liquid showing a marked deflection Water Trichloromethane Propanone Ethoxyethane Nitrobenzene Cyclohexene Ethanol Liquid showing no or small deflection Tetrachloromethane Benzene Cyclohexane .8 Ge 1. 2. In H–Cl molecule. In contrast with electron affinity. fluorine. 4 is assigned to the most electronegative atom. Pauling defined the electronegativity of an atom as the power of that atom in a molecule to attract electrons. Electronegativity 1.7 Be 1. the chlorine being the more electronegative atom.5 In general.5 Si 1.9 B 2.0 Sr 1.IV. electronegativity values increase from left to right across each period and decrease down each group. Polarity in covalent bond 1. In Pauling scale.0 Ba 0. One of the commonly used method is Pauling scale of electronegativity.8 Rb 0.8 I 2. Pauling electronegativity values of some elements (The shaded ones are more electronegative than C) H 2.2 Ca 1.0 P 2.8 Cs 0. D. it is a measure of the power of an single gaseous atom to attract electrons.0 Cl 3.0 Sb 1. the bonding electrons are much closer to the chlorine than to the hydrogen.0 Br 2.5 _ _ N 3.0 Na 0.1 F 4.0 Al 1.5 Se 2.9 K 0. Deflection of a liquid jet by an electric field A dry glass rod rubbed with polythene develops a positive charge. CH3CH2Cl). it introduces some ionic character into the bond. . Cl is said to exert a negative inductive effect. b) Polarity of molecule The overall distortion of charge in molecules.1 D. in HCl. For a molecule with more than one polar bond. is known as polarization. Polarization of a covalent bond represents the existence of some ionic character in the covalent bond.g. Cl attracts electrons more strongly than H does. the center of the positive charge and the center of the negative charge do no coincide and a permanent dipole moment results. Polarization of ions represent the existence of some covalent character in the ionic bonding. the dipole moment of the HCl is 1.34 × 10-30 Cm.IV. Dipole moment Page 5 If two charges of equal magnitude but opposite in sign. 1 D ≈ 3. 2. are separated by a distance d. +q and -q. It represents the departure of the bond from purely covalent. which results from unequal sharing of electrons. then the system is said to have a dipole moment (µ) of magnitude give by: = = Dipole moment (µ) charge × distance between the +ve and -ve centres q×d This dipole moment is usually expressed in terms of Debye (D). HCl. a) Vector quantity of dipole moment Dipole moment is a vector quantity. For example. the dipole moment of the molecule is given by the vector sum of the dipole moments caused by the various polar bonds. Molecules which do not possess a permanent dipole moment are nevertheless polarised to some extent when placed in an electrical field. Bonding intermediate between ionic anc covalent The zero or very small deflection with tetrachloromethane is due to the symmetry of the molecule in which the four dipoles counteract each other exactly. In asymmetric molecules (e. The difference in electronegativity between N(3. Furthermore.IV. NH3 has a stronger dipole moment (1.0) and H(2. But in NF3. (1) Inductive effect (I) This refers to the displacement of electron cloud mainly due to difference in electronegativity.. by resonance. On another hand.0) is greater than that between N(3. if it is attached to a -R group. N is more electronegative than C and N exerts a -I effect. 2. (3) Presence of lone pair Lone pair can be considered as a negative centre and contributes a dipole moment. N. If it is attached to a +R group. But 1. ii. This is transmitted through σ bond. .B. NH2 NH2 NH2 H N H The final dipole moment would be the sum of the two effects.0) and F(4.24D). this results in a much weaker resultant dipole moment. it will behave as a -R group. Benzene ring can serves as a reservoir of electrons in resonance. NF3 should has a stronger dipole moment than NH3. experimentally. For example. it will behave as a +R group. The NF3 has a bond angle smaller than NH3 has. According to these two factors only. the lone pair imposes a dipole moment opposing the dipole moment of N–F. -R effect : electron withdrawing +R effect : electron donating For example. mesomeric effect / resonance effect – delocalization of electron via π bond. in benzenamine. -I effect : electron withdrawing +I effect : electron donating (2) Mesomeric effect / resonance effect (R) This refers to the stabilization due to delocalization of electron through π bond. lone pair on N is donated to the benzene ring and N exerts a +R effect. i. presence of lone pair. ii. iii.46 D) than NF3 (0.1). Bonding intermediate between ionic anc covalent c) Factors affecting dipole moment Page 6 The dipole moment of a molecule is affected by i. the direction of the dipole moment cannot be determined because we do not known which is stronger. inductive effect – a substituent effect on an organic compound due to the permanent polarity or polarizability of groups. Bonding intermediate between ionic anc covalent Page 7 Glossary Past Paper Question polarization of ion polarizing power polarizability Fajans rules electronegativity Pauling scale dipole moment vector inductive effect mesomeric effect / resonance effect 97 2A 1 a i ii iii 98 1A 1 c ii 97 2A 1 a i ii iii 1a i Explain the terms 'dipole' and 'dipole moment'.IV.833 . iii State the effect of an electric field on molecules of the following compounds and explain the effect in terms of dipole moment. using HBr as an example. Compound Theoretical lattice enthalpy Experimental enthalpy NaCl(s) . ii Explain why the dipole moment of HF is greater than that of HI. and the experimental lattice enthalpy of AgCl(s) are given in the table below.905 ii Why is there such a large difference between the theoretical and experimental lattice enthalpies of AgCl(s) ? 1 . Cl Cl Cl 7 Cl 98 1A 1 c ii 1c The theoretical lattice enthalpies of NaCl(s) and AgCl(s).770 ? AgCl(s) . E. ii. 146–147 Chemistry in Context. Valence electrons of individual atoms are contributed to form a sea of electrons.V. Thermal conductivity – the presence of mobile electrons Electrical conductivity – the presence of mobile electrons Malleability and ductility – non-directional electrostatic attraction between nuclei and mobile electrons Shiny lustre – The mobile electron is excited and re-emits the energy in form of lustre. e. Electron Sea model of metal The metal atoms are closely packed (high coordination number) to form a crystal lattice.on top of 2 4s e-. pg. The attraction arises from the electrostatic attraction among the positive ions. packing efficiency of the crystal Transition metal atoms are much smaller than the atoms in Group 1A and 2A.g. (Further reading about band theory of metallic bonding can be found in Modern Physical Chemistry ELBS pg. iii.14 V of Na. 10. From another point of view. Metallic bonding A. iv. 3rd Edition ELBS pg. availability of the valence electron Group 1A metal has 1 valence e-. metallic bond may be considered an delocalization of valence electrons from each atom over the entire metallic crystal. Strength of metallic bond Strength of a metallic bond depends on i. Syllabus Notes General properties of metal i. 74–76 Chemistry in Context. . 158–159) B. ii. E. Transition metal use 3d e. 137 A-Level Chemistry. effective nuclear charge In general. 5th Edition. Metallic bonding Page 1 Topic Reference Reading V. 3rd Edition. Hg has a first I. iii.44 V comparing with 5. However. Thomas Nelson and Sons Ltd. Group 2A metal has 2 valence e-. 121–123 Metallic bonding Electron Sea model Delocalization model V. the metallic bond strength increases with increasing effective nuclear charge on the valence electrons. Metallic Bonding 4. if the effective nuclear charge is too high. The lattice of positive ions is then immersed in the sea of electrons. Ramsden pg.N. mobile electrons and positive ions.9 Modern Physical Chemistry ELBS pg. this will lower the availability of valence electrons and the bond strength will drop. only small amount of the metallic bonds are broken. and account for the trend in each case. where M is a metal? The best answer out of the three is 200-700 kJ mol-1.V. Na. covalent bond and metallic bond Ionic bond (non-directional) Covalent bond (directional) Metallic bond (non-directional) 0 1000 Estimated by Lattice energy Bond energy terms Atomization energy 2000 Range (kJmol-1) -780 (NaCl) – -3791(MgO) 158 (E(F–F)) – 945. N2 (63K/77K). which is the most likely energy range for the change M(s) → M(g). Melting and boiling of metal Upon melting. Metallic bonding C. 1½ 1½ mark ½ ½ mark 98 1A 3 b i ii 3b Sketch the trends for the properties mentioned in (i) and (ii) below.700 800 . This causes a big gap between the melting point and boiling point of metal.2 (V) 3000 4000 kJmol -1 Ionic bond Covalent Metallic Glossary Past Paper Question malleability 92 1A 2 d ii iii 98 1A 3 b i ii ductility lustre 92 1A 2 d ii iii 2d ii Describe the bonding in metallic crystals. This results in a smaller gap between the melting point and boiling point. e.100 200 .1500. Strength of ionic bond. iii Of the three energy ranges in kJmol-1 given below: 5 . Mg and Al 2 .g. Na (371K/1156K). most of the attractions among the molecules have been broken upon melting.g. ‘Free electron’ model or ‘Mobile electron’ description. Na and K 2 ii boiling point of the Period 3 elements. I2 (387K/458K) D. Fe (1809K/3135K) Page 2 In molecular substance. Li.3 (Na) – 514.4 (E(N≡N)) 107. Majority of the metallic bond have to be broken upon boiling. i melting point of the alkali metals. e. V. Metallic bonding Page 3 . H2O and HF 3. B. C. Hydrogen bond A. III. Induced dipole-induced dipole attractions 2. NH3. Biochemical importance of hydrogen bond a) DNA b) Protein Strength of van der Waals' forces and hydrogen bond C. II.CH4. Simple hydrides . Dipole-dipole interactions 3. Structure and physical properties of ice 4.Intermolecular forces Intermolecular Forces I. B. Relationship between structures and properties . Van der Waals’ forces A. Dipole-induced dipole attractions Relative strength of different origins of van der Waals' forces Nature Strength Solubility and hydrogen bond Intramolecular hydrogen bond Examples of hydrogen bonding 1. E. D. F. Discovery of van der Waals' forces Origin of van der Waals' forces 1. Ethanoic acid dimer 2. 114–117 A-Level Chemistry. Van der Waals’ forces A. 116–118 Chemistry in Context. 93–95 Chemistry in Context.g. forces among the molecule. Ideal gas law : PV = nRT where P is the pressure measured V is the volume measured n is the number of mole of gas molecule R is the universal gas constant T is the temperature in Kelvin Instead. Van der Waals’ forces 4. van der Waals'' forces and hydrogen bond. Ramsden pg. an2 van der Waals' equation : ( P + V2 ) (V .g.e. Chemical properties of substances (e. boiling point and solubility) These are depending on the intermolecular forces i. Thomas Nelson and Sons Ltd. 99–103 van der Waals' forces Introduction to intermolecular forces and intramolecular forces Study of chemistry can be divided into 2 main categories : 1. Discovery of van der Waals' forces van der Waals.1 Modern Physical Chemistry ELBS pg.10. stability and reactivity) These are depending on the intramolecular forces i. 3rd Edition. Van der Waals' forces Page 1 Topic Reference Reading I. forces within a molecule – covalent bond. van der Waals' forces Hydrogen bond induced dipole-induced dipole attractions dipole-dipole interactions dipole-induced dipole attractions I. the real gas follows another equation named after him.nb) = nRT where a and b are called van der Waals' constant. was a Dutch physicist. E.g. 5th Edition.I. a is a measurement of the attraction among the particles and b is a measurement of the volume of the gas molecule. pg. Physical properties of substances (e. melting point.N. 3rd Edition ELBS pg.e. In ideal gas equation. He discovered that real gas doesn't obey ideal gas equation perfectly. Johannes Diderik (van der vals). . 2. van der Waals' equation or real gas equation. it is assumed that there is no interaction among the gas molecules and the gas molecules occupy no volume. e. 1837–1923. Intermolecular forces Syllabus Notes There are two type of intermolecular forces : van der Waals'' forces and hydrogen bond. dipole-dipole (permanent dipole-permanent dipole) interactions iii. Van der Waals' forces The values of the empirical constants a and b are depending on the nature of the gas.p.39 3. Dispersion force – it is caused by the unsymmetrical dispersion of electron cloud. 88) has a greater molecular mass than benzene (m. Induced dipole-induced dipole attractions also has many other names i. an unbranched molecule has a larger size than branched molecule. i.m. Butane (b.19 b / dm3mol-1 0. The weak forces of mutual attraction is called van der Waals forces.p. induced dipole-induced dipole attractions ii.0341 1. . hence low polarizability of fluorine atom results in very small induced dipole-induced dipole attractions.5ºC) has a higher boiling point than 2-methylpropane (b. It will induce another temporary dipole on a neighbouring molecule and an attractive force will be established between the two. Gas He N2 CO2 CH3CH2CH2CH3 NH3 a / atmdm6mol-2 0. -12ºC). B. Slight relative displacement of electrons in a molecule will give rise to a temporary electrical dipole. Induced dipole-induced dipole attractions induced dipole-induced dipole attractions is a kind of electrostatic attraction arises from fluctuation of electron cloud.m. dipole-induced dipole attractions 1. -0. In general.0372 Page 2 It can be seen that the attractions among the molecules and the size of the molecules increase from He to butane. ii. CF4 (m.0391 0. London force – it was first explained by scientist London in 1930. Strength of induced dipole-induced dipole attractions depends on the polarizability of the electrons involved. 78) but its boiling point (-128ºC) is lower than that of benzene (80ºC). electrons in a molecule are in constant motion. Temporary induced dipole-temporary induced dipole attractions – dipole is not permanent.0427 0. poly(tetrafluoroethene). The high electronegativity. this gives strong induced dipoleinduced dipole attractions. ii. larger molecular size ⇒ larger electron cloud ⇒ higher polarizability ⇒ stronger induced dipoleinduced dipole attractions A larger molecule means a larger electron cloud and higher polarizability.I. small effective nuclear charge ⇒ more diffuse electron cloud ⇒ higher polarizability ⇒ stronger induced dipole-induced dipole attractions Fluorine is highly electronegative with high effective nuclear charge on the valence electrons. Even at 0 K.123 0. Teflon.63 147 4. Origin of van der Waals' forces van der Waals' forces have 3 origins i. iv. is used to make the non-sticky surface of frying pan.0237 0. Dipole-dipole interactions (permanent dipole-permanent dipole interactions) Page 3 There is dipole-dipole interactions among all polar molecules. This means maximum amount of attractions. polar molecule always has a higher m. The molecules will arrange in a way with lowest potential energy for the system. and b. CH4 < C2H6 < C2H5Cl 1 mark C2H5Cl has one highly polarizable Cl atom and has a permanent dipole moment. while C2H6 is less polarizable than C2H5Cl and has no permanent dipole moment. 4 . than a non-polar molecule with comparable molecular mass. Van der Waals' forces 2. Explain your order by comparing the relative magnitudes and nature of the intermolecular forces. Both CH4 and C2H6 are non-polar but the molecular size and surface area of C2H6 is larger than those of CH4. The oppositely charged ends tend to attract each other and the similarly charged ends tend to repel each other.I. 3. Thus C2H5Cl is expected to have stronger intermolecular 1 mark forces than C2H6.p. This is because on top of dipole-dipole interactions and dipole-induced dipole attractions. C. and a higher boiling point. a polar molecule can also induce another temporary dipole on another molecule. Relative strength of different origins of van der Waals' forces Molecule Ar (non-polar) N2 (non-polar) CH4 (non-polar) CO (polar) HCl (polar) induced dipole-induced dipole attractions 100 % 100 % 100 % ≈ 100 % 81 % dipole-dipole interactions (permanent dipole-permanent dipole interactions) nil nil nil 0. Glossary Past Paper Question van der Waals’ forces dipole-dipole interactions 91 2A 2 a 95 2B 4 a i 98 1A 1 b i temporary dipole induced dipole-induced dipole attractions dipole-induced dipole attractions polarizability 91 2A 2 a 2a Arrange the following substances in order of increasing boiling points: C2H5Cl. Dipole-induced dipole attractions Like a temporary dipole. so 2 marks C2H6 will have a stronger van der Waals’ forces.005 % 15 dipole-induced dipole attractions nil nil nil 0. CH4 and C2H6.08 % 4% Although dipole-dipole interactions and dipole-induced dipole attractions are very weak comparing with induced dipole-induced dipole attractions.p. The attractions caused by this mean is called dipole-induced dipole attractions. there is always induced dipole-induced dipole attractions among polar molecules. with each dot '·' representing an atomic nucleus. illustrate your understanding of the two terms 'covalent radius' and 'van der Waals' radius'. Descending the group. with the increase in relative molecular size. the strength of the instantaneous dipole increases. . Boiling points of halogens depend on the strength of their intermolecular forces (van der Waal’s forces) which is 1 mark related to the strength of the instantaneous dipole of the molecule. (II) Account for the difference between the covalent radius and van der Waals' radius for iodine. Page 4 2 2 (I) Using one or more diagrams of this kind. Hence more energy is needed for the boiling of the higher halogens. 1 mark 98 1A 1 b i 1b i An iodine molecule can be represented by the diagram below. Van der Waals' forces 95 2B 4 a i 4a Explain the following facts: i The boiling points of the halogens increase as the group is descended.I. The size of proton. this will give a strong dipole. 2.g. . O and F are small. 3rd Edition. pg. the bond pair electrons would be attracted towards the latter one. N. naked proton Syllabus Notes δ+ H F δ- δ+ H F δ- intermolecular hydrogen bond bond pair very close to the electronegative atom Since the charge separation is large.II. this facilitates close approach of two dipoles. F. F. Thomas Nelson and Sons Ltd. Ramsden pg. 103–112 Hydrogen bond Strength of hydrogen bond Intramolecular hydrogen bond II. E.N.e. It is then attracted by the lone pair of another very electronegative element. 90–92 Chemistry in Context.10. When a hydrogen is bonded to a very electronegative element e. O. Necessary condition for the formation of hydrogen bond. O. A lone pair on a very electronegative atom i. N. O and F. Hydrogen bond A. Hydrogen bond Page 1 Topic Reference Reading II. A hydrogen atom attaching to a very electronegative atom i. The hydrogen atom without the bonding electron will become a naked proton.e. 5th Edition. 3rd Edition ELBS pg. N. Hydrogen bond 4. 1. 118–123 Chemistry in Context. Nature Hydrogen bond can be considered as a kind of exceptionally strong dipole-dipole interactions. N. 117–126 A-Level Chemistry. The electrostatic attraction between the naked proton and the lone pair is called hydrogen bond.2 Modern Physical Chemistry ELBS pg. 4 °C. -78. this is related to its ability to form hydrogen bond with water molecules. their value may or may not parallel to each other. 138. 25.B. 82. m. Loneliness of the proton Page 2 If the hydrogen is bonded to a more electronegative atom. 138.4 °C.II.9 °C) Arrangement of particles Ordered Attraction among the particles High Random Medium Random Low C.g. 2-methylpropan-2-ol (b. 25. ii. e.p. Hydrogen bond B.p. glucose (C6H12O6) is soluble in water but methane (CH4) is insoluble in water.p. Solubility and hydrogen bond The solubility of the molecule is parallel to its ability to attract with the water molecule.9 °C) CH3 CH2 CH2 CH2 OH If the lone pair is sterically hindered by bulky groups around it.p. e.p.p. m. The solubility is depending on the polarity rather than the size of the molecule.p. the covalent bond will be more polarized. Strength i.1 °C.6 °C). it would be less available to form hydrogen bond. 82. pentan-1-ol (b.g. This causes the proton to be more isolated and the hydrogen bond would be stronger. . m.6 °C) Steric effect – pentan-1-ol (b. -78. Though both melting point and boiling point are depending on the attraction among the particles. In certain extent. Availability of the lone pair on the electronegative atom CH3 CH3 C OH CH3 2-methylpropan-2-ol (b. m.p.1 °C. N. Intramolecular hydrogen bond is formed within the molecule. Intramolecular hydrogen bonding is impossible in 3. VI and VII hydrides. 1 per molecule in NH3 and HF. V.CH4. Boiling point of nitrophenol : H O +N -O H O +N -O O O H O +N -O O 2-nitrophenol (214ºC) 3-nitrophenol (290ºC) 4-nitrophenol (279ºC) 2-nitrophenol has lower boiling point and solubility than the other two isomer. Therefore. 2. 2 per molecule in H2O. H2O and HF CH4 91 109 NH3 195 240 H2O 273 373 HF 190 293 melting point (K) boiling point (K) There is no hydrogen bond among CH4 molecule. Intramolecular hydrogen bond Page 3 Besides intermolecular hydrogen bond.II. i.and 4-nitrophenol. ethanoic acid dimerizes to form a dimer. NH3.e. hydrogen bond may be formed within a single molecule. Furthermore. Simple hydrides . The presence of an eight-membered ring is confirmed by electron diffraction studies. The boiling point of the first member is much higher than the expected because of the presence of hydrogen bond on top of van der Waals' forces. intramolecular hydrogen bond. melting point is also depending on the packing of the crystal. Hydrogen bond D. This is because of the presence of van der Waals' forces only. E. melting process involves breaking of crystal lattice. O H3C C O H O H O C CH3 In aqueous solution. The boiling points of group IV hydrides follow the general trend of increasing boiling point with increasing molecular size. Examples of hydrogen bonding 1. Boiling point of Group IV. VI and VII hydrides . Consider the boiling point of group V. The intramolecular hydrogen bonding prevents formation of intermolecular bonding hydrogen bond between molecules. boiling point is a better indicator of strength of hydrogen bond than melting point because on top of attractions. the molecules of a carboxylic acid link up with water molecules rather than form dimers. Ethanoic acid In gas phase or organic solvent. or between a guanine unit in one chain and a cytosine unit in the other. Including the two covalent bond. the tetrahedrally coordinated oxygen atoms fit into a nonplanar. The ring structure extends in a 3-dimensional way. The two helical nucleic acid chains are held together by hydrogen bonds. This gives ice an open structure and lower density than water. Biochemical importance of hydrogen bond a) DNA (Deoxyribonucleic acids) Hydrogen bonding plays a crucial role in the structures of many biochemical substances. These hydrogen bonds are formed between specific pairs of bases on the DNA chains. 4. In ice.II. such as deoxyribonucleic acids (DNA). The density of water increases with increasing temperature up to 4ºC as the hexagonal structure continues to collapse. the energy absorbed breaks some of the intermolecular associations and the hexagonal crystal lattice collapses. oxygen is tetrahedrally coordinated. six-membered ring of water molecules associated through hydrogen bonding. such as between the adenine unit in one chain and a thymine unit in the other. Hydrogen bond 3. Structure and physical properties of ice Page 4 Each water is capable to form two hydrogen bonds. . When ice melts. The polypeptide forms a helix structure with the amide groups joined together by hydrogen bonds.II. non-polar molecule polar molecule molecule with H-bond induced-induced induced-induced dipole-dipole induced-induced dipole-dipole H-bond Range (kJmol-1) -780 (NaCl) – -3791(MgO) 158 (E(F–F)) – 945. O H H weak hydrogen bond and van der Waals' forces O strong covalent bond H O H H H Glossary hydrogen bond steric effect intramolecular hydrogen bond open structure DNA double helix structure amino acid peptide link amide group polypeptide . covalent bond and metallic bond. van der Waals’ forces have roughly the same strength. Therefore. R O H2N C H α-amino acid O H R O H H R O H H R H C OH C N C C N C C N C chain of polypeptide F.4 (E(N≡N)) 107. Hydrogen bond b) Protein Protein are made up from α-amino acid molecules. The –NH2 group condenses with –COOH group to form a peptide link. This explains why only 100ºC is required to boil water but 8000ºC is required to decompose water into hydrogen and oxygen atoms. For a H-bonded molecule. both H-bond and van der Waals' forces are very weak.2 (V) Ionic bond (non-directional) Covalent bond (directional) Metallic bond (non-directional) Estimated by Lattice energy Bond energy terms Atomization energy Comparing with ionic bond. Strength of van der Waals’ forces and hydrogen bond H-bonds have the strength in the range 5–40 kJmol-1.3 (Na) – 514. there are both hydrogen bond and van der Waals’ force. a H-bonded substance always has a higher boiling point than those not H-bonded. ( O H C N Page 5 ). contains C.p.II.0.9 ºC and 0. Alcohol molecules are held together by hydrogen bond. Indicate the hydrogen bond(s) clearly. ½ mark Methylpropan-2-ol is a tertiary alcohol and butan-1-ol is a primary alcohol. and (II) the boiling point of NH3 is higher than that of PH3. The example must show X–HLY or XLHLX 1 mark iii Explain why (I) the boiling point of CH4 is lower than that of SiH4.0) 7d Which has a higher boiling point.79 gcm-3 respectively.e. (Relative atomic masses : H 1. the molecules are less closely packed and therefore results in lower density. with a relative molecular mass less than 100.van der Waals intermolecular forces NH3 molecules . Intermolecular hydrogen bonding in H2O but not H2S H2S van der Waals forces and permanent dipole-permanent dipole interactions only 2 1 mark 1 mark 92 2C 7 d 7 A carboxylic acid P.p. and O.0. ii Draw a diagram of the structure of a compound which has hydrogen bonds. 55. and those of its isomer 2methylpropan-2-ol are 82. 3 Carboxylic acid has higher b.additional hydrogen bonding 2 1 mark 1 mark 92 2B 6 Bc i 6Bc i Account for the difference in boiling points between H2O and H2S. 7. H atom is situated between 2 small / electronegative atoms. 37.0%. PH3 molecules . A actual example of intra / intermolecular hydrogen bond. 1 mark 1 mark One of these has a lone pair of electrons / bond energy in the range of 2 – 10 kJmol-1. 3º alcohol forms weaker intermolecular hydrogen bonds so its boiling point is lower.81 gcm-3 respectively. >PH3) due to increased intermolecular attraction (hydrogen bonding) 1 mark 2 2 1 2 . ½ mark Due to steric interaction. H. H 1 mark ½ mark 93 1A 3 d 3d The boiling point and density of butan-1-ol are 117. 1 mark (II) Boiling point of NH3 higher than the expected (i. ½ mark 94 1A 2 d i ii iii 2d i Explain the term “hydrogen bonding”. ∴higher b. the carboxylic acid or its methyl ester? Explain your answer. Hydrogen bond Page 6 92 2B 6 Bc i 94 2A 2 b i 98 2B 6 a i 92 2C 7 d 94 2B 4 b Past Paper Question 92 2B 6 Ac i 93 1A 3 d 94 1A 2 d i ii iii 97 1A 1 d i ii 98 1A 2 b 99 1A 1 b iii 6Ac 92 2B 6 Ac i i Account for the difference in boiling points between NH3 and PH3. O 16.8%.2% by mass . Account for these differences. C 12. than its ester ½ mark Hydrogen bonding 1 mark O R C O H O R C O or lone pair on oxygen bonded to acidic proton Need more energy to break off H-bonding. An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H2SO4 gave the desired ester Q but with much of the starting material P unchanged. 1 mark Further.2 ºC and 0. (I) Boiling points is expected to increase with number of electrons in the molecule (CH4 → SiH4) due to increasing van der Waal’s forces. would you expect to have a higher boiling point ? Explain your answer. H2O or F2O. the intermolecular force is H-bond. 1 mark H F H F H F H F H F H F H F (diagram showing H-bonds in HF) 97 1A 1 d i ii 1d Explain why i the boiling point of HF is higher than that of HCl. Hydrogen bond Page 7 94 2A 2 b i 2b Account for each of the following: i Concentrated H3PO4 has a high viscosity. 1 mark 2 98 1A 2 b 2b Which compound. O P O H O H O H ½ mark 94 2B 4 b 4b Explain why hydrogen fluoride is a liquid at room temperature and atmospheric pressure while the other hydrogen 2 halides are gases.6 has a higher tensile strength than poly(ethene). The anomalous boiling point of HF is due to extensive H-bonding between HF molecules and other hydrogen halides have no hydrogen bondings among the molecules. 1 mark H3PO4 has a high viscosity. ½ mark and each H3PO4 molecule is capable of forming 3 H-bonds with its neighbours. With strong intermolecular forces. 99 1A 1 b iii 1b Account for each of the following : iii At 273 K. 2 In H3PO4. 98 2B 6 a i 6a Consider the structures of the two synthetic polymers shown below. CH2CH2 poly(ethene) n HN(CH2)6NHCO(CH2)4CO nylon-6. .II. ice has a smaller density than water.6 n 2 i Suggest an explanation for the fact that nylon-6. ii the boiling point of HI is higher than that of HBr. III...p.11 A-Level Chemistry. insoluble in any solvent except Hg(l) high thermal and electrical conductivities.p. Ramsden pg.0.p. 109 Chemistry in Context. and b.p. ductile. 5th Edition. high m. random arrangement in gas giant ionic structure simple molecular structure / molecular crystal in ice simple molecular structure quite inert quite inert. only conducts electricity in molten state or aqueous state. only decomposes into H and O atom at 8000K reducing agent low thermal and electrical conductivities. & b. pg. soluble in water low thermal and electrical conductivities. 468–469 The relationship between structures and properties of materials In chemistry. chemists intend to explain the chemical and physical properties of substances in term of particles and the interaction among the particles. dense. particles → attractions among particles → arrangement of particles → structure → properties Example Zn(s) Particles Zn atom Attractions among particles strong metallic bond due to delocalized electron weak metallic bond due to less delocalized electron strong ionic bond Arrangement of particles hexagonal close packing random arrangement in liquid face centered cubic Structure giant metallic structure liquid at room temperature giant ionic structure Physical properties high thermal and electrical conductivities. extremely low m. random arrangement in liquid. low thermal and electrical conductivities.1. hard.ion AgCl(s) H2O(l) Ag+ ion & Cl. relatively low m. Thomas Nelson and Sons Ltd. extremely weak intermolecular forces among molecules (van der Waals' forces only) cubic structure H2O is angular in shape. Relationship between structures and properties Page 1 Topic Reference Reading Syllabus Notes III.p.ion H2O molecule H2(g) H2 molecule ionic bond with very strong covalent character strong O–H covalent bond and relatively weak intermolecular forces among molecules (Hydrogen bond and van der Waals' forces) strong H–H covalent bond. insoluble in water low thermal and electrical conductivities.N. 3rd Edition. 402–403. relatively low m. high heat capacity Chemical properties reducing agent very weak reducing agent quite inert Hg(l) NaCl(s) Hg atom Na+ ion & Cl.p. All properties can be explained according to the following framework.p. malleable.p. Relationship between structures and properties 4. open structure in ice H–H is linear in shape. E.0 4. low heat capacity . and b. & b.. weak intermolecular forces among molecules Arrangement of particles C atoms are arranged in hexagonal layers with large gap between layers C atoms are arranged tetrahedral network glucose molecule has a ring structure Structure giant covalent structure Physical properties high thermal and electrical conductivities.p. moderate m. H and O. Glossary particle bonding / attraction arrangement structure properties .p.p. and b.. insoluble in any solvent low thermal and electrical conductivities. extremely high m.p. Relationship between structures and properties Example Graphite Particles C atom Attractions among particles strong covalent bond with delocalized electrons between C atoms.p.p. high m. and b. extremely hard. soluble in water reducing agent chars upon heating. and b. a weak reducing agent. brittle. weak van der Waals' forces between hexagonal layers strong covalent bond between C atoms strong covalent bond between C..III.. insoluble in any solvent Page 2 Chemical properties reducing agent Diamond C atom giant covalent structure simple molecule structure Glucose C6H12O6 molecule low thermal and electrical conductivities. Relationship between structures and properties Page 3 Past Paper Question 90 2A 3 c 91 1A 2 d iii 93 2B 4 b iv 94 2C 9 b iv 99 1A 1 b i 90 2A 3 c 3c Arrange the following substances in order of increasing melting point: NaF.6ºC) < HF(-89.p. the other. NaF would be expected to form ionic solid. 1410 ºC) and lead (m. 1 mark F2(-219. 1 mark Even though HF forms molecular solid but because of the presence of hydrogen bonding between H and neighbouring F atom. HF. C. Hence PVC is more rigid and more durable than PE. = 4 Lead has a close packed metallic structure. The structure of PVC can be represented by 4 1 2 1 The attraction between polymer chains of PVC is dipolar attraction while that between PE polymer is van der Waal's force which is weaker when compared with the former. 1 mark The attraction between F2 molecules is quite weak and due only to van der Waals forces would be expected to have a low melting point. Melting involves the overcoming of forces between the units of which the solid is made.N. = 12 1 mark The Si–Si in silicon is strong compared with the metallic bond in lead and in the liquid state of lead is residual metallic bond remained. The coulombic forces holding an ionic solid together are strong. C. Silicon has a giant covalent network structure. conductivity is low. Explain your order in terms of the bonding involved. which can hold the molecules together more strongly and therefore. 328 ºC). the ions are held together by strong coulombic forces and cannot be easily displaced. In iron. 1 mark 99 1A 1 b i 1b Account for each of the following : i At 298 K and 1 atm pressure.N. F2 would be expected to form molecular solids. 1 mark 94 2C 9 b iv 9b Ethene and chloroethene can undergo polymerization to give polyethene (PE) and polyvinyl chloride (PVC) respectively. Because of lack of mobility of ions. Conductivity is high because electrons can move. the iron atoms occupy the BCC sites and electrons are delocalized. 1 mark 93 2B 4 b iv 4b Account for each of the following facts: iv There is a significant difference in the melting point between silicon (m. carbon dioxide is a gas whereas silicon dioxide is a solid.III. hence NaF would be expected to have high melting point.1ºC) < NaF(993ºC) 91 1A 2 d iii 2d iii Explain why the electrical conductivities of iron and of caesium chloride are different in the solid state. iv Explain why PVC is more rigid than PE. F2. Two of the given solid HF. PVC is more rigid and durable than PE. ∴ silicon has a higher melting point than lead.p. . but incineration of PVC causes a more serious pollution problem. expected to have much 1 mark higher melting point than F2. In CsCl. D.face centred cubic 3. PVT surface Ideal gas equation (Ideal gas law) 1. D. B. Charles' law 2. C. Experimental determination of molar mass of a gas 2. C. Avogadro's law 4. Carbon dioxide (Dry ice) – face centred cubic Gas laws 1.States of Matter I. II. Molar volume Determination of molecular mass 1. Mole fraction 2. Hexagonal close packing 2. Solid state A. Relationship between partial pressure and mole fraction B. Experimental determination of molar mass of a volatile liquid Dalton's law of partial pressure 1. Allotrope of carbon a) Diamond b) Graphite 2. Boyle's law 3. Iodine – face centred cubic 2. Metallic structure 1. Silicon(IV) oxide Molecular crystal 1. Tetrahedral holes and octahedral holes 4. Gaseous state A. . Cubic close packing . Body-centred cubic structure Giant ionic structure Covalent giant structure 1. Thomas Nelson and Sons Ltd. Solid State Unit 1 Page 1 Topic Reference Reading I.I.1 Modern Physical Chemistry ELBS pg. 6 on the same layer. 119–121 Close packing structure Body centred cubic structure I. There are two kinds of close packing . – the number of atoms closest to a particular atom.g. e.c. Coordination no. Solid state 4. Cobalt. each atom is surrounded by 12 atoms.p. pg. of atoms in an unit cell . 147–149 Chemistry in Context. 45. 5th Edition.hexagonal close packing and cubic close packing. 1. 3 on the top and 3 at the bottom.N. Unit cell – smallest portion of the crystal which contains all the fundamentals (information) without repetition. Metallic structure In metal crystal. this is also called abab.. Ramsden pg. 1 12 × 6 1 2×2 3×1 No. coordination is 12. i.e.. E. Titanium and Zinc In hexagonal close packing.12. 3rd Edition. pattern. 3rd Edition ELBS pg. Since the orientation of the first and third layer are the same. Solid state Unit 1 Syllabus Notes A. of atoms in an unit cell 12 at the corner 2 on the face 3 in the cell =2 =1 =3 =6 Total no. Hexagonal close packing (74% of the space is filled) Some of the metal have hexagonal close packing (h.). 134–136 A-Level Chemistry. 154–155 Chemistry in Context. the metal atoms tends to pack with each other closely and has a high coordination no. a face centred cubic unit cell can be constructed. By looking at a four layers unit cell.c.c. Solid State 2.p.c.. of atoms in an unit cell . Examples are Copper. 6 at the second layer.). This is called abcabc. 6 at the third layer and 1 at the fourth layer.I.) is also called face centre cubic packing (f. The first layer has a different orientation with the third layer of atoms. of any atom in the structure is also 12. 1 atom at the first layer. Unit 1 Cubic close packing .face centred cubic (74% of the space is filled) Page 2 Cubic close packing (c.. pattern. Nickel and Calcium The coordination no. No. of atoms in an unit cell 8 at the corner 6 on the face 1 8×8 1 6×2 =1 =3 =4 Total no. Unit 1 Tetrahedral holes and octahedral holes Page 3 Although the crystal is closely packed. It is the space between two layers of triangularly arranged atoms.I. of atoms in an unit cell Glossary coordination no.c.) structure is not a close packing structure. tetrahedral hole hexagonal close packing cubic close packing octahedral hole body centred cubic unit cell . From another angle. No. the coordination no. Tetrahedral hole Tetrahedral hole is surround by 4 atoms. of atoms in an unit cell 8 at the corner 1 at the centre 1 8×8 1×1 =1 =1 =2 face centred cubic Total no. Body-centred cubic structure (68% of the space is filled) e. of any atom is 8. It has a smaller size than octahedral hole.g. Potassium and Chromium Body-centred cubic (b. They are called holes. There are two types of holes – tetrahedral hole and octahedral hole Octahedral hole Octahedral hole is surrounded by 6 atoms. Only 68% of the space is filled comparing with 74% in the two close packing structures. Sodium. there is still some spaces between the atoms.c. it can be seen that the six atoms are arranged in a form of octahedron. 4. In body-centred cubic structure. Solid State 3. The four atoms are arranged in a form of tetrahedron. . arrangement for atoms in different layers. 1½ Cubic closest packed structure. i Which close-packed structure does abcabcabc. 91 1A 2 d i 2d i Iron has a body-centred structure... or Atoms in both close packing arrangements all have the co-ordination number of 12. 2 marks 1 mark Both arrangements have the same packing efficiency..I. Metals having the face-centred cubic close packing has the abcabc. i Draw the unit cell representation of iron. 3 1 1 mark 92 1A 2 d i 2d The arrangement of atoms in metals can be described by the close-packing of spheres.. ii Deduce the number of atoms in one unit cell of iron. Draw a unit cell representation of iron. describe? Indicate on the diagram below one tetrahedral hole (marking it T) and one octahedral hole (marking it O). arrangement for atoms in different layers. iron has a body-centred cubic structure. Solid State Unit 1 90 1A 2 c 91 1A 2 d i 92 1A 2 d i 97 1A 1 a i ii Page 4 Past Paper Question 90 1A 2 c 2c Compare and contrast the two close packing atomic arrangements in metals. ½ mark ½ mark for O and ½ mark for T 97 1A 1 a i ii 1a At room temperature. 2 .. Metals having the hexagonal close packing has the abab. 12. E. 165–166 Chemistry in Context. Ramsden pg. of Na+ ions in the unit cell 1 at the centre 1×1=1 1 12 on the edge 12 × 4 = 3 Total no.ion = 8 Coordination no. 8 6 4 Shape Simple cubic Face centered cubic (Octahedral) Face-centered cubic (Tetrahedral) r+/r0. of S2. This depends on the relative size of the ions in the ionic crystal. Solid state Unit 2 Page 1 Topic Reference Reading I. Unit 2 Syllabus Notes If the attraction forces and the repulsion forces are not balanced. 124–126 Giant ionic structure B.73 and above 0. Thomas Nelson and Sons Ltd. of Cl. pg. of Cl. 3rd Edition.ion = 6 Zinc blende (Zinc sulphide) face centred cubic (tetrahedral) Zn2+ ions occupy half of the tetrahedral holes formed by 4 S2.ion no. of Na+ ions = 4 no. of Cl. of Zn2+ ion = 4 Coordination no.41 Example CsCl NaCl ZnS Caesium chloride Simple cubic + Cs ions occupy the centre of the cubes formed by 8 Cl. of Zn2+ ions = 4 no. the structure will be unstable.ions in the unit cell 1 4 at the corner 4 × 4 = 1 Total no.ion = 4 .73 0. of S2.ions = 4 Coordination no.0.ions = 4 Coordination no. of Cl.ions no.ions = 1 Sodium chloride face-centred cubic (octahedral) Na+ ions occupy the octahedral holes formed by 6 Cl.I.22 .2 Modern Physical Chemistry ELBS pg.0. 138–146 A-Level Chemistry. 159–162. The ions will be rearranged and another more stable structure will be resulted. of Na+ ion = 6 Coordination no. of Cs+ ion = 8 Coordination no. of Cl. of S2.41 . Solid state 4.ions in the unit cell 1 6 on the face 6×2=3 1 8 at the corner 8 × 8 = 1 Total no. of Cs+ ions in the unit cell 1 at the centre 1×1=1 Total no.ions no. 3rd Edition ELBS pg. 149–151 Chemistry in Context. of Cs+ ions = 1 no. 5th Edition.ions in the unit cell 1 6 on the face 6×2=3 1 8 at the corner 8 × 8 = 1 Total no. of Zn2+ ions in the unit cell 4 in the cell 4×1=4 Total no.N. Stable Limiting radius Unstable Stable The arrangement of the ions in an ionic crystal is determined by the ionic radii ratio (r+/r-): Coordination no. Giant ionic structure Ionic crystal is the result of a balance between attractions and repulsions. of Cl. of ions in the unit cell and ratio of coordination no. The lattice points are arranged in a simple cubic manner i. not the position of individual ions.ion as a lattice point. Solid state Unit 2 Page 2 Both the no. consider each pair of Cs+ ion and Cl. i What is the empirical formula of this compound? 1 No. of atom B per unit-cell = 12 × 4 = 3 ∴ Empirical formula of the compound is AB3. Take CsCl as the example. of atom A per unit-cell = 8 × 8 = 1 1 No.I. The position of each formula unit is called a lattice point. . The structure of an ionic crystal is labeled according to the positions of the formula units. Draw the unit cell of caesium chloride.e. Glossary Past Paper Question giant ionic structure 94 1A 1 b i ii 99 2A 2 c i ii zinc blende 94 1A 1 b i ii 1b The crystal structure of a compound AxBy can be described as a simple cubic lattice of A atoms with B atoms at the middle of all the edges. 1 mark ii What are the coordination numbers of an A atom and a B atom respectively? Coordination number of A = 6 ½ mark Coordination number of B = 2 ½ mark 99 2A 2 c i ii 2c i Consider the unit cell of calcium fluoride shown below: 1 1 ii (I) (II) (I) (II) State the respective coordination numbers of each calcium ion and each fluoride ion. 8 lattice point occupy the 8 corners of a cube. Describe the lattice of calcium ions and that of fluoride ions. Describe the lattice of caesium ions and that of chloride ions in caesium chloride. can be used to determine the empirical formula of the ionic compound. This can be explained by the structures of the two allotropes. Giant covalent structure 1. Carbon has two common allotropes – graphite and diamond. 46. The p orbitals perpendicular to the carbon layer are overlapped to form π bond.g. E. The sp3 hybrid orbitals are overlapped to form strong σ covalent bond ⇒ hard. shiny appearance . red phosphorus and white phosphorus. 5th Edition.N.5 gcm-3 does not conduct 3550 ºC Graphite black. transparent crystal hardest natural substance known 3. oxygen and ozone. high melting point and boiling point. Diamond colourless. shiny solid brittle 2.I. ⇒ brittle and low density The π electrons are delocalization above and below the layer in a direction parallel to the layer.12. 123–124 Giant covalent structure C. high density. 138–146 A-Level Chemistry. The two allotropes have different physical properties but similar chemical properties. Allotrope of carbon Unit 3 Syllabus Notes Allotropy – The existence of more than one physical form of the element in the same state. 165–166 Chemistry in Context. 152–154 Chemistry in Context. graphite and diamond. There are only weak van der Waals’ forces between the layers. 3rd Edition. 3rd Edition ELBS pg. e. Thomas Nelson and Sons Ltd. This gives a hexagonal layer structure. ⇒ higher melting point than diamond. The sp2 hybrid orbitals are overlapped to form strong σ covalent bond. Solid state Unit 3 Page 1 Topic Reference Reading I. Properties appearance hardness density electrical conductivity melting point a) Diamond Each carbon is sp3 hybridized.2 gcm-3 conducts in one direction 3730 ºC b) Graphite Each carbon is sp2 hybridized. pg. The bonding is extended to form a 3-dimensional network. ⇒ conducts electricity in one direction. Ramsden pg.3 Modern Physical Chemistry ELBS pg. Solid state 4. 159–162. SiO2 . some of the strong Si-O covalent bonds have to be broken. Glossary Past Paper Question allotrope 90 2B 4 b 92 2A 3 a i ii 98 2A 2 a i ii hexagonal layer structure 90 2B 4 b 4 Account for the following observations. The silicon and oxygen are joined together by strong covalent bond. are shown below. with strong bonds only .0 mark) CO2 . 4 2 marks 2 marks 92 2A 3 a i ii 3a i Describe the bonding and intermolecular forces in ice and in SiO2 solid. With reference to the above structures. Therefore. is an infinite three dimensional network(giant structure) solid in which each Si atom is covalently bonded in a tetrahedral arrangement to four O atoms. whereas CO2 is a gas at room temperature.three dimensional polymeric solid with very strong Si-O bonds. SiO2 solid. In order for SiO2(s) to melt. Ice consists of covalent H2O molecules held together by hydrogen bonding.I. . a lot of energy is required to break the Si–O covalent bond. Each silicon atom is joined to 4 oxygen atoms and each oxygen atom is then connected to another silicon atom. (Si-O bond must be stated. diamond and graphite. Silicon(IV) oxide Unit 3 Page 2 Quartz / Sand / silicon(IV) oxide / silicon oxide / silicon dioxide (SiO2) The structure of quartz is similar to that of diamond.exists as discrete molecules with only weak intermolecular van der Waals’ forces. explain why diamond is hard whereas graphite is soft enough to be used as lubricant. 2 2 6 i ii Comment on the three different carbon-carbon distances as indicated in the above structures. silicon(IV) oxide has a relatively high melting point 1610ºC. 2 marks 98 2A 2 a i ii 2a The structures of the two allotropes of carbon. Quartz. 2 marks ii What type of interactions must be overcome to melt these solid? Hydrogen bonding must be overcome in order to melt ice. 4b SiO2 is a solid with a high melting point. Solid state 2. Upon melting. Fillans pg. The structure of molecular crystals are determined by the shape of the basic molecular units which tend to pack together as efficient as possible. 151–152 Chemistry in Context. Solid state 4. 47. pg. 3rd Edition ELBS pg.12.N. 130–132 Molecular crystal Unit 4 Syllabus Notes D. 1. 165–166 Chemistry in Context.4 Modern Physical Chemistry ELBS pg. 212–220 A-Level Chemistry. E. Iodine – face centred cubic 2. Thomas Nelson and Sons Ltd. Carbon dioxide (Dry ice) .face centred cubic Glossary Past Paper Question molecular crystal . 138–146 Physical Chemistry. 3rd Edition. Molecular crystal Molecular crystal consists of molecules held in lattice sites by weak intermolecular forces such as van der Waals’ forces or hydrogen bonds.I. Ramsden pg. 159–162. Solid state Unit 4 Page 1 Topic Reference Reading I. 5th Edition. pg 155 Unit 1 Assignment A-Level Chemistry (3rd ed. of particle... the volume is directly proportional to the absolute temperature (Kelvin scale) at a given pressure. of mole of particle . Gas Laws 1. Thomas Nelson and Sons Ltd.). the volume of a gas is directly proportional to the number of gaseous particle. Gaseous state A. 137–139 Reading Syllabus Charles' law Molar volume Boyle's law Avogadro's law Ideal gas equation (Ideal gas law) Molar volume of gases II.3. pg. V ∝ T 2. 167 Chemistry in Context.II. Gaseous state 0. Gaseous state Unit 1 Page 1 Topic Reference Reading II. Avogadro’s Law At constant pressure and temperature. pg 163–165.). the volume is inversely proportional to the pressure at 1 a given temperature. Stanley Thornes (Publisher) Ltd.1–0.2 ILPAC 9 (2nd ed. Thomas Nelson and Sons Ltd.5.). Bell & Hyman Ltd. n : no. pg 143–144 Chemistry in Context (4th ed. V ∝ N or V ∝ n N : no.3. 5th Edition. Boyle’s Law For a given mass of gas. V ∝ P 3. Charles’ Law Notes For a given mass of gas.5.. pg 7–19 Modern Physical Chemistry (4th ed. John Murray.). PVT surface for an ideal gas B.314 JK-1mol-1 = 82. Gaseous state 4 PVT surface Unit 1 Page 2 PVT relationship of an ideal gas can be represented by a 2-dimensional surface. R and the relationship becomes PV = nRT Unit P : 1 atm = 101325 Nm-2 (Pa) = 760 mmHg V : 1 m3 = 1000 dm3 = 1000000 cm3 n : no. V ∝ P and V ∝ n ⇒ V ∝ P PV Therefore. an ideal gas equation can be derived. nT 1 By combining V ∝ T.II. of mole T : must be in Kelvin (K) R : 8.05 cm3atmK-1mol-1 . Ideal gas equation (Ideal gas law) By combining the three gas laws. nT = constant The constant is called gas constant. The molar volume at stp can be determined by using ideal gas equation. i.5 dm3 Since the value is only an approximate value. but the number of moles of ideal gas contained in A is only half of that in B. Molar volume Unit 1 Page 3 Since the volume occupied by a gas is mainly space whose size is depending on the motion of the particles only.4 dm3 1 mole × 273 K = 1 mole × 298 K V2 = 24.4 dm3. Gas Helium Hydrogen Carbon dioxide Formula He H2 CO2 Molar volume at stp. Glossary Charles' law absolute temperature (Kelvin scale) Boyle's law Avogadro's law ideal gas equation gas constant / universal gas constant molar volume standard temperature and pressure (stp) / standard condition room temperature and pressure (rtp) / room condition 94 2A 1 a i PVT surface Past Paper Question 94 2A 1 a i 1a Gas container A and B each contain an ideal gas at low pressure and 298 K.262 By using the ideal gas equation.e. all gases have the similar molar volume. usually the molar volume of a gas is assumed to be 24 dm3 at rtp. the molar volume is independent of the nature of the particles.432 22. i Calculate the ratio of the gas pressure in the two containers. PV = nRT P1V1 P2V2 n1T1 = n2T2 1 atm × V2 1 atm × 22. Gaseous state 1.II. the molar volume of a gas at different condition can also be determined. It is found that the molar volume of any gas at 0 ºC (273 K) and 1 atm is about 22. PAVA = nART 1 mark PBVB = nBRT PA n A V B 1 mark P B = n B ( V A) 1 1 1 = 2 (2 ) = 4 1 PA = 4 PB or PB = 4PA 1 mark 3 . Vm / dm3mol-1 22.396 22. For example. The condition 0 ºC and 1 atm is called standard temperature and pressure (stp) or standard condition. The volume of container A is twice that of container B. the condition 25ºC (273K) and 1 atm is called room temperature and pressure (rtp) or room condition. Gaseous State 0. 3rd Edition. PV = nRT m PV = M (RT) m where m is the mass of the gas Mm is the molar mass of the gas m Mm = PV (RT) or m ρ Mm = P (RT) where. Thomas Nelson and Sons Ltd. John Murray. Determination of molecular mass Although molecular mass of a substance is a microscopic properties which cannot be measured directly. Gaseous State Unit 2 Page 1 Topic Reference Reading II. pg. ρ is the density of the gas = V .. pg 158–160 Chemistry in Context. 168. ideal gas equation provides a link between molecular mass of a gas and some measurable quantities. molecular mass can be determined by measuring the mass. 180–181 Reading Syllabus Notes Determination of molecular mass C. PV = nRT where P is the pressure of the gas V is the volume of the gas n is the number of mole of gaseous particle R is the gas constant T is the temperature in Kelvin m The quantity of no. 5th Edition.N. 143–144 Unit 2 Assignment A-Level Chemistry. where m is the mass of the m gas and Mm is the molar mass of the gas.5.II.3 ILPAC 9 (2nd ed.).3. n = M . pressure. Ramsden pg. of mole is related to the molar mass by the relationship. E.). Thomas Nelson and Sons Ltd. volume and temperature of a gas. Since molar mass is numerically the same as molecular mass. pg 21–25 Chemistry in Context (4th ed. 0 757 3 760 atm × 1000 dm Molar Mass of CO2 = = .Mass of volumetric flask (filled with air) (the mass of air is neglected in this step) 152.00 gcm-3 Density of air at 26ºC and 757 mmHg = 0. in K mRT = PV pressure of CO2 × volume of CO2 0. a volumetric flask is used to measure the volume and the mass of the gas.809 g = mass of volumetric flask filled with CO2 .Mass of air = 47.809 g = 0.933 g = 104.8 g 26 ºC 757 mmHg Given values Density of water at 26ºC = 1. in K mRT = PV pressure of CO2 × volume of CO2 Note ! Your should keep all the units in the intermediate steps.189 g Pressure of CO2 = atmospheric pressure Volume of CO2 = volume of the flask mass of CO2 × gas constant × temp.II. In the following experiment.00 gcm-3 = 104. Mass of CO2 Mass of water = = Mass of volumetric flask filled with water .47.124 g Mass of empty volumetric flask Mass of CO2 = Mass of volumetric flask filled with air . the molar mass of the gas can be determined.993 g 47.0821 atmdm3K-1mol-1 × (273 + 26)K = 44.mass of empty volumetric flask = 47.998 g 152.9 g mass of water Volume of the flask = density of water = 1. CO2(g)).g.0. Gaseous State 1.9 g 104.00118 gcm-3 = 0. Measured values Mass of volumetric flask filled with air Mass of volumetric flask filled with CO2 Mass of volumetric flask filled with water Room temperature Atmospheric pressure = = = = = 47.9 cm3 × 0.124 g = 47.933 g .00118 gcm-3 Gas constant (R) = 0. pressure.4 gmol-1 105. volume and temperature of a gas (e.189 g × 0.0821 atmdm3K-1mol-1 Calculation Molar Mass of CO2 = mass of CO2 × gas constant × temp.47.9 cm3 Mass of air = Volume of flask × density of air = 104.8 g . Unit 2 Experimental determination of molar mass of a gas Page 2 By determining the mass.998 g . II. It is determined by comparing the mass of the hypodermic syringe before and after injection. To ensure accurate injection of liquid. the molar mass of the volatile liquid can be determined using the ideal gas equation. the liquid must be vaporized first. Unit 2 Experimental determination of molar mass of a volatile liquid Page 3 The molar mass of a volatile liquid cannot be determined directly according to ideal gas equation. The needle of hypodermic syringe must be long enough to inject the liquid into the inner part of gas syringe where the temperature is higher. 3. The volume of the vapour is measured by the volume displaced by the vapour in the heated gas syringe. the hypodermic syringe should be rinsed with the liquid first and any bubbles in the syringe should be expelled. Precautions : 1. By injecting a known mass of volatile liquid into a heated gas syringe. The hypodermic needle is equipped with a self-sealing rubber to ensure no loss of vapour from the gas syringe. Molar Mass of the volatile liquid = = mass of the liquid × gas constant × temp. Keep the syringe horizontal and avoid touching the piston or warming the barrel with you hand. . The temperature and volume readings must become steady before any record is taken. in K pressure of the vapour × volume of the vapour mRT PV The mass of the liquid is measured by a hypodermic syringe and a balance. Gaseous State 2. either of which could result in loss of liquid. This ensures complete vaporization of liquid. 2. 0821 atmdm3K-1mol-1 1 atm = 760 mmHg Calculation m PV = nRT = M RT m Page 4 = = = = = = 13.12. pressure P.0 cm3 = 61. R = 8.5.0 cm3 . Assuming that the vapor behaves ideally.00 atm × 0.0821 atmdm3K-1mol-1 × 373 K = 135 gmol-1 1.5.00 atm Volume of vapour = Volume of air and vapour in gas syringe .189 -12.0 cm3 = 0. P and T explicitly. the vapour produced by 0.269 g Temperature of vapour = temperature of the steam jacket = 100 ºC = 373 K 762 Pressure of vapour = atmospheric pressure = 760 atm = 1.920 g 100 ºC 762 mmHg 5.0 cm3 66.0610 dm3 self-sealing rubber steam jacket Molar Mass of the volatile liquid = = Glossary Past Paper Question molar mass 95 2A 3 a i 97 2A 1 b 99 1A 1 a i ii gas syringe hypodermic syringe 95 2A 3 a i 3a i Derive an expression for the molar mass M of an ideal gas. Gaseous State Unit 2 Measured values Mass of hypodermic syringe and liquid before injection Mass of hypodermic syringe and liquid after injection Temperature of the vapour Atmospheric pressure Volume of air in gas syringe before injection Volume of air and vapour in gas syringe after injection Given values Gas constant (R) = 0.0 .920 g = 0.189 g 12.31 JK-1mol-1) 2 . calculate the molar mass of the liquid. However.920 g) 762 3 × 0.0610 dm3 mass of the liquid × gas constant × temp.189 g .0 cm3. in terms of its density d.0 (13. some candidates did not express M in terms of d. and absolute temperature T. 97 2A 1 b 1b At 360 K and 101 kPa pressure.0 cm3 66. in K mRT pressure of the vapour × volume of the vapour = PV 0.0821 atmdm3K-1mol-1 × (100 + 273) K Mm 760 atm × ( 1000 dm ) = Mm = 135 gmol-1 Mass of liquid used = mass of hypodermic syringe before injection .Volume of air in gas syringe = 66.mass of hypodermic syringe after injection = 13. PV = nRT ½ mark m n=M where m is the mass of dry air ½ mark m PV = M (RT) m d M = PV (RT) = P (RT) 1 mark C i Generally well answered.II. (1 kPa = 1 × 103 Nm-2) (Gas constant.269 g × 0.226 g of a volatile liquid occupies 85. Gaseous State Unit 2 99 1A 1 a i ii 1a At 433 K.0 cm3 and a pressure of 80. (1 kPa = 1 × 103 Nm-2) ii Deduce the molecular formula of aluminium chloride under the above conditions and suggest a structure consistent with this formula.0 kPa. Page 5 . i Assuming that the aluminium chloride vapour behaves as an ideal gas.569 g of aluminium chloride produces a vapour having a volume of 96. calculate its molar mass. 0.II. . of mole of oxygen molecules mole fraction of oxygen = total no. Mole fraction Mole fraction is the percentage of a particular component in a mixture in term of number. 0. 1.2 mole of oxygen. E. 3rd Edition. Gaseous State 4. Thomas Nelson and Sons Ltd. John Murray.).pn The partial pressure is defined as the pressure the gas would exert if it was contained in the same volume as that occupied by the mixture.3 mole of carbon dioxide weighs 13.4 g. PTotal = p1 + p2 + p3.4 g + 0. Percentage by mass of oxygen = 6. 0. the mole fraction oxygen in the mixture can be calculated by 0.3 mole of hydrogen weighs 0. e.2 no.6 g and 0.25 It is different from percentage by mass.4 g mass of oxygen total mass = 6.3 = 0.3 mole of carbon dioxide. pg 156 Chemistry in Context.2 mole of oxygen weighs 6.). 140 Unit 3 Assignment A-Level Chemistry. pg 33–34 Chemistry in Context (4th ed.4 ILPAC 9 (2nd ed. 169 Reading Syllabus Notes Dalton's law of partial pressure Relationship between partial pressure and mole fraction D.32 = 32% .. Dalton’s law of partial pressure Dalton’s law of partial pressure – The total pressure (PTotal) exerted by a mixture of gases is equal to the sum of the partial pressure (p) of each constituent gas.6 g + 13. of mole of molecules = 0.g.3 + 0.N.2 + 0. In the above example. 5th Edition. pg.13. Gaseous State Unit 3 Page 1 Topic Reference Reading II..2 g. 0.3 mole of hydrogen and 0.2 g = 0. For a mixture containing 0. Thomas Nelson and Sons Ltd. Ramsden pg.II. of gaseous particle or the no.25 Pa 1 mark 3 3 2 . of mole).6 kPa × 8.98 gmol-1 1 mark Let mole fraction of N2 be x 28.324) 2 marks for intermediate answers (partial pressure. Unit 3 Relationship between partial pressure and mole fraction Page 2 According to the ideal gas equation. of mole of that kind of particle.067646 x = 0.02 x + 32.315 . the pressure of a gas is directly proportional to the no. pi ni PT = n = mole fraction of individual component in the mixture The mole fraction of individual component can be calculated from the partial pressure pi and the total pressure PT.20) Pa = 0. In the mixture ½ mark partial pressure of N2 = ¼ × 0. of mole of particles.05 Pa ½ mark partial pressure of O2 = ½ × 0. 95 2A 3 a ii 3a ii At 98. calculate the composition of the sample.5 · 233. Gaseous State 2. Assuming that dry air contains only nitrogen and oxygen and behaves ideally. By combining with Dalton's law of partial pressure.20 Pa = 0. 96 2A 2 a ii 2a ii At 298 K 1. the partial pressure exhibited by individual component in a gaseous mixture should also be proportional to the no. x is mole fraction of propane (2) ÷ (1) T2 P2 1.146 gdm-3 1. 1 : PT ∝ n 2 : pi ∝ ni where PT is the total pressure and n is the total no. Glossary Past Paper Question Dalton's law of partial pressure 91 1A 2c 95 2A 3 a ii 96 2A 2 a ii partial pressure mole fraction percentage by mass 91 1A 2c 2c A container holds a gaseous mixture of nitrogen and propane.20 Pa pressure of mixture = (0.6 kPa and 300 K.146 g dm-3. and n is the no.40 Pa = 0. P ∝ N or P ∝ n where N is the total no. of mole of particle present.x) = 28.5 473. R = 8.759 Composition of dry air 1 mark 75. Calculate the mole fraction of propane in the original gaseous mixture.20 Pa pressure was mixed with 2.9% N2 and 24.0 dm3 of N2 at 0.0 dm3 of O2 at 0.0 dm3 container.15 (1 . of mole of that component Combining 1 & 2.6 × 103 Nm-2 × 8.146 × 103 gm-3 = 98.x) T = P (1 . At -40ºC the propane completely condenses and the pressure drops to 1.0.x) = 4.3235 1 1 1 mark for correct final answer (0.x)RT2 = P2V1 where nt is total number of moles. Assuming that both N2 and O2 behave ideally. the density of a sample of dry air is 1.40 Pa in a 4. of mole. calculate the pressure of the gaseous mixture at 298 K.05 + 0.0 (1 .II.15 = 0.98 1 mark x = 0. partial volume.5 atm. The pressure in the container at 200ºC is 4. ntRT1 = P1V1 (1) (2) nt (1 .5 atm. (Gas constant. no. pi is the partial pressure of a component and ni is the no. equations and method.1% O2 C ii A number of candidates made mistakes in the numerical calculation and in the units used.31 JK-1mol-1 × 300 K = 98. they are related by the expression N = nL where L is Avogadro's constant.31 J K-1 mol-1) Molar mass of dry air 1.31 JK-1mol-1 × 300 K = 28. Effect of temperature change on molecular speeds 3. Order of reaction A. Definition of rate of reaction 1. Second order reaction 3. Graphical presentation of reaction rate 3. First order reaction a) Half life (t½) b) Carbon-14 dating c) Examples of calculation 2.Chemical Kinetics Page 1 Chemical Kinetics I. . Unit of rate of reaction Measuring the rate of reaction 1. By measuring the rate of reaction at different temperature 1 2. Rates of chemical reactions A. By plotting graph of ln rate versus T III. Differential form and integrated form of rate law 2. Zeroth order reaction Collision theory 1. B. Different approaches a) Constant amount approach b) Constant time approach 2. B. Collision theory and activation energy a) Arrhenius equation Determination of activation energy 1. B. By plotting graph of ln k versus T 1 3. Examples of different reaction 1. Interpretation of physical measurements made in following a reaction a) Volume of gas formed b) Colorimetric measurement Rate Law or rate equation 1. Order of reaction a) Experimental determination of order of reaction (1) (2) By measuring the rates of reaction at different reactant concentration By plotting graph of ln rate versus ln [X] II. Distribution of molecular speed (Maxwell-Boltzman distribution) 2. Collision theory and activation energy A. E. F. B. b) Rate determining step 2. Characteristics of catalyst 2. Order of reaction leads to interpretation of reaction mechanism at molecular level a) Multi-stages reaction (1) (2) Alkaline hydrolysis of 2-chloro-2-methylpropane Reaction between bromide and bromate(V) in acidic medium B. Energy profile of reaction A. D.Chemical Kinetics Page 2 IV. Application of catalyst a) Use of catalyst in Haber process and hydrogenation b) Catalytic converter c) Enzymes Concentration Temperature Pressure Surface area Catalyst Light V. C. Energetic stability and kinetic stability Effect of catalyst 1. Transition state 1. . Factors influencing the rate of reaction A. Theory of catalyst a) Homogeneous catalysis b) Heterogeneous catalysis 3. If the reaction is performed in solution. 356–358 Reading Syllabus Definition of rate of reaction Unit of rate of reaction Measuring of rate of reaction I. reactant is converted to product. ii.I.. Reactant → Product Rate of reactions is reflected by either rate of appearing of product or rate of disappearing of reactant.1 A-Level Chemistry (3rd ed. Definition of rate of reaction In every reaction. v. ∆ [product] mol dm-3 . vi. Rates of chemical reactions 5. volume or any physical quantities related to amount. the unit of rate of reaction would be concn time-1 (e. Thomas Nelson and Sons Ltd.g. 352–354. concentration temperature catalyst surface area pressure light Rate of a reaction depends on : 1. Rate of reaction = ∆ [product] = ∆t ∆ [reactant] ∆t [ ] : concentration Note : ∆ [reactant] ∆t because rate of reaction is always positive ∆ [reactant] but the quantity is negative.). the amount can also be expressed in term of concentration. pg. 5th Edition. . 397–399. Stanley Thornes (Publisher) Ltd. 298–300 Assignment Chemistry in Context (4th ed. 401–403 Chemistry in Context. Thomas Nelson and Sons Ltd. Rates of chemical reactions Notes A. iii. Rates of chemical reactions Page 1 Topic Reference Reading I. Unit of rate of reaction If the rate of reaction is expressed as the change in concentration of reactant or product with respect to time.. ∆t negative sign is added for the i. Rate of reaction = change in amount of product = time taken change in amount of reactant time taken The amount can be measured in term of mass.). iv. moldm-3s-1) s ∆t . I. Rates of chemical reactions B. Measuring the rate of reaction By definition, rate is the change in certain quantity with respect to time. For example, speed is the change in distance with respect to time. Rate = 1. ∆ amount ∆ time Speed = ∆ distance ∆ time Page 2 Different approaches i. constant amount approach ii. constant interval approach There are two approaches to measure the rate, a) Constant amount approach Constant amount approach – Like Formula-1 car racing, a fixed no. of laps is set. The one who finishes all the laps first would be the fastest one and would be the winner. e.g. Rate of disproportionation of thiosulphate in acidic medium Sodium thiosulphate and hydrochloric acid solution is put into a beaker. In acidic medium, thiosulphate decomposes into yellow suspension of sulphur. Once the cross on the paper is masked by the yellow suspension, the time is noted. As constant amount of sulphur precipitate is needed to mask the cross and the time taken is noted, the rate of reaction can be estimated. Rate of reaction = Amount of sulphur precipitate formed time taken 1 Rate ∝ time taken Different concentration of thiosulphate and acid can be used to study the kinetic of the reaction. I. Rates of chemical reactions b) Constant interval approach Page 3 Constant interval approach –Like Daytona 24 hours car racing, a fixed interval e.g. 24 hours is set. The one who can finish more laps in 24 hours would be the winner. If the amount of a reactant or a product can be traced at constant intervals throughout the reaction, the rate of reaction can also be determined. For example, the reaction of disproportionation of thiosulphate, S2O32-(aq) + 2H3O+(aq) → SO2(g) + S(s) + 3H2O(l), can also be followed by titrating a fixed amount (e.g. 25.00 cm3) of reacting mixture at regular interval (e.g. 2 minutes) with standard I2(aq) solution using starch indicator. I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq) Before titration, the disproportionation reaction must be quenched (stopped) first. This reaction is catalyzed by acid, therefore, quenching can be done by neutralizing the acid with excess alkali, e.g. sodium hydrogencarbonate (a weak alkali is preferred) or by sudden cooling of the reaction mixture. - d [S2O32-] Rate = dt The rate of reaction at any particular instant can be calculated from the slope. Rate = - slope It is found that the reaction is first order with respect to S2O32-(aq) and zero order with respect to H3O+(aq). Rate = - d [S2O32-] = k [S2O32-] dt In another example, for the catalytic decomposition of H2O2(l) in the presence of MnO2(s) catalyst, it can be quenched by an acidic medium where acid is the negative catalyst to the reaction. The concentration of the H2O2(l) can be determined by titrating with standard MnO4-(aq). 5H2O2(aq) + 2MnO4-(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) I. Rates of chemical reactions 2. Interpretation of physical measurements made in following a reaction Page 4 Besides titration, there are many other methods to follow a reaction. There are many quantities proportional to the amount of a chemical in the reaction mixture. e.g. volume of gas, colour, conductivity and optical activity a) Volume of gas formed In the reaction between magnesium and hydrochloric acid, hydrogen gas is evolved. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) 1 d[HCl(aq)] d[MgCl2(s)] d[H2(g)] = = dt Rate = - 2 dt dt The rate of reaction can be traced by monitoring the volume of the hydrogen gas evolved. N.B. mass All pure solids have a constant density or concentration, equals volume . Therefore, the rate of a reaction is usually depending on the size of the solid particle rather than the amount of the solid present. A graph of volume of hydrogen against time b) Colorimetric measurement If one of the reacting substances or products has a colour, the intensity of this colour will change during the reaction. This method relies on the measurement of intensity of colour, therefore, it is called colorimetry. An example will be the disappearance of the colour of bromine during the oxidation of methanoic acid by bromine: Br2(aq) + HCO2H(aq) → 2Br-(aq) + 2H+(aq) + CO2(g) The change in colour intensity could be followed by a photoelectric device in a colorimeter, as shown in the following figure. Since the colour intensity is directly proportional to the concentration of Br2(aq), therefore, Rate ∝ d (colour intensity of the solution) dt constant time approach initial rate of reaction Glossary rate of reaction constant amount approach quenching colorimetry I. Rates of chemical reactions Page 5 Past Paper Question 93 1A 2 a iii 95 2A 2 b i 96 1B 4 d i ii iii 97 2A 3 a i ii iii 98 2A 3 a iii 99 1B 7 a 93 1A 2 a iii 2a Answer the following questions by choosing in each case one of the species listed below, putting it in the box and giving the relevant equation(s). Al(s), AlCl3(s), AlO2-(aq), Na(s), CO32-(aq), Cu2+(aq), P4O10(s), S(s), S2O32-(aq), Zn2+(aq) iii Which species forms a pale yellow precipitate with dilute hydrochloric acid? S2O32-(aq) + 2H+(aq) → S(s) + SO2(g) + H2O(l) 2 marks S2O32-(aq) C Many candidates picked the wrong chemical species for the inorganic reactions in parts (i) to (iv). 95 2A 2 b i 2b In an aqueous solution, the decomposition of hydrogen peroxide in the presence of manganese(IV) oxide can be represented by the following equation: i 2 C 2H2O2(aq) 2 → 2H2O(l) + O2(g) For a given amount of manganese(IV) oxide, outline how you would use a chemical method to determine the concentration of hydrogen peroxide, at different times, in the reaction mixture. Withdraw a known volume of the reaction mixture ½ mark ½ mark and add to excess dilute H2SO4, 1 mark titrate H2O2 against standard MnO4-. 5H2O2 + 2MnO4- + 6H+ → 2Mn2+ + 8H2O + 5O2 This question was poorly answered. Candidates should brush up their skills in answering questions which involve the description of experimental procedures. MnO 2 96 1B 4 d i ii iii 4d In an experiment to determine the concentration of H2O2 contained in a rainwater sample, 5.0 cm3 of the sample were mixed with an excess of a certain transition metal complex solution, giving a coloured mixture. The absorbance of the mixture was measured by a colorimeter and was found to be 0.0273. When 5.0 cm3 of 1.46 × 10-6 M H2O2 (instead of the rainwater sample) were treated in the same way, an absorbance of 0.0398 was recorded. i Calculate the concentration of H2O2 in the rainwater sample assuming that concentration is directly proportional to 2 absorbance. 1.46 × 10-6 concn. of H2O2 in the sample = 0.0398 × 0.0273 1 mark -6 -6 1 mark = 1.00 × 10 M / 1.0 × 10 M (Deduct ½ mark for answers given in 1 × 10-6 M or 10-6 M) ii Why is the method of titration not suitable for the determination of the concentration of H2O2 in the rainwater 1 sample ? The concentration of H2O2 in rainwater is too low to be determined by titrimetric method. 1 mark iii Why is it not suitable to collect the rainwater sample for this experiment in an iron container ? Fe can be oxidized by H2O2 / transition metal ions can catalyse the decomposition of H2O2. 1 1 mark I. Rates of chemical reactions 97 2A 3 a i ii iii 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + H+(aq) + I-(aq) 3a i The progress of the reaction can be monitored by a titrimetric method. Outline the experimental procedure. ii State how the initial rate of the reaction can be determined from the titrimetric results. iii Suggest another method to monitor the progress of the reaction. 98 2A 3 a iii 3a The table below lists the rate constants, k, at different temperatures, T, for the first order decomposition of a dicarboxylic acid, CO(CH2CO2H)2, in aqueous solution : CO(CH2CO2H)2(aq) → CH3COCH3(aq) + 2CO2(g) T/K 273 293 313 333 353 k / s-1 2.46 × 10-5 4.75 × 10-4 5.76 × 10-3 5.48 × 10-2 ? iii Suggest a method to monitor the progress of the reaction. 99 1B 7 a 7a In a chemical kinetics experiment, samples of the reaction mixture are removed at regular time intervals for titrimetric analysis. Suggest TWO methods by which the reaction in the samples removed can be stopped or stowed down. Page 6 6 9 II. Order of reaction Page 1 Topic Reference Reading II. Order of reaction 5.2 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 96–99, 103–106 A-Level Chemistry Syllabus for Secondary School (1995), Curriculum Development Council, 197–198 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 403–410 Experiment – Determination of the order of reaction of disproportionation of thiosulphate in acidic medium Assignment A-Level Chemistry (3rd ed.). Stanley Thornes (Publisher) Ltd., 302–308 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 358–365 Reading Syllabus Notes Order of reaction Rate Law II. Order of reaction A. Rate Law or rate equation Rate Law or rate equation – A mathematical expression relates the concentration of a species with the rate of reaction or time. There are three different forms of rate law Examples of rate law (ordinary form) e.g. H2(g) + Br2(g) → 2HBr(g) Rate = k’[H2][Br2] ½ [HBr] 1+k’’( [Br ] ) 2 i. ordinary form ii. differential form iii. integrated form e.g. 2NO(g) + O2(g) → 2NO2(g) Rate = k[NO]2[O2] e.g. 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) Rate = k[NO]2[H2] N.B. The rate equation has no direct relationship with the stochiometric coefficient. It can only be determined by experiment. According to collision theory, the rate of reaction always increases with concentration of reactant. concentration of reactant ↑ ⇒ rate of reaction ↑ In some reactions, the rate is directly proportional to the concentration. i.e. double the concentration of a reactant, only double the rate of reaction. But in some reactions, double the concentration of a reactant may cause 4 folds increase in reaction rate. The relationship needs not be linear. e.g. H2(g) + I2(g) → 2HI(g) For the above reaction, it is found experimentally that Rate ∝ [H2] Rate ∝ [I2] Rate ∝ [H2][I2] Rate = k [H2][I2] Rate law or rate equation (an overall second order reaction) The reaction is first order with respect to H2(g) The reaction is first order with respect to I2(g) k, the proportionality constant, is known as rate constant which depends on temperature only. d [H2] = k’[H2] dt d[H2] . Order of reaction Page 2 Rate has the unit (mole dm-3 s-1) and concentration has the unit (mole dm-3). depends on actually what the rate equation is.II. k = [H ][I ] 2 2 mole dm-3 s-1 -1 3 -1 unit of k = = mole dm s mole dm-3 · mole dm-3 e.d [reactant] = dt dt (differential form of rate law) For the reaction. e.ln [H2] 0) = k’t [H2] 0 ln [H ] = k’t 2 at t = 0 at t = ∞ or [H2] 0 k't [H2] = e (Integrated form of rate law) [H2] 0 = [H2] [H2] 0 [H2] = ∞ ⇒ [H2] = 0 . Differential form and Integrated form of rate law (The derivation of the integrated form of rate law is not required in A-Level) Rate of reaction = ∆ [product] = ∆t ∆ [reactant] ∆t By using notation of calculus Rate of reaction = d [product] . the reaction is first order with respect to H2(g) .(ln [H2] . k = [NO]2[H ] 2 mole dm-3 s-1 -2 6 -1 unit of k = = mole dm s (mole dm-3)2 · mole dm-3 1. Therefore.d [H2] .[H ] = k’dt 2 f ⌠.d[H2] = k' ⌡dt ⌠ ⌡ [H2] 0 [H2]0 [H2] .g. Rate For the reaction. k.d [I2] 1 d [HI] dt = dt = 2 dt . Rate = k[NO]2[H2]. 2H2(g) + 2NO(g) → 2H2O(g) + N2(g). H2(g) + I2(g) → 2HI(g) Rate = .[ ln [H2] ][H ] = k' t 2 0 [H2] [H2] is the concentration at time = t [H2] 0 is the concentration at time = 0 .d [H2] dt = k[H2][I2] and Rate = k[H2][I2] ∴ Rate = Keep the [I2] constant by using excess I2 where k’ = k [I2]. Rate = k [H2][I2].g. the unit of rate constant. Rate In the rate equation. II.d [I-] 1 . Because it is in large excess.d [H2O2] 1 . Order of reaction 2.d [NO2] = 2 dt = dt Rate = 2 dt ∆ [product] ∆t Notation of calculus is preferable to or - ∆ [reactant] since the former one reflects the rate ∆t of reaction at a particular instant instead of average rate over a period of time. . ∆ [product] = ∆t ∆ [reactant] ∆t Rate of reaction = - Using notation of calculus Rate = d [I2] . Graphical presentation of reaction rate Page 3 For the reaction H2O2(aq) + 2I-(aq) + 2H+(aq) → 2H2O(l) + I2(aq) The completeness of the reaction can be monitored by following the concentration of one of the reactant or product. For the reaction of decomposition of NO2(g) at 300 ºC 2NO2(g) → 2NO + O2(g) 1 d [NO] d [O2] 1 . it is also the solvent.d [H+] = 2 dt = 2 dt dt = dt Note the rate of disappearing of I-(aq) is twice of the rate of disappearing of H2O2(aq). Although H2O(l) is one of the product. its concentration would be rather constant. Order of reaction 3. A zeroth order means that the rate of reaction is independent of the concentration of that reactant. . Order of reaction Page 4 Order of individual reactant – Overall order of reaction – The order of a reaction with respect to a given reactant is the power of that reactant’s concentration in the experimentally determined rate equation.II. The (overall) order of the reaction is the sum of the powers of the concentration terms in the experimentally determined rate equation. A second order means that the rate of reaction is directly proportional to the square of the concentration of that reactant. In general. A first order means that the rate of reaction is directly proportional to the concentration of that reactant. 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) Rate = k[NO]2[H2] The reaction is second order with respect to NO(g) The reaction is first order with respect to H2(g) The overall order of the reaction is 3 or the reaction is a third order reaction.g. the higher the order of a reactant. the more dependent will be the rate of reaction on the concentration of that reactant. e. the order of a reaction (i. rate) with different concentrations of the reactant (i. while the concentration of Y is kept constant. [Y]b will be a constant. The order of reaction can be obtained from the slope of the graph with (ln rate) plotting against ln [X].e. Since more than 2 readings are used to determine the order. [X[). a) can be determined by measuring the rate of reaction with different concentration of X. Keep [Y] constant by using excess Y.e. the order determined will be more accurately than the above method. X) can be determination by measuring the rates of reaction (i. For the same reaction X + Y → Z rate = k[X]a[Y]b rate = k'[X]a ln rate = ln (k'[X]a) ln rate = ln k' + a ln [X] According to the above equation. rate1 = k[X]1a[Y]b rate2 = k[X]2a[Y]b (1) ÷ (2) [X]1 a rate1 k[X]1a[Y]b rate2 = k[X]2a[Y]b = ( [X]2 ) [X]1 rate1 ln rate = a ln ( [X] ) 2 2 (1) (2) [X]1 rate1 a = ln rate ÷ ln ( [X] ) 2 2 (2) By plotting graph of ln rate versus ln [X[ The order of reaction with respect to X can also be determined by graphical method. .e.II.e. a) with respect to a reactant (i. since b is also a constant.e. Order of reaction a) Experimental determination of order of reaction (1) By measuring the rates of reaction at different reactant concentrations Suppose a reaction X+Y→Z has the rate law rate = k[X]a[Y]b Page 5 The order of reaction with respect to X (i. After 5 half-lives. half life (t½). This is the only reaction which is independent of temperature. I0 ln I = kt or I0 kt I =e I0 : abundance of radioactive isotope at time = 0 I : abundance of radioactive isotope at time = t By definition.B. of isotope It can be seen that the half life (t½) of a first order reaction is related to the decay constant (k). Order of reaction B. First order reaction Page 6 First order reaction – the rate of reaction is directly proportional to the concentration of the reactant. of isotope ∝ Rate of decay Rate of decay ∝ I -d I Rate = k I = dt By integration according to time.II.: when t = t½ then I0 = 2 I ln 2 0. the fraction remaining will be = ½×½×½×½×½ = 0. Since the rate of decay is directly proportional to the amount of isotope remains. Radiation ∝ No. This form an exponential decay curve.693 t½ = k = k I : abundance of radioactive isotope ∝ no. Radioactive decay is so special that the rate of reaction is even independent of temperature. since the rate is proportional to the number of radioactive isotopes present. a) Half life (t½) The rate of radioactive decay reaction is directly proportional to the abundance of the radioactive isotope. is the time taken by a given amount of radioactive isotope to decay to half of the original amount.125% Radioactive decay is said to be a first order reaction.03125 = 3. . Examples of different reaction 1. I0 ln I = kt 2I ln I = kt½ = ln 2 N. The 14 6 C is present as 14 6 CO2 . This 14 6 CO2 enters plants via photosynthesis and animals via the food chain.693 t½ = k = k . I0 ln I = kt 32 ln 2 ln 8 = t · t ½ ln 2 ln 4 = 5730 · t t= ln 4 · 5730 = 2 × 5730 = 11460 years ln 2 and ln 2 0. the atmosphere contains a constant concentration of 14 6 C . Order of reaction b) Carbon-14 dating In nature the. At the same time. all living things have a constant proportion of their carbon in form of or plant dies.II. is transmutated to 14 6 C by the neutron from the cosmic ray. c) Example of calculation 14 6 Half life (t½) of C = 5730 years If 32 counts per minute per gram of carbon is emitted by living organisms. it is possible to estimate the age of the specimens. C .e. 14 7 Page 7 N in the air. By comparing the content of C in archaeological specimens with that in similar living materials living at the present time. calculate the age of the fossil. it undergoes beta decay which is a first order reaction. 14 6 14 7 C isotope is radioactive. N +1 n → 6 C +1 H 0 1 14 i. It is assumed that the concentration of 14 6 CO2 in atmosphere remains unchanged throughout million of years. The number of counts is proportional to the abundance of the radioactive isotope. replacement of 14 6 C ceases while decay of 14 6 C continues. 14 6 C → 7 N + 01 e − 14 6 14 As a result of this simultaneous formation and decay of 14 6 C . when the animal 14 6 Therefore. However. while a fossilized bone gives 8 counts per minute per gram of carbon. 1 at t = t½. Consider the reaction A(l) + B(l) → C(l) + D(l) which is first order with respect to A.[B] [B] [B]0 . where k’ = k [A] Rate = k’[B]2 = d[B] . Since k’ = k [A].[B] 0 0 2 1 1 k' t½ = [B] .[B] = [B] 0 0 0 1 t½ = k' [B] 0 N.[B] 0 1 1 [B] = k' t + [B]0 1 By plotting a graph [B] versus t. the slope will equal to k’.k' t = [B] .k' t = .d[B] dt 1 1 . half life of a second order reaction is depending on the initial concentration of the reactant. In contrast to the constant half life of a first order reaction.[B] 0 1 1 . the rate constant (k) can be determined. then [A] will remain a constant throughout the reaction.k' dt = ⌠ [B]2 ⌡ 0 [B]0 t 1 .[B] . Rate = k’[B]2 the reaction becomes second order with respect to B. Rate = k[A][B]2 if we keep A in large excess.k' t = . Half life (t½) of second order reaction 1 1 [B] = k' t + [B]0 1 1 ½[B]0 = k' t½ + [B]0 1 1 k' t½ = ½[B] . Second order reaction Page 8 Second order reaction – the rate of reaction is directly proportional to the square of the concentration of the reactant..B. [B] = 2 [B] 0 . but second order with respect to B.II.k’dt = [B]2 [B] d[B] ⌠ ⌡. Order of reaction 2. it is found that the reaction is zero order with respect to both N2(g) and H2(g).kt + [A]0 1 2 [A] 0 = . Actually.kdt = d[A] t 0 [A] [A]0 [A] [A]0 ⌠-k dt = ⌠d[A] ⌡ ⌡ -kt = [[A]] . Zeroth order reaction Page 9 Zeroth order reaction – the rate of reaction is independent of the concentration of the reactant. The availability of the surface of Fe is the limiting factor of the reaction.d[A] dt .II. Order of reaction 3. This is because Fe catalyst is used in the synthesis. there is no reaction with an overall order equals zero. In synthesis of ammonia by Haber process.[A]0 [A] = . But zero order reaction with respect to individual reactant is commonly found in catalytic reaction.kt½ + [A] 0 1 1 kt½ = [A] 0 .kt = [A] . For a general zero order reaction. A → product Rate = k[A]0 = k(1) = k Rate = k = .2 [A] 0 = 2 [A] 0 [A] 0 t½ = 2 k Glossary order of reaction rate law / rate equation rate constant differential form of rate law integrated form of rate law overall order of reaction half-life carbon-14 dating decay constant . N2(g) + 3H2(g) d 2NH3(g).kt + [A]0 Half life of zero order reaction [A] = . Both N2(g) and H2(g) are in large excess comparing with Fe catalyst. 2 ln 1 = 1. The candidates should have the ability to take In and to recognize equations in the form of a straight line. if not -½) 1½ mark ∴t= 15369 × 10 −10 . ln[A] = ln[A]0 .231 : 1. ii C . i Starting from this equation. Which can be rearranged to give an equation in the form of a straight line (y = mx +b) 1 mark ln [A] = -kt + ln[A]0 ½ mark Therefore.51 × 109 years. ½ mark This is followed by taking the natural logarithm to give ln(0. A k → product . k can be determined from the slope of the plot ln[A] vs t.231 The amount of U at time t (I) = 1 ½ mark ln1231 .231 : 1. Calculate the age of the mineral if the I0 = kt where I0 is the initial I Pb : 238 U is 0. 92 1A 2 a i ii 2a For the first order reaction.51 × 109 The ratio of Pb : U = 0. i Give a rate expression for this reaction 1 2 Rate = k[A][B] 1 mark 91 2A 1 a iii 1a The radioactive decay of that all 206 82 238 92 U may be represented by 238 92 238 92 U→ 206 82 Pb + α + β.693/k. No knowledge of calculus was required in answering this question.II.000. an experimental straight line plot can be obtained by taking the ln of the ½ mark integrated rate equation. Therefore. 92 The integrated form of a first order reaction my be expressed as ln C concentration and I is the concentration at time t. t½ = 0.000 + 0. where [A] and [A] 0 are the concentrations of A at time = t and time = 0 respectively.35 × 109 years (no / wrong unit -½) (3 to 4 sig. [A] = [A]0/2. ½ mark Substituting into the integrated rate law give [A]0/2 = [A]0 e-kt. Many candidates were not able to establish the correct ratio of I0/I. Order of reaction Page 10 Past Paper Question 90 1A 2 d i 91 2A 1 a iii 92 1A 2 a i ii 93 2A 2 b i ii 94 1B 4 e i ii iii 95 1A 1 d i ii 96 1A 1 b i ii 97 2A 3 b i ii c 98 2A 2 b i ii 99 2A 3 a i ii 92 2A 2 a ii iii 95 2A 2 a ii 96 2A 3 c i ii 98 2A 3 a ii 95 2A 2 b ii 90 1A 2 d i 2d Consider the reaction A(l) + B(l) → C(l) + D(l) which is first order with respect to A.kt.231 = 1. 1 The half life of the reaction is defined by the concentration condition. fig. show how you would determine the rate constant k from a set of experimental 2 measurements of concentration at various times. 206 82 iii The age of a mineral containing comparison with that of mole ratio of Note : 206 82 238 92 U and 206 82 Pb can be estimated from the mole ratio. A good example to demonstrate that most candidates lack the most basic mathematical application / manipulation skills. and hence failed to obtain the correct answer. Without using the half-life method. = 1. but second order with respect to B. Few candidates were able to score full marks in this question.000 ½ mark The amount of U at the beginning (I0) = 1. the integrated form of the rate equation is [A] = [A] 0 e − kt .5) = -kt½. It is assumed 4 92 Pb has arisen only by the decay of the uranium and that all the subsequent decays are small in U which has a half-life of 4. The decay constant k can be calculated from the half-life. Pb : 238 U .5369 × 10-10 year-1 1½ mark k= 4. derive the relationship between the half-life t½ of the reaction and the rate constant k. From the integrated rate equation. 0 × 10-3 = k (0.20 1.0 × 10 −5 ( 0.08) ( 015) ( 0.10 y 3.15 0. Order of reaction 92 2A 2 a ii iii 2a In acid solution. Since the rate of reaction 1 mark Rate = k[ClO3-]l [Cl-]m [H+]n n −5 l m n 10 × 10 .16 0. 1 2 1 = ⇒ = ⇒ − log 4 = n log ⇒ n = 2 4 4 2 4. Similarly l = 1 and m = 1.08)( 015)( 0.08 0.20 0.10)2 2 marks k = 300 (mol dm-3)-4 s-1 (1 mark for answer.7 × 10-2 = 9 Rate = k[C]3[D]0[E]2 = k[C]3[E]2 1 mark ii Calculate the rate constant.2) . Rate = k[ClO3-][Cl-][H+]2 mol dm −3 s −1 10 × 10 −5 .10 0. ( 0.0 × 10-3 = 1 -3 1 rate 1 0.10 2.20 )x = 2.30 ) = 2.10)3(0. Rate = k[C]x[D]y[E]z 3. rate k= = = 2. 4 2 .0 × 10-5 0.20 0.7 ? 10 i Deduce the rate law of the above reaction. Using results of experiment 1 3.15 0.40 4.0 ? 10 -2 2 0.10 z 3.0 × 10 ∴z=2 1 mark rate 4 = (0.10 0.0 × 10-5 ii Determine the order of the reaction with respect to each reactant.08 0. chlorate(V) ions.20 2. slowly oxidize chloride ions to chlorine. with working max.4) n .30 0.10 0. 1 mark for correct unit) C Some candidates could not do the calculation of the rate constant.4 ? 10 -3 3 0.40 8.4 × 10-2 8 rate 1 0. [Cl-]/mol dm-3 [H+]/mol dm-3 Initial rate/mol dm-3 [ClO3-]/mol dm-3 -5 0.0 × 10-3 1 rate 1 0.08 × 10 −2 mol −3 dm9 s −1 2 + 2 ( 0.2) ( mol dm −3 ) 4 .10 0.10 3.20 ) = 3.0 ? 10 -2 4 0.15 0.08 0.10 3.0 × 10-3 ∴y=0 1 mark rate 3 = (0.10 = ∴x=3 1 mark Since rate 2 = (0. The following kinetic data are obtained at 25°C. 2 1 mark 1 mark no unit -½ mark 93 2A 2 b i ii 2b Given the following data for the reaction at 298 K 2C + 3D + E → P + 2Q -3 -3 -3 -3 -1 Experiment [C] / mol dm [D] / mol dm [E] / mol dm initial rate / mol dm s -3 1 0. ClO3-.0 × 10 0. [ClO 3 ][Cl ][H ] C Many candidates gave wrong units for the rate constant.II. due to their unfamiliarity in handling exponential values.10 0.08) l ( 015) m ( 0.10 0. 3 marks without working: all correct 2 marks or wrong 0 mark Page 11 3 iii Determine the rate constant at this temperature.0 × 10-5 0.30 2. indicating the measurements you would take. In some cases. the experimental nature of the rate results was not appreciated.0 10.0 5. (Density of CH3COCH3 = 0. Write the rate expression for the reaction. 1½ mark ½ mark Measure temperature of solution / set thermostat to 298 K ½ mark Add a fixed volume of acid to Na2S2O3 (using measuring cylinder) Start stop clock ½ mark Note time. while others used a colorimeter or filtering and weighing to quantitate sulphur.0 2.05 M I2 in KI Initial rate / mol dm-3 s-1 B 10.96 × 10 k = 0.0 10. merely stating that 'rate increases with increased concentration'. ½ mark for unit) C Many candidates were unable to deduce the quantitative relation between reaction rate and change in concentration of the reactants.00 M HCl 0.45 × 10-6 E 10.0 30.789 gcm-3) 100 × 0.078 × 4.789 rate = 4.II. The temperature control was omitted in most cases. t. 2 2 .04 × 10-6 D 5.0 50.96 × 10-6 C 10. Order of reaction Page 12 94 1B 4 e i ii iii 4e In an experiment to determine the rate constant at 298 K for the decomposition of sodium thiosulphate by a large excess of dilute hydrochloric acid.1] [ 58.0 65.0 10. the progress of the reaction was followed by colorimetric measurement. use B and D : rate ∝ [propanone] ½ mark H +. but none of these answers were acceptable. and many candidates did not give a method for measuring the time taken for the formation of a certain amount of sulphur. t.0 20.789 = 3.96 × 10-6 = k [0. taken for a certain amount of sulphur to appear was measured. the time. calculate the rate constant for the reaction at the temperature of the experiment. Some candidates had difficulty in calculating the concentrations of the reactants from the data given.47 × 10-6 i Determine the effects of the changes in concentration of each of the reactants (iodine and propanone) and the catalyst (hydrochloric acid) on the reaction rate. Some utilized titration. The results are tabulated below. use B and C : rate independent of [I2] ½ mark propane. 95 1A 1 d i ii 1d The iodination of propanone is catalysed by hydrogen ions. Under these conditions: Rate = k’[S2O32-]. 1 mark iii What quantities would you plot on a graph for the determination of the rate constant k’ at 298 K? 1 1 -1 -3 21 or 0 mark t / s and [S2O3 (aq)] / mol dm C Many candidates have previous experience of this experiment. 4 Make standard [S2O32-] solution of different molarity and place equal volume of these solutions in conical flasks.0 65.1 × 100 × 0.0 20. C. but nonetheless their answers were poor.0 20. so that the order with respect to iodine was calculated to be non-zero or fractional. i Write a balanced equation for the reaction between hydrochloric acid and sodium thiosulphate. use B and E : rate ∝ [H+] ½ mark + ½ mark rate = k [H ] [CH3COCH3] ii For mixture B. The overall equation is: CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + HI(aq) Using four mixture B.0 4. candidates did not know that rate is inversely proportional to reaction time.0 2. In (iii).65 × 10-5 mol-1dm3s-1 1 mark (½ mark for numerical answer. 1 S2O32-(aq) + 2H+(aq) → H2O(l) + SO2(g) + S(s) or Na2S2O3 + 2HCl → 2NaCl + S + SO2 + H2O 1 mark ii Describe how you would carry out this experiment. when a black “X” marked on a cardboard becomes invisible when looking through the flask.078 ] 1 mark -6 58. D and E.0 5. I 2. Composition by volume of mixture /cm3 Mixture propane water 1.0 60. 1 mark titrate H2O2 against standard MnO4-. 88 Page 13 ii 228 228 A sample containing 0. Order of reaction 95 2A 2 a ii 228 2a Ra decays by the emission of β particles. (Note : The integrated form of the rate expression for radioactive decay can be represented by the following 3 .II. 1 mark This question was poorly answered.297 g 95 2A 2 b i ii 2b In an aqueous solution. slope = -k. Nt = N0e-kt. Withdraw a known volume of the reaction mixture ½ mark ½ mark and add to excess dilute H2SO4. emitting β-particles.50 g of 88 Ra is kept in a closed container.+ 6H+ → 2Mn2+ + 8H2O + 5O2 How would you use the results obtained in (i) to show graphically that the decomposition is first order with respect to H2O2 and to find the rate constant of the decomposition ? Plot [H2O2] against t to obtain the following graph MnO 2 4 2 marks ln 2 for first order reaction t½ = constant and k = t ½ OR Plot ln [H2O2] against t to obtain a straight line 2 marks 1 mark [H2O2] ln [H O ] = . 14 6 1 1 mark 14 12 C→ 0 −1 e+ 14 7 N or 14 6 C→β+ 14 7 N ii A charcoal sample from the ruins of an ancient settlement was found to have a C/ C ratio 0. The half-life for the decay is 6. i Write an equation for the decay of 14C. (1) Explain why the 14C/12C ratio in the charcoal sample is smaller than that in living organisms. 96 1A 1 b i ii 1b Carbon-14.kt 2 2 0 ⇒ ln[H2O2] . (2) Given that the half-life for the decay of 14C is 5730 years.104 (5) 1 mark t 1 mark Nt = 0. outline how you would use a chemical method to determine the concentration of hydrogen peroxide.5 ln N = 0. Calculate the mass of 88 Ra remaining after 5 3 years.60 times that found in living organisms. calculate the age of the charcoal sample. is radioactive. where N0 and Nt respectively represent the initial number and number at time t of the radioisotope. and k is the decay constant. in the reaction mixture.) ln 2 k = t½ = 0. at different times. the decomposition of hydrogen peroxide in the presence of manganese(IV) oxide can be represented by the following equation: i ii 2H2O2(aq) 2 → 2H2O(l) + O2(g) For a given amount of manganese(IV) oxide. 14C. (The decay of a radioisotope can be described by the equation.104 year-1 1 mark 0.67 years. 5H2O2 + 2MnO4. Candidates should brush up their skills in answering questions which involve the description of experimental procedures.ln[H2O2]0 = -kt 2 marks C for first order reaction. 95 0.12 1.0 0. decay of 14C continues while 12C remains constant ∴ 14C / 12C ratio drops 1. t½ is the half-life for the decay) (1) In the piece of charcoal intake of C stops.6 = 5730 ⇒ t = 4223 ± 3 year (Deduct ½ marks for wrong /no unit) Page 14 ½ mark ½ mark 1 mark + 1 mark 96 2A 3 c i ii 3c The following equation represents the decomposition of N2O5(g) : 2N2O5(g) → 4NO2(g) + O2(g) The progress of the above decomposition can be followed by measuring the partial pressures of N2O5(g).04 0.43 0.10 ln(PN2O5) 1.slope = k = 1. Order of reaction equation : N0 0.77 1.39 2.99 × 10-3 minute-1 (2.19 1.4 0. t.301 log10 ( N ) = t t t ½ where N0 is the initial number of radioactive nuclei.4 800 2. at 2 5 different times.3 100 10. find the order of the decomposition. Nt is the number of radioactive nuclei at time t.8 × 10-3 min-1) (1 mark for numerical answer.301 t (2) log 0. PN O .II.97 1.59 2.48 0.7 1250 1.86 0.77 0.2 400 5. From the plot of ln(PN2O5) against t. .99 0.9 200 8. The table below lists the results of such an experiment : t / minute PN O / kPa 2 5 0 13.9 300 7. 1 mark for correct labelling of the axes) the reaction is 1st order Using your graph in (i). determine the rate constant of the decomposition.2 × 10-3 – 1.1 3 i By plotting a suitable graph.9 550 4. ½ mark for unit) 1 mark 2 1 mark 1 mark . t 0 100 200 300 400 550 800 1250 2.04 or log(PN2O5) Plotting the following graph 2 marks ii (1 mark for the straight line. 0 × 10-2 4. One of the assumptions is that all 40Ar present in the rock sample is derived from the decay of 40K. in aqueous solution : CO(CH2CO2H)2(aq) → CH3COCH3(aq) + 2CO2(g) T/K 273 293 313 333 353 k / s-1 2.76 × 10-3 5.5 × 10-5 2.0 × 10-3 -4 -4 -1 1.4 × 10-5 For this reaction.0 × 10-3 -5 -4 -1 3. Give one other assumption relating to 40Ar.0 × 10 12. CO(CH2CO2H)2. calculate the rate constant. and 9 i ii .0 × 10 5.27 × 109 years.0 × 10 6. ii The above method of estimation is based on several assumptions. k. 3c Suppose that the reaction takes place in a buffer solution of pH 4.5 × 10-4 2. Order of reaction 97 2A 3 b i ii c 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + H+(aq) + I-(aq) 3b The following initial rates and initial concentrations were obtained in an experiment at 298 K: Initial concentration / mol dm-3 -3 -1 Initial rate / mol dm s [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.0 × 10 4. T. deduce its rate equation.0 × 10-1 5.0 × 10 8. Page 15 5 4 98 2A 2 b i ii 5 2b Potassium-40 is radioactive and decays to give the stable isotope.4 × 10 2.48 × 10-2 ? ii Estimate the rate constant of the reaction at 353 K and hence calculate the half-life of the reaction at the same temperature. argon-40.5 × 10 4.46 × 10-5 4.4 × 10-5 -2 -2 8. 99 2A 3 a i ii 3a Consider the following data for the reaction A + B → products Initial concentration / mol dm-3 [A] [B] Initial rate / mol dm-3s-1 4. the ratio of 40K to 40Ar is 1 to 9.75 × 10-4 5.0 × 10 1. for the first order decomposition of a dicarboxylic acid. ii Calculate the rate constant for the reaction at 298 K.5 × 10 2.0 × 10-3 i Deduce the rate equation for the reaction.5 × 10 1. at different temperatures. The half-life of the decay is 1.8 × 10-5 -2 -2 4. Estimate the age of the rock sample.0 × 10-2 -5 -4 -1 7. On the basis of your results in (b) deduce the half-life of the reaction at 298 K.0 × 10-2 6.5 × 10 4. 98 2A 3 a ii 3a The table below lists the rate constants. i In a rock sample.0 × 10 5.II.0 × 10 2. Collision theory and activation energy 5.III. The kinetic energy of the gas increases with increasing temperature. 369 Notes For a sample of gas at a specific temp. Thomas Nelson and Sons Ltd. some particles gain energy from the others and some particles loss energy to the others. 410–412 Reading Syllabus Distribution of molecular speed Effect of temperature change on molecular speeds Collision theory and activation energy Arrhenius equation III.. Effect of temperature change on molecular speeds Maxwell-Boltzmann distribution As temperature increases. Only very few have very high or very low speed (extreme speeds).3 Unit 1 Assignment A-Level Chemistry (3rd ed. it is found that most particles possess speeds about average speed. . 266–367. pg. Collision theory 1. 310–311 Chemistry in Context (4th ed. Maxwell-Boltzmann distribution is different from normal distribution in the way that it is not symmetrical. Origin of Maxwell-Boltzmann distribution Gas particles are in constant random motion. different particles will have different speed. 5th Edition. This makes the height of the peak drops at higher temperature.. Collision theory and activation energy Unit 1 Page 1 Topic Reference Reading III. Stanley Thornes (Publisher) Ltd. This causes a wider spread in particle speed. After a series of collisions.). During collisions. 172–173.). 2. This gives a Maxwell-Boltzmann distribution. of particle and should remain constant and independent of the temperature. so the peak of most probable speed shifts to the right. collisions among the particles become more vigorous. Collision theory and activation energy A. 140–142. Distribution of molecular speed (Maxwell-Boltzmann distribution) Chemistry in Context. Thomas Nelson and Sons Ltd. The area under the curve represents the total no. Meanwhile. Reaction is something more than just collision. Different reaction has different value of EA. every 10 K increase in temperature will double the reaction rate. where = = shaded area total area no. Unit 1 Page 2 Collision theory and activation energy To simplify the discussion. it is assumed that a reaction will only take place if the collision has an energy exceeds the activation energy EA. Z∝c Conclusion 1 – Reaction takes place as a result of collision. all reacting particles must get in contact with other for reaction to take place. an increase in 10 K at room temperature will only cause 1. The collision frequency (Z) is proportional to the velocity of the particle (c). The simple collision theory is contradictory to the experimental finding. Rate ∝ Z and therefore Rate ∝ e − EA RT − EA RT Rate ∝ Ze Z : collision frequency R : (universal) gas constant = 8.III. − EA RT increases exponentially with In order to explain the expontential increase of reaction rate with temperature. by mathematical approximation. of particles e − EA RT The value e temperature.6 % increase in particles velocity. Except in self decomposition. For most reactions. a sample of ideal gas is considered first. the fraction of the shaded area is represented by e EA is called the activation energy. Experimental finding finds that reaction rate increases exponentially with increasing temperature. Collision theory and activation energy 3. of particles having energy greater than EA total no. . activation energy. − EA RT . Refering to the Maxwell-Boltzmann distribution of molecular speed.314 JK-1mol-1 EA : activation energy T : temperature in Kelvin Conclusion 2 – Reaction only occurs when the collision exceeds certain threshold energy. III. Concentration of reactants Glossary Maxwell-Boltzmann distribution Gas constant (R) threshold collision theory activation energy Arrhenius constant (factor) collision frequency (Z) . Steric factor & Orientation factor) 3. rate constant (a constant depends on temperature only) Arrhenius constant activation energy : activation state factor (fraction of particle has energy greater than EA at temperature T) Rate of reaction = k [Reactants]n = Ae − EA RT [Reactants]n From the above equation. it can be seen that the rate of a reaction is depending on 1. Correct orientation of collision a) Arrhenius equation The equation k = Ae − EA RT is called Arrhenius equation. thus rate of reaction. Exceed activation energy iii. Temperature 2. It is a factor reflects the steric factor and the fraction of collisions with effective orientation. Conclusion 3 – Reaction only occurs when the colliding particles are correctly oriented to one another. Collision theory and activation energy Unit 1 Page 3 Since the rate of reaction is directly proportionally to the rate constant. P is found to be a factor always smaller than 1. Rate ∝ k k ∝ Ze − EA RT Z : collision frequency P : steric factor / orientation factor (a proportionality constant) where A = PZ A is called Arrhenius factor k = PZe k = Ae − − EA RT EA RT Through experiment. k: A: EA : e − EA RT This equation relates the temperature and activation energy with rate constant. i. a reaction will take place only if the following 3 conditions are satisfied and the collision is then called an effective collision.e. Collision ii. Nature of the reaction (Activation energy. According to collision theory. i. k. Plot of Boltzmann distribution curve at two temperatures T1 and T2. Collision theory and activation energy Unit 1 Page 4 Past Paper Question 96 2A 1 b i ii 98 2A 3 b i ii 96 2A 1 b i ii 1b i Draw the Maxwell-Boltzmann curves for the distribution of molecular speeds at two different temperatures for an ideal gas. force exerted by the collision of molecules on the container wall / the chance in momentum of molecules upon collision / frequency of collision increases. where T2 > T1 2 ii (½ marks for correct shape of the curves. at two different temperatures. ½ marks for indicating T2 > T1) Use your curves in (i) to explain why. 1 mark Hence. As temperature increases. i Sketch curves to show the distribution of molecular kinetic energy of the reactant. the pressure increases as the temperature is raised. explain why the rate of reaction (1) increases with temperature. . fraction of molecules with high average speed / kinetic energy increases.III. for a fixed mass of an ideal gas at constant volume. 1 mark 2 98 2A 3 b i ii 3b The exothermic reaction E(g) → E'(g) (1) is a single stage reaction. ii With reference to your answer in (i). E(g). . the activation energy of a particular reaction can be determined experimentally. can be determined. . The activation energy.4 A-Level Chemistry Syllabus for Secondary School (1995).RT ln e EA ln k = ln A . Collision theory and activation energy 5. 100–101 Chemistry in Context. Collision theory and activation energy Unit 2 Page 1 Topic Reference Reading III. 367–368 Determination of activation energy B. 199–200 Unit 2 A-Level Chemistry (3rd ed.RT where ln e = 1 1.. k = Ae − EA RT Syllabus Notes ln k = ln Ae − EA RT ln k = ln A + ln e − EA RT EA ln k = ln A . T2 can be obtained. T1 and Rate2.RT EA ln k2 = ln A . Stanley Thornes (Publisher) Ltd. 2 EA EA ln k1 . pg.RT ) 1 2 1 2 EA 1 1 k1 ln k = .).(ln A .T ) 2 1 2 EA 1 1 Rate1 ln Rate = . 365. EA.R ( T .ln k2 = (ln A . R = 8. Curriculum Development Council. Thomas Nelson and Sons Ltd. Determination of activation energy By rearranging the Arrhenius equation. 2 set of data – Rate1. Thomas Nelson and Sons Ltd. By measuring the rate of reaction at different temperature Rate ∝ k Rate1 k1 Rate2 = k2 By measuring the rate of reaction at 2 different temperature.).T ) 2 1 2 By putting the Rates and temperature determined.R ( T .314 J-1K-1mol-1. 5th Edition.RT ) . EA ln k1 = ln A . 2 Eq.. 1 Eq.Eq. 1 .RT Eq. 413–414 Advanced Practical Chemistry. 312 Reading Assignment Chemistry in Context (4th ed.III. John Murray (Publisher) Ltd. the slope of the graph will equal to . By plotting graph of ln rate versus 1 T 1 Actually. Since Rate ∝ k Rate ∝ Ae − and EA RT k = Ae − EA RT Rate = rAe Rate = qe − EA RT r : a proportionality constant q : another proportionality constant = rA − EA RT E − A RT ln rate = ln qe ln rate = ln q + ln e − EA RT EA ln rate = ln q .R . (Refer to order of reaction and rate constant). To determine the Ea. where ln e = 1 . (Refer to order of reaction and rate constant).R T EA 1 By plotting a graph ln rate versus T .III. many sets of values of rate / concentration at different temperature have to be measured. 3. A and EA can then be determined by plotting a graph with ln k versus T k at different temperature can be determined by measuring the rates of a reaction at different temperatures according to the integral form of rate equation. a set of values of rate / concentration have to be determined first. it is very time consuming to use plotting of ln k versus T to determine the activation energy.RT EA 1 ln rate = ln q . This provides a more convenient way to determine the activation energy if the determination of the Arrhenius constant is not required.RT ln e EA ln rate = ln q . Collision theory and activation energy 2. By plotting graph of ln k versus 1 T Unit 2 Page 2 1 To be more accurate. In order to determined a value of k at a certain temperature. Calculate i the activation energy. as the temperature changes. . ½ mark The poor performance to the first part of this question was disappointing.91 × 10 −4 8.91 × 10 (wrong unit -½) 2 marks k723 = 1. E ln k = ln A − a RT kT Ea 1 1 ln = ( − ) 1 mark kT R T1 T2 T1 = 273 + 370 = 643 K ½ mark T2 = 273 + 470 = 743 K −2 E 4.fig. 2 1 4 2 92 1A 2 e 2e The diagram below gives the Maxwell-Boltzmann energy distribution of a system of molecules at two temperatures: 2 C What do the shaded areas of the curves represent and why are they different at different temperatures? These curves plot the distribution of molecular kinetic energies at different temperatures.III.2 × 103 1 1 ( ) ln( )= − −4 8. Collision theory and activation energy Unit 2 Page 3 Glossary Past Paper Question Arrhenius equation 91 2A 3 a i ii 92 1A 2 e 93 2A 2 a 94 2A 1 c iii 96 1A 1 b iii 97 2A 3 d 98 2A 3 a i 91 2A 3 a i ii 3a For the reaction 2XY(g) → X2(g) + Y2(g). ½ mark The increase of the reaction rate with temperature corresponds closely to the ratio of the corresponding areas ½ mark where Ea represents the minimum collision energy necessary for the reaction to occur.05 × 10 1 1 ) ln ½ mark = a ( + 3. the rate constant is 3.91 × 10-4 mol-1dm3s-1 at 370ºC and 4.77 × 10-2 mol-1dm3s-1 C i Some candidates were not familiar with calculations involving the use of natural logarithm. Generally the rate constant of a reaction is related to the temperature by k = A exp(-Ea/RT).31 643 743 Ea = 184.31 JK-1mol-1) 450 ºC ≡ 723 K k 723 184. R = 8. It is intended to reflect ½ mark that the most sensitive parameter which affects the reaction rate is the activation energy.31 643 723 3. the area under the distribution curve for E ≥ Ea changes. The shaded area under either one of these curves corresponds to E ≥ Ea and is approximately equal to the fraction of molecules that collide with kinetic energy ≥ Ea. ii Some weaker candidates failed to realize that the absolute temperature should be used in the calculation.2 kJmol-1 (wrong unit -½) (not 3-4 sig.05 × 10-2 mol1 dm3s-1 at 470ºC. Therefore. -½) 2 marks ii the rate constant at 450ºC (Gas constant. Ea. yet candidates continued to fail to clearly explain the meaning of the various features of the Maxwell-Boltzmann distribution curve. This is a standard question which has appeared almost on a yearly basis. III. in terms of the Arrhenius equation. T. is radioactive. The rate of reaction is normally increased by raising the temperature. in aqueous solution : CO(CH2CO2H)2(aq) → CH3COCH3(aq) + 2CO2(g) T/K 273 293 313 333 353 k / s-1 2. Comment on the effect of temperature upon the rate of decay.7 kJ (no unit -½ mark) 96 1A 1 b iii 1b Carbon-14. 1 mark 94 2A 1 c iii 1c The table below lists the concentration of the reactant C as a function of time at 298 K for the following reaction.R ( 298 ) Ea 1 (2) ln k + ln 2 = C . According to Arrhenius Equation. k. 2 1 2 9 . for the first order decomposition of a dicarboxylic acid. Determine the activation energy of the reaction.6 10.1 (Gas constant R = 8.46 × 10-5 4.1 13. C→D Time / s 0 60 120 180 240 300 -2 -3 [C] /10 mol dm 20. the effect of temperature on the rate of a reaction. at different temperatures. Arrhenius Equation Ea ln k = C . emitting β-particles.48 × 10-2 ? i Determine the activation energy for the reaction by plotting an appropriate graph.31 JK-1mol-1) iii The rate constant of the above reaction is found to be doubled when the temperature is raised from 298 K to 306 K. 14C.R ( 306 ) Ea 1 1 (2) . 1 mark 97 2A 3 d 3 The reaction of iodine with propanone in acidic solutions can be represented by the following equation I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + H+(aq) + I-(aq) 3d For a given set of initial concentrations the initial rate doubles when temperature is increased from 298 K to 308 K.306 ) 1 mark 1 mark Ea = 65.76 × 10-3 5.RT Ea 1 (1) ln k = C . Collision theory and activation energy Unit 2 93 2A 2 a 2a Discuss. hence ln k (k) increases. Ea Arrhenius Equation: ln k = -RT + C or k = Ae − Ea RT Page 4 2 1 mark 1 Ea Increase in T makes T smaller and -RT larger. CO(CH2CO2H)2. 98 2A 3 a i 3a The table below lists the rate constants. rate of decay is independent of temperature. iii All radioactive decay has zero activation energy.2 9.75 × 10-4 5.5 11. Calculate the activation energy of the reaction.0 16.(1) ln 2 = R ( 298 . Energy profile of reaction 5. Stanley Thornes (Publisher) Ltd. the reacting molecules/ions.IV. Transition state Transition state is an activated complex at the top of the potential curve. Rate = k[Br-][BrO3-][H+]2 Syllabus Notes The theory of collision and activation energy cannot account for the order of the reaction determined. Energy profile of reaction Unit 1 Page 1 Topic Reference Reading IV. immediately decompose to form the desired products.. 373–375 Energy profile Transition state Reaction mechanism Rate determining step Consider the reaction between bromide and bromate(V) ion 5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) Kinetic studies show that this is an overall fourth order reaction. the transition state is a pentavalent trigonal bipyramidal complex.. The reaction would then continue to proceed as the transition state. pg. as it proceeds via the following mechanism: A. . Thomas Nelson and Sons Ltd. The transition state is highly unstable having 10 valence electrons. 417–420 Chemistry in Context. Thomas Nelson and Sons Ltd. which is so unstable and can never be isolated. 313–315 Reading Assignment Chemistry in Context (4th ed. The idea of transition state (activated complex) comes from an assumption that in any reaction.5 Unit 1 A-Level Chemistry (3rd ed. if acquired sufficient energy to overcome the energy barrier. Energy profile of reaction During a reaction. In the hydrolysis of 1-bromobutane. 5th Edition. IV. bonds are in the process of making and breaking.). Consider a simple reaction of hydrolysis of 1-bromobutane in aqueous alkaline solution.). would first form a transition state. which is energetically unstable. Activation energy (Ea) – the energy difference between the reactant and the transition state (the threshold energy level leading to a reaction) The energy profile for the reaction can be depicted as follows. the step 1 would be slower and known as the rate determining step. it is possible to isolate the intermediate from the reaction mixture. The reaction. since the formation of the transition state involves 2 molecules. Molecularity is the number of the molecule involved in the formation of the transition state in a step of the reaction.particle and 1 CH3CH2CH2CH2Br molecule.particle and a CH3CH2CH2CH2Br. the rate of reaction is directly proportional to the concentration of OH-(aq) ion and concentration of 1-bromobutane. the rate of formation of the transition state is directly proportional to [OH-] and [CH3CH2CH2CH2Br].B. a) Multisteps reaction (1) Alkaline hydrolysis of 2-chloro-2-methylpropane Alkaline hydrolysis of 2-chloro-2-methylpropane is found to be a first order reaction. While. . It is an intermediate of this reaction. unlike the highly unstable transition state. it is a multisteps reaction consists of 2 steps Step 1 (slow) CH3 CH3 C Cl CH3 CH3 CH3 C + + ClCH3 Step 2 (fast) CH3 CH3 C+ CH3 -OH CH3 CH3 C OH CH3 As formation of the transition state for step 1 involves a higher activation energy Ea’.ion is not involved in the formation of the transition state. N. The reaction is called unimolecular nucleophilic substitution (SN1 reaction) because only 1 molecule is involved in the formation of the transition state of the slowest step. order of reaction is the sum of the power of the concentration terms in the rate equation. Unit 1 Page 2 Order of reaction leads to interpretation of reaction mechanism at molecular level Hydrolysis of 1-bromobutane is a second order reaction. They may or may not have the same value but their meaning are indeed different. Energy profile of reaction 1. Since the formation involves only 1 OH. CH3 CH3 C+ CH3 is called a carbocation or carbonium ion. Rate ∝ [OH-][CH3CH2CH2CH2Br] Rate = k[OH-][CH3CH2CH2CH2Br] It is proposed that the transition state is formed upon the collision between a OH. therefore. Indeed. Hydrolysis of 1-bromobutane is called a bimolecular nucleophilic substitution reaction (SN2 reaction). The molecularity of the reaction is said to be 2. Molecularity is not the same as order the reaction. Because it is at the bottom of a potential well. CH3 CH3 C Cl + OH(aq) CH3 CH3 CH3 C OH + Cl-(aq) CH3 Rate = k[(CH3)3CCl] It is very clear that the reaction is not a SN2 reaction and OH. is only first order with respect to (CH3)3CCl.IV. B. Energy profile of reaction Unit 1 Page 3 (2) Reaction between bromide and bromate(V) in acidic medium For the reaction. 5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) Rate = k[Br-][BrO3-][H+]2 If the transition state is formed by collision between Br-. The rate of Step 4 and 5 would only be the same as Step 3. For the concentration of HBr. Rate of reaction ∝ [HBr][HBrO3] [HBr] ∝ rate of formation of HBr ∝ [H+][Br-] [HBrO3] ∝ rate of formation of HBrO3 ∝ [H+][BrO3-] Therefore rate of reaction ∝ [HBr][HBrO3] ∝ ([H+][Br-])([H+][BrO3-]) ∝ [Br-][BrO3-][H+]2 Rate = k[Br-][BrO3-][H+]2 N. Since the rate of the reaction is independent of the other steps. Since Step 4 and 5 is following the rate determining step. . It is proposed that the reaction between bromide and bromate(V) in acidic medium involves 5 steps. the rate of disappearing of HBr in Step 4 and 5 need not to be considered.→ HBrO3 HBr + HBrO3 → HBrO + HBrO2 HBrO2 + HBr → 2HBrO HBrO + HBr → H2O + Br2 Fast Fast Slow Fast Fast Since Step 3 is the slowest step.and 2 H+ particles. BrO3. b) Rate determining step For multisteps reaction.IV. the reaction rate would be extremely slow. The probability of having four particles collide at the same time is extremely low. The step with the slowest rate is called the rate determining step.→ HBr H+ + BrO3. the rate of reaction is only controlled by the step with the slowest rate. it would be the rate determining step. Step 1 Step 2 Step 3 Step 4 Step 5 H+ + Br. EA = 668 kJmol-1. N2(g) + 3H2(g) → 2NH3(g) ∆H = -92 kJmol-1. ↑ temp. E(N≡N) E(H–H) 945. A mixture of N2(g) and H2(g) is said to be energetically unstable but kinetically stable. C iii Compare the effect of increasing temperature on the rate of reaction in the system. However. the % increase in rate will be large for system X Many candidates discussed only one system when the question asked for the effect on each system. Unit 1 Page 4 Energetic stability and kinetic stability The reaction N2(g) + 3H2(g) → 2NH3(g) is exothermic. N2(g) does not react with H2(g) at room temperature.4 kJmol-1 435. Energy profile of reaction 2. will increase the rate of reaction of both systems. This is because the reaction involves a very high activation energy.9 kJmol-1 Glossary energy profile transition state (activated complex) reaction mechanism molecularity unimolecular nucleophilic substitution (SN1) bimolecular nucleophilic substitution (SN2) carbocation / carbonium ion intermediate rate determining step energetic stability kinetic stability 91 1A 2 b iii 93 1A 1 c i ii 94 2A 2 a iii 95 2B 4 c ii 99 2A 3 a i ii iii Past Paper Question 91 1A 2 b iii 2b The energy profile of the reaction A(g) + B(g) d C(g) under two different catalysis X and Y are represented below.IV. The high activation energy is assoicated with the high bond energies of N≡N and H–H bond. 2 1 mark 1 mark . diamond or graphite. 99 2A 3 a i ii iii 3a Consider the following data for the reaction A + B → products Initial concentration / mol dm-3 . HI is the most unstable hydrogen halide with respect to decomposition to its elements. i.8 +367 H–I +26. is energetically more stable? Explain why carbon does not convert from the less stable allotrope to the more stable one at room temperature. From the ∆Hf values.1 +298 ii At temperatures above 400 K. Is the reaction endothermic or exothermic? Explain.8 +430 H–Br -36. ∆Hocomb. 2 2 marks 2 1 mark 1 mark 94 2A 2 a iii 2a Given the following thermochemical data.395. hydrogen chloride and 4 hydrogen bromide do not decompose. However.5 H2(g) + ½O2(g) →H2O(l) . This indicated that their concept of 'energetic stability' was quite confused and/or that they could not relate the sign of enthalpy terms to the stability of compounds.75. It is the intermediate step with highest activation energy. ∴ Conversion does not occur at room temperature. Which allotrope of carbon. ∆H is negative. the activation energy for its decomposition is lowest. Reaction ∆Ho98 / kJ mol-1 2 C (graphite) + 2H2(g) → CH4(g) .393. 2 marks And therefore it is the most easily decomposed HX. 1 mark Conversion from diamond to graphite involves the rearrangement of atoms in a giant covalent network would require a high activation energy. but.IV. hydrogen iodide decomposes to produce violet fumes. 1 mark 1 mark 2HI(g) → H2(g) + I2(g) C ii Based on the given information candidates should have been able to point out that HI decomposes readily at moderately low temperatures because it has a small bond dissociation energy corresponding to a low activation energy for decomposition.e. 1 mark C Many candidates erroneously said that diamond was the more stable allotrope of carbon. Exothermic since the potential energy of the products is lower than that of the reactant. HI has the smallest bond dissociation energy. Step 3.4 kJmol-1 ∆Hocomb.4 kJ mol-1.4 +562 H–Cl -92.0 C (graphite) + O2(g) → CO2(g) .285.[C(diamond)] = . HBr and HCl do not decompose at temperature > 400 K. Page 5 i ii Which is the rate-determining step? Explain.393. Energy profile of reaction Unit 1 93 1A 1 c i ii 1c Consider the energy diagram shown below for a certain reaction which takes place with three steps. The positive and therefore the largest enthalpy change of formation amongst the hydrogen halides means that HI is the least stable hydrogen halide with respect to decomposition into its elements.[C(graphite)] = . 2 95 2B 4 c ii 4c Consider the data given below for the hydrogen halides and answer the questions that follow. Briefly explain this difference and write a balanced equation for the decomposition of HI. Bond dissociation energy / kJmol-1 Standard enthalpy change of formation / kJmol-1 H–F -269. ∴ it is the slowest step.5 kJmol-1 ∴ Graphite is the relatively more stable allotrope.9 iii The enthalpy change of combustion of diamond at 298K is -395. 4 × 10-5 Page 6 For this reaction.IV.4 × 10-5 12.0 × 10-2 [B] 4. Energy profile of reaction [A] 4.0 × 10-2 4.0 × 10-2 8. i deduce its rate equation.0 × 10-2 4. (6 marks) . ii calculate the rate constant.0 × 10-2 Unit 1 Initial rate / mol dm-3s-1 6.0 × 10-2 8. and iii sketch a possible energy profile.8 × 10-5 6. 5th Edition.). pg. catalytic reaction has a faster rate of reaction.). Energy profile of reaction 5. 102 Chemistry in Context. John Murray (Publisher) Ltd. There are 2 ways to make more particles having an energy higher than activation energy. Thus.IV. Energy profile of reaction Unit 2 Page 1 Topic Reference Reading IV. 370–373 Effect of catalyst Application of catalysts B. thus increase the reaction rate.6 Unit 2 A-Level Chemistry (3rd ed. This can be done by using of catalyst. Stanley Thornes (Publisher) Ltd. 1. 414–416 Advanced Practical Chemistry. Characteristics of catalyst Catalyst – a regenerated reagent which modify the rate of reaction. . 2. Effect of catalyst According to collision theory.. The following examples can be used to illustrate this: Using of catalyst provides an alternative multisteps reaction pathway. It can thus be used repeatedly without undergoing any permanent changes. 296. only the collision with an energy exceeding the activation energy may lead to a reaction. It is consumed in one reaction step and regenerated in a subsequent step.. Thomas Nelson and Sons Ltd.. Increase the temperature of the particle ⇒ more particles possess an energy greater than Ea. Syllabus Notes 1. In general. Lower the activation energy ⇒ more particles at the original temperature posses an energy greater than the new Ea'. a catalyst is chemically involved in a reaction. A lower activation energy is involved in the rate determining step of the catalyzed reaction. Thomas Nelson and Sons Ltd. 315–317. 446 Reading Assignment Chemistry in Context (4th ed. thus.e. e. R' C O H R + O H charge separation involved O. it may be physically changed after the reaction. .d. 1.g. f R C O H + O R' fR O C OR' + H2O + H+ In the acid catalyzed pathway. the proton possessed by the alcohol is shifted to the hydroxyl group to convert it to a better leaving group. the equilibrium constant K of the reaction is not changed. As ∆H remains unchanged. Note : It can be proved that ln K = constant ∆H RT Catalyst speeds up a reaction but has no effect on the yield of the reaction. the carbonyl oxygen is first protonated. no reaction intermediate with charge separation is involved in the acid catalyzed mechanism and a much lower activation energy would result.s. This activates the carbonyl group by increasing the polarity of the C=O bond and make the carbonyl carbon more positive and more vulnerable to the attack of the nucleophile.d. 2. alcohol. It speeds up both forward and reverse reaction for the same extent. Theory of catalyst According to the physical state. Furthermore.IV. Acid Catalyzed Esterification H O R C OH H O H O R' H O H R C OH +O H R' no charge separation involved f H+ +O R C OH f r. a rather slow rate. This makes the reaction have a high activation energy. Then. a particular catalyst usually can catalyze only one reaction. water H2O. In contrast .H R' C O+ H R O O R' C O ester R + H2O water R O H alcohol In esterification. Heterogeneous catalyst – the catalyst has different physical state from the reaction mixture a) Homogeneous catalysis Example 1 Acid-catalysed esterification (will be further elaborated in organic chemistry) Esterification O carboxylic acid R' C O H Or. esterification in the presence of any aqueous acid is found to be catalyzed. from granule form to powder form. Homogeneous catalyst – the catalyst has the same physical state as the reaction mixture 2. Energy profile of reaction Unit 2 Page 2 Catalyst is usually highly specific. The catalyst does not change the equilibrium position of a reaction.s. i. the nucleophilic oxygen atom on alcohol is added to the carbonyl carbon. catalysts can be classified into 2 categories. It can be seen that an intermediate with charge separation in the structure is involved. Since catalyst is involved in the formation of reaction intermediate. in the presence of acid. Beside the volume of the oxygen. H2O2(aq) + 2I-(aq) + 2H+(aq) → 2H2O(l) + I2(aq) or 5H2O2(aq) + 2MnO4-(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) I2(aq) + 2S2O32-(aq) → 2I-(aq) + 2S4O62-(aq) Although the MnO2(s) granules used are chemically unchanged after the reaction.IV. b) Heterogeneous catalysis Example 1 Decomposition of hydrogen peroxide in the presence of manganese(IV) oxide (MnO2(s)) catalyst Without the presence of catalyst. N. This is probably because the reaction involves collision among similarly charged particles. the reaction will be slowed down. This reaction is usually used in the laboratory in the preparation of oxygen. hydrogen peroxide will only decompose slowly into water and oxygen. MnO2(s). . Addition of Fe3+(aq) or Fe2+(aq) into the reaction mixture speed up the reaction considerably. The amount of H2O2(aq) can then be determined by back titrating with potassium iodide and sodium thiosulphate or directly with KMnO4(aq). 2H2O2(aq) → 2H2O(l) + O2(g) The rate of reaction can be traced by monitoring the volume of oxygen evolved. 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq) 2Fe2+(aq) + S2O82-(aq) → 2Fe3+(aq) + 2SO42-(aq) Fe3+(aq) is consumed Fe3+(aq) is regenerated The catalyzed reaction has a lower activation energy since the alternative pathway involves only collision between oppositely charged particle. However. The acid is said to be a negative catalyst. the reaction can also be traced by quenching the reaction using excess acid. In the presence of catalyst.B. the rate of reaction will be increased tremendously. Please note that the presence of the catalyst only affect the rate of the reaction but it has no effect on the amount of oxygen obtained eventually. Oversimplified mechanism of the catalyzed reaction – Fe3+(aq) or Fe2+(aq) are served as a medium of e.transfer. it will be physically changed into powder form. Energy profile of reaction Unit 2 Page 3 Example 2 Reaction between aqueous solutions of iodide and persulphate ions: 2I-(aq) + S2O82-(aq) → I2(aq) + 2SO42-(aq) The reaction rate of the above redox reaction is rather low. Therefore. That's why fat is a solid and oil is a liquid at room temperature. carbon dioxide. . N2(g) + H2(g) d 2NH3(g) Hydrogenation of unsaturated oil All lipids are triesters of glycerol (propan-1.2. the new activation energy is much lower than in the uncatalyzed reaction. 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) The catalysts are coated on a honeycomb support to increase the surface area for better action.g. nitrogen and water) with the use of metal catalysts such as platinum (or palladium) and rhodium. carbon monoxide. O CH2 CH CH2 O O O O O from propan-1. The hydrogenated oil formed is called margarine. some pollutants from car exhaust (e. With the Fe catalyst.3-triol) and fatty acid (long chain carboxylic acid).2.IV.g. Energy profile of reaction 3. b) Catalytic converter All the cars imported into HK after 1 January 1993 are installed with catalytic converter. This is done by bubbling H2(g) through oil under heat and pressure with the presence of Ni(s) catalyst. In the catalytic converter. Saturated carbon chain is larger than the unsaturated one with the same length. This makes the direct combination of N2(g) and H2(g) viable. nitrogen monoxide and unburnt hydrocarbons) are converted into relatively harmless substances (e. The difference between oil and fat are only the degree of unsaturation of the fatty acids in the molecule.3-triol from different fatty acids Oil can be converted to fat by hydrogenation of the unsaturated carbon chain. the van der Waals' forces among the saturated carbon chain is stronger. Applications of catalysts Unit 2 Page 4 a) Use of catalyst in Haber process and hydrogenation Haber process Iron is used as the catalyst in the Haber process to speed up the formation of ammonia. Oil contains more unsaturated fatty acid that fat does. enzymes are used in various aspects such as in the manufacture of biological washing powders. In part (iv) some candidates related the order of reaction to activation energy.61 g of ethanol in the presence of a catalyst. i Name a suitable catalyst for this reaction in the forward direction. Enzymes obtained from yeast have long been used in the production of alcohol by fermentation. sweat as well as fats.01 g of ethanoic acid are treated with 4. 2 2 marks 95 1A 1 e i 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12. Energy profile of reaction c) Enzymes Unit 2 Page 5 Enzymes are proteins that catalyse specific biochemical reactions. These washing powders contain enzymes which can break down stains caused by blood. They have the advantage of removing stains even at normal temperature.04 g of ethanoic acid are found to have reacted. egg. iv C Why could the use of a different catalyst change the order of the reaction? different reaction paths or different reaction mechanisms is involved with the catalyst added. They are often called biological catalysts. The fermentation of glucose to form ethanol can be represented by the following equation : C6H12O6 → 2C2H5OH + 2CO2 enzyme Nowadays. 5.IV. Concentrated sulphuric(VI) acid / hydrochloric acid / hydrogen chloride gas 1 mark 1 . When the reaction reaches equilibrium at 298 K. Glossary catalyst regenerated reagent Haber process hydrogenation 91 1A 2 b iv 95 1A 1 e i 98 2A 3 b iii homogeneous catalysis catalytic converter heterogeneous catalysis palladium rhodium adsorption enzyme Past Paper Question 91 1A 2 b iv 2b The energy profile of the reaction A(g) + B(g) d C(g) under two different catalysis X and Y are represented below. reaction (1) will proceed at a faster rate via the following mechanism : E(g) + C(g) → EC(g) EC(g) → C(g) + E'(g) (EC is the reaction intermediate.IV. C.) Sketch labelled energy profiles for the conversion of E(g) to E'(g). Page 6 . Energy profile of reaction Unit 2 98 2A 3 b iii 3b The exothermic reaction E(g) → E'(g) (1) is a single stage reaction. with and without the catalyst. Explain why reaction (1) proceeds faster in the presence of the catalyst. iii In the presence of a catalyst. Factors influencing the rate of reaction 5. This will result in higher velocities and hence more collision per unit time. A larger surface area would lead to more collisions per unit mass of the solid reactant and higher reaction rate. 2 marks Catalyst: A catalyst that increases a reaction rate does so by providing a new mechanism. A larger fraction of particles will have energy greater than EA. Glossary Past Paper Question 90 2A 1 c 90 2A 1 c 1c Give. then the rate is increased. 522 Chemistry in Context. 4 Temperature: An increase in temperature increases the number of particles having sufficient energy to produce fruitful collision. 399–401 Reading Laboratory experiments for general. Factors influencing the rate of reaction Page 1 Topic Reference Reading V. a finer particles means a higher surface area to volume ratio. 354–356 Factors influencing the rate of reaction V.7 Assignment A-Level Chemistry (3rd ed. 187–196. 293–296 Chemistry in Context (4th ed. Light Beside heat. Thomas Nelson and Sons Ltd. pressure is equivalent to concentration for solution. this makes a larger fraction of particles having an energy greater than EA. Surface area For solid reactant. Syllabus Notes F. Concentration The collision frequency among the particles increases with increasing concentration. D. Factors influencing the rate of reaction A.). Stanley Thornes (Publisher) Ltd.. organic & biochemistry (2nd ed. pg. two factors that would increase the rate of a reaction. A gas under pressure means that there are more gaseous particle in a given volume. 2 marks Concentration: An increase in concentration will give more particles per unit volume and hence more collision per unit time. If the activation energy of the limiting step of the new mechanism is lower than the activation energy of the limiting step of the uncatalyzed reaction. 5th Edition. light is also another source providing energy for the reaction. C Many candidates failed to point out explicitly that only in gas-phase reactions would an increase in pressure lead to an increase in reaction rate.. Saunders College Publishing. C. 2 marks Maximium of 4 marks for any 2 correct answers. Catalyst Use of catalyst lower the activation energy EA. the collision frequency also increase. E. Temperature Increase in temperature causes increase in energy of the particles. with explanations.). Pressure For gaseous reactant.). Furthermore. B.V. and also that the average kinetic energy will be increased. Thomas Nelson and Sons Ltd. . Concentration 2. Examples of calculation C. Dynamic nature Examples of equilibrium 1. Kw. KIn.Chemical Equilibria Page 1 Chemical Equilibria I. Kp. Bromine water in acidic and alkaline medium 2. Equilibrium constant (Kc and Kp) 3. Relationship between rate constant and equilibrium constant Effect of change in concentration. pressure and temperature on equilibria (Le Chaterlier's principle) 1. Pressure 3. Nature of equilibrium A. D. Temperature ∆H a) Equation : ln K = constant T 4. Kd 2. Hydrolysis of bismuth chloride Equilibrium law 1. Potassium dichromate in acidic and alkaline medium 3. . Effect of catalyst on equilibria 5. Concentration of solid 5. Ka. Introduction to different equilibrium constant (K) – Kc. Kb. Determination of equilibrium constant (Kc) a) Esterification b) Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) 6. Degree of dissociation (α) 4. B. Dissociation constant (Ka and Kb) a) Dissociation of polybasic acid (1) Charge effect 3. Titration using indicator a) Choosing of indicator 4. Conductimetric titration a) Strong acid vs Strong base b) Weak acid vs Strong base c) Weak acid vs Weak base Calculation involving pH. C.Chemical Equilibria Page 2 II. Thermometric titration 5. Ka and Kb a) Relationship between Ka and Kb (pKa and pKb) b) Relationship between pH. Difference between equivalence point and end point 2. G. Use of indicator (1) Strong and weak acids/bases 1. Ionic product of water pH and its measurement 1. pOH and pKw c) Some basic assumptions applied Experimental determination of Ka . F. Measuring of pH or conductivity of acid / base 2. Bronsted-Lowry definition 3. Arrhenius definition 2. D. 4. Measurement of pH a) Use of pH meter (1) Calibration of pH meter Colour of indicator B. Acid-base Theory 1. H. Titration using pH meter 3. Lewis definition Strength of acid 1. b) E. Buffer 1. Acid-base Equilibria A. Principle of buffer action a) pH of an acidic buffer b) pH of an alkaline buffer 2. Definition of pH 2. Calculation of buffer solution Theory of Indicator Acid-base titration 1. Temperature dependence of pH 3. Leveling effect Dissociation of water 1. Cell diagrams (IUPAC conventions) Electrode potentials 1. Redox reaction 1. Electrochemical series 4. Standard hydrogen electrode 2.Chemical Equilibria Page 3 III. Use of salt bridge a) Requirement of a salt bridge b) An electrochemical cell does not need salt bridge 3. Use of electrode potential a) Comparing the oxidizing and reducing power b) Calculation of e. C. Lead-acid accumulator 2.m. Electrochemical process of rusting 2. Calculation of mass liberated in electrolysis Electrochemical cells 1.f. a) Potentiometer b) High impedance voltmeter / Digital multimeter 2. b) By the change in oxidation no. Prevention of rusting a) Coating b) Sacrificial protection c) Alloying . Redox Equilibria A. Secondary cell and fuel cell 1. Hydrogen-oxygen fuel cell Corrosion of iron and its prevention 1.f. Relative electrode potential (Standard reduction potential) a) Meaning of sign and value 3. Balancing of redox reaction a) By combining balanced half-ionic equation (1) (2) Steps of writing balanced half-ionic equation Combining half-ionic equations B. Faraday and mole 3. 2. E. of a cell c) Prediction of the feasibility of redox reaction (1) Disproportionation reaction D.m. Measurement of e. Acid-equilibrium and redox equilibrium are 2 examples of chemical equilibrium. pg. 3. we are going to study chemical equilibrium. Stanley Thornes (Publisher) Ltd. acid-base equilibrium.). Heinemann. Solid phase. Aqueous phase and organic phase.. Thomas Nelson and Sons Ltd. In A-Level. This can be done by running at the same speed of the escalator but at a different direction.I. 5th Edition. 289–296 Nature of equilibrium Dynamic nature I. 11–12 Part 1 Assignment A-Level Chemistry (3rd ed. For examples 1. There are two kinds of equilibrium : Static equilibrium and Dynamic equilibrium. Nature of equilibrium Part 1 Page 1 Topic Reference Reading I.).0 Practical Chemistry (3rd ed. 2. Static equilibrium Dynamic equilibrium The positions of the two boys remain unchanged because they are not in motion. 231 Chemistry in Context (4th ed.. 83–86. liquid phase and gaseous phase.). They are in a state of dynamic equilibrium. Nature of Equilibrium 6.0 – 6. Moving phase and stationary phase Static equilibrium Equilibrium Dynamic equilibrium Chemical equilibrium Phase equilibrium Acid-base equilibrium Redox equilibrium . Phase equilibrium is the study of equilibrium involving interconversion of phases. All equilibria studied in chemistry are dynamic in nature. The position of the boy remains unchanged but both the escalator and the boy are in motion. 228–229. Chemical equilibrium includes all equilibria involving any chemical reaction. Nature of equilibrium Syllabus Notes A. 325–329 Reading Chemistry in Context. Dynamic nature Equilibrium means a state of balance. redox equilibrium and phase equilibrium. Thomas Nelson and Sons Ltd. They are in a state of static equilibrium. A boy can also maintain a fixed position on a moving escalator.1. The amounts of reactant and product will remain constant and an equilibrium is established. the rate of the forward reaction will decrease and the rate of the backward reaction will increase. Nature of equilibrium Part 1 Page 2 All chemical equilibria are dynamic in nature. forward rate = backward rate When the increasing backward rate reaches the decreasing forward rate. there will be no net conversion between product and reactant. Reactant forward reaction backward reaction Product Rate of forward reaction = k1 [Reactant] Rate of backward reaction = k-1 [Product] Concentrations of reactant and product Concentration / mol dm -3 Concentration of reactant Concentration of product 0 Time At the beginning. However. there is always a constant conversion between the reactant and the product. Therefore. Since the rate of reaction increases with concentrations. . Consider the following reversible reaction. the equilibrium is said to be dynamic in nature. there is only reactant but no product.I. the forward reaction and backward reaction do not stop after the equilibrium has established. the concentration of the reactant drops and the concentration of the product rises. Once the reaction proceeds. Rate of reaction Rate / mol dm s -3 -1 Rate of forward reaction Rate of backward reaction 0 Time At equilibrium. This system doesn’t involve any chemical change but only interconversion between liquid phase (water) and gaseous phase (steam). This is an example of phase equilibrium.I. . In the above example. no equilibrium will exit if the cover of the pressure cooker is removed. the amounts of water and steam will become constant. Eventually. At the same time. H2O(l) vaporizes to steam H2O(g). Nature of equilibrium Consider another example Part 1 Page 3 If H2O(l) is heated in a sealed pressure cooker at a constant temperature. All equilibria only exist in close systems. steam also condenses back to water. The equilibrium can be disturbed by either addition of acid or alkaline. And the concentrations of all species will remain constant. the concentration of H+(aq) increases suddenly accompanying with a sudden increase in backward reaction rate. Once the equilibrium is established. the forward rate of reaction will be the same as the backward rate of reaction. Nature of equilibrium B. The position of the equilibrium is said to be shifted to the left by the addition of acid. the concentration of product will be lower that the original concentration while the concentration of reactant will become higher. Addition of acid Br2(aq) + orange H2O(l) d colourless H+(aq) + colourless Br-(aq) + colourless HBrO(aq) colourless Upon addition of acid. . And the colour of the equilibrium mixture will become darker. Br2(aq) is orange in colour and all the other species are colourless. The concentration of product drops and the concentration of the reactant rises. the equation may be rewritten as follow : Br2(aq) + orange H2O(l) d colourless H+(aq) + colourless Br-(aq) + colourless HBrO(aq) colourless In this equilibrium. Part 1 Page 4 Bromine water in acidic and alkaline medium Br2(aq) + orange H2O(l) d colourless HBr(aq) + colourless HBrO(aq) colourless Since HBr(aq) is a strong acid while HBrO(aq) is a weak acid. Examples of chemical equilibrium 1. At the new equilibrium position. A new equilibrium position will be established when the concentrations of product and reactant reaches a level that the forward rate and the backward rate become the same again. More product is converted back to reactant.I. the equilibrium position can be shifted to the right and becomes orange. the equilibrium position can also be shifted to the right by addition of alkaline. Nature of equilibrium Addition of alkali Br2(aq) + orange H2O(l) d colourless Part 1 Page 5 H+(aq) + colourless Br-(aq) + colourless HBrO(aq) colourless Conversely. Questions : What if NaBr(s) is added ? 2. The colour intensity of the equilibrium mixture will become lighter as the concentration of Br2(aq) decreases.I. . While upon addition of alkali. the equilibrium will adjust to a new position. Potassium dichromate in acidic and alkaline medium 2CrO42-(aq) yellow O O - + 2H+(aq) colourless O - d Cr2O72-(aq) orange O + H2O(l) colourless O O - - Cr O O + O Cr O O + H+ + H+ O Cr O Cr O O + H2O chromate(VI) ion dichromate(VI) ion Upon addition of acid. Upon addition of alkaline. H+(aq) + OH-(aq) → H2O(l) The decrease in concentration of H+(aq) slows down the backward reaction. the hydrogen ions are neutralized. Eventually. the equilibrium position can be shifted to the left and becomes yellow. state the expected observation and write the relevant balanced equation(s).+ H2O 1 mark Candidates should take care in providing a 3-D diagram. And consider an irreversible reaction only an equilibrium lying completely onto one side. Upon addition of drops of HCl(aq). 1 98 2B 8 b i 8b For each of the following. the white disappears and the white ppt. Glossary equilibrium / equilibria static equilibrium dynamic equilibrium chemical equilibrium acid-base equilibrium redox equilibrium phase equilibrium reversible reaction forward rate backward rate close system shifting in equilibrium position chromate(VI) ion dichromate(VI) ion bismuth chloride bismuth oxychloride 92 2B 4 b i ii 98 2B 8 b i Past Paper Question 92 2B 4 b i ii 4b i Draw the structures of the chromate(VI) and dichromate(VI) ions. 2CrO42. i Dilute sulphuric(VI) acid is added dropwise to a solution of potassium chromate(VI). Some candidates stated that the shape of the chromate(VI) ion is square-planar.+ 2H+ d Cr2O72. 6 . it would be beneficial to consider them all reversible. 2 ii C i ii 1 mark each Write an equation showing the action of aqueous acid on the chromate(VI) ion.I. Hydrolysis of bismuth chloride BiCl3(aq) colourless + H2O(l) colourless d Part 1 Page 6 BiOCl(s) white ppt. which shows the correct bond orders and charge(s). as stated by some candidates. the chromate(VI) . of bismuth oxychloride reappear upon addition of drops of water. a white suspension of bismuth oxychloride (BiOCl) is formed. + 2HCl(aq) colourless When solid bismuth chloride (BiCl3) is mixed with water. Although not all reactions are reversible. Nature of equilibrium 3.dichromate(VI) equilibrium is not a redox reaction. The expression is called equilibrium constant. Stanley Thornes (Publisher) Ltd. denoted by the symbol K or Keq. 230–238 Reading rd A-Level Chemistry (3 ed. 232–233.. because the whole reaction is in equilibrium. 201–202 Chemistry – Experimental foundation. Kp. the rate of reaction is no longer depending on the rate determining step only. pg. Kd Considering the following reversible reaction aA+bBdcC+dD [C]c[D]d At equilibrium.1 A-Level Chemistry Syllabus for Secondary School (1995). Kw. Moreover. 296–303 Equilibrium law Equilibrium constant (Kc and Kp) C. The equation is also known as Law of equilibrium or Equilibrium law. Nature of equilibrium Part 2 Page 1 Topic Reference Reading I. Prentice Hall Inc. Stanley Thornes (Publishers) Ltd. 5th Edition.I.). Nature of Equilibrium 6. all intermediate steps of the reaction are in equilibrium. the rate of reaction is depending on the concentration and stoichiometric coefficients. Equilibrium law 1. 51–55 Part 2 Assignment Advanced Practical Chemistry. [C]c[D]d K or Keq = [A]a [B]b This equation describes the relationship between the concentrations of reactant and product.). If the reaction is not at equilibrium. Curriculum Development Council. KIn. Therefore. 333–340 Chemistry in Context. = = where k1 is the rate constant of forward reaction k-1 is the rate constant of backward reaction Syllabus Notes Backward rate at equilibrium k-1[C]c[D]d -1 This is applicable only if the reaction is at equilibrium. the rate of forward reaction equals the rate of backward equation. The unit of K is depending on the values of stoichiometric coefficients.. a) Relationship between rate constant and equilibrium constant At equilibrium. it is found that the expression [A]a [B]b is a constant at a given temperature. Thomas Nelson and Sons Ltd. 236–238 Chemistry in Context (4th ed.B. Forward rate at equilibrium = k1[A]a[B]b Backward rate at equilibrium = k-1[C]c[D]d Forward rate at equilibrium k1[A]a[B]b k1 [C]c[D]d k-1 = [A]a[B]b = K k1 ∴ K=k N. Kb. Ka. Instead. .. 53–56 Calculations for A-Level Chemistry. Introduction to different equilibrium constant (K) – Kc. Thomas Nelson and Sons Ltd.1. 86–87.. 229–231. the rate equation must be determined experimentally. John Murray (Publisher) Ltd. there are many different ways to express concentration. Ka. Type Chemical Chemical Meaning Equilibrium constant in term of concentration. Given : density of water at 25ºC is 1. a) Concentration of pure solid and liquid Molarity is only one way of expressing concentration. for a gas. concentration is expressed in term of molarity. In general. Kp Ionic product of water. concentration is expressed in term of partial pressure (which is related to mole fraction). Kw. However. Concentration of pure solid / liquid = amount of solid / liquid mass of solid / liquid = volume occupied = density volume occupied Example : Calculation of the concentration of water in term of molarity at 25ºC.0 gmol-1 = 55. Kp.00 gdm-3.6 moldm-3 = = mass / molar mass volume occupied density molar mass .g. Kw Acidity constant. the concentration of a pure solid / liquid is the equivalent to its density which is a constant at a given temperature.00 gcm-3 = 1000. Kd. Equilibrium constant (Kc and Kp) The equilibrium constant K or Keq describes the relationship between the concentrations of reactants and products. for a solution. For a pure solid / liquid. Kb Indicator constant. Concentration of pure water in molarity = = = no.I. Kb.]2[H+ ]2 (aq) 4 (aq) pNH32 Kp = p · p 3 N2 H2 Kw = [H+(aq)][OH-(aq)] [H3O+(aq)][CH3COO-(aq)] [CH3COOH(aq)] [NH4+(aq)][OH-(aq)] Kb = [NH3(aq)] [H3O+(aq)][In-(aq)] KIn = [HIn(aq)] [I2(hexane)] Kd = [I ] 2(aq) Ka = Chemical Acid-base Acid-base Acid-base Phase 2. Nature of equilibrium Part 2 Page 2 There are many different kinds of expression of equilibrium constant depending on the type of equilibrium concerned. Therefore. concentration is a very general concept concerning the relationship between the amount of a substance and the volume occupied by that substance. e. Kd Example 2CrO42-(aq) + 2H+(aq) d Cr2O72-(aq) + H2O(l) N2(g) + 3H2(g) d 2NH3(g) H2O(l) d H+(aq) + OH-(aq) CH3COOH(aq) + H2O(l) d H3O+(aq) + CH3COO-(aq) NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) HIn(aq) + H2O(l) d H3O+(aq) + In-(aq) I2(aq) d I2(hexane) Equilibrium Law [Cr2O72-(aq)] Kc = [CrO 2. of mole of water volume occupied volume × density / molar mass volume occupied 1000. KIn.00 gdm-3 18. Ka Bascity constant. KIn Partition coefficient / distribution coefficient. Kc. Indeed. the amount is usually measured in term of mass. Kc Equilibrium constant in term of partial pressure. Therefore. It has a significant physical meaning. water is also the solvent in this equilibrium. This fact should be reflected in the equilibrium law. In the writing of any equilibrium constant. [Fe3+(aq)] Kc = [Fe2+ ][Ag+ (aq) (aq)] . Consider a saturated sugar solution. Since the concentration of water is constant. The exclusion of concentration of pure solid / liquid in an equilibrium law is not only doctrinal. the equilibrium position is not depending on the amount of solid silver whose concentration is constant.I. Kc Consider the following equilibrium 2CrO42-(aq) + 2H+(aq) d Cr2O72-(aq) + H2O(l) [Cr2O72-(aq)][H2O(l)] K = [CrO 2. Nature of equilibrium Part 2 Page 3 b) Equilibrium constant in term of concentration. Fe2+(aq) + Ag+(aq) d Fe3+(aq) + Ag(s) Equilibrium mixture with less Ag(s) suspension Equilibrium mixture with more Ag(s) suspension It is found that the addition of extra Ag(s) into the mixture has no effect on the concentrations of Fe2+(aq). it may be incorporated into the equilibrium constant and the expression will become [Cr2O72-(aq)] K = [CrO 2.]2[H+ ]2 = Kc [H2O(l)] (aq) 4 (aq) The expression excludes the concentration of pure solid and liquid is known as Kc.]2[H+ ]2 (aq) 4 (aq) Besides being a product. the concentration of the solution is not depending on the amount of sugar remains undissolved. it can be considered as a pure liquid in here. equilibrium constant in term of concentration. sugar(s) + aq d sugar(aq) A saturated sugar solution A saturated solution with more undissolved sugar crystal Consider another equilibrium. Because it is in large excess comparing with the solutes. Ag+(aq) and Fe3+(aq). the concentration of a pure solid or liquid in the equilibrium law should be excluded. the equilibrium constant is known as Kp. N2(g) + 3H2(g) d 2NH3(g) [NH3(g)]2 Kc = [N ][H ]3 2(g) 2(g) or pNH3(g)2 Kp = p 3 N2(g) pH2(g) . Kp Consider the thermal decomposition of CaCO3(s). Nature of equilibrium c) Part 2 Page 4 Equilibrium constant in term of partial pressure.I. CaCO3(s) d CaO(s) + CO2(g) K= [CaO(s)][CO2(g)] [CaCO3(s)] [CaCO3(s)]K [CaO(s)] = [CO2(g)] = Kc Or if the concentration of CO2(g) is expressed in term of partial pressure. Kp = pCO2(g) Consider the Haber process. I. Nature of equilibrium 3. Degree of dissociation (α) Part 2 Page 5 Sometimes, the equilibrium positions of dissociation reactions can be described by degree of dissociation. Degree of dissociation is the percentage of molecules that undergoes dissociation and represented by α. Consider the dissociation of carboxylic acid dimer in gaseous state, H H C H C O H O O H O H C C H H H 2 H C H C O H O (CH3COOH)2(g) d 2CH3COOH(g) Consider 1 mole of dimer, the mole no. would be numerically the same as no. in percentage (i.e. mole fraction). Initial no. of mole no. of mole at equilibrium Partial pressure at equilibrium 1 1-α p(CH3COOH)2(g) 0 2α pCH3COOH(g) The equilibrium law in term of partial pressure can be expressed as pCH3COOH(g)2 Kp = p Let P be the total pressure, P = p(CH3COOH)2(g) + pCH3COOH(g) ni pi Since mole fraction is related to partial pressure by the equation, mole fraction = n = P ni pi = n P = mole fraction × P 1-α 1-α P= P (1 - α) + 2α 1 +α 2α 2α P= P pCH3COOH(g) = mole fraction of CH3COOH(g) × P = (1 - α) + 2α 1+α p(CH3COOH)2(g) = mole fraction of (CH3COOH)2(g) × P = 2α 2 2 pCH3COOH(g)2 (1 + α ) P 4α2 Kp = p = = P 1 - α2 1-α (CH3COOH)2(g) P 1+α In this equilibrium, unlike equilibrium constant Kp, dissociation constant (α) is also depending on the total pressure of the gaseous mixture on top of temperature. For example, T/K Kp / Nm-2 298 72.9 303 121.1 308 182.7 313 302 (CH3COOH)2(g) At 303 K and pressure of 1400 Nm-2 4α2 × 1400 Nm-2 121.1 Nm-2 = 1 - α2 At 303 K and pressure of 200 Nm-2 4α2 × 200 Nm-2 121.1 Nm-2 = 1 - α2 α = 0.145 α = 0.363 Thus, more carboxylic acid dimers will dissociate at a lower pressure. I. Nature of equilibrium 4. Part 2 Page 6 Determination of equilibrium constant (Kc) The equilibrium constant Kc can be determined by measuring the concentrations of different species in the equilibrium mixture. Indeed, sometimes it is not necessary to measure the concentrations of all the species. By knowing the initial concentration of the reactant and the equilibrium concentration of the reactant or product, the concentration of the other species can be determined according to the law of conservation of mass. Consider an equilibrium mixture with initial concentrations of 1 mol dm-3 Fe3+(aq) and 2 mol dm-3 of NCS-(aq) respectively. + Fe3+(aq) iron(III) ion pale yellow e.g. Initially At equilibrium NCS-(aq) thiocyanate ion colour d FeNCS2+(aq) thiocyanato iron(III) ion dark red complex + NCS-(aq) 2 mol dm-3 d FeNCS2+(aq) 0 mol dm-3 FeNCS2+(aq) x mol dm-3 Fe3+(aq) 1 mol dm-3 + Fe3+(aq) (1 - x) mol dm-3 NCS-(aq) d (2 - x) mol dm-3 By knowing the equilibrium concentration of either Fe3+(aq), NCS-(aq) or FeNCS2+(aq), the equilibrium constant can be determined. [FeNCS2+(aq)] x Kc = [Fe3+ ][NCS- ] = (1 - x) (2 - x) dm3 mol-1 (aq) (aq) I. Nature of equilibrium a) Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) Part 2 Page 7 First of all, equilibrium constant is a value affected by temperature. It is better to keep the temperature constant by immersing the solution to be studied in a water bath. For the equilibrium, Fe3+(aq) + NCS-(aq) d FeNCS2+(aq), the initial concentrations of Fe3+(aq) and NCS-(aq) can be determined by prepared corresponding standard solution of Fe(NO3)3(aq) and KNCS(aq). The problem is how to determine the equilibrium concentration of either Fe3+(aq), NCS-(aq) or FeNCS2+(aq). If titration is used to determine the concentration, the reaction must be quenched first. Otherwise, when the chemical is being titrated, the equilibrium position will also be shifted. Since FeNCS2+(aq) is deep red in colour, its concentration can be determined by colorimetric measurement without quenching. Firstly, a calibration curve which relates the concentration of FeNCS2+(aq) and the colour intensity must be constructed. This can either be done by using a calorimeter or visual comparison. In the construction of calibration curve, samples of standard FeNCS2+(aq) with various concentrations are prepared. These are prepared by adding excess NCS-(aq) into standard Fe3+(aq). It is assumed that all Fe3+(aq) ions are converted to FeNCS2+(aq). Therefore, the concentration of FeNCS2+(aq) can be determined from concentration of Fe3+(aq). Equilibrium mixture with known initial Fe3+(aq) and NCS-(aq) concentrations are prepared. The equilibrium concentration of FeNCS2+(aq) in the mixture is determined by a colorimeter or by visual comparison of the [FeNCS2+(aq)] intensity of the samples prepared before. Hence, the equilibrium constant Kc = [Fe3+ ][NCS- ] can be (aq) (aq) determined. Several trials with different concentrations should be done to obtain an average value. I. Nature of equilibrium b) Esterification CH3COOH(aq) + CH3CH2OH(aq) Part 2 Page 8 d CH3COOCH2CH3(aq) + H2O(l) Both esterification and hydrolysis of ester are very slow, even in the presence of acid catalyst, it takes 48 hours for the system to reach a state of equilibrium. Hence, the equilibrium concentration of CH3COOH(aq) can be determined by titration even without quenching. In this example, water is not the solvent and is not in large excess. It cannot be considered as a pure substance and its concentration must be reflected in the equilibrium law. [CH3COOCH2CH3(aq)][H2O(l)] Kc = [CH COOH ][CH CH OH ] 3 (aq) 3 2 (aq) Since aqueous acid is required to catalyze the reaction, it would be more convenient to approach the problem by treating it as a hydrolysis problem. This reduces the number of measurement of from 3 liquids (ethanoic acid, alkanol and aqueous acid) to 2 liquids at the preparatory stage. CH3COOCH2CH3(aq) + H2O(l) d CH3COOH(aq) + CH3CH2OH(aq) [CH3COOH(aq)][CH3CH2OH(aq)] 1 Kc = Kc' = [CH3COOCH2CH3(aq)][H2O(l)] Procedure An accurately known volume of standard HCl(aq) is pipetted into a test tube containing known mass of CH3COOCH2CH3(l). The solution is shaken occasionally for 48 hours to ensure equilibrium is reached. The initial amount of water is determined by subtracting the mass of HCl from the mass of HCl(aq). The initial amount of CH3COOCH2CH3(l) is determined by weighing. The total no. of mole of HCl(aq) and CH3COOH(aq) at equilibrium are determined by titrating with standard NaOH(aq). Experimental result Mass of the empty test tube = 12.01 g Mass of the empty test tube with CH3COOCH2CH3(l) = 16.52 g Mass of the empty test tube with HCl(aq) and CH3COOCH2CH3(l) = 5.20 g Volume of HCl(aq) added = 5.00 cm3 Concentration of HCl(aq) = 1.953 moldm-3 Calculation I. Nature of equilibrium Part 2 Page 9 Glossary equilibrium law / law of equilibrium equilibrium constant (K/Keq) equilibrium constant in term of partial pressure (Kp) equilibrium constant in term of concentration (Kc) degree of dissociation (α) dimer thiocyanate ion thiocyanato iron(III) ion calibration curve 90 2A 1 a i ii 94 2A 1 b 95 1A 1 e ii 96 1A 1 f i ii 97 1A 2 a i 98 1A 2 c 99 2A 4 a i ii Past Paper Question 90 2A 1 a i ii 1a Consider the following equilibrium at constant pressure: A2(g) + 3B2(g) d 2AB3(g) A mixture of 4.0 mol of A2(g) and 12.0 mol of B2(g) was placed in a vessel of volume 20.0 dm3, and heated to 565K. When the system had reached equilibrium, it was found that 4.0 mol AB3(g) was present. i Calculate the concentration of each species at equilibrium. A2(g) + 3B2(g) d 2AB3(g) Since 4 moles of AB3 will consume 2 moles of A2 and 6 moles of B2 ∴ at equilibrium: 12 - 6 6 [B2] = 20 = 20 moldm-3 ½ mark 4-2 2 [A2] = 20 = 20 moldm-3 ½ mark 4 [AB3] = 20 moldm-3 ½ mark ii Calculate the equilibrium constant of this reaction at 565K. [AB3]2 4 20 20 K = [A ][B ]3 = (20)2 ⋅ ( 2 ) ⋅ ( 6 )3 = 14.82 dm6mol-2 (wrong unit -½) 2½ mark 2 2 C i Some candidates mistakenly used the initially given concentration rather than the equilibrium concentration in their calculation of equilibrium constant. ii Some weaker candidates failed to give the correct unit for the equilibrium constant. 1½ 2½ I. Nature of equilibrium Part 2 94 2A 1 b 1b A colorimetric method can be used to provide data for the determination of the equilibrium constant of the following reaction. Fe3+(aq) + NCS-(aq) d FeNCS2+(aq) Outline such a method for the determination of the equilibrium constant Kc of the above equilibrium. FeNCS2+(aq) ions absorb visible light in the electromagnetic spectrum / hence [FeNCS2+(aq)] can be determined by measuring the absorbance of the solution. 1 mark Calibrate a colorimeter by measuring the absorbance of several solutions which contained known concentrations of FeNCS2+(aq). (In these solutions, the NCS- ions are in excess, so that all Fe3+ ions present can be considered as 1 mark existing in the complex form, FeNCS2+) By mixing comparable volumes of standard Fe3+(aq) and NCS-(aq) solutions and measuring the [FeNCS2+] formed with the colorimeter, [Fe3+(aq)], [NCS-(aq)] and [FeNCS2+(aq)] at equilibrim can be found. 1 mark [FeNCS2+(aq)] Hence, the equilibrim constant, Kc = [Fe3+ ][NCS- ] , can be determined. 1 mark (aq) (aq) C Poorly-answered. This indicated that most candidates were not familiar with the use of a colorimetric method in the determination of equilibrium constants. A large number of candidates did not mention the calibration of the colorimeter. 95 1A 1 e ii 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12.01 g of ethanoic acid are treated with 4.61 g of ethanol in the presence of a catalyst. When the reaction reaches equilibrium at 298 K, 5.04 g of ethanoic acid are found to have reacted. ii Calculate the equilibrium constant, Kc , for the reaction at 298 K. CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) initial no. of moles 12.01 60.052 = 0.2 12.01 - 5.04 60.052 = 0.116 0.116 V 4.61 46.068 = 0.1 0.1 - 0.084 = 0.016 0.016 V 0.084 V 0.084 V – – Page 10 4 3 no. of mole at equilibrium 0.084 0.084 2 marks concentration at equilibrium C 0.084 0.084 V × V 1 marks Kc = 0.116 0.016 = 3.80 (no unit) × V V A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. ) 97 1A 2 a i 2a Consider the following dissociation reaction: PCl5(g) d PCl3(g) + Cl2(g) At 400 K and 101 kPa pressure. Kp = 1 = P O2 3. the equilibrium constant for the following reaction is 1. ii Calculate Kc for the reaction at 723 K.25 1. i Calculate the respective concentrations of nitrogen.5 0. i Write an expression for Kp for the reaction of SO2(g) and O2(g) to give SO3(g).I. 4 98 1A 2 c 3 2c At 4200 K.25 × 373 (kPa) = 0.25 × 373 kPa) 0.5 total = 3. 25. one half of the SO2(g) had been converted into SO3(g). N2(g) + O2(g) d 2NO(g) 1.5 PSO3 = PSO2 (= 3. (PSO3)2 (PSO3) Kp = (P )2(P ) or (P )(P )½ 1 mark SO2 O2 SO2 O2 ii Calculate Kp for the above reaction under the given conditions.0% of the original nitrogen was consumed. 99 2A 4 a i ii 4a In the Haber process.19 (kPa) Page 11 1 2½ ½ mark ½ mark ½ mark 1 mark (1 mark) (½ mark for numerical answer. When equilibrium was attained at 373 kPa pressure. hydrogen and ammonia in the equilibrium mixture. Nature of equilibrium Part 2 96 1A 1 f i ii 1f SO2(g) and O2(g) were mixed in the mole ratio of 3 : 1 at 1000 K in the presence of a catalyst. ½ mark for unit) (The answer and the expression (part (i)) for Kp should be consistent with each other.50 mol dm-3 respectively.25 Kp = (P )2(P ) = P = 0.25 × 373 kPa-1 = 0.50 mol dm-3 and 1. N2(g) + 3H2(g) d 2NH3(g) In a simulation of the process. a mixture of nitrogen and hydrogen was placed in a closed container.2 × 10-2.035 kPa-1 SO2 O2 O2 or. i Calculate Kp for the reaction at 400 K. Calculate the concentration of N2(g).25 × 373 kPa (PSO3)2 1 3.0 mol of N2(g) are allowed to react in a 2. The initial concentrations of nitrogen and hydrogen were 0.0 dm3 closed container. ammonia is synthesized by the exothermic reaction of nitrogen and hydrogen at around 723K. in the equilibrium mixture at 4200 K. .25 PO2 = 3. in moldm-3.25 -½ -½ 0. 2SO2 + O2 d 2SO3 initial 3 1 0 at equilibrium 1.0 mol of O2(g) and 2. the percentage dissociation of PCl5(g) is 86%.25 1. When the equilibrium was attained at 723 K. pressure and temperature on equilibria (Le Chaterlier's principle) Effect of catalyst on equilibria Le Chaterlier's principle D. upon addition of Na3PO4(aq) or NaHPO4(aq). However. . 160 Part 3 Assignment A-Level Chemistry (3rd ed.1.1. pg. some Br2(aq) added will be converted to HBr(aq) and HBrO(aq). 87. the solution will turn darker. Curriculum Development Council. [Br2(aq)] [HBr(aq)][HBrO(aq)] Therefore. Concentration Br2(aq) + orange H2O(l) d colourless HBr(aq) + colourless HBrO(aq) colourless Notes Upon addition of extra Br2(aq) into the equilibrium system. Nature of equilibrium Part 3 Page 1 Topic Reference Reading I.. Thomas Nelson and Sons Ltd.). pressure and temperature on equilibria (Le Chaterlier's principle) 1. 231–232 Chemistry in Context (4th ed.I. Stanley Thornes (Publisher) Ltd.).2. Nature of Equilibrium 6. Because more products are formed upon addition of the reactant. the colour will get paler. Effect of change in concentration. upon addition of Fe3+(aq) or NCS-(aq).0–6. 305–309 Effect of change in concentration. the position of the equilibrium is said to be shifted to the right upon the addition of Br2(aq). Thomas Nelson and Sons Ltd. the equilibrium will adjust in a way to restore the expression back to the original [Br2(aq)] K c. PO43-(aq) ions form complex with free Fe3+(aq) and lower its concentration. the net change in concentration of Br2(aq) is always less than the actual amount added. 5th Edition.2. An increase in [Br2(aq)] will make the value of the expression + Fe3+(aq) iron(III) ion pale yellow NCS-(aq) thiocyanate ion colour d FeNCS2+(aq) thiocyanato iron(III) ion dark red complex [FeNCS2+(aq)] Kc = [Fe3+ ][NCS2+ ] (aq) (aq) Similarly.. It shift the equilibrium position to the right. 343–354 Reading Syllabus Chemistry in Context.4 AS-Level Chemistry Syllabus for Secondary School (1991). This shifts the equilibrium position to the left. Kc = [HBr(aq)][HBrO(aq)] [Br2(aq)] [HBr(aq)][HBrO(aq)] smaller than the original Kc. Therefore. At equilibrium. However. Hence a new equilibrium position is established eventually. forward rate = backward rate and forward rate = k1[NO2(g)]2 backward rate = k-1[N2O4(g)] The forward reaction is a second order reaction but the backward reaction is only a first order reaction. . the colour of the gas gets darker because of an immediate increase of the concentration of brown NO2(g). pressure has only effect on the equilibrium involving gaseous species.I. 2NO2(g) brown d N2O4(g) colourless Once the plunger of the gas syringe is pressed. Nature of equilibrium 2. Pressure Part 3 Page 2 Since only gas is compressible. The increase in pressure increases the forward rate more than the backward rate. the gas turns paler gradually. 0 kJmol-1 When the tube containing the mixture of NO2(g) and N2O4(g) is immersed in hot water. the colour of the gas gets paler. Conversely. . ∴ An increase in temperature will increase the values of k1 and k-1 for different extent. it can be seen that the forward reaction and backward reaction involve different activation energy. the reaction with higher activation energy will have a higher percentage increase in rate. The expression K = k will become smaller.I. a higher -1 temperature will shift the equilibrium position of an exothermic reaction to the left.e. Moreover. Temperature 2NO2(g) brown N2O4(g) colourless Part 3 Page 3 d ∆H = -58. Ea1 < Ea-1 Once the temperature is increased. The forward reaction has a lower activation energy Ea1. Nature of equilibrium 3. Therefore. from the diagram of distribution of molecular speed. when it is put inside icy water. This is an exothermic reaction. i. thus the equilibrium k1 constant and equilibrium position. the rate of both forward reaction and backward reaction are increased. From the energy profile of the reaction. the colour of the gas gets darker. RT + R ∆H ln K = .T∆S = . It is a constant for a specific reaction. the expression . ∆H is negative.RT less positive. The actual derivation is not required in A-Level but you are expected to use the equation to explain the effect of temperature on an equilibrium. ∆S is entropy change of the system. which indicates the total free energy change of the system.RTln K = ∆H . Free energy change and entropy change are not required in A-Level.RTln K where ∆G is Free energy change. Nature of equilibrium a) Equation : ln K = constant ∆H RT Part 3 Page 4 Or the effect of the temperature can be interpreted from another point of view. ∆H + constant RT ∆G = ∆H . which is a measure of the change in degree of disorderness. Conversely. This shifts the equilibrium position to the reactant side.RT + RT ∆H ∆S ln K = . N.I.RT + constant (1) Derivation of ln K = - By rearranging the equation N.T∆S ∆H T∆S ln K = .B. It is a constant for a specific reaction. ∆H For an exothermic reaction.RT is positive. the equilibrium position of an exothermic reaction will be shifted to the product side if the temperature is lowered. therefore.RTln K . hence K will become smaller.T∆S = . T is the temperature in Kelvin. ∆H is enthalpy change of the system. An increase in ∆H temperature will make the expression . .B. It can be proved that the equilibrium constant is related to the ∆H according to the equation ∆G = ∆H . the decrease in pressure can be reduced by converting 1 N2O4(g) molecule to 2 NO2(g) molecules. Le Chaterlier's principle states that when a change is imposed on an equilibrium system. the changes in forward rate and backward rate must be considered.0 kJmol-1 The effect of decrease in temperature can be reduced by converting some NO2(g) to N2O4(g) because the reaction is exothermic. the net decrease of concentration of Fe3+(aq) will be reduced and the equilibrium position is shifted to the left. It only increases the forward rate and backward rate for the same extent and shortens the time required to attain equilibrium. Therefore. called Le Chaterlier's principle. a decrease in temperature will shift the equilibrium position of an exothermic reaction to the right. some Fe3+(aq) added with combine with NCS-(aq) to form FeNCS2+(aq). Le Chaterlier's principle The effect of a change in condition on the position of an equilibrium can be generalized as a principle. some FeNCS2+(aq) will dissociate to give more Fe3+(aq). If the pressure is decreased. the system will response in a way to minimize the effect of the change. As a result the net increase of concentration of Fe3+(aq) will be reduced and the equilibrium position is shifted to the right. Effect of pressure 2NO2(g) d N2O4(g) brown colourless Any increase in pressure can be reduced by converting 2 NO2(g) molecules to 1 N2O4(g) molecule Therefore. 5. Effect of concentration + NCS-(aq) Fe3+(aq) pale yellow colourless FeNCS2+(aq) dark red d Upon addition of FeCl3(s). Le Chaterlier's principle is only a rule which helps to predict the outcome of the change. Therefore.I. Effect of catalyst on equilibria Part 3 Page 5 Addition of catalyst has no effect on concentrations nor equilibrium constant. Consequently. In order to explain the change. it has no effect on the position of an equilibrium. Nature of equilibrium 4. increase in pressure will shift the equilibrium to the right. If PO43-(aq) is added to reduce the concentration of Fe3+(aq). Glossary equilibrium position shifting of equilibrium position disorderness Le Chaterlier's principle free energy change entropy change . While a increase in temperature will shift the equilibrium position to the left. Effect of temperature 2NO2(g) d N2O4(g) brown colourless ∆H = -58. 1½ mark What is the effect of decreasing pressure on the equilibrium of each system? ↓ pressure will shift the equilibrium to the left for system X and system Y. (Relative atomic masses : H 1. O 16. i. 94 2A 2 b ii 2b Account for each of the following: ii The melting point of ice decreases with an increase in pressure.0.e. H2SO4 or other drying agent C Some candidates misunderstood the question and gave alternative synthetic routes from acid to ester. increase in pressure. = 4. 1 mark 3 . H. the molecules are packed closer together ∴ water has a higher density than water. C. A number of candidates failed to point out that esterification is reversible and the yield of product depends on the equilibrium position of the reaction.0) 7c Suggest. of ice decreases with increase in pressure.I.N.8%. an open structure. 1½ 1½ 92 2C 7 c 7 A carboxylic acid P. Nature of equilibrium Part 3 Page 6 Past Paper Question 91 1A 2 b i ii 92 2C 7 c 94 2A 2 b ii 95 1A 1 e iv 96 1A 1 f iii 97 1A 2 a ii 95 2B 4 a iii 97 2B 8 a iii 97 2B 8 b i 91 1A 2 b i ii 2b The energy profile of the reaction A(g) + B(g) d C(g) under two different catalysis X and Y are represented below. two ways which would make the esterification go towards completion. the intermolecular attraction is H-bond. will shift the equilibrium to the left for system X and system Y.2% by mass .p.0%. i ii C What is the effect of increasing temperature on the equilibrium of each system? ↑ temp. Some candidates confused equilibrium and rate of reaction thinking that by changing reaction conditions to increase the rate of reaction would make the reversible reaction go to completion. 1½ mark Many candidates discussed only one system when the question asked for the effect on each system. and O. contains C. with a relative molecular mass less than 100. 55. Many candidates did not distinguish kinetics from chemical equilibria. An attempt to convert P to its methyl ester Q by prolonged refluxing of P with methanol in the presence of aqueous H2SO4 gave the desired ester Q but with much of the starting material P unchanged. 37. each H2O molecule is tetrahedrally surrounded by 4 other H2O 1 mark molecules. C 12. shifts the equilibrium to the right ∴ m. with explanations. 1 mark For the reaction. 7. In ice. H2O(s) d H2O(l). In liquid water. 4 The reaction is reversible 1 mark 1 mark RCOOH + CH3OH d RCOOCH3 + H2O 1 mark Use large excess of CH3OH. mass action to shift equilibrium 1 mark Remove H2O as it is formed to shift equilibrium to the right ½ mark Use conc.0. one half of the SO2(g) had been converted into SO3(g).01 g of ethanoic acid are treated with 4. ½ 4 6 7 . I2(s) + KI(aq) d KI3(aq) or I2(s) + I-(aq) d I3-(aq) Page 7 1 1 2 1 mark 1 mark 96 1A 1 f iii 1f SO2(g) and O2(g) were mixed in the mole ratio of 3 : 1 at 1000 K in the presence of a catalyst. the percentage dissociation of PCl5(g) is 86%. iii dinitrogen tetraoxide. In each case. will the value of Kp increase. 1 mark C A common mistake in the answers was the omission of the concentration of water from the equilibrium expression. catalyst can affect only the rate of (forward and backward reaction. yield does not change. decrease or remain the same ? remains the same ½ mark 97 1A 2 a ii 2a Consider the following dissociation reaction: PCl5(g) d PCl3(g) + Cl2(g) At 400 K and 101 kPa pressure. Therefore.61 g of ethanol in the presence of a catalyst. 97 2B 8 a iii 8a Suggest how the following nitrogen oxides can be prepared in the laboratory.with KI in solution. Concentrated sulphuric(VI) acid / hydrochloric acid / hydrogen chloride gas 1 mark iv Would the addition of more of the same catalyst affect the value of Kc? Explain. Nature of equilibrium Part 3 95 1A 1 e i iv 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12. 95 2B 4 a iii 4a Explain the following facts: iii Iodine is more soluble in aqueous potassium iodide solution than in water. When the reaction reaches equilibrium at 298 K. ii State the effect of an increase in pressure (I) on Kp. No. 5. N2O4 97 2B 8 b i 8b The synthesis of ammonia using the Haber Process involves the following: N2(g) + 3H2(g) d 2NH3(g) ∆Ho = -92 kJ mol-1 i State the effect of a change in temperature on the reaction at equilibrium. and (II) on the percentage dissociation of PCl5(g).I. 1 mark OR Catalyst does not affect position of an equilibrium. iii If the above reaction takes place in the absence of the catalyst. state the reactant(s) used and the reaction conditions. When equilibrium was attained at 373 kPa pressure. but other conditions remain unchanged.04 g of ethanoic acid are found to have reacted. i Name a suitable catalyst for this reaction in the forward direction. and write balanced equation(s) for the reaction(s) involved. I2 appears more soluble. I2 forms soluble complex I3. Nature of equilibrium 6.00 g of H2(g) are heated to equilibrium at 470ºC.181 . Stanley Thornes (Publisher) Ltd.9. 232–238 Chemistry in Context (4th ed.0. 335–338 Reading Syllabus Notes 5.0075 mol 0. the equilibrium mixture contains 1. 2. of mole of H2(g) no.0 g initial no.174 = 0. of mole of I2(g) ( (a) 1. of mole of H × no.I.). (a) How may moles of each gas are present in the equilibrium mixture? (b) Compute the equilibrium constant. (Given : Relative atomic masses: I.9 gmol-1 × 2 = 0. Nature of equilibrium Part 4 Page 1 Topic Reference Reading I. of mole of I2(g) at equilibrium = 126. 46. of mole of HI(g)2 no. from initial concentrations.0 g of I2(g) and 1.181 mol 0.174 × 2 = 0.500 mol 0 produced used up 0.90 g no. of mole of HI(g) 2 ) V no.0075 mol = 50 2(g) 2(g) .326 mol 0 + 0.1.5 Part 4 Assignment A-Level Chemistry (3rd ed.00) [HI(g)]2 Kc = [H ][I ] = 2(g) 2(g) = 2.). of mole of H2(g) = 1.326 mol × 0.0075 = 0. 1.174 mol 0. Thomas Nelson and Sons Ltd. 126. Write an equilibrium law(s) to describe the system. Example 1 H2(g) + I2(g) d 2HI(g) When 46. of mole of H2(g) × no.348 mol)2 Kc = no.500 .348 mol I2(g) H2(g) HI(g) (b) 3 0. of mole of I2(g) × V V no. 1. stoichiometric coefficients and law of conservation of mass.174 mol at equilibrium 0.0075 mol H2(g) + I2(g) d 2HI(g) initial 0. Figure out the relationship among the equilibrium concentrations of different species. no.9 g I 2. all calculations must be based on the concentration ( ) instead of the absolute V amount (n). of mole of I2(g) = 126. Solve the equation(s).9 gmol-1 × 2 = 0.2. Examples of calculation Since the equilibrium law describes the relationship of the concentrations of different species in an n equilibrium system.00 gmol-1 × 2 = 0. of mole of HI(g)2 (0.. H.348 mol no. of mole of I = 0. 3.00 g initial no. N.B.181 mol 1.348 = 0..500 mol 1.0. 2. Using the value of Kc from the Example 1.0 dm3 evacuated chamber to 470ºC.00 . of mole of I2(g) consumed.x) = 50 4x2 1.00) mol = 2. of mole of I2(g) consumed. no.00 mol 1.93 or 2. of mole of I 2(g) 2(g) Let x mole be the no.I.2.0.0821 atm mol-1K-1 × 743 K = 4. x mol × 2 (2x)2 Kc = (1.07 mol 4z2 = 100 – 150z + 50z2 46z2 – 150z + 100 = 0 used up z mol z mol at equilibrium (1.22 mol 4x2 = 50 .00 .0821 atm mol-1K-1) (a) 1.00 + 1. no. Nature of equilibrium Example 2 Part 4 H2(g) + I2(g) d 2HI(g) Page 2 If 1.00x + x2 = 50 x = 0.00 mol of gas remains at equilibrium.00 mole each of H2(g) and I2(g) are heated in a 30.3. of mole I2(g) remaining = (1.0. (2z)2 Kc = (2. of mole of HI(g)2 Kc = no.z) = 50 4z2 2. of mole of I 2(g) 2(g) Since the final equilibrium position is not depending on when the additional H2(g) is introduced. initial produced I2(g) 1. (c) Now if one additional mole of H2(g) is introduced into this equilibrium system. 2.z)(1. of mole of HI(g)2 Kc = no.x) mol (1.00 .00z + z2 = 50 z = 0. of mole of H × no.07 atm P= V = 30. Let x mole be the no.3 (rejected) no.00 – z) mol (2.4 (rejected) no.00 mol × 0.x)(1.00 .00 mol 0 produced used up x mol x mol at equilibrium (1.78 or 1.00 mol HI(g) 0 z mol × 2 3. it can be assumed that it is introduced with the original H2(g). determine (a) how many mole of I2(g) remain unreacted when equilibrium is established.00 .00 .78) mol = 0. (b) the total pressure in the chamber.00 . initial 1.100x + 50x2 46x2 . of mole I2(g) remaining = (1.00 .x) mol 2x mol I2(g) H2(g) HI(g) 3.93) mol = 0. of mole of H × no. PV = nRT nRT 2.100x + 50 = 0 (b) The number of moles of gas does not change as the reaction proceeds at 470ºC.00 . how many mole of the original iodine will remain unreacted? (Given : Universal gas constant (R) = 0.00 – z) mol 2z mol .00 mol H2(g) (1. hence 2.0 dm3 (c) 1.00 . 01 g 46. 5. of moles 4.068 gmol-1 60.052 = 0.0.5.82 dm6mol-2 (wrong unit -½) 2½ mark 2 2 iii When the volume of the vessel was increased and the system allowed to come to a new equilibrium at the same temperature. for the reaction at 298 K. Nature of equilibrium Part 4 stoichiometric coefficients law of conservation of mass Page 3 Glossary Past Paper Question equilibrium concentration 90 2A 1 a i ii iii 95 1A 1 e ii iii 90 2A 1 a i ii iii 1a Consider the following equilibrium at constant pressure: A2(g) + 3B2(g) d 2AB3(g) A mixture of 4.116 V 0.016 V 0. i Calculate the concentration of each species at equilibrium. Some weaker candidates failed to give the correct unit for the equilibrium constant. When the system had reached equilibrium.04 60. A2(g) + 3B2(g) d 2AB3(g) Since 4 moles of AB3 will consume 2 moles of A2 and 6 moles of B2 ∴ at equilibrium: 12 .016 0. it was found that 4.0 mol of B2(g) was placed in a vessel of volume 20.01 g of ethanoic acid are treated with 4.82 × 93 × 3 = 90 dm3 1 mark 4 Some candidates mistakenly used the initially given concentration rather than the equilibrium concentration in their calculation of equilibrium constant.116 0.0 mol of A2(g) and 12.01 .0 mol AB3(g) was present. Let V be the new volume 9 ½ mark [B2] = V 4-1 3 [A2] = V = V ½ mark 2 [AB3] = V ½ mark 2 V V K = 14.100 mol 12.084 V – – 3 no.084 2 marks concentration at equilibrium . When the reaction reaches equilibrium at 298 K.04 g of ethanoic acid are found to have reacted. 1½ 2½ 3 95 1A 1 e ii iii 1e The reaction between ethanoic acid and ethanol can be represented by the following equation: CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) 12.100 . Calculate the new volume. At new equilibrium: 9 moles of B2 is present. CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) initial no.61 g 12.052 gmol-1 = 0. of mole at equilibrium 0. Kc .0 dm3. 9.0 mol of B2(g) was found to be present. ii Calculate the equilibrium constant.82 = (V)2 ⋅ ( 3 ) ⋅ ( 9 )3 ½ mark V= C i ii 14.6 6 [B2] = 20 = 20 moldm-3 ½ mark 4-2 2 [A2] = 20 = 20 moldm-3 ½ mark 4 [AB3] = 20 moldm-3 ½ mark ii Calculate the equilibrium constant of this reaction at 565K.084 = 0.0840.I. and heated to 565K. [AB3]2 4 20 20 K = [A ][B ]3 = (20)2 ⋅ ( 2 ) ⋅ ( 6 )3 = 14.2000 mol= 0.084 V 0.61 g of ethanol in the presence of a catalyst. 60 g of ethanoic acid ? 0.010 mole CH3COOH(l) + C2H5OH(l) d CH3COOC2H5(l) + H2O(l) no.0219 mole ½ mark Need to add 0.016 = 3.084 0.I.106 x ∴ x = 0.74 g ½ mark iii A common mistake in the answers was the omission of the concentration of water from the equilibrium expression.0106 x 0.80 (no unit) × V V iii What additional mass of ethanol would be required in order to use up a further 0. Few candidates were able to complete part (iii). of mole at equilibrium 0.0942 3.094 0.084 V × V 1 marks Kc = 0.016 mole of ethanol = 0. Nature of equilibrium Part 4 0.80 = 0.116 0.60 g of ethanoic acid ≡ 0. Page 4 2 C .094 1 mark 0. Acid-base Equilibria 6. 2. 369–370 Reading Chemistry in Context (5th ed. H+(aq). H2SO4(l) + H2O(l) → H3O+(aq) + HSO4-(aq) HSO4-(aq) + H2O(l) → H3O+(aq) + SO42-(aq) Examples of Arrhenius base e.. Arrhenius definition Syllabus Notes B rφ n A rr h s ef wis d initio Le Lowry d n e d te nius defi f e Common Sense n it io in tion ni Arrhenius definition is based on ionization. produces hydrogen ions. 326 Acid-base Equilibria Acid-base Theory Dissociation of water II. Acid-base Theory At the early stage of development of acid-base theory.. They are called Arrhenius definition.g. Acid is restricted to hydrogen-containing species and base is restricted to oxide or hydroxide. And base is only a substance which neutralizes the sour taste of an acid. The coverage of the latter definition is boarder than the former one. acid was only described as a sour substance. Brønsted Lowry definition and Lewis definition.).II.1–6. 365–367. The theory is only applicable to aqueous medium where a lot of acid-base reaction takes place in the absence of water.2. 1. . they developed a series of definitions to describe acid and base. 198–201.). NaOH(aq) + HCl(aq) → Na+Cl-(aq) + H2O(l) CuO(s) + HCl(aq) → Cu2+Cl-2(aq) + H2O(l) This is the definition used in certificate level. Examples of Arrhenius acid e. 255–258 Chemistry in Context (4th ed. Acid-base Equilibria Part 1 Page 1 Topic Reference Reading II.). Nevertheless. According to these properties. Base – a substance which reacts with an acid to give salt and water only.. when dissolved in water. Swedish chemist Svante Arrhenius proposed that Acid – a hydrogen-containing compound that.2 Part 1 Assignment A-Level Chemistry (3rd ed.2. Thomas Nelson and Sons Ltd.g. Arrhenius theory is criticized that 1. 219–222. In 1884. Acid-base Equilibria A. Stanley Thornes (Publisher) Ltd. Thomas Nelson and Sons Ltd. scientists find that acids and bases bear some other properties. e. it can behave as both proton acceptor and proton donor.g. Stronger a proton donor. All Arrhenius acids and bases can be classified into Brønsted-Lowry acid and base accordingly. in which Acid – a proton donor Base – a proton acceptor e. . H2O(l) + proton acceptor (base) H2O(l) d proton donor (acid) H3O+(aq) + proton donor (conjugate acid of H2O(l)) OH-(aq) proton acceptor (conjugate base of H2O(l)) The above reaction is also known as self-dissociation / self-ionization of water.II. Water is a very special example. Therefore. H3N:(g) + BF3(g) → H3N→BF3(s) electron electron donor acceptor (Lewis base) (Lewis acid) Most of the chemical reaction is caused by redistribution of electrons which leads to rearrangement of atoms and formation of a new substance. Lewis definition The American chemist Gilbert N. weaker will be the conjugate acid. it is a Brønsted-Lowry acid and a Brønsted-Lowry base. It does not require formation of a salt or an acid-base conjugate pair. Similarly.g. Lewis (1923) proposed an even broader definition for acid and base. weaker will be the conjugate base. It works with solvents other than water. when a Brønsted-Lowry base accepts a proton.g. it becomes a potential proton donor and is called conjugate acid of the base. it becomes a potential proton acceptor and is called conjugate base of the acid. iii. Brønsted-Lowry definition Part 1 Page 2 In 1923. virtually. Acid-base Equilibria 2. NH3(aq) + proton acceptor (base) e. He proposed that Acid – an electron acceptor Base – an electron donor This definition offers many advantages. Since all chemical species are potential electron acceptor or electron donor. e. HCl(aq) + proton donor (acid) H2O(l) d proton donor (acid) H2O(l) d proton acceptor (base) NH4+(aq) + proton donor (conjugate acid of NH3(aq)) H3O+(aq) + proton donor (conjugate acid of H2O(l)) OH-(l) proton acceptor (conjugate base of H2O(l)) Cl-(aq) proton acceptor (conjugate base of HCl(aq)) When a Brønsted-Lowry acid donates a proton. The acids are not limited to compounds containing hydrogen. all kinds of reaction can be considered as Lewis acid-base reaction. all chemical species are either Lewis acid or Lewis base. ii. Stronger a proton acceptor. Therefore. 3. Danish chemist Johannes Brønsted and British chemist Thomas Lowry proposed a boarder definition. including i.g. Concept of Lewis acid-base is very useful in describing reaction mechanisms. pH = . oxidizing agent + e. Ka can also be expressed in a negative log scale. more HA(aq) molecules will dissociate into H3O+(aq) ion and A-(aq) ion and give a larger value for K a. B. Ka is defined as Ka = [H3O+(aq)][A-(aq)] = Keq[H2O(l)] [HA(aq)] where [H2O(l)] is a constant since water is the solvent in large excess. Part 1 Page 3 In the description of an acid-base equilibrium in A-Level. For an acid. Relative strength of selected acids and their conjugate bases Stronger an acid. A strong acid or base is less stable than a weak acid or base. Ka. Stronger a base.d reducing agent In redox reaction. However. only strong acid react with strong base. in the description of reaction mechanism. A strong acid tends to react with a strong base to form a weak conjugate base and weak conjugate acid.II. e.→ NH2. NH2. acid base reaction can be considered as a competition for proton.B.+ HC≡CH . Acid-base Equilibria N.log Ka A low pH means a high concentration of H+ ion.+ HC≡CH → NH3 + HC≡Cbut not NH3 + HC≡C. N. Lewis definition is used. the Brønsted-Lowry definition is used. [H3O+(aq)][A-(aq)] Equilibrium constant. HA(aq) in water HA(aq) + H2O(l) d H3O+(aq) + A-(aq) Acidity constant. only strong oxidizing agent reacts with strong reducing agent. Keq = [HA ][H O ] (aq) 2 (l) For a stronger acid.g. The strength of an acid or a base is also an indicator of their stability. weaker will be its conjugate base.B. conjugate base + H+ d conjugate acid In acid base reaction. A low pKa means a larger value for Ka and a stronger acid. Strength of acid Strength of an acid can be measured by an equilibrium constant called acidity constant or acid dissociation constant. Similarly. Similar to pH. weaker will be its conjugate acid.log [H+] pKa = . Redox reaction can be considered as a competition for electron. K [H2O(l)] = [H+(aq)][OH-(aq)] = Kw Kw is known as ionic product of water. C. thus [H2O(l)] is a constant and can be incorporated in the equilibrium constant. [H+(aq)][OH-(aq)] [H2O(l)] Ionic product of water or simply H2O(l) d H+(aq) + OH-(aq) H2O(l) d H+(aq) + OH-(aq) K= [H+(aq)][OH-(aq)] [H2O(l)] In aqueous medium. liquid NH3(l)) must be used. The water will convert all those acid molecules into H3O+(aq) ions. Like other equilibrium constant. the ions are from self-dissociation / self-ionization of water molecule. H2SO4(aq) and HNO3(aq) are strong acid though they don't have the same strength. H2O(l) is in large excess.e. it is a quantity depending on temperature and has the value of 1. For example. Dissociation of water Pure water is found to be a poor conductor of electricity but not an insulator. Pure water is neutral because the concentration of H+(aq) is the same as the concentration of OH-(aq). [H+(aq)] = [OH-(aq)] Acidic Neutral Alkaline . Acid-base Equilibria 1. i. do not show any difference in acidity in water. H2O(l) + H2O(l) d H3O+(aq) + OH-(aq) K= 1.00 × 10-14 mol2dm-6 at 25ºC. a base stronger than hydroxide ion does not exist in aqueous medium. NaNH2(s) is a very strong base. HCl(aq) + strong acid HBr(aq) + strong acid H2O(aq) → H3O+(aq) + strong weak base acid H2O(aq) → H3O+(aq) + strong weak base acid Cl-(aq) weak base Br-(aq) weak base This is why we consider all HCl(aq). a non-aqueous solvent (e. NaNH2(s) + H2O(l) → Na+(aq) + OH-(aq) + NH3(aq) This is known as the leveling effect of solvent. The water will convert all those base molecules into OH-(aq) ions. Indeed. Therefore.g. it must contain a very small amount of ions.II. Similarly. Leveling effect Part 1 Page 4 Acids stronger than hydroxonium ion H3O+(aq). [H+(aq)] 1. but few candidates gave the correct equation for it acting as a base. i Identify all Brφnsted acids and all Brφnsted bases in the solution. 97 2A 4 c i 4c A solution is formed by mixing equal volumes of 0. ii The equation for nitric(V) acid acting as an acid was given correctly. but a base in liquid hydrogen fluoride.000 ( 1.00 × 10-7 moldm-3 Recall that pure water has the concentration of 55.000.II.80 × 10 [H2O(l)] = 1 This means that only 1 out of every 556.6 moldm-3.80 × 10-9 ) water molecule will dissociate into a pair of H+(aq) and OH-(aq).dissociation / self-ionization of water ionic product of water (Kw) 94 1A 1 f i 97 2A 4 c i 99 1A 4 a 94 1A 2 c i ii Glossary Past Paper Question 94 1A 1 f i 1f i Write an equation to show that HPO42-(aq) can act as a Brφnsted base in water.20 M CH3CO2Na(aq).6 moldm-3 = 1. 99 1A 4 a 4a Write all the Bronsted acids present in aqueous ammonia. acid-base equilibria Arrhenius acid-base Brønsted Lowry acid-base Lewis acid-base conjugate acid-base proton donor / acceptor electron acceptor / donor pH pKa leveling effect acidity constant / acid dissociation constant (Ka) self.20 M CH3CO2H(aq) and 0. HPO42-(aq) + H2O(l) d H2PO4-(aq) + OH-(aq) 1 1 mark 94 1A 2 c i ii 2c i What is a “Brφnsted acid”? “Brφnsted acid” is a proton donor 1 mark ii Write equations to show that nitric(V) acid is an acid in water. This explains the extremely low conductivity of pure water but it is not an insulator.00 × 10-7 moldm-3 -9 55. 1 2 . HNO3 + H2O d H3O + NO31 mark + 1 mark HNO3 + HF d H2NO3 + F C i Many candidates thought that an aqueous medium was required for the proton donation. Acid-base Equilibria For pure water at 25ºC Part 1 Page 5 Kw = [H+(aq)][OH-(aq)] = [H+(aq)]2 = 1.00 × 10-14 mol2dm-6 [H+(aq)] = 1. 30 × 10-14 0. water at 50 ºC has a pH 6. For example. thus the Kw.63 Therefore. 268 Chemistry in Context (4th ed. 2.1 mol dm-3 0. N..47 × 10-14 mol2dm-6 = 2.63. pH and its measurement 1. pure water has pH 7 at 25ºC.01 mol dm-3 0. 264.3 Part 2 Assignment A-Level Chemistry (3rd ed. Kw = 5.log [H+(aq)] Because it is a log scale.B.). 258–259. Thomas Nelson and Sons Ltd.34 × 10-7 moldm-3 pH = . Temperature dependence of pH Pure water has pH 7 only at 25ºC because Kw is temperature dependent.00 × 10-14 mol2dm-6 [H+(aq)] = 1.log[H+(aq)] = . increases with increasing temperature.. not because of the pH.34 × 10-7 = 6.47 × 10-14 51.68 × 10-14 1.II.2.47 × 10-14 mol2dm-6 [H+(aq)] = 5.3 × 10-14 A solution is said to be neutral because the concentration of H+(aq) is the same as the concentration of OH-(aq). Kw = [H+(aq)][OH-(aq)] = [H+(aq)]2 = 1.00 × 10-7 = 7. Definition of pH pH 0 1 2 3 4 [H+(aq)] 1 mol dm-3 0. Find the pH of a 0.11 × 10-14 0. Acidic Neutral Alkaline .47 × 10-14 mol2dm-6 Kw = [H+(aq)][OH-(aq)] = [H+(aq)]2 = 5. Thomas Nelson and Sons Ltd.log 1. pH = . 325–329 pH and its measurement Temperature dependence of pH D. at 50ºC.. For pure water at 25ºC.00 × 10-14 5.).0001 mol dm-3 Syllabus Notes pH is defined as the negative log of the concentration of H+(aq) in molarity.00 × 10-7 moldm-3 pH = . Find the pH of a 10-8 M HCl(aq) solution. Acid-base Equilibria 6.00 Therefore.1 M HCl(aq) solution. 367–369 Reading Chemistry in Context (5th ed. Acid-base Equilibria Part 2 Page 1 Topic Reference Reading II. A neutral solution has the pH 7 only at 25 ºC. Stanley Thornes (Publisher) Ltd. The degree of dissociation of water.log 2. every 1 unit increase in pH means ten folds decrease in the concentration of H+(aq). Temperature /ºC 0 10 20 25 50 100 Kw / mol2 dm-6 0.001 mol dm-3 0.).log[H+(aq)] = . it is a weak acid or a weak base whose conjugate acid and conjugate base have different colours. Some candidates erroneously pointed out that the low pH of water at 323 K was caused by the dissolution of some acidic gases. (1) Calibration of pH meter pH meter is calibrated by immersing the electrode into a buffer solution with known pH and adjusting the knob on the pH meter. blood is a buffer solution which has a constant pH of about 7. ∴ pH of pure water at 323 K is less than 7. (1) Colour of indicator Indicator litmus methyl orange phenolphthalein 0 | | | 1 red 2 red 3 4 5 | pH value 6 7 8 purple | 10 11 12 13 14 blue | yellow | |pale pink| red | 9 |orange| colourless Universal indicator / pH indicator is a mixture of several indicators. Like other testing paper. is an endothermic process. 99 1A 4 d 4d Constant boiling hydrochloric acid contains 20. Increase in temperature. 2 + 1 mark The dissociation of water.0. pH meter needs to be calibrated before use. 2H2O(l) d H3O (aq) + OH (aq).4 which is independent of the acidity of our diet. Calculate the mass of constant boiling . Acid-base Equilibria 3. the pH of pure water is less than 7. a) Use of pH meter pH meter is an electronic instrument which gives the pH value of a solution directly. Usually. indicator and pH paper are commonly used to determine the pH of a solution. Measurement of pH Part 2 Page 2 pH meter. it should be dipped into the solution to be tested and removed immediately. However. b) Use of indicator Indicator is a chemical which has different colour at different pH. Buffer solution is a solution whose pH is insensitive to the addition of small amount of acid or base. the dye will diffuse into the solution and the colour observed would be paler than the standard. 1 mark C i Many candidates were not aware of the fact that the dissociation of water is an endothermic process. In pure water. 94 2A 3 b i 99 1A 4 d pH meter calibration of pH meter buffer solution indicator universal 94 2A 3 b i 3b Account for each of the following: i At 323K. the universal indicator gives a series of colour change at different pH instead of just 2. Glossary Past Paper Question pH scale indicator. [H3O(aq)+] at 323 K is higher than that at 298 K.2% by mass of HCl.II. pH paper is a filter paper is soaked in universal indicator. If the paper is immersed in the solution. For example. the equilibrium shifts to the right and hence [H3O+(aq)] increases. As a result. Page 3 .0 at 298 K.00 dm3 of HCl(aq) of pH 2.II. Acid-base Equilibria Part 2 hydrochloric acid required to prepare 1. 1 M NaOH(aq) 1 M NaOH(aq) pH 0 1 3 7 11 13 14 Notes e.. Dissociation constant (Ka and Kb) For the equilibrium. 369 Reading Chemistry in Context (5th ed. a strong acid (HCl(aq)) has lower pH than a weak acid (CH3COOH(aq)). the conductivity of a strong acid / base is also higher than that of a weak acid / base.). Strong and weak acids/bases An acid or a base is said to be strong if the degree of dissociation is high. Thomas Nelson and Sons Ltd. Thomas Nelson and Sons Ltd. HA(aq) + H2O(l) d H3O+(aq) + A-(aq) [H3O+(aq)][A-(aq)] K = [HA ][H O ] (aq) 2 (l) Since H2O(l) is in excess. with the same concentration.g.II. 2. this is measure in term of pH..1 M HCl(aq) 0. 258–262. Usually. Stanley Thornes (Publisher) Ltd. John Murray (Publisher) Ltd. HA(aq) + H2O(l) d H3O+(aq) + A-(aq) [H3O+(aq)][A-(aq)] K = [HA ][H O ] (aq) 2 (l) K[H2O(l)] = [H3O+(aq)][A-(aq)] = Ka [HA(aq)] Solution 1 M HCl(aq) 0. Measuring of pH or conductivity of acid / base Obviously.4 Advanced Practical Chemistry. 1. its concentration can be incorporated into the equilibrium constant and the new dissociation constant is called acidity constant Ka. NH3(aq)). 273 Chemistry in Context (4th ed. 283 Part 3 rd Assignment A-Level Chemistry (3 ed.2.g. Owing to the higher ionic concentration.1 M NH3(aq) 0. NaOH(aq)) has higher pH than a weak base (e. Bell and Hyman.). Ka and Kb D.. 62–64 Modern Physical Chemistry. Acid-base Equilibria 6. CH3COOH(aq) + H2O(l) d CH3COO-(aq) + H3O+(aq) Ka = [CH3COO-(aq)][H3O+(aq)] = 1.. the conductivity of the solution or the dissociation constant.g. 329–330 Syllabus Strong and weak acids/bases Measuring of pH or conductivity of acid / base Dissociation constant (Ka and Kb) Calculation involving pH.1 M CH3COOH(aq) pure water 0.7 × 10-5 mol dm-3 [CH3COOH(aq)] .). And a strong alkali (e. Acid-base Equilibria Part 3 Page 1 Topic Reference Reading II. the first dissociation constant Ka1 is always larger than the second dissociation constant Ka2.g. in a solution of H3PO4(aq).] [ H3PO4(aq)] [ H2PO4-(aq)] 4 (aq) + 3 3[H (aq)] [ PO4 (aq)] = [ H3PO4(aq)] (1) Charge effect For a polybasic acid. When the second proton is removed from the acid molecule. The first dissociation constant Ka1 is low. Acid-base Equilibria Part 3 Page 2 Similarly. the dissociation constants of a tribasic acid can be expressed as : H3PO4(aq) d 3H+(aq) + PO43-(aq) Ka(overall) = K a1 × K a2 × K a3 [H+(aq)][ H2PO4-(aq)] [H+(aq)][ HPO42-(aq)] [H+(aq)][ PO43-(aq)] = × × [ HPO 2. it is removed from an electrically neutral molecule. This is because when the first proton is removed from the acid molecule. The effect of the charges on the dissociation of acid molecule is known as charge effect.9 × 10-3 6. Acid H3PO4(aq) H2PO4-(aq) HPO42-(aq) Ka(298K)/mol dm-3 7. B(aq) + H2O(aq) d HB+(aq) + OH-(aq) [HB+(aq)][OH-(aq)] K = [B ][H O ] (aq) 2 (l) K[H2O(l)] = [HB+(aq)][OH-(aq)] = Kb [B(aq)] e. it dissociates in water through 3 consecutive steps. it is removed from a negatively charged species which is energetically less favorable. . NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) Kb = [NH4+(aq)][OH-(aq)] = 1. the second dissociation constant Ka2 is lower than the first one and the third dissociation constant Ka3 is even lower. Moreover. Relative abundance of different species H2O(l) >> H3PO4(aq) > H+(aq) > H2PO4-(aq) > HPO42-(aq) > PO43-> OH-(aq) Furthermore. it consists of mainly water and H3PO4(aq) molecules.8 × 10-5 mol dm-3 [NH3(aq)] a) Dissociation of polybasic acid Consider the dissociation of tribasic acid H3PO4(aq) in water.2 × 10-8 4.II.] 4 (aq) K a1 = Phosphoric acid is a tribasic acid which is not very strong. Only 25% of the molecules dissociate in water. the dissociation constant of a base (basicity constant Kb) can also be defined in a similar way. Therefore.4 × 10-13 Equilibrium H3PO4(aq) d H+(aq) + H2PO4-(aq) H2PO4-(aq) d H+(aq) + HPO42-(aq) HPO42-(aq) d H+(aq) + PO43-(aq) [H+(aq)][ H2PO4-(aq)] [ H3PO4(aq)] [H+(aq)][ HPO42-(aq)] K a2 = [ H2PO4-(aq)] + [H (aq)][ PO43-(aq)] Ka3 = [ HPO 2. pOH and pKw Kw = [H3O+(aq)][OH-(aq)] = 1 × 10-14 mol dm-3 -log Kw = -log ([H3O+(aq)][OH-(aq)]) -log Kw = -log [H3O+(aq)] .log Kw pKa + pKb = pKw = 14 b) Relationship between pH. Ka and Kb a) Relationship between Ka and Kb (pKa and pKb) Sometimes the value of Ka and Kb can also be expressed in negative log scale known as pKa and pKb.log Kb = .log [OH-(aq)] pKw = pH + pOH = 14 . pKa = -log[Ka] pKb = .II. Acid-base Equilibria 3.log[Kb] For the hydrolysis of a conjugate acid-base pair in water.log Kw -log Ka . Part 3 Page 3 Calculation involving pH. HA(aq) + H2O(l) d H3O+(aq) + A-(aq) Ka = [H3O+(aq)][A-(aq)] [HA(aq)] A-(aq) + H2O(l) d HA(aq) + OH-(aq) Kb = [HA(aq)][OH-(aq)] [A-(aq)] [H3O+(aq)][A-(aq)] [HA(aq)][OH-(aq)] × = [H3O+(aq)][OH-(aq)] = Kw [HA(aq)] [A-(aq)] K a × Kb = Ka × Kb = Kw = 1 × 10-14 mol dm-3 OR -log (Ka × Kb) = . x must be very small comparing with the original concentration (0.x ≈ 0. Calculate the pH of 0.30 × 10-3) = 2.29 × 10-3) = 2.100 .log x = .100 = 1.x mol dm-3 = 1. Acid-base Equilibria c) Some basic assumptions Part 3 Page 4 Depending on whether the small quantity is omitted.89 It can be seen that no matter the small quantity is omitted or not. The dissociation constant of any weak acid with Ka less than 1 × 10-4 mol dm-3 can be neglected in the expression : (original concentration – x) M.x d CH3COO-(aq) + H+(aq) x x [CH3COO-(aq)][H3O+(aq)] x·x = 0. .100 .100 .70 × 10-5 mol dm-3.x d CH3COO-(aq) + H+(aq) x x [CH3COO-(aq)][H3O+(aq)] x·x = 0.30 × 10-3 pH = .x mol dm-3 = 1. The equation is solved with the small quantity omitted. This is applicable to most weak acids. The equation is solved accurately concentration at equilibrium (M) Ka = CH3COOH(aq) 0.log x = .100 M CH3COOH(aq). there may be two possible answers for a single question. i.e.100. 0. Ka of CH3COOH(aq) is 1. they give similar results.70 × 10-5 mol dm-3 [CH3COOH(aq)] or -1.70 × 10 x = 1.31 × 10-3 (rejected) x = 1. 1.II.1M). x·x -5 0.100 .log (1.100 . This is because the degree of dissociation of ethanoic acid is only about 1%.89 2.29 × 10-3 pH = .70 × 10-5 mol dm-3 [CH3COOH(aq)] Since CH3COOH(aq) is a weak acid.log (1. concentration at equilibrium (M) Ka = CH3COOH(aq) 0. 1. in acidic medium. When NH4Cl(s) is dissolved in water NH4Cl(s) → NH4+(aq) + Cl-(aq) water NH4+(aq) + H2O(l) d NH3(aq) + H3O+(aq) [NH4+(aq)] >> [NH3(aq)] The amount of NH4+(aq) hydrolyzed is considered negligible comparing with the amount of NH4+(aq) from the ionization of salt (a strong electrolyte). H3O+(aq) + NH3(aq) → NH4+(aq) + H2O(l) Hydrolysis of the salt is very minimal in acidic medium NH4+(aq) + H2O(l) d NH3(aq) + H3O+(aq) [NH4+(aq)] >> [NH3(aq)] . e. weak alkali will also be neutralized.g. If x is very small. weak acid will be neutralized.1 · x ≠ 0.II. CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) And because the solution is alkaline. Acid-base Equilibria 3. but not multiplied/divided to another number. 3. hydrolysis of CH3COO-(aq) is very minimal. In alkaline medium.1 2. Basic assumptions Part 3 Page 5 Attention must be taken that the small quantity can be omitted only if it is added/subtracted to another number. All salts are strong electrolyte and completely ionize in water.1 but 0. 0. only very small percentage is hydrolyzed.x ≈ 0.1 . CH3COO-(aq) + H2O(l) d CH3COOH(aq) + OH-(aq) [CH3COO-(aq)] >> [CH3COOH(aq)] Similarly. 70 × 10-5 mol dm-3 and Kw of water is 1.100 M CH3COO-Na+(aq) solution.100 0.x ≈ 0.100 x = 7.log 7.100 .88 .x x·x 0.67 × 10-6 pOH = . Acid-base Equilibria 4.67 × 10-6 = 5. CH3COO-(aq) + Inital concentration (mol dm-3) Equilibrium concentration (mol dm-3) [CH3COOH(aq)][OH-(aq)] [CH3COO-(aq)] 0. Ka of CH3COOH(aq) is 1.log [OH-(aq)] = . therefore 0. CH3COO-Na+(aq) → CH3COO-(aq) + Na+(aq) A very small percentage of CH3COO-(aq) hydrolyzes in water and makes the solution alkaline.70 × 10-5 0.12 = 8.5.00 × 10-14 x2 = 1. Answer 1 CH3COO-Na+(aq) is a strong electrolyte which completely ionizes in water.100.100 .100 .II.x = Ka Since x << 0.pOH = 14 .100 1.00 × 10-14 mol2dm-6. Example Part 3 Page 6 Some Useful Relationships Kw = [H+(aq)][OH-(aq)] = Ka × Kb = 1 × 10-14 mol2 dm-6 pKw = pH + pOH = pKa + pKb = 14 Calculate the pH of 0.12 pH = 14 .100 -x = [CH3COOH(aq)] + [CH3COO-(aq)][H+(aq)] × [H (aq)][OH (aq)] = 1 K a × Kw H2O(l) d CH3COOH(aq) + OH-(aq) 0 x 0 x Kc = = = [CH3COOH(aq)][OH-(aq)] [H+(aq)] × [H+ ] [CH3COO-(aq)] (aq) Kw x·x 0. x 0 x 0 x Since the dissociation constant of the conjugate acid-base pair is related by the relationship Kw = Ka × Kb. Acid-base Equilibria Anwser 2 Part 3 Page 7 Alternatively.12 pH = 14 .88 × 10-10 = [CH3COOH(aq)][OH-(aq)] x·x = 0.67 × 10-6 (rejected) If the equation is solved exactly. CH3COO-(aq) + H2O(l) d CH3COOH(aq) + OH-(aq) Inital concentration (mol dm-3) Equilibrium concentration (mol dm-3) 0. x = 7.88 4.pOH = 14 .67 × 10-6 pOH = .100 0.log [OH-(aq)] = .II.x [CH3COO-(aq)] or .7.88 × 10-10 a 5. the problem can be interpreted as hydrolysis of the base CH3COO-(aq) which is the conjugate base of CH3COOH(aq).100 .00 × 10-14 Kb = K = 1.100 . 1 1 mark .70 × 10-5 = 5. Experimental determination of Ka By half-neutralization By other methods Glossary Past Paper Question conductivity charge effect 91 1A 3 b ii 92 2A 3 c ii 93 1A 1 e 94 1A 1 f ii 95 2A 2 c i ii iii 96 2A 1 c i 97 1A 2 b dissociation constant acidity constant Ka pKb Kw pH pOH pKa basicity constant Kb polybasic acid pKw hydrolysis of salt 94 2A 3 c i 91 1A 3 b ii 3b Write equation(s) to describe the reaction of ii NaHCO3(s) with water. NaHCO3(s) → Na+(aq) + HCO3-(aq) HCO3-(aq) + H2O(l) d H2CO3(aq) + OH-(aq) C ii Most knew the hydrolysis of HCO3-(aq) to form H2CO3(aq) and OH-(aq). Kw 1.12 = 8.67 × 10-6 = 5.5.log 7. Ka = = = 18 × 10 −5 [ HA] a−x Since HA is a weak acid a . (III) the concentration of the base. 7.8 × 10-5 = C ii .2589 × 10-9 10 −14 10 -14 and [HA] = [OH-] = = = 7.008847 12589 × 10 −9 . Calculate (I) the initial concentration of the acid. × × 1000 −6 (40.] (40.0 + V) / 1000 0.943 × 10-6 [H + ] . a pure salt solution is formed. the solution is basic.89 cm3 2 marks or V = 37.0 + V) / 1000 -5 Ka = = 1.2589 × 10-9 at equivalence point.943 × 10-6 (40. (II) the volume of the base added to reach the equivalence point. 12589 × 10 −9 Also the amount of A. A.995 × 10-3 Let a be the concentration of the weak acid HA d H+ + Aa-x x x [ H + ][ A − ] x ⋅ x .89 cm3 0. HA(aq).2212M 2 marks (II) Let V be the volume (in cm3) of MOH added.008847 mole Total no. The total volume of the solution at the equivalence point = 40.70 ⇒ -log[H+] = 2.0 + V) = 77. Thus. before the addition of base.008847 × 1000 = 0.x ≈ a ∴ Concentration of weak acid = a = 0. MOH(aq).008847 1.90. the [HA] = [OH-] at equivalence point Since pH = 8. of significant figures in the calculation) Few candidates obtained all three answers correctly.943 × 10 −6 0.8 × 10-5 mol dm-3 (I) pH = 2. HA d H+ 0. It was because of hydrolysis. but marks were awarded for all intermediate steps. Page 8 5 1. of mole of HA present = 1000 At equivalence point. The initial pH.008847 1.+ H2O d HA + OHi.89 (Comment : Wrong no. (The dissociation constant of the weak acid and the ionic product of water at 298K are respectively Kw = 1. at 298K. the pH was 8. was 2.2589 × 10 −9 × [H + ][A .943 × 10 (40.8 × 10 = [HA] 7. was titrated with a strong base.e. the amount of acid is the same as the amount of base. At the equivalence point of the titration.at equivalence point can be assumed to be equal to the amount of HA in the original solution because HA is a weak acid and the solution is alkaline.2212 = 0. + ATherefore.0 cm3 of an aqueous solution of a weak acid. Acid-base Equilibria Part 3 92 2A 3 c ii 3c ii 40.70 ⇒ [H+] = 1.0 + V) 7.II.0 × 10-14 mol2 dm-6 ) Ka = 1.90 at equivalence point ⇒ [H+] = 1.2335 M 1 mark (III) Concentration of MOH = 37.0 + V (cm3) 40 × 0.70. 10 M disodium hydrogenphosphate(V) solution at pH 10.and H+.0 × 10-14 (mol dm-3)2 (Assume the extent of dissociation of HPO42.1 [H+] = 1. Write chemical equations for four equilibrium reactions. ∴ Solubility of Ag3PO4 = 1.II.indicating that the candidates were not familiar with the interplay between Ka and Kb of the respective conjugate acid and conjugate base. [H+][PrCOO-] Ka = [PrCOOH] Assumming [H+] = [PrCOO-]. ½ for value) for any 4 of the above equilibria 1 mark for correct dimension of equilibrium constants C Most candidates could get the two equilibria involving HCO3.225 × 10-3 M pH = 2. Give that at 298 K: K1 = 4.12 × 10-4 1 K2 2HCO3 (aq) + OH (aq) d CO3 (aq) + H2O(l) K = K = 4.84 × 10-22 ∴ [Ag+(aq)]3 = 2.8 × 10-11 moldm-3 for H2CO3 Kw = 1.84 × 10-22 (mol dm-3)5. that occur in the solution and calculate the value of the equilibrium constant for each reaction. Given : Ionic product of water at 298 K = 1.05 × 10-6 moldm-3 Q 1 mole of Ag3PO4 produces 3 moles of Ag+(aq) ions.91 Page 9 5 3 2 1 mark 1 mark . in mol dm-3.5 × 10-5 = 0. each involving the hydrogencarbonate ions.ions at pH 10 is negligible. of silver phosphate(V) in 0.33 × 10 mol dm K2 HCO3-(aq) + HCO3-(aq) d CO32-(aq) + H2CO3(aq) K = K = 1.017 × 10-6 moldm-3 3 marks 94 2A 3 c i 3c Given: Ka for CH3(CH2)2COOH = 1. but failed to realize the two equilibria involving HCO3.1 [H+]2 1.5 × 10-5 moldm-3 at 298K Calculate the pH of i an aqueous solution of 0. Ag3PO4(s) + H2O(l) d 3Ag+(aq) + HPO42-(aq) + OH-(aq) Calculate the solubility.84 × 10-17 [Ag+(aq)] = 3.8 × 10-11 moldm-3 HCO3-(aq) d H+(aq) + CO32-(aq) Kw HCO3-(aq) + H2O(l) d H2CO3(aq) + OH-(aq) K = K =2. K2 = 4.10M CH3(CH2)2COOH. the equilibrium constant Kc = 2.0 × 10-14 mol2dm-6 for H2O K2 = 4. 94 1A 1 f ii 1f ii For the following equilibrium at 298 K. [PrCOO-] = 0. At pH = 10 [OH-(aq)] = 10-4 moldm-3 [HPO42-(aq)] = 10-1 moldm-3 Kc = [Ag+(aq)]3[HPO42-(aq)][OH-(aq)] = 2.and OH. Acid-base Equilibria Part 3 93 1A 1 e 1e A solution is prepared by dissolving potassium hydrogencarbonate in water at 298 K.3 × 10-7 moldm-3.33 × 10-8 moldm-3 1 1 + 6 -1 3 HCO3 (aq) + H (aq) d H2CO3(aq) K1 = 2.8 × 103 mol-1dm3 w 4 marks(½ for equation. 050 M CH3CH2COOH is 3.+ H+ [CH3CH2COO-][H+] Ka = [CH CH COOH] 3 2 Assuming [CH3CH2COO-] = [H+] (10-3.00 1 mark The performance of candidates in this question was poor.25 – 1.5. Acid-base Equilibria Part 3 95 2A 2 c i ii iii 2c The following reversible reaction occurs in an aqueous solution of ammonium ethanoate : (I) CH3COO-(aq) + NH4+(aq) d CH3COOH(aq) + NH3(aq) i Write an expression for the dissociation constant.94 × 102 mol dm-3 HSO4-(aq) + H2O(l) d H3O+(aq) + SO42-(aq) Ka = 0.10 M solution of ammonium ethanoate at 298 K.10) = 1.100 .76 × 10-5 = 3.18 × 10 1 mark x = 5.59 × 1010 mol dm-3 respectively. the dissociation constants of ethanoic acid and ammonium ion are 1.II. (1) Let x = [NH3] = [CH3COOH] 1 mark [NH4+] = [CH3COO-] = 0.10)2 Ka = (0. CH3CH2COOH d CH3CH2COO.76 × 10-5 mol dm-3 and 5. and (2) the pH of the solution. calculate (1) the concentration of ammonia.x) x 1 mark y = 9.05 .96 × 10-8 M 1 mark pH = -log [H+] = 7. NH4+ d NH3 + H+ [NH3][H+] Ka = [NH +] 1 mark 4 ii Calculate the equilibrium constant of reaction (I) at 298 K.5.100 . Calculate the Ka of CH3CH2COOH at 298 K.10.00 OR y (x) Ka(NH4+) = (0.60 × 10-4) 5. of ammonium ion.60 × 10-4) y (5.59 × 10-10 = K (CH COOH) = 1.x)2 = 3.60 × 10-4 M (2) Let y = [H+] Ka(CH3COOH) = y (100 .60 × 10-4) 1 mark y = 9.59 × 10-10 = (0. (At 298 K.x x2 -5 K = (0.93 × 10-8 M pH = 7.10-3.30 × 10-5 mol dm-3) (Deduct ½ mark for no / wrong unit) 97 1A 2 b 2b Account for the difference in Ka values given in the following equilibrium reactions: H2SO4(aq) + H2O(l) d H3O+(aq) + HSO4-(aq) Ka = 7.10 mol dm-3 2 ½ mark ½ mark 1 mark .d NH3 + CH3COOH [NH3][CH3COOH] K = [NH +][CH COO-] 1 mark 4 3 Ka(NH4+) 5.100 . Page 10 1 2 4 C 96 2A 1 c i 1c i At 298 K the pH of 0.100 .26 × 10-5 mol dm-3 (1.) NH4+ + CH3COO. indicating that they were not able to apply what they had learnt to a slightly varied situation. Ka .18 × 10-5 1 mark a 3 iii For a 0. Buffer must be a mixture of a conjugate acid-base pair of a weak acid or weak base.. Thomas Nelson and Sons Ltd. it must be a mixture of an acid and a base. This involves the equilibrium of CH3COOH(aq) d CH3COO-(aq) + H+(aq) Upon addition of H+(aq). Buffer Buffer is a solution which resists to pH change upon addition of small amount of acid or alkali. CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Therefore. Stanley Thornes (Publisher) Ltd. Blood is an example of buffer solution. Acid-base Equilibria Part 4 Page 1 Topic Reference Reading II.II. John Murray (Publisher) Ltd. Thomas Nelson and Sons Ltd. However. CH3COO-(aq) + H+(aq) → CH3COOH(aq) Upon addition of OH-(aq). 65–67 A-Level Chemistry (3rd ed. or A mixture of NH3(aq) (base) and NH4+(aq) (conjugate acid) Syllabus Notes For the buffer consists of CH3COOH(aq) and CH3COO-Na+(aq).. Principle of buffer action Since buffer is capable to absorbs both H+(aq) and OH-(aq).g. ..). A mixture of CH3COOH(aq) (acid) and CH3COO-(aq) (conjugate base). 271–272 Reading Chemistry in Context (4th ed.5 Part 4 Assignment Advanced Practical Chemistry. 374–377 Chemistry in Context (5th ed. e. 334–337 Buffer E. they neutralize each other.2. Acid-base Equilibria 6.4. 1. the pH of the solution can be maintained. H+(aq) will be removed by the base CH3COO-(aq)..).). which has a constant pH of about 7. OH-(aq) will be removed by the acid CH3COOH(aq). strong acid and strong base cannot coexist. II.log [CH COO.pOH [base] [base] pH = pKw .( pKb . of mole of salt / V no.log no. CH3COOH(aq) and CH3COO-Na+(aq)). of mole of salt Since pKw = pH + pOH = 14. CH3COOH(aq) d CH3COO-(aq) + H+(aq) Ka = [CH3COO-(aq)][H+(aq)] [CH3COOH(aq)] Ka [CH3COOH(aq)] [CH3COO-(aq)] [H+(aq)] = pH = [CH3COOH(aq)] -log Ka .log [NH + ] 4 (aq) pOH = pKb . of mole of acid ) = pKa . of mole of salt The pH is depending on the nature of the acid i.log ( no.g.log [acid] [salt] pH = In a buffer solution.e. of mole of base no. pH = pKw .g.log no. therefore. pKa and the ratio of concentrations of the acid and its salt. NH3(aq) and NH4+(aq)) NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) [NH4+(aq)][OH-(aq)] [NH3(aq)] Kb = Kb[NH3(aq)] [OH-(aq)] = [NH + ] 4 (aq) [NH3(aq)] pOH = pKb .pKb + log [salt] ) . the higher will be the buffering capacity of the solution because it will be able to absorb more base or acid.] (aq) 3 pKa . acid and its salt share the same volume ∴ pH = pKa . Furthermore. of mole of acid / V no.log [salt] ) = 14 . Acid-base Equilibria a) pH of an acidic buffer Part 4 Page 2 For a buffer consists of an acid and its conjugate base (e.log [base] [salt] The base and salt share the same solution. pOH = pKb . the more concentrate the acid and its salt. b) pH of an alkaline buffer For an buffer solution involving a weak base and its conjugate acid (e. 16 × 10-5 or .58 × 10-6 = 0 x = 3.50 . [NH4+(aq)] = (0.log x = .100 M 2.log pKw = pH + pOH = 14 a) pH of buffer solution Calculate the pH of the buffer solution consists of 0.0500 · x = 0.pOH = 14 .100 .0500 + x) M ≈ 0.50 = 9.0500 M [NH4+(aq)][OH-(aq)] 0. of mole of salt no.16 × 10-5) = 4.log no.50 B.100 M NH3(aq) and 0. of mole of salt pOH = pKb .log x = .100 .log [base] [salt] OR pOH = pKb .50 pH = pKw . A.16 × 10-5) = 4.0500 x .100 . Kb = [NH4+(aq)][OH-(aq)] (0.100 = 1. ∴ [NH3(aq)] = (0.log (3.log [acid] [salt] OR pH = pKa .4.II.4.0500 + x) x = 0.100 0.50 = 9.0500 + x + OH-(aq) 0 x Like the example mentioned in the last unit. Calculation of buffer solution Part 4 Page 3 Important relationships pH = pKa . The equation is solved accurately. It can be assumed that the 1.0.0500 0.58 × 10-5 mol dm-3.58 × 10-5 [NH3(aq)] Kb = x = 3. Kb of NH3(aq) 1.58 × 10-5 [NH3(aq)] x2 + 0.x) M ≈ 0.1. the equation can be solved accurately or approximately. Initial concentration (mol dm ) Equilibrium concentration (mol dm-3) -3 NH3(aq) + 0.16 × 10-5 pOH = .0500 (rejected) pOH = . of mole of acid no.log (3.0500 M NH4+(aq) respectively. The equation is solved with the approximation applied.x = 1. Acid-base Equilibria 2. hydrolysis of ammonium ion is also very minimal in the presence of NH3(aq). hydrolysis of ammonia is very minimal in the presence of NH4+.pOH = 14 .50 pH = pKw .x H2O(l) d NH4+(aq) 0. of mole of base no. This means that the approximations are valid.0830 = 1.200 = 1.pOH = 14 .50 [NH3(aq)] Practically.050 = 4. of mole of CH3COOH(aq) = (0.120 mol NaOH is a strong alkali which will neutralize 0. they all give the same result (pH = 9.200 mol dm-3 × 0.log 0.58 × 10-5) . Ka of CH3COOH(aq) is 1. b) Effect of addition of acid / alkali into a non-buffered solution 1.0500 = 4.84 × 10-3) = 2. In the following examples. Calculate the pH of a 0.08 .100 pOH = -log (1. No. the solution will become very acidic.0500 mole of HCl(g) is dissolved in 600 cm3 of 0.74 2.0.log [H+(aq)] = .4.070 pH = pKa .x) M ≈ 0. Calculate the pH of the solution if 0.0500 mol [H+(aq)] = [HCl(aq)] = 0.log [NH + ] can only be used with the approximations applied.log (1.log 0.200 M = 0.0500 mole of solid NaOH(s) is added into 600 cm3 of 0.0500 mole of CH3COOH(aq).0500 + x) M ≈ 0. At equilibrium d CH3COO-(aq) + H+(aq) CH3COOH(aq) 0.600 dm-3 = 0.600 dm3 = 0.0500 mol no.70 × 10 pH = .50 = 9. The effect of change in volume can be neglected when HCl(g) is being dissolved.200 M CH3COOH(aq) solution.50 pH = pKw .log (1.70 × 10-5 mol dm-3.log no.120 . Since HCl(aq) is a strong acid.200 M CH3COOH(aq). In the equilibrium mixture. of mole of acid 0.B. the equation pOH = pKb . of mole of CH3COO-(aq) = no. of mole of salt = .100 .200 .0830 mol dm-3 pH = .log 0. 0. of mole of CH3COOH(aq) in 600 cm3 of 0.0500) mol = 0. No matter which method is used. [NH3(aq)] pOH = pKb .II.100 M [NH4+(aq)] = (0. no.0500 M 0. of mole of CH3COOH(aq) neutralized = 0.84 × 10-3 x·x -5 0.200 M CH3COOH(aq). This is 4 (aq) because a quadratic equation cannot be set up without considering the expression of Kb. Acid-base Equilibria Part 4 Page 4 C.200 x x x = 1.6 3. In acidic medium. Calculate the pH of the solution if 0.070 mol no.70 × 10-5) .50).x ≈ 0. Solving by applying the equation with approximations applied.log [NH + ] 4 (aq) [NH3(aq)] = (0. N. approximation is used whenever applicable. the ionization of CH3COOH(aq) can be neglected comparing with the ionization from HCl(aq). 21 × 10-3 mol dm-3. no.180 + 0. of mole of CH2ClCOOH(aq) in 600 cm3 of the buffer = 0.0500) mol = 0. no.300 M × 0. of mole of CH2ClCOOH(aq) = (0.08 2. Part 4 Page 5 Effect of addition of acid / alkali into a buffer solution Calculate the pH of a buffer solution with 0. [acid] 0. Acid-base Equilibria c) 1.0500 mole of CH2ClCOO-(aq) will be produced.0.300 M × 0. no.600 dm3 = 0. of mole of CH2ClCOOH(aq) in 600 cm3 of the buffer = 0. of mole of CH2COO-(aq) = (0.40 d) Comparison of the action of acid and alkali Non-buffered Acid Buffer solution Original pH 2.0500 mole of CH2ClCOOH(aq) will be produced and 0.21 × 10-3) .log no.200 M × 0. 0.0500 mole of solid NaOH(s) is added into 600 cm3 of the buffer.180 .170 mol no.II.log 0.0500 mole of CH2ClCOO-(aq) will be consumed.40 Comparatively.0500) mol = 0.130 pH = pKa .21 × 10-3) . In the equilibrium mixture.0500 mole of HCl(g) is dissolved. of mole of salt = .log no.74 2. 0.0500 mole of CH2ClCO2H(aq) will be neutralized and 0. no.03 Acid added 1. For a non-buffered acid.0500) mol = 0. with [acid] the acid alone.200 = 2.600 dm3 = 0.120 mol When 0.300 pH = pKa . This causes a relatively big change in pH. [acid] i.log (1. Calculate the pH of the solution if 0. Calculate the pH of the solution if 0.log 0.300 M of CH2ClCO2H(aq) and 0.log (1.130 mol no. of mole of CH2ClCOO-(aq) in 600 cm3 of the buffer = 0. Furthermore. addition of alkali will cause a big change in [salt] .600 dm3 = 0.180 mol no.21 × 10-3) . it contains no base to absorb the extra H+(aq) ions if acid is added.03 3.600 dm3 = 0.log [salt] = .230 pH = pKa . In the equilibrium mixture. of mole of acid 0.log (1. of mole of salt = . pH = pKa . of mole of CH2ClCOOH(aq) = (0. of mole of CH2COO-(aq) in 600 cm3 of the buffer = 0.0.120 .74 2. of mole of acid 0.230 mol no.180 mol no.200 M × 0.200 M of CH2ClCO2-Na+(aq) respectively.120 mol CH2ClCOOH(aq) d CH2ClCOO-(aq) + H+(aq) When 0.170 = 3.6 3.070 mol no. of mole of CH2COO-(aq) = (0.0500 mole of NaOH(s) is added.e.log 0.120 + 0.0500) mol = 0. the pH of a buffer solution is more resistant to addition of alkali or acid.0500 mole of HCl(g) is dissolved in 600 cm3 of the buffer.74 Alkali added 4.log [salt] . Ka of CH2ClCO2H(aq) is 1.070 = 2. 50 mol of MCl in 1. of mole of salt can be adjusted to prepare a solution of desired pH.log 1 = pKa This method is used to prepared a buffer solution with the pH equals pKa of the acid. Solution S is made up of 0. i Calculate the pH of solutions S and T. assuming (0. a buffer solution is prepared by mixing a weak acid / base with its salt.1 . 0. MOH d M+ + OHSolution S: Since MOH d M+ + OHinitial 0.1 0 0 at eqm.25 mole no. Acid-base Equilibria 3.5 and 0.x .x ≈ 0.x x x x2 -5 ∴ Kb = 2 × 10 = 0.1 1 mark ∴ x = 2 × 10-6 = 1.0 dm3. of mole of acid to no.10 mol of MOH and 0.5 M CH3COOH(aq) requires 0.5 -log (4 × 10-6) = 5.1 . 0.40 pH = 14 . and a second solution T has 0.1 .1 0. of mole of acid Since pH = pKa .85 = 11.2.1 = 4 × 10-6 M 1 mark x = 2 × 10-5 × 0. of mole of salt = pKa .x 0.25 mole ≈ 0.5 0 at eqm. For any pH not the same as pKa. Or it can be prepared by halfway titration.1 .5 + x x x × (0. Preparation of buffer solution Part 4 Page 6 Usually.15 1 mark Solution T: Since MOH d M+ + OHinitial 0.5 + x ≈ 0.5 + x) -5 with 0.1 -x) ≈ 0. 0.5 mole of NaOH(aq) for complete neutralization.1 0.x = 2 × 10 0.414 × 10-3 M pOH = -log(1. the ratio of no. Glossary Past Paper Question buffer hydrolysis half-way titration 91 2A 2 c i ii 94 2A 3 c ii iii 96 2A 1 c i ii iii 97 2A 4 c ii 98 2A 1 a i ii 99 2A 4 c ii 91 2A 2 c i ii 2c A weak base MOH has an ionization constant Kb = 2.II.40 = 8.5.log no.60 1 mark 4 . where Kb = [M+][OH-]/[MOH].44 × 10-3) = 2. For example.25 mole ≈ 0 mole CH3COO-(aq) + H2O(l) ≈ 0 mole ≈ 0.85 pH = 14 .1 . Initial amount Final amount d CH3COOH(aq) + OH-(aq) ≈ 0. if 1dm3 of 0.25 mole of NaOH(aq) can be added so that only half of the CH3COOH(aq) will be neutralized.0 × 10-5 mol dm-3.10 mol of MOH in 1.0 dm3.5 mole ≈ 0. 09 .05 M Ka = [H+] ∴ pH = pKa 1 mark = 4.5 × 10-5 = (0.x = 2 × 10 9 × 10-5 = 3.05 M CH3CH2COOH ½ mark [CH3CH2COO-] = 5.05 .050 + 1.26 × 10-5 mol dm-3 (1.09 . 94 2A 3 c ii iii 3c Given: Ka for CH3(CH2)2COOH = 1.0 × 10-14 mol2dm-6) Solution T: + OHSince MOH d M+ Before addition of acid and hydrolysis 0.5 × 10-5 M pH = 4.10-3.x ≈ 0.90 (4.051 M [H+] (0. calculate that for solution T.25 – 1.0 × 10-3 moles solid NaOH [PrCOOH] = 0.44 × 10-5 M pH = 4.0 × 10-3 mol of solid NaOH.051) 1.01 0. CH3CH2COOH d CH3CH2COO.050 M CH3CH2COOH and (2) 100 cm3 of the solution in (ii) above. (1) For 100 cm3 of 0. it is generally known as a buffer.53 × 10-6 1 mark x=2× 51 -log (3.30 × 10-5 mol dm-3) (Deduct ½ mark for no / wrong unit) ii Calculate the pH at 298 K of a solution which is 0. The new pH value for solution S is 10.0 × 10-4 mol of solid NaOH were added separately to (1) 100 cm3 of 0.34 2 1 mark 1 mark 3 1 mark 1 mark 1 mark 96 2A 1 c i ii iii 1c i At 298 K the pH of 0.51 + x ≈ 0.050 .01 mol of a strong acid HX is added separately to the solutions S and T.0 × 10-3 = 0.050 M 0.5 × 10-5 moldm-3 at 298K Calculate the pH of ii an aqueous solution of 0.050 M with respect to CH3CH2COOH and to CH3CH2COONa. Acid-base Equilibria ii Part 4 Page 7 4 C i 0.II.10.0 × 10-3 = 0.+ H+ [CH3CH2COO-][H+] Ka = [CH CH COOH] ½ mark 3 2 + ½ mark Assuming [CH3CH2COO ] = [H ] (10-3.5 + 0.050 M [PrCOO-] = 0.53 × 10-4) = 5. After addition of 1.09 .1 0.0.050 M CH3CH2COOH is 3.50 [H+] = 1.10) 1 mark = 1. [CH3CH2COO-] = [CH3CH2COOH] = 0.26.049 M [PrCOO-] = 0.049) [H+] = 1.51 + x x 1 mark x × (0. 1 mark Most candidates were not able to work out the final pH of solution T.01 0 Upon hydrolysis at eqm. Comment on the difference in the pH change of the two solutions.0 × 10-3 M 2 2 5 .88 – 4.x 0.1 . What conclusion can you draw from this result? (Kw = 1.51 + x) -5 with 0.51 and 0.5 Upon addition of acid 0.90) 1 mark iii 5.1.09 0. and Assuming [PrCOOH] = 0. Assuming that the change in volume upon the addition of NaOH is negligible calculate the pH at 298 K of the solution in each case.050M CH3(CH2)2COONa and 0. 0. In the solution.10)2 Ka = (0.0 dm3 of the solution in (ii) after the addition of 1.55 1 mark Solution T can resist the change in pH to a greater extent than solution S.050M CH3(CH2)2COOH. Calculate the Ka of CH3CH2COOH at 298 K.82 iii 1.50 1.5 × 10-5 = [H+] × 0.45 pH = 8. 5 × 10-2 M [H+] (5.76 × 10-5 mol dm-3 at 298 K) Page 8 ½ mark 1 mark ½ mark ½ mark 1 mark 1 mark (1 mark) 7 98 2A 1 a i ii 1a Solution A is 0.1 M aqueous HCO2H.1 M aqueous NaOH. approximately 0.38 × 10-4 moldm-3.15 M lactic acid.03 × 10-5 mol dm-3 (1.II. i Calculate the pH of A and of B at 298 K.15 × 10-4 mol dm-3) pH = 3. Compare the effect of such action on the pH of the two solutions.5 × 10-2 M [CH3CH2COOH] = 4.20 M CH3CO2Na(aq). describe a method to determine the dissociation constant. Acid-base Equilibria Part 4 [CH3CH2COOH] = 4.95 (3. of HCO2H. you may assume that the volume changes are negligible. the solution in (ii) is a buffer ∴ pH change is very small 97 2A 4 c ii 4c A solution is formed by mixing equal volumes of 0. [CH3CH2COO-] = 5.5 × 10-2 + [H ] = 1.0 × 10-3) Ka = 4. 99 2A 4 c ii 4c ii Given the following materials and apparatus. Explain your answer.10 M aqueous sodium hydroxide solution. titration apparatus. and a calibrated pH meter 9 . approximately 0. and solution B is a mixture of equal volumes of 0. (2) For (ii) and (iii). ii A few drops of dilute hydrochloric acid are added to 50.5 × 10-2 M [H+] (5. present in the solution at 298 K.13 × 10-4 mol dm-3 (1. (Ka of CH3CO2H = 1. Ka.05 M CH3CH2COOH is not a buffer ∴ pH changes significantly or.05 × 10-5 mol dm-3) pH = 4.99 (4.93 – 3.0 cm3 of A and B.20 M CH3CO2H(aq) and 0.5 × 10-2) Ka = 4.00) 0.97 – 5.5 × 10-2 + [H ] = 1. phenolphthalein indicator. Note : (1) Lactic acid is a monprotic acid and its Ka at 298 K is 1. ii Calculate the concentration of each chemical species.30 M lactic acid and 0.96) (2) For 100 cm3 of the solution in (ii). excluding H2O. Theory of Indicator Acid-base indicator is a weak acid which has different colour at different pH. 331–334 Indicators Acid-base titration F.3) OH O- HO C O C O + 3H2O HO C OH C O + + 2H3O O Conjugate acid (colourless) Conjugate base (red) . they change colour at different pH.7) H3C H3C Conjugate acid (red) O N N N S O OH + H2O H3C H3C Conjugate base (yellow) O N N N S O O + H3O+ - In acidic medium. 68–69 Experiment – Acid-base titration using method of double indicator Reading A-Level Chemistry (3rd ed.. Because methyl orange and phenolphthalein have different Kin. 264–270 Chemistry in Context (4th ed. Therefore. Acid-base Equilibria Part 5 Page 1 Topic Reference Reading II.2. methyl orange is a weak acid. Stanley Thornes (Publisher) Ltd. the conjugate acidbase pair of the acid have different colour.). The conjugate acid of methyl orange is red in colour while the conjugate base of methyl orange is yellow. John Murray (Publisher) Ltd. Methyl orange (pKin of methyl orange = 3.7 Part 5 Assignment Advanced Practical Chemistry.).2. Phenolphthalein is also a weak acid. Thomas Nelson and Sons Ltd.6–6. 371–374 Chemistry in Context (5th ed. Actually. Acid-base Equilibria 6. For a similar reason. the concentration of H+(aq) is high and the equilibrium position is lying on the left.. it is yellow in colour in alkaline medium. Thomas Nelson and Sons Ltd.).II. it will be red in colour in acidic medium. Its conjugate acid is colourless while the conjugate base is red. HIn(aq) + Colour I H2O(l) d H3O+(aq) + In-(aq) Colour II Kin = [H3O(aq)+][In-(aq)] [HIn(aq)] Syllabus Notes For example... Phenolphthalein (pKin of phenolphthalein = 9. [salt] = 10 [acid] pH = pKin . G.e.7 to 4. The equivalence point is determined as an end point. So that the pH of the solution will not be affected. For methyl orange. It cannot notice any difference unless the intensity of one colour is ten times of another. a small amount at a time until just sufficient has been added to complete the reaction. It has pKin 9.7 . temp. only minimal amount of indicator should be added during a titration. a naked eye cannot distinguish colours accurately.7 + 1 = 4.log 10 = 3.7 HMe(aq) + H2O(l) d H3O+(aq) + Me-(aq) Kin = [H3O(aq)+][Me-(aq)] [HMe(aq)] [acid] Therefore.e. Difference between equivalence point and end point Equivalence point – the stage in a titration when the reactants and products are present in equivalent amounts according to the stoichiometry of the reaction.log [salt] = pKin .3. colour.II. . [salt] = 10 [acid] 1 pH = pKa . Titrations may be carried out by hand from a burette or automatically.log [salt] = pKin . Working range of a pH indicator = pKin ± 1 Phenolphthalein works in the same way as methyl orange.3. it will become red at pH = .g. i. therefore. N. The accuracy of a titration is depending on how close the end point (experimental value) is to the equivalence point (theoretical value). Acid-base Equilibria Part 5 Page 2 However.7 The pH range from 2. pH.. pKin = 3.3 to 10. One solution is added to the other.7 [acid] 1 The solution will become yellow at pH = .log 10 = 3. conductivity changes) is observed indicating the end of the titration. End point – the stage in a titration when a change (e. Because indicator is actually an acid. Acid-base titration Titration is a process for determining the volume of one solution required to react quantitatively with a given volume of another. 1.B. i.7 is called the working range of methyl orange.1 = 2. the working range of phenolphthalein is from pH 8. plotting of a pH graph is not applicable because there is no abrupt pH change throughout the course. . the end point can be determined. the solution shows a great pH change upon addition of small amount of acid or base. The graphy is usually called titration curve. By plotting a graph with the pH of the solution against the volume of titrant added. Acid-base Equilibria 2. Titration using pH meter Part 5 Page 3 About the equivalence point of most acid-base titration. It is hard to decide where is the end point. For the titration between a weak acid and a weak base.II. Therefore. Titration using indicator Part 5 Page 4 a) Choosing of indicator Choosing of indicator has a great impact on the accuracy of titration. Conversely. A wrongly chosen indicator will cause a large error. If phenolphthalein is used. And a large error will be caused. in the titration of weak acid versus strong base. A-(aq) + H2O(aq) d HA(aq) + OH-(aq) Furthermore. the colour will change with a volume very close to the equivalence point. the equivalence point will be at pH 7 because the salt formed will not hydrolyse in water. Acid-base Equilibria 3. the colour will change far before the equivalence point is reached. The indicator is chosen so that the abrupt change on the pH curve will fall across the working range of the indicator. The value of pKin should be close to the value of the pH of the solution at the end-point.II. because of hydrolysis of the salt. only phenolphthalein can be used in the titration between a weak acid and a strong base. the equivalence point will be higher than pH 7. the vertical portion of the pH curve will also be shifted to the higher pH range. In the titration of strong acid versus strong base. if methyl orange is used. However. both methyl orange and phenolphthalein can be used in the titration between a strong acid and a strong base. Although neither methyl orange nor phenolphthalein changes colour at pH 7. . so that the colour change occurs as closely as possible to the equivalence point. Therefore. the end point will be very close to the equivalence point. The selection is depending on the shape of the pH curve of titration. In a titration of a weak acid and a strong base. It is only used in the form of litmus paper where a white background is available. by plotting a graph with the temperature versus the volume of the titrant added.II. 4. The graph is composed of 2 straight lines. One representing the exothermic reaction and another represents the cooling. N. the temperature of the solution will increase until the equivalence point is reached. According to the theory behind thermometric titration. no indicator is appropriate because there is no abrupt pH change throughout the course. The end point of a titration between a weak acid and a weak base can neither be detected by any indicator nor a pH meter. the addition of any cold excess titrant into the warm solution will cause a drop in temperature. the points on the graph should not be joined together by a smooth curve. . However. Acid-base Equilibria Part 5 Page 5 Litmus is usually not used in titration because the colour change is not as obvious as methyl orange and phenolphthalein. In a titration of a strong acid and a strong base. either methyl orange or phenolphthalein can be used. pH value Indicator 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 methyl orange | red |orange| yellow | phenolphthalein | colourless |pale pink| red | For the titration between a weak acid and a weak base. The end point is where the two straight lines intersect each other. In a titration of a strong acid and weak base. Thermometric titration Since neutralization is exothermic.B. According to this principle. The end points are usually determined by thermometric titration or conductimetric titration. the end point can be determined. methyl orange should be used. phenolphthalein should be used. if the titration is between a weak acid and a weak base HA(aq) + BOH(aq) → H2O(l) + A-(aq) + B-(aq) Two covalent molecule reacts to form 1 water molecule and 2 ions. thus the conductivity. Sometimes. the solution contains only H+(aq) and Cl-(aq). in the titration between NaOH(aq) and HCl(aq). Thus. Furthermore. the conductivity / conductance of the solution will also change throughout the titration process. a pair of H+(aq) and OH-(aq) ions is converted to a H2O(l) molecule. NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) At the beginning. the conductivity will increase even further upon the addition of excess strong base. Beyond the end point. Beyond the end point. the addition of the weak acid will only dilute the solution and the conductivity will decreases. Conductimetric titration Part 5 Page 6 Depending on the nature of the acid. c) Weak acid vs Weak base However. The increases in ionic concentration will cause an increase in conductivity. For example. the product will be water and the salt. addition of a strong base (a strong electrolyte) increases the ionic concentration. base and the salt formed. The end point is determined by extrapolating the two sessions of the graph with 2 straight lines. Acid-base Equilibria 5.II. the acid solution contains mainly undissociated acid molecules and the conductance is minimal. the excess OH-(aq) ions are more conductive. The addition of a strong base will convert the molecular acid molecules to ionic salt. the conductivity will increase. b) Weak acid vs Strong base At the beginning. Beyond the end point. Glossary indicator methyl orange phenolphthalein Kin naked eye working range titration equivalence point end point titration curve (pH curve) indicator thermometric titration conductimetric titration conductivity / conductance ionic concentration extrapolating . N. Conductivity of a solution depends on i) ionic concentration ii) mobility of the ions a) Strong acid vs Strong base If a strong acid is titrated against a strong base. Upon neutralization. Therefore. Consequently H+(aq) in the acid is replaced by Na+(aq) which has lower conductivity. the conductivity will drop before the end point. it is difficult to predict the actual shape of the graph but a sudden change in slope can be treated as the existence of end point.B. thus they change colour over different pH range. Acid-base Equilibria Part 5 Page 7 Past Paper Question 90 1A 3 a 92 2A 3 c i 94 2A 3 b ii 95 1B 4 e i ii 98 2A 1 a iii 99 1A 4 b i ii 99 1A 4 c 90 1A 3 a 3a Explain why phenolphthalein turns pink in a solution of sodium carbonate. but remains colourless in a solution of sodium hydrogencarbonate. The end point is the point in a titration where a particular indicator changes colour which occurs at pH 2 marks close to the value of pKa for the indicator. ½ mark ½ mark In a solution of Na2CO3. Equivalence point is the point in the titration where moles of acid equals moles of base. and that at the higher pH.30 M lactic acid and 0.0 cm of A and of B are titrated with a strong base.38 × 10-4 moldm-3. corresponding to a salt solution.+ H2O d HCO3. CO32.II.takes place to a different extent. (2) For (ii) and (iii). 99 1A 4 b i ii c 4b The graph below shows the variation of pH when 25. 10 M NaOH(aq). Explain your answers. If an indicator is required. you may assume that the volume changes are negligible.15 M lactic acid. 2½ 2 3 1 1 . In(red) + H+ d InH+(colourless) lies to the left and a pink colour is exhibited. while that of phenolphthalein is above 7. 92 2A 3 c i 3c i Explain the difference between the equivalence point and the end point of a titration. Compare the pH of the two solutions at the respective equivalence points of the titrations. 95 1B 4 e i ii 4e For each of the volumetric analyses (i) to (iii). [H+] is sufficiently high for the equilibrium to shift to the right. 1½ mark C ii Very few candidates mentioned the fact that an acid-base indicator is a weak acid or weak base. HIn and the alkaline form In½ mark (aq) which are of different colours.and HCO3. while phenolphthalein shows its acidic colour. They also failed to relate the colour of the indicator to its pH range. phenolphthalein exists in a form which has a resonance structure and the solution becomes pink. whose pH is also above 7. in a solution of pH 7.and HCO3. the indicator methyl orange shows its alkaline colour (yellow). ∴ at pH 7.+ OHresulting in a pH sufficiently high ( and [H+] sufficiently low) 1 mark such that the equilibrium. 94 2A 3 b ii 3b Account for each of the following: ii At 298K. and solution B is a mixture of equal volumes of 0. The pH range of methyl orange is below 7. potassium dichromate(VI) solution.0. methyl orange shows its alkaline colour.10 M aqueous sodium hydroxide solution. 10 M HCl(aq) is titrated against 0. select the appropriate one from those given below: litmus. while phenolphthalein shows its acidic colour (colourless). Note : (1) Lactic acid is a monprotic acid and its Ka at 298 K is 1.hydrolyse to a different extent. CO32. methyl orange. phenolphthalein.0 cm3 of 0. Acid-base indicators are weak acids / bases. The dissociation constant Ki of different acid-base indicators are different. The dissociation of which can be represented by 1 mark HIn(aq) + H2O(l) d H3O+(aq) + In-(aq) The colour of an indicator depends on the relative concentrations of the acidic form. C Few candidates were able to answer that a solution of Na2CO3 has a higher pH value than that of a solution of NaHCO3 because the hydrolysis of CO32. starch solution i ethanedioic acid titrated with sodium hydroxide phenolphthalein 1 mark ii sulphuric(VI) acid titrated with aqueous ammonia methyl orange 1 mark 98 2A 1 a iii 1a Solution A is 0. state whether an indicator is required. ½ mark In a solution of NaHCO3. 3 iii 50. 6 99 1A 4 c 4c Given an aqueous solution of Na2CO3 and NaHCO3. 10 M CH3CO2H(aq) is titrated against 0. . 10 M NaOH(aq). Indicator pH range of colour change bromocresol green 3.3 . Explain your choice.8 . From the table below. suggest how to determine the concentrations of the two substances by titrimetric method using indicators.0 cm3 of 0.II.10.0 .7.4 bromothymol blue 6.5.6 thymolphthalein 8. choose an appropriate indicator for the titration in (i). Acid-base Equilibria Part 5 Page 8 i ii On the above graph. sketch a curve to represent the variation of pH when 25. Redox Equilibria 6. A. of electrons donated by reducing agent. An acid-base equilibrium can be considered as a competition of donation and acceptance of proton. 21–29 Reading Syllabus Notes Redox Equilibria Faraday and mole III. Oxidizing agent + + Mg2+(aq) electron 2e-(aq) d d Reducing agent Mg(s) Upon donation of electron. Conjugate base NH3(aq) + + proton H+(aq) d d Conjugate acid NH4+(aq) Upon donation of proton. e.g. a balanced redox reaction can be constructed in two ways a) By combining balanced half-ionic equation b) By the change in oxidation no.3. Redox reaction A complete redox reaction consists of two systems of redox equilibria. in a redox reaction. Therefore. Thomas Nelson and Sons Ltd. of atoms and the no. Ag+(aq) + e-(aq) d Ag(s) Cu(s) d Cu2+(aq) + 2e2Ag+(aq) + Cu(s) d 2Ag(s) + Cu2+(aq) . Reduction (acceptance of electron) Oxidation (donation of electron) Overall redox reaction 1. a conjugated acid will be converted to a conjugate base. pg. Curriculum Development Council.. 5th Edition.III. the no.). a reducing agent will be converted to an oxidizing agent. Balancing of redox reaction In a balanced equation. 74–80 Chemistry in Context (4th ed.g. of electrons accepted by oxidizing agent must be the same as the no. 188–189 Part 1 Assignment A-Level Chemistry (3rd ed. thus two half ionic equation. the no. One half ionic equation represents the reduction and another represents oxidation. Redox Equilibria Part 1 Page 1 Topic Reference Reading III.1 A-Level Chemistry Syllabus for Secondary School (1995). 186–187 Redox equilibrium is similar to Acid-base equilibrium. Stanley Thornes (Publisher) Ltd.. Thomas Nelson and Sons Ltd. e. of charges must be balanced at the two sides of the equation. Furthermore. Redox Equilibria Chemistry in Context. Redox equilibrium can also be considered as a competition of donation and acceptance of electron.). of Cr) 2Cr2O7 (aq) → 2Cr3+(aq) + 7H2O(l) (Add H2O(l) for lack of O) 14H+(aq) + Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l) (Add H+(aq) for lack of H) 3. Adding H+(aq) for lack of H (from dissociation of water. Reduction half ionic equation Ag+(aq) + e.lost by Cu(s).+ Cu(s) → 2Ag(s) + Cu2+(aq) + 2e(left and right must have the same no.(Reducing agent: Cu(s)) Since the no. Adding H2O(l) for lack of O ii. equation may be balanced by i. which can only be memorized. of e. of e-) 2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq) 2Ag+(aq) + 2e- Overall ionic equation .to balance the charge → 2Cr3+(aq) + 7H2O(l) 14H+(aq) + Cr2O72-(aq) (14 × +1) + (-2) = +12 (2 × +3) + (0) = +6 14H+(aq) + Cr2O72-(aq) + 6e→ 2Cr3+(aq) + 7H2O(l) (14 × +1) + (-2) + (6 × -1) = +6 (2 × +3) + (0) = +6 4. i.→ Ag(s) (Oxidizing agent: Ag+(aq)) Oxidation half ionic equation Cu(s) → Cu2+(aq) + 2e.gained by Ag+(aq) must equal the no. Note: (2) Combining half-ionic equations In a redox reaction. Redox Equilibria Part 1 Page 2 a) By combining balanced half-ionic equation (1) Steps of writing Balanced Half ionic equation 1. of e. the ionic equation can be reconstructed. an oxidizing agent must react with a reducing agent. (First of all. H2O d H+ + OH-) 2Cr2O7 (aq) → 2Cr3+(aq) (Balance the no.e.III. ∴ 2 Ag+ are needed for each Cu. Reduction half ionic equation Oxidation half ionic equation → 2Ag(s) Cu(s) → Cu2+(aq) + 2e–––––––––––––––––––––––––––––––––––––––– 2Ag+(aq) + 2e. By combining an oxidation half equation and a reduction half equation. 2.) For example. Write down the reactant and the product Cr2O72-(aq) → Cr3+(aq) Mass balance – by adding something already existed. Oxidizing agent does not react with another oxidizing agent. Since the reaction is usually conducted in water medium. Check once again → 2Cr3+(aq) + 7H2O(l) 14H+(aq) + Cr2O72-(aq) + 6eMass: 14H + 2Cr + 7O 2Cr + 14H + 7O Charge: (14 × +1) + (-2) + (6 × -1) = +6 (2 × +3) + (0) = +6 Balancing of half-ionic equation can only be started with correct reactant and product. Charge balance – Add e. the half ionic equations must be balanced. MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) again mass 1 Mn 4O 5 Fe 8H charge (-1) + 5×(+2) + 8×(+1) = +17 mass 1 Mn 4O 5 Fe 8H charge (+2) + 5×(+3) = +17 No. so add some No. Charges (Q) = Current (I) × Time (t) If the current is measured in ampere (A) and the time is measured in second (s). mole is used as the counting unit and method of counting by weighing is used. of O is not balanced. of proton associated with an atom. Redox Equilibria b) By the change in oxidation no. Similarly. Faraday and mole In counting the no. so add some Check the no.N. is the difference between the no. MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l). of electron. Charges (Q) = 1C = 1C = Current (I) × Time (t) 1A×1s 1 As . the unit of charge will be coulomb (C). 2.02 × 1023 F ≡ 6.60 × 10-19 C By measuring the size of the current and the time taken. MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) H+(aq). Oxidizing agent MnO4-(aq) +7 → Mn2+(aq) +2 Each Mn gains 5 electrons Each Fe loses 1 electron Reducing agent Fe2+(aq) +2 → Fe3+(aq) +3 By the fact that the no. of atoms and charges The equation. a decrease in O. the equation will become MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) H2O(l). In counting the no. This method is particularly useful in answering M. 6. of atom.02 × 1023 C 1 electron charge ≡ 1. Therefore. of H is not balanced. method of counting charges is used.02 × 1023 electron charges ≡ 1 F of charges ≡ 96500 C 96500 1 1 electron charge ≡ 6. an increase in O.N. question where the completely balanced equation is usually not required. of electron gained by the oxidizing agent must be the same as the no.N.III. by 1 is equivalent to gaining of 1 e-.C. each MnO4-(aq) must reacts with 5 Fe2+(aq). The amount of charges carried by 1 mole of electrons is called 1 Faraday of charges (1 F). by 1 is equivalent to losing of 1 e-. is balanced. the amount of charges flowing through a circuit can be measured.66 × 10-24 F ≡ 1. of electron and no. Part 1 Page 3 Since O. of electron lost by the reducing agent. 1 Faraday of charges is found to be 96500 Coulombs (96500 C). No.49 × 10-3 mol = = 6.25 × 10-3 mol × 63.49 × 10-3 mol = = 1.0140 dm3 = 14. the amount of Cu(s) and O2(g) can be calculated. if a beaker of CuSO4(aq) is electrolyzed at 200 mA for 20 minutes. no. of mole of O2(g) formed = Volume of O2(g) at s. of mole of electron 2. All volume is measured at standard temperature and pressure.III. no.p.→ Cu(s) 4OH-(aq) → O2(g) + 2H2O(l) + 4e- Page 4 By the relationship Q = It. of mole of Cu(s) × molar mass of Cu = 1.0794 g Similarly. For example. Part 1 Calculation of mass liberated in electrolysis In the electrolysis of CuSO4(aq) using graphite electrodes.00 ) Molar volume of gas at s.25 × 10-3 mol 2 2 Mass of Cu(s) deposited = no. therefore.23 × 10-4 mol × 22. of mole of oxygen × molar volume of gas at s.t.t. is 22. By measuring the size of current and the time taken of electrolysis. no. what will be the mass of Cu(s) deposited on the cathode and the volume of oxygen evolved at the anode.55. of mole of Cu(s) formed = no.t.p.4 dm3mol-1 = 0. O : 16. (Given : Relative atomic mass Cu : 63. rheostat and ammeter must be installed to adjust and measure the current. Cu(s) will deposit on the cathode and O2(g) will evolve at the anode.49 × 10-3 mol Formation of each mole of Cu(s) requires 2 moles of electron. Redox Equilibria 3. of mole of electron flowing through the current = Faraday constant = 96500 C = 2. formation of each mole of O2(g) requires losing of 4 mole of electron. of mole of electron 2.4 dm3 At cathode At anode Cu2+(aq) + 2e. of coulombs flowing through the circuit = Current in ampere × time in second 200 = 1000 A × (20 × 60) s = 240 C 240 charges in coulomb No.p. In the circuit.55 g mol-1 = 0. = 6.0 cm3 reducing agent conservation of mass and charge Glossary redox equilibrium oxidizing agent faraday (F) coulomb (C) .23 × 10-4 mol 4 4 = no. slowly oxidize chloride ions to chlorine.ions react with S2O32.0 × 10-5 i Write the balanced equation for this reaction.50M Na2CrO4 solution is needed to react completely with 40.16 0. 1 mark CrO42.40 4. 3H2O2(aq) + 2Cr(OH)4-(aq) + 2OH-(aq) d 2CrO42-(aq) + 8H2O(l) 1½ mark C A surprisingly large number of candidates could not do the balancing of equation for redox reactions which should be basic work for A-Level candidates. 93 1A 1 d i ii 1d Write balanced equations for: i the reaction between MnO2(s) and PbO2(s) in acidic solution to give MnO4-(aq) and Pb2+(aq). 1 1½ 1½ . ii Most candidates were not able to write balanced equations for the half-reactions which take place in basic medium.08 0.+ 4OHS2O32. The following kinetic data are obtained at 25°C.0 cm3 of 0.20M Na2S2O3 solution 2 in basic medium? No.+ 6H+ → 3Cl2 + 3H2O 1 mark C i Weaker candidates were not able to give the balanced equation. chlorate(V) ions.20 1. Redox Equilibria Part 1 Page 5 Past Paper Question 91 2A 2 b i ii 92 2A 2 a i 93 1A 1 d i ii 93 2B 6 b 94 1A 1 d e i ii 95 1A 2 b i 97 1A 1 b 99 2A 4 b i ii 93 2A 3 a i ii 94 2B 4 e i ii 95 2B 4 b i ii 91 2A 2 b i ii 2b i CrO42.+ 5H2O + 8e1 mark overall equation : 1 mark 8CrO42.+ 10OH.0 × 10-5 0.0 × 10-5 0.15 0.30 0. The balanced equation for this reaction ClO3. of moles of CrO4 = 0. of mole of S2O32. [Cl-]/mol dm-3 [H+]/mol dm-3 Initial rate/mol dm-3 [ClO3-]/mol dm-3 0.= 0.+ 2OHii What volume of 0.20 2.III.+ 17H2O → 8Cr(OH)4.08 0.→ Cr(OH)4.08 0.40 8.+ 5Cl.04267 dm3 = 42.67 cm3 1 mark 0.ions in basic medium. Write balanced equations for 3 each half reaction and for the overall reaction.→ 2SO42. ClO3-.0 × 10-5 0.15 0.50 (no unit -½) (more than 2 decimal places -½) C i Some candidates failed to recognize that the given reaction takes place in basic medium.+ 3S2O32.15 0. 2MnO2(s) + 3PbO2(s) + 4H+(aq) d 2MnO4-(aq) + 3Pb2+(aq) + 2H2O(l) 1½ mark ii the reaction between H2O2(aq) and Cr(OH)4-(aq) in alkaline solution to give CrO42-(aq).and SO42.+ 6SO42.+ 4H2O + 3e.2 × 40 × × 10 3 8 0.2 × 40 × 10-3 Since 3 moles of S2O32.2 × 40 × × 10 −3 3 Volume of Na2CrO4 = = 0.≡ 8 moles of CrO421 mark 8 -3 2∴ no. 92 2A 2 a i 2a In acid solution.ions to form Cr(OH)4. n) Overall equation: 35H2O + 10Crn+ + (12 .+ (12 .per Crn+ is (6 .00 A for 2.50 × 60 × 60 no.2n = 4 n=2 1 mark C This part was unsatisfactorily answered.+ 10H+ 2 mark Chlorine will react similarly as bromine because Cl2 is a stronger oxidising agent when compared with Br2 and +6 (in SO42-) is the maximium oxidation state for S. No. Redox Equilibria Part 1 93 2A 3 a i ii 3a i Write balanced half equations for the redox reaction of Crn+ and MnO4. 93 2B 6 b 6b In aqueous solution. 1 mark Equation for the reaction between I2 and S2O32.50 hours.50 × 10-3 mol of Crn+ is titrated with 0. Explain your prediction. of moles of Ti liberated = 47.is 1000 1 mark 5 moles of Crn+ ≡ 4 moles of MnO410 5 1 mark 12 .00150 48. The equivalence point of the reaction is reached after 48.2n)MnO4. of moles of MnO4.2n)8H+ → 5Cr2O72.is I2 + 2S2O32.2n)Mn2+ + (13 . of moles of e.+ 2SO42.e.989 ≈ 4 1 mark 1e Write a balanced equation for each of the following reactions. 1 mole of Ti produced requires n moles of e5.III.+ 8H+ + 5e.663 × 10 ∴ n = 1. i The reaction between BrO3-(aq) and Br-(aq) in acidic solutions to give Br2(aq).0250 = 0.663 × 10-1 1 mark 96550 -1 4.169 × 10-1 1 mark 5. whereas 4 moles of bromine reacts with 1 mole of thiosulphate ions.+ (26 .+ 7OH.passed = = 4. 1½ mark BrO3-(aq) + 5Br-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) ½ mark for completing the equation 1 mark for balancing the equation ii The reaction between MnO4-(aq) and SO2(g) in alkaline solution to give MnO2(s) and SO42-(aq). Some candidates suggested that chlorine could oxidize thiosulphate ions to peroxodisulphate(VI). In terms of n. 1½ mark 2MnO4-(aq) + 3SO2(g) + 4OH-(aq) → 2MnO2(s) + 3SO42-(aq) + 2H2O(l) ½ mark for completing the equation 1 mark for balancing the equation C Generally well-answered. 2 marks C Few candidates were able to construct the equation for the reaction of bromine and thiosulphate ions.00120 1 mark No. or that chlorine was a stronger reducing agent.in acidic solution. 94 1A 1 d e i ii 1d An aqueous solution of titanium (Ti) salt was electrolysed by passing a current of 5.169 × 10-1 = 3.+ 5H2O → 8Br. As a result. Many candidates started with wrong equations so that they were not able to obtain the correct final answer.+ (12 . Calculate the value of n.2n)Mn2+ + (12 -2n)4H2O or 1 mark 10Crn+ + (12 .→ Ti i.+ (12 . of moles of e. Page 6 5 4 5 3 1½ 1½ .→ 2I. 1 mole of iodine reacts with 2 moles of thiosulphate ions.16n)H+ → 5Cr2O72.60 no.60 g of metallic Ti were deposited at the cathode. are involved in the oxidation reaction? Write the balanced overall chemical equation for the above reaction.0250 M KMnO4 solution.8n)H2O ii An acidified solution containing 1. 5.+ S4O62for the reaction between 4Br2 and S2O32-.2n)e2 marks MnO4.90 = 1. The balancing of equations for the redox reactions should be given more attention. Write balanced ionic equations for the two reactions. The half reactions 7H2O + 2Crn+ → Cr2O72. 4Br2 + S2O32. Deduce the charge on the Ti ion in the solution. Give : 1 faraday = 96500 Cmol-1 Tin+ + ne.00 × 0.→ Mn2+ + 4H2O 1 mark 1 mark No.2n)MnO4. except that some candidates still put electrons in the overall equation.00 × 2.00 cm3 of KMnO4 has been added. Predict the reaction between chlorine and thiosulphate ions in aqueous solution.+ 14H+ + (12 . of moles of Crn+ is 0. how many moles of electrons per mole of Crn+. Balance the equation in each case. 1 2 1 mark 1 mark 2 1 mark 1 mark 97 1A 1 b 1b In a nickel-plating experiment.should be released from each mole of NH2NH2. the oxidation state of nitrogen in the reaction product. of N in the reaction product is 0.→ ICl + 3H2O NH2NH2 → N2 + 4H+ + 4e1 mark IO3. after passing a current of 5. Assuming that the current efficiency is 100%. 3 1 mark IO3. O. i One of the cells contained an aqueous solution of silver(I) nitrate(V) and 2. 99 2A 4 b i ii 4b In an electrolysis experiment. Redox Equilibria Part 1 Page 7 94 2B 4 e i ii 4e In concentrated HCl. and hence suggest what the colourless gas might 2 be.+ 5H+ + HCl + 4e.+ 5H+ + HCl + 4e. Q NH2NH2 : IO3. Deduce the oxidation state of gold in the gold salt. They just wrote that a redox reaction involves reduction and oxidation. 1 mark C i Generally well answered.+ HCl + NH3NH32+ → ICl + N2 + 3H2O + H+ 1 mark C This question demanded application of knowledge to an unfamiliar situation and was poorly-answered. 95 1A 2 b i 2b i What is the essential feature of a "redox reaction" ? “Redox reaction” : a reaction involving the transfer of electron(s) from one reactant to another 1 mark OR A reaction involving a change in oxidation state / oxidation no. except that some candidates did not point out the essential feature of a redox reaction.S. inert electrodes were used and a constant current of 1. there are 2 moles of N atoms. 1 mark In 1 mole of NH2NH2.= -1 2S2O32. 4 moles of e. ½ mark ½ mark The colourless gas in probably N2. ∴ O.= +5 / V Oxidation state of I in I2 = 0 5I-(aq) + IO3-(aq) + 6H+(aq) → 3I2(aq) + 3H2O(l) C ii Many candidates could not give the correct oxidation state of iodine in IO3-.23 g of gold was deposited on the cathode.50 g of metallic nickel was deposited at the cathode. i S2O32. i Deduce.0 A through an electroplating bath containing a nickel compound for 5. ii The other cell contained an aqueous solution of a gold salt and 1.) 3 .(aq) + I2(aq) → S4O62.5 minutes. Some candidates gave I+. instead of ICl.00 A was passed for 1800 s through two electrolytic cells connected in series.→ ICl + 3H2O NH3NH32+ → N2 + 6H+ + 4e1 mark IO3.02 g of silver was deposited on the cathode. as the product in the reduction of IO3-. 95 2B 4 b i ii 4b Give the oxidation states of iodine in the two equations below. (Assume 100% current efficiency in this experiment. 0.III. deduce the oxidation state of nickel in the compound.= 1 : 1 In the reduction of IO3-. Deduce the Faraday constant and the charge carried by an electron. 1 mole of hydrazine H2NNH2 reacts with 1 mole of potassium iodate(V) to give 1 mole of iodine monochloride ICl with the evolution of a colourless gas.(aq) + I2(aq) → S4O62.(aq) + 2I-(aq) ii I-(aq) + IO3-(aq) + H+(aq) → I2(aq) + H2O(l) Oxidation state of I in I. ii Write balanced half-equations and a balanced overall equation for the reaction.+ H+ + HCl + NH2NH2 → ICl + N2 + 3H2O 1 mark OR 1 mark IO3. of I decreases from +5(IO3-) to +1(ICl) ∴ In the oxidation of NH2NH2.(aq) + I-(aq) Oxidation state of I in I2 = 0 Oxidation state of I in I. Only a few candidates worked out the identity of the colourless gas from the change in oxidation states of nitrogen.S.= -1 Oxidation state of I in IO3. the electrons from the reducing agent flow to the oxidizing agent directly.).).). Use of salt bridge Cell diagrams (IUPAC conventions) B. the electrons from the reducing agent will flow through the external circuit. The chemical energy of the two reagents will be converted to heat energy.W. pg.g. John Murray (Publisher) Ltd. . 188.m.. e. In this situation. 280–281 Chemistry in Context (4th ed. P. Redox Equilibria Part 2 Page 1 Topic Reference Reading III. Electrochemical cells Chemistry in Context.f. Redox Equilibria 6. 259–263 Part 2 Assignment A-Level Chemistry (3rd ed.2 Advanced Practical Chemistry. Thomas Nelson and Sons Ltd.III.. the chemical energy is converted to electrical energy. if a reducing agent is connected to an oxidizing agent through an external circuit and completed with a salt bridge.3. Thomas Nelson and Sons Ltd. Zn(s) strong reducing agent + Cu2+(aq) strong oxidizing agent → Zn2+(aq) weak oxidizing agent + Cu(s) weak reducing agent However. 5th Edition. 278–280 Notes If a reducing agent is mixed with an oxidizing agent. Stanley Thornes (Publisher) Ltd. And an electrochemical cell is constructed. 209 Reading Syllabus Electrochemical cells Measurement of e. 108–109 Physical Chemistry (3rd ed. Atkins.. f. a) Potentiometer Potentiometer is the best device to measure the e.m. e. NH4Cl(aq) salt bridge should not be used with Ag(s) | Ag+(aq) system. Criteria of a salt bridge 1. The electrolyte must have no reaction with the chemicals in the cell. Ag+(aq) + Cl-(aq) → AgCl(s) 2. owing to the internal resistance of a cell. both of them have very high resistance and the current flowing through them will be very minimal.m. otherwise precipitation will occur. b) High impedance voltmeter / Digital multimeter High impedance voltmeter or digital multimeter is not as good as potentiometer because the current flowing through the cell would not be zero.f. If the current is not zero. The cation and anion in the salt bridge should have the same migration speeds.f. The sliding contacted is slided along the potentiometer wire until zero reading is attained on the galvanometer. of the accumulator × XY However. of a cell is defined as the potential difference across a cell when the circuit is open. of the cell will be proportional to the potentiometer wire and given by the equation XB e. the setup and operation of a potentiometer is not very convenient. Part 2 Page 2 e.f. Although voltmeter and multimeter are not the ideal instruments for e.f.f. If they have different speeds.m. The error will be within an acceptable level. a) Requirement of a salt bridge Not all kind of strong electrolytes can be used as a salt bridge.f.m. That means the current is zero. For this reason. A standard accumulator is used as the reference in the potentiometer and the circuit is connected as the diagram shown.f. Moreover. Use of salt bridge Salt bridge is made by a thread of cotton soaked in saturated KNO3(aq) or NH4Cl(aq) solution.m. It is used to complete the circuit when the two half cells are connected together. 2. Redox Equilibria 1. For example.m.III.m.f. of a cell because the current flowing through the cell to be measured would be exactly zero. KNO3(aq) and NH4Cl(aq) are the only two commonly used salt bridges. The e. of the cell will be affected by the choice of salt bridge. Measurement of e.m. of the cell = e. A voltmeter or multimeter may be used instead. the operations of them are very simple. measurement. . the potential difference measured will be lower than the e.m. Redox Equilibria Part 2 Page 3 b) An electrochemical cell does not need salt bridge Salt bridge can be replaced by a porous partition if accurate supply of e. liquid or gaseous phase) salt bridge porous partition used to separate different species in the same phase used to enclose different species in the same chemical system used between different species in the same chemical system The rule of writing of cell diagram is not very strict. neither salt bridge nor porous pot is required.B. Cell diagrams (IUPAC conventions) Instead of drawing the actual diagram of an electrochemical cell.m. IUPAC has an agreed convention for writing an notation. For the redox reaction 2AgCl(s) + H2(g) d 2Ag(s) + 2HCl(aq) Reduction Oxidation AgCl(s) + e. some electrochemical cells even do not require the used of a porous partition. In the notation. the two electrolytes may mix together and an undesirable reaction may take place. 2.d Ag(s) + Cl-(aq) H2(g) d 2H+(aq) + 2e- Overall reaction 2AgCl(s) + H2(g) d 2Ag(s) + 2Cl-(aq) + 2H+(aq) In this example the two half cells share the same electrolyte. Daniel cell invented in 1837 Moreover. therefore. . called cell diagram. The tiny holes allows the ions to diffuse slowly but prevent the solutions from mixing. to represent a cell. After a long period of time. 3. 1.III. it is not as good as salt bridge for two reasons. it may be possible to construct more than 1 diagram for 1 electrochemical cell. The difference in diffusion speed of the ions in the two partitions affects the e.f. phase boundary (boundary between solid.m. is not required. Although the setup is more convenient than the one with the salt bridge. of the cell. HCl(aq).f. [] + N. | || or MM M . of a cell is given by the polarity of the electrode at the right in a cell diagram. of the cell = + 1. Example 2 For the Daniel cell with salt bridge Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) e.f. Therefore.m.1.m. the e. of the cell = + 1. Redox Equilibria Example 1 Part 2 Example 2 Example 3 Page 4 Zn(s) | Zn2+(aq) M Cu2+(aq) | Cu(s) Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Pt(s) | H2(g) | HCl(aq) | AgCl(s) | Ag(s) Example 1 For the Daniel cell with the porous partition Zn(s) | Zn2+(aq) M Cu2+(aq) | Cu(s) e.10 V .m.m. the sign of e.10 V.III.f. if the cell diagram is written as Cu(s) | Cu2+(aq) M Zn2+(aq) | Zn(s).f.f of the cell will become .10 V By convention. m. the cell diagram is written in this way (Pt) | RedL. For the system doesn't involve metal. " is used to used to separate different species in the same phase " [ ] " is used to enclose used to enclose different species in the same chemical system Glossary electrochemical cell e. " can be used interchangeably depending on how do you define a phase.III. Fe3+(aq) || [Mn2+(aq) + 4H2O(l)] .d Mn2+(aq) + 4H2O(l) Fe2+(aq) d Fe3+(aq) + e- The cell diagram is written as C(graphite) | Fe2+(aq) . Redox Equilibria Example 3 Pt(s) | H2(g) | HCl(aq) | AgCl(s) | Ag(s) Part 2 Page 5 e. Example 4 Electrochemical cell is not limited to metal | metal ion system.f potential difference potentiometer potentiometer wire accumulator galvanometer impedance multimeter salt bridge migration speed porous partition (pot) Daniel cell IUPAC cell diagram convention phase boundary inert electrode half-cell half-ionic equation 90 2A 1 b i ii iv 95 2A 3 c i ii iii 96 2A 3 a ii 97 2A 4 a ii iii Past Paper Question . Therefore it is written as above. OxR | (Pt) Red : Reduced form Ox : Oxidized form The two half-ionic equations involved in the cell on the right are Reduction half equation Oxidation half equation MnO4-(aq) + 8H+(aq) + 5e.B.f of the cell = + 0. an inert electrode is required. [MnO4-(aq) + 8H+(aq)] | C(graphite) RedL Fe(II) N.22 V Sometimes " | " and " .m. OxL || RedR. AgCl(s) and Ag(s) can be considered as two different phases because they are different substances. For those system doesn't involve metal. OxL Fe(III) RedR Mn(II) OxR Mn(VII) " . Graphite and platinum are the two commonly used inert electrodes. i Write half equations for the reactions at the anode and at the cathode. and the equation for the overall reaction that 3 occurs in the dry cell. The basic concept of the half cell was not clearly understood by most candidates. [Mn2O3(s) + 2NH3(aq)] | C(s) (1 mark for the correct species at cathode and anode ½ mark for using vertical solid lines to indicate phase difference. the electrode potential will also drop. with the two solutions being connected by a salt bridge. (1) If a current is drawn for some time. 3 (1) after it has been used for some time. the equilibrium will shift to the left. 95 2A 3 c i ii iii 3c The electromotive force of a new zinc-carbon dry cell is 1. the following changes occur at the two electrodes: Anode : Zn(s) reacts to give Zn2+(aq). 1½ mark leading to a drop in electrode potential.5 V. NH4+ which is an acid will react with Zn Zn(s) + 2NH4+(aq) → Zn2+(aq) + 2NH3(aq) + H2(g) ½ mark + + 2+ or Zn(s) + 4NH4 (aq) → Zn(NH3)4 (aq) + 4H (aq) + 2e ½ mark will decrease in [NH4+(aq)]. ½ mark for using vertical double line (dotted line) to represent salt bridge or using vertical single line to represent porous partition) iii Explain why the electromotive force of the dry cell drops. 1 From anode to cathode 1 mark iv represent the cell using the I. With reference to this cell: i write balanced equations for the reactions occurring at the cathode and the anode.→ Mn2O3(s) + OH-(aq) + NH3(g) Overall equation 1 mark Zn(s) + 2MnO2(s) + 2NH4+(aq) → Zn2+(aq) + Mn2O3(s) + 2NH3(g) + H2O(l) or Zn(s) + 2MnO2(s) + NH4+(aq) → Zn2+(aq) + Mn2O3(s) + NH3(g) + OH-(aq) ii Write the cell diagram for the dry cell.P. 2 Zn(s) | Zn2+(aq) | MnO2(s) | Mn2O3(s) | C(s) 2 marks or Zn(s) | Zn2+(aq) M NH4+(aq) | MnO2(s) | Mn2O3(s) | C(s) or Zn(s) | Zn2+(aq) | [MnO2(s) + NH4+(aq)]. and another platinum electrode dipped into a separate beaker containing FeSO4(aq). using the IUPAC convention. NH3(g) will accumulate at the cathode.→ 2Cr3+(aq) + 7H2O(l) 1 mark Oxidation occurs at the anode 1 mark Fe2+(aq) → Fe3+(aq) + eii state the direction in which electrons move when the electrodes are connected externally.[Cr2O72-(aq). 96 2A 3 a ii 3a ii Consider the following electrochemical cell : 6 .Fe3+(aq)||Cr3+(aq).H+(aq)]|Pt 2 mark C i Many candidates did not give state symbols in the half equation. hence they were not able to write the correct half equations and the equation for the overall reaction.III. (2) after it has been stored for a long time without being used. the equilibrium will shift to the left. making the electrode potential of Zn | Zn2+ less negative. Zn2+(aq) will accumulate at the anode. OR If a current is drawn for some time. convention 2 Pt|Fe2+(aq). Many candidates could not write the IUPAC cell diagram for the zinccarbon dry cell. 2 Reduction occurs at the cathode Cr2O72-(aq) + 14H+(aq) + 6e. Cathode: MnO2(s) and NH4Cl(aq) react to give Mn2O3(s) and NH3(g). When it is producing an electric current.A.U. Redox Equilibria Part 2 Page 6 90 2A 1 b i ii iv 1b A cell based on the reaction: 6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) consists of one platinum electrode dipped into a beaker containing an acidified solution of K2Cr2O7(aq). C As in previous years. [Mn2O3(s) + OH-(aq)] | C(s) or Zn(s) | Zn2+(aq) | [MnO2(s) + 2NH4+(aq)]. 1 mark At anode Zn(s) → Zn2+(aq) + 2e+ At cathode 2MnO2(s) + 2NH4 (aq) + 2e → Mn2O3(s) + 2NH3(g) + H2O(l) 1 mark or 2MnO2(s) + NH4+(aq) + 2e. this type of question on electrochemistry was poorly answered.C. (2) ½ mark If the cell is allowed to stand for some time. (2) In beaker E.III. using the IUPAC convention. (2) Comment on the change in concentration of Cu2+(aq) ions in beakers D and E as electrochemical reaction occurs. [Cu2+(aq)] remains almost constant because Cu2+(aq) ions produced in the electrochemical reaction combine with OH-(aq) ions in the buffer to form Cu(OH)2(s) 1 mark (3) Cu(s) | Cu(OH)2(s) | OH-(aq) || Cu2+(aq) | Cu(s) or Cu(s) | Cu(OH)2(s) | OH-(10-4 M) || Cu2+(aq) | Cu(s) or Cu(s) | Cu2+(aq) (2. iii Write the cell diagram for the electrochemical cell.0 M) | Cu(s) (Deduct ½ marks for each minor mistake) 1 mark (1 mark) (1 mark) 97 2A 4 a ii iii 4a The diagram below shows an electrochemical cell connected to a digital voltmeter. the negative terminal of the electrochemical cell. write the cell diagram for the electrochemical cell. with explanation. (1) The left hand electrode / Cu electrode in beaker D is the negative terminal ½ mark 1 mark [Cu2+(aq)] on LHS is lower than that on RHS / 1. [Cu2+(aq)] decreases because Cu2+ is reduced to Cu or Cu2+(aq) +2e-→ Cu(s) ½ + 1 mark ½ mark In beaker D.0 M Oxidation occurs at left hand electrode / electrons flow from left hand electrode to right hand electrode ½ mark so that [Cu2+(aq)] in the beaker E decreases / the difference in [Cu2+(aq)] of the two beakers becomes smaller.0 × 10-11 M) || Cu2+(aq) (1.83 6 V was recorded at 298K. An electromotive force of 0. (3) Using the IUPAC convention. ii Write the overall equation for the electrochemical reaction. Redox Equilibria Part 2 Page 7 (1) Identify. . 298 K with all solution having molar concentration. of a cell is depending on both electrodes. Thomas Nelson and Sons Ltd.. H2(g) will be converted back to H+(aq). H+(aq) will be converted to H2(g). if electron is taken from the electrode. Redox Equilibria Part 3 Page 1 Topic Reference Reading III. It would be more convenient if a standard condition is agreed and a reference electrode is chosen for comparison. 209–217 Chemistry in Context. 176–177 A-Level Chemistry (3rd ed. Redox Equilibria 6. 281–285 Reading Chemistry in Context (4th ed.f. The electrode is enclosed in a hydrogen jacket at 1 atm and immersed in a 1M H+(aq) solution. 5th Edition. the concentrations of electrolytes and temperature. Cell diagram of hydrogen electrode : Pt(s)H2(g) | H+(aq) Syllabus Notes .3 Part 3 Assignment Advanced Practical Chemistry. Or. Standard condition is defined as 1 atm.).d H2(g) If electron is supplied to the electrode.m. a redox equilibrium is set up and can be described by a half ionic equation.III.3. It consists of an inert platinum electrode coated with platinum black (platinum powder) which increases the area of contact. Electrode potentials 1. Eventually. Thomas Nelson and Sons Ltd. Stanley Thornes (Publisher) Ltd. Hydrogen electrode is chosen by the scientist as the standard of reference. pg. Standard hydrogen electrode Since the actual e. 110–112. John Murray (Publisher) Ltd.).. 189–195 Electrode potentials Standard hydrogen electrode Relative electrode potential (Standard reduction potential) Electrochemical series Use of electrode potential C. 2H+(aq) + 2e.. For example when the half-cell Pt|Cl2(g)|Cl-(aq).f. hydrogen electrode is chosen arbitrarily as the standard of comparison.36 V Furthermore.f.m. the relationship between the overall e.d H2(g) reducing agent Eo = 0 V At standard condition. Reduction half equation 2H+(aq) oxidizing agent + 2e.f. the polarity of the right-hand electrode is assigned to the e. a cell with the hydrogen electrode as the left-hand electrode and the unknown electrode as the right-hand electrode is set up. The e.m.d 2Cl-(aq) Eo = + 1. value to do the job.f. However. Standard hydrogen electrode is assigned an (relative) electrode potential of 0 V. the e.f. Part 3 Relative electrode potential (Standard reduction potential) Page 2 In order to compare the ease of losing or gaining electron of different chemical systems (different half cells). By definition.f.m.36 V. The values are also called reduction potentials because the half equations are always expressed in reduction form when listed in the electrochemical series.m. the electrode potential measured is also called standard reduction potential.m. of the cell is +1.f. Cl2(aq) + 2e. Eocell = Eoright . is determined to be 1. electrode potential of an electrode is given by the e.m. of a cell and the reduction potential of the 2 half cells is given by Eocell = Eoright . of the cell if the LEFT electrode in the cell diagram is a hydrogen electrode.36 V Cell diagram Pt | H2(g) | H+(aq) || Cl-(aq) | Cl2(g) | Pt Eo = + 1. the relative affinity to electron).e. is used as the right-hand cell and the hydrogen electrode is used as the left-hand cell.III.Eoleft When hydrogen electrode is chosen as the left-hand electrode. in this example. the e.m. Therefore.36V . a sign should be assigned to the e. of the cell is then determined. Redox Equilibria 2. Determination of relative electrode potential (standard reduction potential) In the determination of the standard reduction potential of an unknown half cell. the value of Eoleft will be zero. by definition.Eoleft = Eoright Therefore. value to denote the direction of flow of electrode (i. electrode potential serves as a very good indicator of the oxidizing and reducing power of different species. the list is called electrochemical series. Eo = .14 V Page 3 Eo = . .0. we are referring to the cell diagram instead of the position of the actual set up. a) Meaning of sign and value A positive electrode potential means that the electrode accepts electrons more readily than a hydrogen electrode.14 V When we are discussing the position of the electrode. As the sign and value of the electrode potential are depending on the readiness of a system to accept / donate electrons. Cell diagram Pt | H2(g) | H+(aq) || Sn2+(aq) | Sn(s) Sn2+(aq) + 2e.B.III. Redox Equilibria Part 3 Consider another example where the positions of the cell diagram and the physical setup are different. When the reduction equations are listed according to the electrode potential.d Sn(s) N.0. Strong reducing agent K(s) Pb(s) + SO42-(aq) Sn(s) H2(g) H2SO3(aq) + H2O(l) Ag(s) + Cl-(aq) PbO(s) + 2OH-(aq) 4OH-(aq) H2O2(aq) Ag(s) N2O4(g) + 2H2O(l) Mn2+(aq) + 2H2O(l) 2Cr3+(aq) + 7H2O(l) 2Cl-(aq) Pb2+(aq) + 4H2O(l) Mn2+(aq) + 4H2O(l) PbSO4(s) + 2H2O(l) MnO2(s) + 2H2O(l) Weak reducing agent Eo / V d d d d d d d d d d d d d d d d d d . 4.68 + 0.70 The values are only correct at standard conditions.80 + 0.28 + 0.51 + 1.36 . Reduction potential of 1M H2SO4(aq) 4H+(aq) + SO42-(aq) + 2e.0.69 + 1. Redox Equilibria 3. concentrated sulphuric acid has a concentration of about 18 M instead of 1 M. Electrochemical series Part 3 Page 4 Reduction equation Weak oxidizing agent K+(aq) + ePbSO4(s) + 2eSn2+(aq) + 2e2H+(aq) + 2e4H+(aq) + SO42-(aq) + 2eAgCl(s) + ePbO2(s) + H2O(l) + 2eO2(g) + 2H2O(l) + 2e2H+(aq) + O2(g) + 2eAg+(aq) + e2NO3-(aq) + 4H+(aq) + 2eMnO2(s) + 4H+(aq) + 2eCr2O72-(aq) + 14H+(aq) + 6eCl2(aq) + 2ePbO2(s) + 4H+(aq) + 2eMnO4-(aq) + 8H+(aq) + 5ePbO2(s) + SO42-(aq) + 4H+(aq) + 2eMnO4-(aq) + 4H+(aq) + 3eStrong oxidizing agent N.2.d H2SO3(aq) + H2O(l) Eo = +0. .46 + 1.B. e. while dilute sulphuric acid is not. Use of electrode potential a) Comparing the oxidizing and reducing power On the electrochemical series.14 0 + 0.0. MnO4/H+(aq). If the condition is not standard.III.17 + 0.g.17 V Concentrated sulphuric acid has a much more positive electrode potential and can be considered as a strong oxidizing agent. The electrode potentials are measured at standard condition. another value would be expected. the chemical species are arranged according to their electrode potentials.17 + 0. Those oxidizing agents low on the list accept electron readily and are strong oxidizing agents.36 + 1. For example.33 + 1. And those reducing agents high on the list donate electron readily and are strong reducing agents. e. K(s).23 + 1.g.92 .40 + 0.80 + 1. m.d Ag(s) Eo = .f. of cell. In contrast. Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s) Or consider a cell diagram. of electrons) flowing through the circuit.f of a cell can be calculated from the values of the electrode potentials. RedL + OxR → RedR + OxL Sn(s) + Ag+(aq) → 2Ag(s) + Sn2+(aq) Eo = + 0.94 V Eocell = Eoright . of the cell can be calculated by using the equation. of a cell is depending on the nature of the two electrodes only.94 V Furthermore.f. of electrons conserve. of electrons involved. Redox Equilibria b) Calculation of e.0.d Ag(s) Eo = .f is depending on the direction of reaction of a cell.14 V) + (+ 0.d Sn(s) Ag+(aq) + e.(.80 V Eo = (+ 0.f.Eoleft Sn2+(aq) + 2e. Eocell = Eoright .m.d Ag(s) Eo = + 0.m.m.Eoleft = (+ 0.0. e.III. the half equations are multiplied whenever necessary to make the no.14 V) = + 0. of a cell Part 3 Page 5 The e.m.94 V Since the sign of the e.14 V Eo = + 0. Sn(s) | Sn2+(aq) || Ag+(aq) | Ag(s) The e. Consider an overall cell reaction. in the balancing of a redox reaction.d Ag(s) ------------------------------------------Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s) .m.0.14 V Eo = + 0. In the above example Oxidation half ionic equation Reduction half ionic equation Overall ionic equation Sn(s) d Sn2+(aq) + 2e×2 Ag+(aq) + e.d Sn(s) Ag+(aq) + e.14 V Eo = + 0. the following convention should be observed. It is not depending on the current (or no. the electrode potentials are only simply added together and are not multiplied by the no.80 V The electrode potential is reversed when the direction of reaction is reversed. Sn(s) d Sn2+(aq) + 2eAg+(aq) + e.80 V) = + 0.80 V) .80 V The overall electrode potential can then calculated by adding the 2 values together. when a cell diagram is converted to an equation or vice versa. Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s) It is comprised two half cells presented by 2 half equations Sn2+(aq) + 2e. In the calculation of e.f. 80 V) = + 0. of mole of e.0.f.m.T∆S = . oxidizing agent reacts with reducing agent.80 V) = .46 V Eocell = (+ 0. For example. of a cell is related to the ∆G by the expression. both Cu(s) and Ag(s) are reducing agents while Cu2+(aq) and Ag+(aq) are oxidizing agents. ∆Go = -nFEocell where n is the no. Relationship between ∆G and Eo is not required in A-Level) When 1 e. Its value is depending on the total no. only strong oxidizing agent reacts with strong reducing agent. N.60 × 10-19 J The e.60 × 10-19 C × 1 V = 1.34 V Eo = + 0. This means an energetically feasible reaction. 1 eV = 1.f.B.RT ln K = .0.m.46 V Page 6 Cu(s) + 2Ag+(s) → Cu2+(aq) + 2Ag(s) Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(s) (energetically feasible) (energetically infeasible) The reaction will only be energetically feasible if the sum of electron potentials is positive.jumps across a potential difference of 1V.0.d Cu(aq) Ag+(aq) + e.d Ag(s) Eo = + 0. F is the Faraday constant Eocell is the e. Note : ∆G = ∆H . According to the expression ∆Go = -nFEocell . a positive Eocell means a negative ∆Go. In contrast. the energy acquired by the electron is called 1 eV (electron volt). of the cell ∆Go is the total free energy change of the reaction.34 V) + (. Cu2+(aq) + 2e. Eocell is a reflection of the electron pressure whose value is only depending on the nature of the two electrode. Redox Equilibria c) Part 3 Prediction of the feasibility of redox reaction In general.III.80 V Eocell = (.involved. But only Cu(s) reacts with Ag+(aq) but Ag(s) doesn't reacts with Cu2+(aq). of mole of electron passing through the cell. Indeed.nFE .34 V) + (+ 0. g. Glossary standard hydrogen electrode e.m. e. Two half-cell potentials cannot be added to give a reduction potential for a different half-cell.→ Cu(s) Cu+(aq) → Cu2+(aq) + eEo = . E3 ≠ E1 + E2. Eo = + 0.f.0. i. Redox Equilibria (1) Disproportionation reaction Part 3 Page 7 Disproportionation – A process in which the same element undergoes oxidation and reduction simultaneously. What conditions should be maintained when a hydrogen electrode is used as a standard hydrogen electrode? State how the conditions can be met in the laboratory. it must be noticed that the overall reduction potential of Cu2+(aq) to Cu(s) is not the sum of the reduction potentials of the two steps. Deduct 1 mark for each wrong/missing label Deduct ½ mark if ‘Platinized’ omitted Conditions: Maintaining the stated conditions: 298K/25ºC thermostat/water-bath 1 mark pressure gauge 1 mark H2 gas at 1 atm pressure the use of standard acid solution 1 mark [H+] = 1 M Diagrams of a hydrogen electrode were generally not well drawn. C 3 marks for correctly labelled diagram.e. reference electrode standard condition platinum electrode relative electrode potential (standard reduction potential) electrochemical series feasibility disproportionation 90 1A 2 a 92 2A 1 c i 95 2B 6 c iii 96 2B 6 d i ii e 97 1A 2 c i ii 99 2A 1 b i ii 90 2A 1 b iii 90 2B 6 a ii Past Paper Question 97 2A 4 a i 90 1A 2 a 2a A convenient laboratory method for determining electrode potentials makes use of a standard hydrogen electrode.III. 6 Draw a labelled diagram of a hydrogen electrode. . i. E2 > E1.37 V (energetically feasible) 2Cu+(aq) Furthermore.15 V ---------------------------------------------------------------------------------------→ Cu(s) + Cu2+(aq) Eo = + 0.e.52 V Cu+(aq) + e. 2Cu+(aq) → Cu2+(aq) + Cu(s) A chemical species will disproportionate if the reduction potential to the next lower oxidation state is more positive than that from the next higher state. and even fewer considered the depth of its immersion in the solution containing hydrogen ions. Only about a third of the candidates mentioned the use of platinized platinum or platinum black. 33 . With reference to this cell: iii calculate the e.35 V 2Cr3+(aq) + Zn(s) → Zn2+(aq) + 2Cr2+(aq) is also favourable because of the positive Eo value. However.33 = 2.33V. 96 2B 6 d i ii e 6d You are provided with the following standard reduction potentials : Eo/V + 0. with the two solutions being connected by a salt bridge. of the cell.76) + 1.f. giving a balanced equation.→ [Ti(H2O)6]3+(aq) +0. 1½ mark 92 2A 1 c i 1c Given the following standard reduction potentials.f.→ 2Cr3+(aq) + 7H2O(l) ------------------------------------------------------------------------------Eo = -(-0. SI6 is therefore very unstable with respect to disproportionation. predict. 1 mark At standard conditions. Eo : 1.d Cu+(aq) + 0.41V Cr3+(aq). iii The violet solution formed should be kept in a sealed vessel or handled in an inert atmosphere.76) + (-0. given that the standard reduction potentials of Fe3+/Fe2+ and acidified Cr2O72-/Cr3+ are respectively +0. in each case. why this is so.15 Cu2+(aq) + e. most of them failed to do so.33V (Cr2O72-(aq)+14H+(aq)). Using the 4 following data.III.23 .11) V = +1. and another platinum electrode dipped into a separate beaker containing FeSO4(aq). 1 mark C iii Candidates were expected to predict that if [Ti(H2O)6]3+ is not kept in a sealed container. what will happen to the violet solution if it is not kept in a sealed vessel or not handled in an inert atmosphere.76V Zn2+(aq)|Zn(s) i Predict the products of the reaction of zinc with dichromate(VI) solution.09 V Cr2O72-(aq) + 14H+(aq) + 3Zn(s) → 3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l) is favourable because of the large positive Eo value. O2 is a stronger [O] agent than TiO2+.12 V. In aqueous acid solution the overall reaction: Zn(s) → Zn2+(aq) + 2eCr2O72-(aq) + 14H+(aq) + 6e. Explain your prediction. Explain.34 Cu2+(aq) + 2e.56V 1 mark 90 2B 6 a ii 6a The following species are either impossible to prepare or very unstable.(-0.11 O2(g) + 2H2O(l) + 4e. Redox Equilibria Part 3 90 2A 1 b iii 1b A cell based on the reaction: 6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) consists of one platinum electrode dipped into a beaker containing an acidified solution of K2Cr2O7(aq). [Ti(H2O)6]3+ will be oxidized to TiO2+(aq) / the violet solution will turn colourless. Page 8 1 1½ 3 95 2B 6 c iii 6c In aqueous solutions.m.77V and +1. in the presence of oxygen. Eo / V 3+ 2+ [Ti(H2O)6] (aq) + e → [Ti(H2O)6] -0.41) = 0. 2 marks 1 mark Therefore.23 From the Eo values.(+ 0.37 TiO2+(aq) + 5H2O(l) + 2H+(aq) + e.(2Cr3+(aq)+7H2O(l))|Pt(s) -0.12 V 1 mark 4Ti3+ + O2 + 2H2O → 4TiO2+ + 4H+ 1 mark Therefore.m. the products are Zn2+(aq) and Cr2+(aq). the Eo of the following reaction is +1.→ 4OH-(aq) +0. the reaction 4[Ti(H2O)6]3+ + O2 → 4TiO2+ + 4H3O+ + 18H2O takes place because the overall Eo for the reaction is +1. TiO2+ is colourless. ii SI6 S(VI) has a high redox potential and would oxidize I. The overall reaction: → Zn2+(aq) + 2eZn(s) 3+ 2e + 2Cr (aq) → 2Cr2+(aq) ------------------------------------------------Eo = . C i Candidates were not able to use all the given electrode potential data to predict the products : Zn2+ and Cr2+.40 O2(g) + 4H+(aq) + 4e. It can be reduced to give a violet solution containing [Ti(H2O)6]3+.(+0.77) = 0.d Cu(s) . and e.to I2. = +1.Cr2+(aq)|Pt(s) -0.→ 2H2O(l) +1. 92 . Write a balanced equation for the 2 reaction.(+0.0.d 2I-(aq) Eº = + 0.15) = + 0. Eo/ V +1. Redox Equilibria Cu+(aq) + e.76 V Fe3+(aq) + e. 1 mark 4 97 2A 4 a i 4a The diagram below shows an electrochemical cell connected to a digital voltmeter. An electromotive force of 0.54 V I2(aq) + 2e.→ 2CuI + I2 1 mark has a Eº value of (0.d Fe2+(aq) Explain the following observations. and 2 In aqueous solutions. the brown colour fades.) 97 1A 2 c i ii 2c Consider the standard reduction potentials listed below: Eº = + 0. ½ mark VO2+(aq) + 2H+(aq) + e. Give a balanced equation for any reaction that takes place.33 Cr2O72-(aq) + 14H+(aq) + 6e. i Write half-equations for the reaction at the anode and at the cathode. ii When concentrated KF(aq) is added to the resulting solution in (i) above.37 V predict what will be observed when a potassium iodide solution is added to a copper(II) sulphate(VI) solution.d 2Cr3+(aq) + 7H2O(l) +1.d Cu(s) I2(s) + 2e.54) = + 0.→ VO2+(aq) + H2O(l) +) SO2(g) + 2H2O(l) → SO42-(aq) + 4H+(aq) + 2e½ mark --------------------------------------------------------------------------------------------- 2VO2+(aq) +SO2(g) → 2VO2+(aq) + SO42-(aq) (Award FULL marks if only the overall equation is given. i When Fe(NO3)3(aq) is added to KI(aq).d CuI(s) i Part 3 + 0.d [FeF6]4-(aq) Eº = + 0. VO2+(aq) can be reduced by sulphur dioxide to give VO2+(aq). giving a balanced equation in each case.52 .d 2I-(aq) Cu2+(aq) + I-(aq) + e. 99 2A 1 b i ii 1b i What is 'standard reduction potential' ? ii With reference to the standard reduction potentials listed below. 3 Explain your prediction and write a balanced equation for the reaction involved.51 MnO4-(aq) + 8H+(aq) + 5e. A brown solution and a (white) precipitate are formed because the reaction ½ + ½ mark 2Cu2+ + 4I.40 V [FeF6]3-(aq) + e.54 + 0. a brown solution is formed.d Mn2+(aq) + 4H2O(l) . and give their corresponding standard electrode potentials.III.36 Cl2(g) + 2e.92 Page 9 ii 6e Using the above information explain why copper(I) compounds are unstable in aqueous solutions.83 6 V was recorded at 298K.d 2Cl-(aq) +1.52 + 0. Cu+ disproportionates to give Cu(s) and Cu2+(aq) ½ + ½ mark 2Cu+(aq) → Cu(s) + Cu2+(aq) (1 mark) 1 mark because the above reaction has a +ve Eº of + 0. explain why hydrochloric acid can be used to acidify potassium dichromate(VI) solution but cannot be used to acidify potassium manganate(VII) solution.38 V 1 mark In acidic conditions. 4 Part 4 Assignment A-Level Chemistry (3rd ed. Comparing with the other secondary cell. Redox Equilibria Part 4 Page 1 Topic Reference Reading III. 288–290 Chemistry in Context (4th ed. . 5th Edition. Thomas Nelson and Sons Ltd. However. Lead-acid accumulator Chemistry in Context. Question : What would happen if a Daniel cell is recharged ? PbO2 cathode (Reduction) PbO2(s) + 4H+(aq) + SO42-(aq) + 2e. When the battery is charged up. it produces a much larger current. pg. Cell diagram Pb(s) | PbSO4(s) | H2SO4(aq) | PbSO4(s) | PbO2(s) | Pb(s) OxL RedR OxR RedL Pb(0) Pb(II) Pb(II) Pb(IV) No salt bridge or porous partition is required because the two electrodes share the same electrolyte. H2SO4(aq) is consumed and H2O(l) is produced.. Upon discharging. the 36% H2SO4(aq) has the density 1.→ PbSO4(s) + 2H2O(l) Pb anode (Oxidation) Pb(s) + SO42-(aq) → PbSO4(s) + 2eOverall reaction Eo = + 1. the reaction is reversible. Redox Equilibria 6. PbSO4(s) will be converted back to PbO2(s) and Pb(s).. Stanley Thornes (Publisher) Ltd. The condition of charge in the cell can be determined by measuring the density of the acid with a hydrometer.III.36 Pb(s) + PbO2(s) + 2H2SO4(aq) d 2PbSO4(s) + 2H2O(l) Eocell = + 2. Notes The two electrodes are lead alloy grids filled with spongy PbO2(s) (cathode) and spongy Pb(s) (anode) immersed in the electrolyte of dilute H2SO4(aq).3.69 Eo = + 0. 217–219 Reading Syllabus D. Thomas Nelson and Sons Ltd. the density of the electrolyte will decrease.).). H2SO4(aq). thin layers of PbSO4(s) will be coated on both the cathode and anode. Upon discharge. Secondary cell and fuel cell 1.27 gcm-3.05 V In a fully charged lead-acid accumulator. 196–198 Lead-acid accumulator is commonly used in vehicle to start the engine. → 4OH-(aq) 2H2(g) + O2(g) d 2H2O(l) Eo = 0 V Eo = + 0. i Write balanced equations for the half reactions and for the overall reaction in the cell. Redox Equilibria 2. H2 Anode (Oxidation) O2 Cathode (Reduction) Overall Reaction H2(g) → 2H+(aq) + 2eO2(g) + 2H2O(l) + 4e.III.g. if a single arrow) -½ 2 2 marks . Glossary Past Paper Question Secondary cell lead-acid accumulator hydrogen-oxygen fuel cell 91 1A 2 a i ii iii 93 2A 1 c i ii iii 98 2A 4 a i ii iii alloy grid spongy hydrometer 91 1A 2 a i ii iii 2a An electrochemical cell containing an oxygen cathode and a hydrogen anode is shown below : The pistons above the gas chambers are frictionless.40 V On top of electrical energy.d H2O anode: H2 d 2H+ + 2eoverall: H2 + ½O2 → H2O wrong coefficient -½ reverse labelling of cathode and anode reaction -½ wrong species in each reaction -½ wrong direction of reaction (e.40 V Eo = + 0. Hydrogen-oxygen fuel cell Part 4 Page 2 Fuel cell is the battery used in spaceship to produce electricity. the fuel cell also provide the warmth and water for the astronaut in the spaceship. cathode: 2H+ + ½O2 + 2e. should be used to denote a phase boundary. (II) Write the overall equation for the electrochemical reaction and hence determine the standard electromotive force (e.23 + 1.) of the cell. Redox Equilibria ii Part 4 Page 3 C 1 How does the concentration of H2SO4 affect the equilibria of the half reactions ? ↑[H+] will shift equilibrium of 2H+ + ½O2 + 2e.d Pb(s) O2(g) + 4H+(aq) + 4e. 2 C Labelled diagram 1 mark Direction of electron flow 1 mark ii Write a balanced equation for the reaction at each electrode. Many candidates were not aware that in writing cell diagrams.0. A fuel cell and an electrolytic cell could be the same problem if viewed from the equilibrium principles. 93 2A 1 c i ii iii 1c i Draw a labelled diagram of a cell that makes use of the reaction: 2H2SO4(aq) + PbO2(s) + Pb(s) → 2PbSO4(s) + 2H2O(l) Indicate the direction of electron flow on your diagram.OxL||RedR. instead of a comma ".+ H+ or Pb(s) + SO42-(aq) → PbSO4(s) + 2e1 mark iii Using the IUPAC convention.as the product in one or both of the half reactions.d Pb(s) + SO42-(aq) Pb2+(aq) + 2e.83 . They also confused the anode and the cathode and the direction of overall reaction.". how would the reading of the voltmeter change ? Explain 2 your answer.d H2O to right.0. General format: (Pt)|RedL.d H2(g) + 2OH-(aq) PbSO4(s) + 2e. 2 2 98 2A 4 a i ii iii 4a A lead-acid rechargeable cell is formed by dipping a lead plate coated with PbO2(s) and another lead plate coated with PbSO4(s) in H2SO4(aq). write the cell diagram for the above cell.13 + 1. At PbO2 (cathode) PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e. voltage reading increase 1 mark ½ mark pressure in O2 and H2 chamber increase or [O2] and [H2] increase equilibrium of H2 + ½O2 → H2O is shifted to right and favours the discharge of battery ½ mark Surprisingly.to left. and ½ mark each shift equilibrium of H2 d 2H+ + 2e.→ PbSO4(s) + 2H2O(l) or PbO2(s) + 4H+(aq) + SO42-(aq) + 2e.m.36 .d PbSO4(s) + 2H2O(l) Eo/V .OxR|(Pt) Pb(s)|PbSO4(s)|H2SO4(aq)|PbSO4(s)|PbO2(s)|Pt(s) or Pb(s)|PbSO4(s)|H2SO4(aq)|PbO2(s)|PbSO4(s)|Pt (without physical state -1) 2 marks or PbSO4(s)|PbO2(s)|H2SO4(aq)|PbSO4(s)|Pb(s) ii Many candidates had difficulties in writing balanced equations. i You are provided with the following standard reduction potentials at 298 K. iii If weights are added to the pistons of both chambers. .d 2H2O(l) PbO2(s) + SO42-(aq) + 4H+(aq) + 2e.69 8 (I) Write half-equations for the reaction at the anode and at the cathode of the cell.III.0.→ PbSO4(s) + 2H2O(l) 1 mark At Pb (anode) Pb(s) + HSO4-(aq) → PbSO4(s) + 2e.f. a vertical line " | ". iii Few candidates gave the right cell diagram. Over half of the candidates gave OH. many candidates could not apply knowledge of chemical equilibria to an electrochemical reaction. 2H2O(l) + 2e. III. Redox Equilibria Part 4 ii Write the cell diagram in accordance with the IUPAC convention. iii Explain why (I) the voltage of the cell drops upon discharge: (II) the cell is rechargeable. Page 4 . 271–272 Formation of rust is a complicate process.40 V) = + 0.84 V (energetically feasible) Fe2+(aq) precipitates with OH-(aq) ion in water to form solid iron(II) hydroxide. pg. of the cell = (+ 0. Initially.3. . Anode (oxidation) Fe(s) → Fe2+(aq) + 2eEo = + 0. Thomas Nelson and Sons Ltd. 309–310 Reading Syllabus Notes Corrosion of iron and its prevention E. the iron(III) hydroxide changes to a reddish brown solid – rust. Thomas Nelson and Sons Ltd.40 V The overall e. iron rusts much faster in sea water (sodium chloride solution) than in distilled water. Since Fe2O3·nH2O(s) has a very loose structure. 516–518 Chemistry in Context (4th ed. Redox Equilibria Part 5 Page 1 Topic Reference Reading III. it flakes off from the iron surface and exposes the inside of the iron block to further corrosion.. Stanley Thornes (Publisher) Ltd.e. Because rusting involves flowing of charges.44 V) + (+ 0.44 V The dissolved oxygen in water accepts the electrons and becomes hydroxide ion. 5th Edition.).5 Part 5 Assignment A-Level Chemistry (3rd ed..m. Redox Equilibria 6. Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s) Iron(II) hydroxide is then further oxidized by the dissolved oxygen to iron(III) hydroxide.III. Corrosion of iron and its prevention 1. Rusting can be considered as an electrochemical process.). Acidic condition also accelerates rusting by reacting with iron metal. Cathode (reduction) O2(aq) + 2H2O(l) + 4e. Fe2O3·nH2O(s).. Rusting only occurs when iron gets in contact with dissolved oxygen in water. 4Fe(OH)2(s) + O2(aq) + 2H2O(l) → 4Fe(OH)3(s) Upon standing.f. i. Fe(s) atom loses 2 electrons and becomes Fe2+(aq) ion. rusting would be faster in the presence of an electrolyte.→ 4OH-(aq) Eo = + 0. Electrochemical process of rusting Chemistry in Context. the center is the anodic area. 2. Grease or oil is used in the moving part of machinery where the paint will be scratched. After a long time. the tin-plated can will rust even faster than an ordinary iron can. However. The process is called galvanising or galvanization.. A.Potassium hexacyanoferrate(III) Since Fe2+(aq) ion is the one of the initial product in rusting. . a layer of tin is coated on the surface of iron can to exclude the air and water. Redox Equilibria Part 5 Rust indicator . [Fe(CN)6]3-(aq) + Fe2+(aq) yellow → [FeFe(CN)6]-(aq) deep blue Page 2 When a drop of solution containing phenolphthalein and potassium hexacyanoferrate(III) is dropped onto an iron surface. And. Instead. it is not used in protecting the food can from rusting. where Fe2+ and OH. Prevention of rusting a) Coating Since rust is only formed in the presence of oxygen and water. This make it more resistant to rusting. is a yellow chemical which turns deep blue in the presence of Fe2+(aq) ion. Paint or grease Painting is the most cost effective method for large iron object but if the painted surface is scratched. zinc oxide has a very tight structure and protects the zinc underneath from further corrosion. Potassium hexacyanoferrate(III). Galvanising (zinc dipping) Iron is dipped into a tank of molten zinc or molten zinc is spray onto iron so that the surface of the iron is coated with a layer of zinc. iron can be protected from rusting by putting a coating on the surface. This shows that the rim is the cathodic area where the concentration of dissolved oxygen is higher. iron will be vulnerable. formation of rust can be monitored by detecting the presence of Fe2+(aq) ion. The body of most automobiles are galvanized before the paint is applied. the rim of the drop will turn pink and the center of the drop will turn blue gradually. K3Fe(CN)6(aq). C.ion meet. B. Tin-plating Since zinc is toxic. Although zinc also reacted with oxygen in the air to form zinc oxide. if the tin-plating is scratched and the iron underneath is exposed.III. The iron is called galvanised iron. most rust will form somewhere between the cathodic area and anodic area. This is because iron is more reactive than tin and iron will lose electrons through tin quickly. It serves as a very good indicator of rust. iron pipe under the ground. (7 marks) . it will acquire the electrons from the more reactive metal and changes back to Fe. For the same reason. Consequently. i galvanized iron ii tin-plated iron 99 2A 1 b iii 1b iii Explain why cathodic protection can be used to prevent the corrosion of iron. Zn(s) also protects iron by sacrificial protection because Zn(s) is more reactive than Fe(s). iron is protected from corrosion but the more reactive metal is sacrificed instead. hull of ship. Most water taps are electroplated with chromium. iron will become stainless steel which is more resistant to rusting. e. chromium and nickel are added. This make the car body negatively charged and less vulnerable to rusting. Glossary rusting (corrosion of iron) rust (hydrated iron(III) oxide) anodic area rust indicator (potassium hexacyanoferrate(III)) coating galvanising sacrificial protection tin-plating electroplating alloying 97 2A 4 b i ii 99 2A 1 b iii cathodic area zinc dipping Past Paper Question 97 2A 4 b i ii 4b Discuss and write relevant equations for the electrochemical processes involved when broken surfaces of the following are exposed to moist air. the body of a car is always connected to the negative terminal of the car battery. iron can be protected from rusting. If the formation of iron(II) ion can be suppressed. By connecting iron to a more reactive metal. Once a Fe2+(aq) ion is formed. formation of Fe2+(aq) can be retarded. But if appropriate amounts of carbon. Chromium is frequently used because of its silvery appearance. c) Alloying Pure iron rust easily in moist air.g. b) Sacrificial Protection Iron is first changed to iron(II) ion in the rusting process.III. Sacrificial protection by galvanization Besides isolating iron from air and water.g. Redox Equilibria Part 5 Page 3 D. This method is preferable for the iron object which is hard to be repainted e. Fe(s) d Fe2+(aq) + 2eA reactive metal loses electron more readily than a less reactive metal. Electroplating of other metal Chromium is another metal which is coated on iron to prevent rusting. Raoult's law 2.Phase Equilibria I. Composition of the liquid mixture comparing with composition of the vapour 3. Solvent extraction a) Calculation 2. B. II. Two components system A. Isotherm of water 3. Boiling point-composition diagram a) Definition of boiling point b) Boiling point-composition diagram c) Fractional distillation Non-ideal solution 1. Three components system A. PVT surface of water Carbon dioxide 1. III. Paper chromatography a) Rf value b) Separation of amino acids B. Deviation from Raoult's law a) Positive deviation b) Negative deviation c) Enthalpy change of mixing 2. Isotherm of carbon dioxide Ideal system 1. . Phase diagram of water 2. One component system A. Elevation of boiling point and depression of freezing point by an involatile solute 3. Boiling point of two immiscible liquid a) Steam distillation Partition of a solute between two phases 1. Water 1. Phase diagram of carbon dioxide 2. Diatomic oxygen and ozone is another pair of example.. volume and temperature.1 Unit 1 A-Level Chemistry (3rd ed. Different regions of the surface represents different phases of a substance.Phase Equilibria Unit 1 Page 1 Topic Reference Reading Assignment Reading Syllabus Phase Equilibria 7. the same substance may have different physical state or phase. There are 3 physical states : solid. the same substance with the same physical state may not be exactly the same. By measuring the physical properties of a substance at different pressure. Stanley Thornes (Publisher) Ltd.). . Moreover. e.g. liquid and gas. the two different phases of solid carbon. a PVT surface can be constructed. graphite and diamond are two allotropes of carbon. One component system Notes One component system is a system consists of one substance. PVT surface is solely a summary of experimental findings and can only be obtained through experiment. At different condition. 239–240 One component system PVT surface Phase diagram Isotherm I. it can be seen that the melting point decreases with increasing pressure. It is the only condition at which solid. how is this related to the volume of ice ? This can be explained by Le Chatelier's principle. Phase diagram of water BD – melting/freezing curve AB – sublimation curve BC – vaporization/condensation curve EB – supercooling curve Water is a very special substance which expands upon freezing. activation energy is provided). It is an unique condition for a specific substance.e. . The liquid water which has a temperature lower than 0ºC is called supercooled liquid. gaseous phase can be further divided into vapour and gaseous phases. a gas cannot be compressed into liquid without cooling. For example. which means highly unstable. the system will response in a way so that the effect of increased pressure can be reduced. It would be easier to investigate a 2-dimensional projection of the 3-dimensional PVT surface along the temperature or volume axes. The curve EB represents the supercooling curve of water. The point where the 3 solid lines joined together is called triple point. From the negative slope of the freezing curve.e. the water will freeze immediately. Once the pressure is removed. The temperature at which solid and gas coexist is called sublimation point. the open structure will be broken and ice will be converted to water. melting point of ice at different pressure. The temperature at which liquid and gas coexist is called boiling point. i. The temperature at which solid and liquid coexist is called melting point. Supercooled water is energetically less stable than ice. The projection along the volume axis is called phase diagram. If the supercooled water is disturbed (i. Comparatively. The existence of water extends into the ice region. Water 1. However. volume and temperature altogether. it will melt because the freezing point of water is lowered.Phase Equilibria Unit 1 Page 2 Depending on the properties. Like all kind of change. though it is very small comparing with other chemical process. Phase diagram of water A solid line on the diagram represents the condition at which two phases coexist. it is rather difficult to study pressure. Supercooled water is a metastable state.. water will freeze again. when a high pressure is exerted onto ice. Water (liquid) d higher density (smaller volume) Ice (solid) lower density (larger volume) When pressure is exerted onto ice. it may reach a temperature below 0ºC without freezing. This is done by converting ice to water which has lower volume. Its freezing curve (BD) has a negative slope. The freezing curve (BD) represents the conditions at which water and ice are at equilibrium. If a beaker of water is cooled calmly. carbon dioxide has a freezing curve with positive slope. If pressure is exerted onto a cube of ice. freezing process also involves activation energy.16 K for water 5. 6. Conversely. Vapour is defined as a gas which can be liquefied by compression alone. At molecular level.03 × 10-3 atm and 273. But.6 K for carbon dioxide. A.11 atm and 216. liquid and gaseous phases can coexist in equilibrium. The projection along the temperature axis is called isotherm. liquid water is compressed and the volume change is very minimal. at which the vapour and liquid coexist. This has important implication in the manufacturing of liquid air. vapour can be compressed into liquid. ice expands a little bit. Isotherm of water Unit 1 Page 3 Isotherm is the curve representing the relationship between the pressure and volume of a substance at the same temperature. It is only depending on temperature. Along de. Along no. Along ab. The vapour pressure. Along ef. At high temperature. Along rs. If the temperature is too higher. Along qr. Along op. the gaseous molecules are moving at great speed. At a temperature higher than the critical temperature. Saturated vapour pressure is a quantity independent of volume. Along pq. If the volume of the system is decreased. Water vapour d Liquid water 3. water is heated up and expands. The isotherm has the shape of a hyperbola only at high temperature. PVT surface of water . ice is compressed and the volume gets a little bit small.e. This makes a gas behave as an ideal gas. Ice and water coexist at melting point. This means that the pressure is inversely 1 proportional to volume (i. ice starts to melt and contract. Along bc. the air can never be compressed into liquid no matter how high the pressure is. it becomes a gas. the vapour is compressed. more vapour will condense into liquid and the saturated vapour pressure will remain unchanged. The curve ghjk represents the compression of water vapour to liquid water at room temperature. is called saturated vapour pressure. the vapour sublimes directly into ice. At a temperature beyond critical temperature. Along cd. The curve nopqrs represents the compression of water vapour at a temperature lower than triple point temperature.Phase Equilibria 2. At a temperature lower than critical temperature. The intermolecular attractions among particles become insignificant comparing with thermal motions. ice melts because of the increasing pressure. P α V which is stated by Boyles' Law). Water and water vapour coexist at boiling point. water boils and turns to water vapour. a gas cannot be liquefied by compression alone. PVT surface of water The curve abcdef represents the change of water when it is warmed at 1 atm pressure. the temperature and volume of water vapour increase. Carbon dioxide Unit 1 Page 4 Liquid carbon dioxide doesn't exist at atmospheric pressure. Carbon dioxide behaves like many other substances. When dry ice is warmed. Critical point ½ mark At points on isotherm above P. It contracts upon freezing and has a freezing curve with a positive slope. Name the point P and comment on its significance. the triple point pressure (5. 1 mark 1½ . Phase diagram of carbon dioxide 2. Isotherm of carbon dioxide Glossary one component system sublimation phase allotrope PVT surface vapour gas projection phase diagram isotherm triple point freezing curve sublimation curve vaporization curve supercooled liquid supercooling curve metastable state hyperbola ideal gas critical point vapour pressure (saturated) vapour pressure 90 1A 2 e i ii iii 92 1A 2 g i ii iii iv 98 1A 2 a i ii iii Past Paper Question 90 1A 2 e i ii iii 2e i Each of the graphs A to D in the above diagram shows the variation of pressure with volume for a particular substance at different temperatures T1.1 atm) is higher than atmospheric pressure. it sublimes directly to carbon dioxide gas. 1. T3 and T4.Phase Equilibria B. the gaseous state of the substance cannot be liquefied by the increased in pressure alone. From the phase diagram of carbon dioxide. T2. 1½ Graph A ½ mark 1 mark Q it has the shape of a hyperbola or Q it represents the relationship that P is inversely proportional to 1/V. 1 mark Or The horizontal portion signifies the liquefaction (vaporization) of the gaseous (liquid) state of the substance. 1 mark 1 1 mark 98 1A 2 a i ii iii 2a The phase diagram for carbon dioxide is shown below.Phase Equilibria ii Unit 1 Page 5 C What is the significance of the horizontal portion of graph D? 1 Along the horizontal portion. Point X . Suggest the meaning of the curve YW. 1 1 1 iv Point B is the boiling point. . Z iii Mark on the diagram. using the symbol B. the gaseous and liquid states of the substance can co-exist. iii Which graph provides experimental evidence that PV = nRT? Explain your answer. Metastable state or Supercooling occurs. liquid and vapour are all at equilibrium) ½ mark ii Which two points in the diagram define the sublimation curve? Y. 92 1A 2 g i ii iii iv 2g For the phase diagram below: i Identify the point X and Y. the normal boiling point. iii Candidates were expected to know that Graph A provides experimental evidence that PV = nRT because the graph gives the relationship in the form of a hyperbola in which P is inversely proportional to V (at a given temperature).critical temperature (substance can no longer exist in the liquid state beyond this temperature) ½ mark Point Y .triple point (solid. (I) solid CO2 (dry ice) (II) liquid CO2 1 1 2 .Phase Equilibria Unit 1 Page 6 i Identify points B and C. ii What is the physical meaning of point C ? iii Starting with a sample of CO2(g) at room temperature and atmospheric pressure. suggest how the following can be obtained. Two components system Two components system is a system consists of 2 different substances. 372–377 Unit 2 Assignment A-Level Chemistry (3rd ed. 181–186 Reading Syllabus Two components system Ideal system Raoult's law II. However. if two volatile liquids are mixed.1 Modern Physical Chemistry ELBS pg. It will also be depending on the abundance of individual component.Phase Equilibria Unit 2 Page 1 Topic Reference Reading Phase Equilibria 7. If a system is comprised of one more substance. some vapour will be compressed into liquid and the (saturated) vapour pressure will remain a constant. the physical properties of a mixture is also depending on the relative abundance (mole fraction) of individual component. Therefore. At a given temperature in an one component system. the saturated nature of the vapour is understood. In the presence of a liquid. the higher will be the (saturated) vapour pressure at a given temperature.. the more volatile a liquid. If the volume of the mixture is decreased. the (saturated) vapour pressure is a constant independent of volume.2. It would be easier if the investigation is limited to a given temperature. Liquid d Saturated vapour In general. Stanley Thornes (Publisher) Ltd. A 4-dimensional space will be required to represent the experimental findings. On top of pressure. the physical behavior will be more complicate. the (saturated) vapour pressure will no longer depend on temperature only. The saturated vapour pressure of a liquid is also commonly known as the vapour pressure of the liquid. the effect of temperature can be excluded. PAº = vapour pressure of pure A (more volatile) PBº = vapour pressure of pure B (less volatile) Notes .). The situation can further be simplified by considering the liquid and vapour phases only. volume and temperature. 1.e. N. Meaning of superscript º : pure substance o : standard state Vapour pressure-composition diagram By Dalton's law of partial pressure.Phase Equilibria By Dalton's law of partial pressure. the vapour will tend to condense to form more liquid. the (saturated) vapour pressure will be jointly proportional to the mole fractions of the two components. i. Total vapour pressure PT = = = PT = partial pressure of vapour A PA mole fraction of component A × + + vapour pressure of + pure A + partial pressure of vapour B PB mole fraction of × component B XB × PBº vapour pressure of pure B XA × PAº A vapour pressure-composition curve can also be considered as a kind of phase diagram. the partial vapour pressure of any volatile component of an ideal solution is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution. The ideal behavior of the solution is further confirmed by the nearly zero enthaply change of mixing of the two components. Rauolt's Law The system in which the total (saturated) vapour pressure is jointly proportional to the mole fractions of the two components is called an ideal system. ALB and BLB are all similar. .B. Or in words. At an external pressure higher than the equilibrium vapour pressure. the intermolecular forces ALA. Ideal system = = Unit 2 Page 2 partial pressure of vapour A PA + + partial pressure of vapour B PB It is found that if the intermolecular forces among all molecules in a mixture have similar magnitude. Total vapour pressure PT A. the liquid will tend to vaporize to form more vapour. And if the external pressure is lower than the equilibrium vapour pressure. The relationship PA = XA × PAº is called Raoult's law. Vapour pressure-composition diagram Page 3 By repeating the calculation for the mixtures with different compositions. This requires the liquid composition of the mixture and the (saturated) vapour pressure of individual component at the given temperature. Quantitatively.0 × 103 Pa Calculation Determine the composition of the vapour of the equimolar mixture of A and B at 300 K. Unit 2 Composition of the liquid mixture comparing with composition of the vapour Will the composition of the vapour be the same as the composition of the liquid ? Obviously. Mole fraction of A in the liquid = XA Mole fraction of B in the liquid = XB = (1. the more volatile component vaporizes more easily and will have a greater contribution in the total vapour pressure (which is related to concentration) at a certain temperature.XA) PBº Mole fraction of A in the vapour (according to Dalton's law of vapour pressure) X A P Aº X A P Aº PA = = = X A' = PA + PB XA PAº + XB PBº XA PAº + (1 .XA) Partial vapour pressure of A = PA = XA PAº Partial vapour pressure of B = PB = XB PBº Vapour pressure of the liquid mixture (according to Rauolt's law) = PT = XA PAº + XB PBº = XA PAº + (1. a vapour composition curve can be constructed on the same vapour pressure-composition diagram. the vapour will be enriched by the vapour of the more volatile component. when an equal molar solution of 2 liquids evaporates. the composition of the vapour of an ideal solution at a given temperature can be calculated according to Rauolt's law.XA )PBº Data (Saturated) vapour pressure of A at 300 K = 1.0 × 104 Pa (Saturated) vapour pressure of B at 300 K = 3.Phase Equilibria 2. . Therefore. 24 0.1 0.7 0.5 85.20 100.0 86.0 109. thus it has a comparatively lower boiling point.70 87. Boiling point composition diagram Unit 2 Page 4 a) Definition of boiling point A liquid is said to be boiling when the (saturated) vapour pressure reaches the atmospheric pressure. It only provides the relationship between vapour pressure and mole fraction of the liquid.3 1 Actually. A B C D E F G H I J K Mole fraction of propan-1-ol Boiling point of the mixture Boiling point of the distillate Mole fraction of propan-1-ol in the liquid / ºC / ºC in the vapour (prepared experimentally) (measured) (measured) (read from the graph) 0 109.90 84.3 0.50 92.40 94.2 99.4 91. a boiling composition diagram can be constructed.60 89.Phase Equilibria 3. A more volatile liquid has a higher (saturated) vapour pressure at a given temperature.4 0.41 0. The diagram on the right is the boiling pointcomposition diagram of an ideal system consisting of propan-1-ol and 2-methylpropan-1-ol.3 84.) .3 83. b) Boiling point-composition diagram By measuring the boiling pointsof mixtures with different composition and the composition of the vapour. Rauolt's law does not provide any relationship between the vapour pressure and the temperature.30 97. It would be more convenient to determine the boiling point of the distillate (condensed from the vapour) and determine the composition of the vapour using the liquid composition curve as the calibration curve.74 0.3 0.98 1 83.9 94.0 0.10 105.90 0.2 83.53 0. (The boiling point composition curve cannot be constructed from the Rauolt's law.2 0.0 0 0.3 88.82 0. it is quite difficult to determine the composition of the vapour chemically.94 0.65 0. The first one is the boiling point of the mixture with different composition.8 0. The second one is the composition of the vapour when the mixture boils. A boiling point composition diagram also consists of 2 lines.80 85.6 84.6 0. Glossary two components system 4-dimensional space (saturated) vapour pressure Rauolt's law ideal system/solution vapour pressure-composition diagram vapour composition curve boiling point composition diagram boiling point fractionating column glass bead boiling process condensation process 91 2A 3 b i ii 93 2A 1 a i ii 94 2A 2 c i 95 2A 1 c i 96 2A 2 c i 99 2A 2 a i ii jointly proportional liquid composition curve fractional distillation liquid film Past Paper Question 91 2A 3 b i ii .42 mole 2-methylpropan-1-ol. It should be noted that boiling is a constant temperature process. As a result. Instead of separate distillations. (Line DE) If the 0. thus fractional distillation.63 mole 2-methylpropan-1-ol mixture is distilled again. the distillate obtained will have 0. In general. from the curve. Unit 2 Page 5 While more propan-1-ol is removed.Phase Equilibria c) Fractional distillation The boiling point composition curve can be used to predict the purity of the distillate.63 mole of 2-methylpropan-1-ol. in which the liquid mixture and the vapour will have the same temperature (Line CD).2 mole propan-1-ol and 0. the vapour will have 0. the liquid mixture vaporizes and condenses on the glass beads when it moves further from the heat source. If the distillation is continued. On the other hand. the final distillate will be 100% propan-1-ol. The thin liquid film condensed on the glass bead will evaporate again while more hot vapour is coming up and the distillation will continue.37 mole of propan-1-ol and 0.58 mole propan-1-ol and 0. in the fractionating column. For example. when a mixture of 0. the composition will change when a liquid mixture is vaporized but the composition will not change when a vapour mixture is condensed. condensation of the vapour is accompanied with a decrease in temperature.37 mole propan-1-ol and 0. the boiling point of the liquid will increase gradually. the percentage of the less volatile 2-methylpropan-1-ol remain in the flask will increase.8 mole 2-methylpropan-1-ol is boiled. successive distillations can be accomplished by using a fractionating column. The distillate will have the same composition as the vapour. Upon heating. from the curve. 84 + 35. PA = PAºXA 1 mark PB = PBºXB Partial pressure of A = 24 × 0. are 24. measured at 350 K.25 Total vapour pressure of second distillation = 0. Calculate the partial vapour pressure of each component. = 32 mol. C. Page 6 3 4 3 4 .75 = 0. O. of moles of C2H5OH = =4 46 1 Mole fraction of CH3OH = = 0.14 C ii Many candidates were not able to get the final answers correct because they did not realise that the ratio of partial pressures of A to B in the first distillate was the same as the mole ratio of A to B in the condensed phase.52 mmHg ½ mark 17.0 g of methanol and 184.0 kPa respectively.0.84 mmHg 5 4 Partial pressure of C2H5OH = × 44. (Relative atomic masses : H.= ½ mark 3 3 i.75 × 24 + 0. calculate the composition of the first droplet of the second distillate. forming ideal solutions. 1.2 5 4 Mole fraction of C2H5OH = = 0.0) CH3OH mol.8 1 mark 5 By Raoult’s Law PA = x PAº 1 mark 1 Partial pressure of CH3OH = × 89.0. = 46 C2H5OH 32 =1 No.2 = 0.86 = 0. If the vapour pressure of the solute is higher than the vapour pressure of the solvent then the total vapour pressure is higher and vice versa.0. Composition of distillate (1st time) PA 14.2 = 17.6 mm Hg respectively. This indicated that the underlying principles of fractional distillation were not well understood by these candidates. wt.6 = 14.4 XA = P = 19. Comment on the vapour pressure of the solution as compared with that of the pure solvent. i Calculate the vapour pressure of a liquid mixture consisting of 60 mole % A and 40 mole % B at 350 K.4 = 4. wt. the total vapour pressure will be the sum of the partial pressures of the solute and solvent. If this distillate is distilled for a second time at 350 K. 12.2 mm Hg and 44.68 mmHg 1 mark 5 Total vapour pressure = 17. If a volatile solute is added to a solvent.25 × 12 = 21 kPa 1 mark Composition of second distillate 0. 16.8 kPa 1 mark Total vapour pressure PT = 14. The vapour pressures of pure A and pure B.75 × 24 = 0. A mixture containing 32. of moles of CH3OH = 32 184 No.4 + 4.53 3 1 2 The composition of C2H5OH in the vapour = 1 .0 g of ethanol behaves as an ideal solution. ii Careless mistakes in the calculation of mole fractions were common.8 = 19. ratio of CH3OH : C2H5OH is 1 : 2 C i Most candidates were not able to point out that the vapour pressure of the solution could be higher or lower than that of the pure solvent.6 = 35.2 kPa (wrong unit -½) 1 mark ii The mixture in (i) is distilled at 350 K and a small amount of the distillate is collected.e.75 T 1 mark XB = 1 .4 kPa Partial pressure of B = 12 × 0.Phase Equilibria Unit 2 3b i A liquid is added as a solute to a solvent forming an ideal solution..0. 93 2A 1 a i ii 1a Liquid A is miscible with Liquid B over the whole composition range.0 kPa and 12.68 = 53. the total vapour pressure of the mixture and the composition of the vapour.86 1 mark XA = 21 1 mark XB = 1 . 3 marks ii The vapour pressures of pure methanol and ethanol at 25ºC are 89.84 1 = The composition of CH3OH in the vapour = 53. The boiling point of A is higher than that of B.) 99 2A 2 a i ii 1 mark 4 . Boiling point-composition curve 3 C 1 mark for labelling the x.6p0 = 1. calculate (1) the vapour pressure of the mixture.8p0 95 2A 1 c i 1c Two miscible liquids A and B form ideal solutions upon mixing. the vapour pressures of pure A and pure B are 32 kPa and 16 kPa respectively. i Sketch a graph of vapour pressure versus mole fraction for the above solutions at 298K. i Draw the boiling point-composition graph at 1 atm for mixtures of A and B and label the graph in detail.8 (2p0) = 1.6p0 PE = 0. At 298 K.8 From the graph. 1 (1) Vapour Pressure = 4 × 32 + 4 × 16 1 + 1 marks = 20 kPa 1 mark (Deduct ½ mark for wrong / no unit) 8 2 (2) Mole fraction of A = 20 = 5 (Deduct ½ mark for including a unit in the answer. (2) the mole fraction of A in the vapour which is in equilibrium with the mixture.Phase Equilibria Unit 2 94 2A 2 c i 2c Two miscible liquids E and F form ideal solutions on mixing. determine its vapour pressure. the vapour pressure E is p0 while that of pure F is 2p0. in terms of p0.8 p0 OR BY CALCULATION PF = 0. For a mixture of E and F with mole ration E to F = 1 : 4. Page 7 3 4 4 Mole fraction of F = 4 + 1 = 5 or 0.and y-axis1 mark for the liquid curve 1 mark for the vapour curve (deduct 1 mark if TB > TA) This question was well answered except that some candidates mistook the boiling point-composition diagram as the vapour pressure-composition diagram 96 2A 2 c i 2c i Two miscible liquids A and B form ideal solutions when mixed. At 298 K.2p0 ∴ vapour pressure of mixture = PE + PF = 0. For a mixture of 1 mol of A and 3 mol of B at 298 K.2p0 + 1. vapour pressure of the mixture = 1. at 298K. (At 298 K. Suppose that the mixture is an ideal solution. and ii the concentration of methanol.5 kPa and 9. in mol%. Calculate.) i the vapour pressure of the mixture. (3 marks) Page 8 .9 kPa respectively.Phase Equilibria Unit 2 2a A liquid mixture consists of 70.0 mol % of methanol and 30.0 mol% of ethanol. the vapour pressures of methanol and ethanol are 12. at 298 K. in the vapour phase at equilibrium. The minimium vapour pressure in the liquid composition curve represents a mixture which is less volatile than either component. 1. Slightly deviated Highly deviated . 186–187 Reading Syllabus Non-ideal system Deviation from Raoult's law Enthalpy change of mixing B. Curriculum Development Council. 78–80 A-Level Chemistry (3rd ed. If the solution is highly deviated. a curve with a maximium point will result. This is called positive deviation from Raoult's law. Non-ideal system However.2–7.. For an ideal solution..Phase Equilibria Unit 3 Page 1 Topic Reference Reading Phase Equilibria 7. Stanley Thornes (Publisher) Ltd. the curve will curve up a little bit. the liquid composition curve is linear in shape.2. it will be easier for the particles to escape from the liquid mixture and give a (saturated) vapour pressure higher than the expected at a given temperature. Positive deviation If the intermolecular forces among the two kinds of particles (ALB) is weaker the original forces in pure components (ALA & BLB). This is called negative deviation from Raoult's law. Slightly deviated Highly deviated Notes Negative deviation Similarly if the intermolecular forces among the two kinds of particles (ALB) is stronger. If the solution is slightly positively deviated. most solutions do not behave ideally. it will give a (saturated) vapour pressure lower than the expected at a given temperature.2.). Some of them will give a (saturated) vapour pressure higher than the expected and some of them will give a lower value. 203–204 Unit 3 Assignment Advanced Practical Chemistry. John Murray (Publisher) Ltd.3 A-Level Chemistry Syllabus for Secondary School (1995). This means that the mixture with that particular composition is even more volatile than either component. Deviation from Raoult's law This can be interpreted as the change in the strength of intermolecular forces. If a 0. This explains why absolute ethanol is much more expensive than 97% ethanol. Boiling point-composition curve for ethanol / water . At this point. It is a largely deviated system. the first distillate will contain 0. eventually. If the distillate is distilled continuously. This mixture is known as azeotropic mixture or azeotrope.55 mole fraction of ethanol distillate will be obtained. Azeotropic mixture is a constant boiling mixture and cannot be separated by fractional distillation.20 mole fraction of ethanol is distilled. both liquid and vapour have the same composition. The remaining 3% water can only be removed by drying agent e. calcium oxide followed by distillation.39 mole fraction of ethanol.g.Phase Equilibria a) Positive deviation Unit 3 Page 2 A mixture of ethanol and benzene is a positive deviation from Raoult's law.94 mole fraction ethanol or 97% by mass of ethanol.55 mole fraction of ethanol. The mixture has a minimium boiling point at 0. Boiling point-composition curve for ethanol / benzene Ethanol and water is another example of highly positive deviation. only a 0. Fractional distillation can only purify the mixture up to 0. Boiling point-composition curve for trichloromethane / propanone Similar if a 0. a constant boiling mixture of 0. the vapour will contain less trichloromethane (0.30 mole fraction of trichloromethane is distilled. Hydrochloric acid is also a highly negative deviation from Rauolt's law. The distillate obtained will be pure propanone. But the distillate obtained will be pure trichloromethane. Its azeotropic mixture has a composition of 0.Phase Equilibria b) Negative deviation Unit 3 Page 3 Trichloromethane and propane is a system of highly negative deviation.65 mole fraction of trichloromethane. Eventually.20 mole fraction) and more propanone. And the liquid remain in the flask will contain more trichloromethane. Boiling point-composition curve for hydrogen chloride / water . the solution remain in the flask will be 0. The mixture has a maximium boiling point at 0.65 mole fraction of trichloromethane will be left in the flask and cannot be separated into trichloromethane and propanone.11 mole fraction of HCl. If a 0.65 mole fraction of trichloromethane.80 mole fraction of trichloromethane is distilled. bond breaking is an endothermic process. 1 mark As a result.e. When the components of a ideal system is mixed. The attractive forces between A and B are weaker than those between molecules of the same kind. the temperature increases. Enthalpy change of mixing Unit 3 Page 4 The change in strength of intermolecular forces can further be confirmed by the enthalpy change of mixing of components. This means that some intermolecular attractions are weakened. there is an increased tendency to escape form solution and that the vapour pressure of each of the 1 mark components is greater than that predicted by Raoult’s law. Q are the vapour pressure of pure A and pure B respectively. bond formation is an exothermic process. (i. This means no special intermolecular attractions are formed or broken. When the components of a system showing positive deviation is mixed. This means that some intermolecular attractions are strengthened. the temperature change is almost zero. (i. the temperature decreases.e. The intermolecular attraction between molecule A and molecule B is similar to that between molecule A and molecule A and that between molecule B and molecule B.) Glossary Past Paper Question non-ideal system / solution positive deviation from Raoult's law negative deviation from Raoult's law azeotropic mixture / azeotrope / constant boiling mixture enthalpy change of mixing 92 1A 2 f i ii iii 93 2A 1 b 94 2A 2 c i ii 95 2A 1 c i ii 96 2A 2 c ii 97 2A 1 c 99 2A 2 b i ii 92 1A 2 f i ii iii 2f The vapor pressure versus mole fraction diagram for a two component system (A and B) is shown below: i C State the reason(s) which give(s) rise to the shapes of the curves in the diagram.) When the components of a system showing negative deviation is mixed. ii What do points P and Q represent? P. 1 mark iii What assumption(s) would have to be met for the curves to be linear? Give an equation to represent this behaviour.Phase Equilibria c. ½ mark Ideal behaviour (Obeys Raoult’s Law) OR PA=XAPAº ½ mark P=XAPAº+XBPBº iii Some candidates confused ideal gas law with Raoult's Law for ideal solution. 2 1 1 . 2p0 + 1. Weaker (stronger) intermolecular attraction means molecules have greater (smaller) chance to escape. hence vapour pressure of the system is higher (lower) than that expected by Raoult’s Law.6p0 PE = 0.2p0 ∴ vapour pressure of mixture = PE + PF = 0.8 p0 OR BY CALCULATION PF = 0.8p0 In case of negative (positive) deviation. The molecules has a smaller (greater) chance to escape. i Sketch a graph of vapour pressure versus mole fraction for the above solutions at 298K. P > 1. If deviation is negative. 3 4 4 Mole fraction of F = 4 + 1 = 5 or 0. determine its vapour pressure. Vapour pressure – composition diagrams for negative and positive deviations from Rauolt’s Law are : Page 5 4 C (no dotted line -½) If the attraction between the molecules A & B is weaker (stronger) than the average intermolecular attraction in pure A & B. if only one of the cases is discussed. P < 1. 2 marks. positive (negative) deviation results. thus 2 marks resulting in a lower (higher) vapour pressure.8p0 ½ mark ½ mark If deviation is positive. For a mixture of E and F with mole ration E to F = 1 : 4. the interaction betwee E and F is stronger (weaker) than the average of interaction between E and E and between F and F.] ii 3 . in what possible ways could the vapour pressure of the above mixture deviate from the value determined in (i)? Account for the deviation(s).Phase Equilibria Unit 3 93 2A 1 b 1b Draw vapour pressure versus composition graphs of two types of non-ideal solutions formed by mixing two miscible liquids. [Max. 94 2A 2 c ii 2c Two miscible liquids E and F form ideal solutions on mixing. Many ca`ndidates were not aware of the relation between intermolecular interactions and non-ideal behaviour of real solutions.8 (2p0) = 1. the vapour pressure E is p0 while that of pure F is 2p0.6p0 = 1. at 298K. Interpret these graphs in terms of intermolecular attractions. vapour pressure of the mixture = 1. in terms of p0.8 From the graph.8p0 If E and F do not form ideal solution on mixing. At 298 K. If a solution with composition X1 is heated to T1. ½ mark In the pure. . In the mixture C2H5OC2H5 forms H-bond with CHCl3 Thus.composition curves for two miscible liquids A and B at one atmosphere pressure are 7 shown below. the gaseous phase in equilibrium with it has the composition X2 which is richer in B. i Draw the boiling point-composition graph at 1 atm for mixtures of A and B and label the graph in detail.Phase Equilibria Unit 3 95 2A 1 c i ii 1c Two miscible liquids A and B form ideal solutions upon mixing. ½ mark ½ mark (2) Mixtures of C2H5OC2H5 and CHCl3 will show negative deviation from Raoult's Law. Explain your prediction. the intermolecular attraction is H-bond. 1 mark for indication on the graph) This question was well answered except that some candidates mistook the boiling point-composition diagram as the vapour pressure-composition diagram 96 2A 2 c ii 2c ii Predict whether the following pairs of liquids when mixed. liquids the intermolecular attraction is dipole-dipole interaction. Vaporization of this liquid leads to a new vapour with a composition X3. by fractional 4 distillation. the attraction between the molecules is mainly van der Waals' force. (1 mark for explanation. would give solutions showing positive deviation or negative deviation from Raoult's Law. The boiling point of A is higher than that of B. ½ mark Weakening of the intermolecular force causes an increase in vapour pressure / escaping tendency of molecule.and y-axis 1 mark for the liquid curve 1 mark for the vapour curve (deduct 1 mark if TB > TA) Using your graph. Boiling point-composition curve Page 6 3 ii C 1 mark for labelling the x. ½ mark 97 2A 1 c 1c Three types of boiling point . explain how liquid B can be separated from a mixture containing A and B. CH3 4 (1) CH3CH2OH and (2) C2H5OC2H5 and CHCl3 (1) Mixtures of ethanol and methylbenzene will show positive deviation from Raoult's Law. In the mixture EtOH molecules are surrounded by methylbenzene. ½ mark ½ mark In pure EtOH. the mixture will have a lower vapour pressure than the pure liquids. which is richer in B than that with a composition X2 and finally pure B is obtained. 1 mark for indication on the graph) This vapour is cooled to T2 and condenses to liquid with a composition of X2. (1 marks for explanation. (5 marks) . the boiling points of pure nitric(V) acid and water are 359 K and 373 K respectively. ii Describe all changes in the fractional distillation of an aqueous solution of 20 mol % of HNO3. At this pressure. nitric(V) acid and water form an azeotrope which boils at 394 K. 99 2A 2 b i ii 2b i At 1 atm pressure.3 mol %of HNO3.vapour composition relationship. and (II) the boiling point .liquid composition relationship. Sketch a labelled phase diagram for the nitric(V) acid-water system at 1 atm pressure showing (I) the boiling point . The azeotrope consists of 35.Phase Equilibria Unit 3 Page 7 Account for the characteristics of these curves. ). the solute will not be soluble. McGraw-Hill. the forces among the solute and solvent particles (ALB) must be stronger than the solvent-solvent interaction (ALA) and solute-solute interaction (BLB). For example. this must be a negative deviation from Rauolt's law.Phase Equilibria Unit 4 Page 1 Topic Reference Reading Phase Equilibria 7. Normally. curve AB shows the vapour pressure of pure water at different temperature. Elevation of boiling point by an involatile solute A volatile solvent (A) has weak intermolecular forces among the solvent particles (ALA). pure water boils at 100 ºC and freezes at 0 ºC. A solute (B) is involatile only if the intermolecular forces among the particles (BLB) are strong. And a higher temperature is required to make the solution boil. Notes Since the solute-solvent interaction is the strongest. Recall the phase diagram of pure water. a higher temperature will be required to rise the vapour pressure of the mixture to atmospheric pressure. If salt is dissolved in water. when an involatile solute is dissolved in a volatile solvent. the presence of involatile solute increases the boiling point of the solvent. the vapour pressure is always lower than that of pure water. Curve CD is the vapour pressure of the aqueous salt solution.187–189 Reading Syllabus Elevation of boiling point and depression of freezing point by an involatile solute Boiling point of two immiscible liquids Steam distillation 2. Otherwise. If an involatile solute (B) is soluble in the volatile solvent (A). its vapour pressure will reach atmospheric pressure at 100 ºC. Therefore.2. . At any particular temperature. The vapour pressure of the mixture will be lower than the pure solvent at any temperature. Stanley Thornes (Publisher) Ltd..4 Chemistry (Structure and Dynamics). 331–334 Unit 4 Assignment A-Level Chemistry (3rd ed. For this result. This will lead to a lower triple point temperature as well as a lower melting point. the vaporization curve will meet the sublimation curve at a lower temperature.) Page 2 3. The degree of elevation of boiling point and depression of freezing point depends on the nature of the solvent and the concentration of the solute. PT = PAº + PBº Comparing with an ideal solution : PT = XA PAº + XB PBº Since the total vapour pressure will be higher than the vapour pressure of either component at any temperature. Nitrobenzene and water have boiling points 211 ºC and 100 ºC respectively. the two components will vaporize independently. if the intermolecular forces between the two components (ALB) are very small comparing with the original forces (ALA & BLB). the vapour pressure will be the sum of the vapour pressures of the two pure components and independent of the mole fraction of individual component. independent of the relative abundance. (Surprisingly. . This is an extremely case of positive deviation from Rauolt's law. the two liquids will be immiscible with each other. it is not depending on the nature of the solute. Nitrobenzene and water are two immiscible liquids. the mixture boils at 98 ºC. salt is sprayed onto the icy road to melt the ice in winter. the boiling point of the mixture will also be lower than either component. Under this circumstance.Phase Equilibria Unit 4 If the lowering of vapour pressure is extended to the lower temperature range. When they are mixed together. Boiling point of a mixture of two immiscible liquids For a two components system. Most organic compounds are immiscible with water.Phase Equilibria a) Steam distillation Unit 4 Page 3 According to the above principle. This is done by passing steam through the organic mixture to be separated or heat the mixture in the presence of large amount of water. water in the distillate can be separated by a separating funnel. a heat sensitive organic compound can be separated by steam distillation. water can be added to lower the boiling point of the compound. ordinary distillation cannot be used to purify heat sensitive organic compound. the boiling point must be lowered. Thus. Sometimes. Steam provides both the heat and water for the distillation. Glossary elevation of boiling point depression of freezing point separating funnel vaporization curve immiscible liquids sublimation curve steam distillation triple point temperature steam generator safety tube . This is because some organic compounds decompose well before they boil. In order to separate them. Steam distillation Since water and the organic compound are immiscible. The method is called steam distillation. 0 or m = P M = 27 × 123. Thus the total vapour pressure exerted by an immiscible mixture of liquids will reach atmospheric pressure at a temperature below the boiling-point of the heat sensitive / most volatile constituent. of mole of E is proportional to PG as PV = nRT. Let mF and mG be the masses of F and G in the system mF PF nF MF PG = nG = mG MG mf PFMF 733 × 18. the vapour pressures of F and G are 733 mmHg and 27 mmHg respectively.97 m = 0.97 mG 1 mark C ii mG mG mG 1 mark Percentage by mass of G = m + m = 3. 1 mark Similarly. the no.0 = 3. of mole of F is proportional to PF. candidates had to realize that in the vapour phase.97 G G G ⇒ mf = 3. At this temperature.1% f G G G G To answer the questions. (II) A mixture of two immiscible liquids F and G boils at 98°C.Phase Equilibria Unit 4 92 2A 2c ii Page 4 Past Paper Question 92 2A 2 c ii 2c ii The total vapour pressure exhibited by a mixture of immiscible liquids is equal to the sum of their individual 4 saturated vapour pressures. Calculate the percentage of G by mass in the vapour when the mixture boils.201 or 20.0 and 123.0 respectively.) (I) A liquid boils when its vapour pressure reaches the value exerted by external pressure. 1 mark (II) PF = vapour pressure of F = 733 torr PG = vapour pressure of G = 27 torr Since the no. (I) Explain why steam distillation enables some heat-sensitive organic compounds to be distilled at a lower temperature than their normal boiling points.97 m + m = 4. (The relative molecular masses of F and G are 18. the number of moles of F and G are proportional to their vapour pressures and then the candidates needed to set up the relationship PG / PE = nG / nE = (m/M)G / (m/M)E . gcm-3 or gdm-3. the speed of diffusion from one solvent to another is the same as the reverse speed. 205–206 A-Level Chemistry (3rd ed. the concentration of solute in the less dense solvent (upper layer) will be used as the numerator in the equilibrium law. Syllabus Notes . By convention. the concentrations of the solute in both solvents will remain constant.).Phase Equilibria Unit 5 Page 1 Topic Reference Reading Phase Equilibria 7. it dissolves in both of them. A.3 Unit 5 Assignment A-Level Chemistry Syllabus for Secondary School (1995).B.. Three components system The three components systems that we are going to study consist of 1 solute and 2 solvents. Thomas Nelson and Sons Ltd.189–193 Reading Chemistry in Context (4th ed. the choice of unit has no effect on the value of Kd.. At equilibrium. Curriculum Development Council.). Partition of a solute between two phases When a solute is added into a mixture of two immiscible solvents. 330–333 Three components system Partition of a solute between two phase Solvent extraction Paper chromatography III. N. Stanley Thornes (Publisher) Ltd. concentration in the upper layer [Solute(upper layer)] Kd = concentration in the lower layer or [Solute (lower layer)] Kd can also be expressed as [Solute(upper layer)] molar mass of the solute concn in gram per dm3 in the upper layer [Solute(lower layer)] × molar mass of the solute = concn in gram per dm3 in the lower layer Since the numerator and denominator always have the same dimension. The concentration can be expressed in molarity. Kd always has no unit. Solute in lower layer d Solute in upper layer And the equilibrium constant of the system is known as partition coefficient or distribution coefficient. Kd. Therefore. CCl4 and CHCl3 are the only two common organic solvents denser than water. Solvent extraction Unit 5 Page 2 Solvent extraction is an useful technique in organic chemistry.32 g = 8. Calculate the mass of butanoic acid extracted by shaking 100 cm3 of water containing 10 g of butanoic acid with 100 cm3 of ether.5 The total mass extracted by the two 50 cm3 portions of ether = 6. It allows partial removal of solute from one solvent into another immiscible one.78 g of butanoic acid will be extracted.x) / 100 = 3. x / 100 (10 .36 x2 / 50 (3.5.64 . . This is done by shaking the sample to be purified with an immiscible solvent in a separating funnel. Ether can never be evaporated by a naked flame because it is highly immflammable.5 (aqueous phase) Let x g be the mass of acid extracted by 100 cm3 of ether. CH3CH2CH2COOH between ether and water is 3.36 g + 2.mass extracted in the first extraction (10 .B. Ether is a slightly polar organic solvent lighter than water. the partition coefficient of butanoic acid. In the first extraction. Once the solute is extracted. calculate the mass of butanoic acid extracted if two portions of 50 cm3 ether are used.36) g = 3.64 g x2 = 2.68 g Although the same total amount of ether is used. 7.32 x1 = 6. High temperature is also undesirable because ether forms explosive peroxide with oxygen at high temperature.5 x = 7. ether can be evaporated at reduced pressure to obtain the pure sample. the mass of acid extracted in two portions is larger. Let x1 g and x2 g be the masses of acid extracted in the two portions. For this reason.Phase Equilibria 1. N. x1 / 50 (10 .x2) / 100 = 3.6. = = The mass of acid remaining in the aqueous phase original mass . [Solute(ethereal phase)] Kd = [Solute ] = 3.78 Therefore. solvent extraction is usually done with a small amount of solvent at a time. It is commonly used in solvent extraction because it is immiscible with water and volatile.5 In the second extraction. a) Calculation Example 1 At 291 K. Example 2 For the same aqueous solution of butanoic acid.x1) / 100 = 3. In a mixture of dyes. different dyes can be separated. They must be developed first. a) Rf value The colour pattern obtained from paper chromatography is called paper chromatogram. The solute partitions between the stationary phase and the mobile phase. b) Separation of amino acids A mixture of unknown amino acids can be separated and identified by means of paper chromatography.Phase Equilibria 2. Paper chromatography Unit 5 Page 3 Partition of a solute between two phases is not limited to 2 immiscible liquids only. The positions of the amino acids in the chromatogram can be detected by spraying with ninhydrin (a developer). Glossary three components system partition partition coefficient / distribution coefficient separating funnel explosive peroxide paper chromatography stationary phase chromatogram characterize ninhydrin developer adsorbed Rf value solvent extraction mobile phase . different dyes have different partition coefficients between the moving phase and stationary phase. It always has a value smaller than 1. Each spot on the chromatogram is identified by a value called Rf value. However. The dye with larger partition coefficient will move faster on the paper when the solvent is soaking up. [Solute(moving phase)] Kd = [Solute ] (stationary phase) In paper chromatography. And. all amino acid are colourless. moving phase is the solvent used to carry the solute along the paper. which reacts with amino acids to yield highly coloured products. It is defined as : distance moved by the coloured spot Rf = distance moved by the solvent front Rf is for a particular compound depends upon the solvent used and the temperature. stationary phase is the layer of water adsorbed on the natural cellulose fibres of filter paper. It is possible to characterize a particular compound separated from a mixture by its Rf value. Paper chromatography is also a kind of distribution equilibrium. Because of these differences. ∴ α = 12β and 100α + 50β = 4 50α =4 100α + 12 48 α= = 0.15 × 0. 96 1A 1 e i ii 1e At 298 K. was separated by paper chromatography using an appropriate solvent. Decide whether (i) or (ii) is the better extraction method. whereas 50. The chromatogram obtained is shown below: .0384 1250 The mass of X in liquid D = 100α = 3.0 cm3 of the aqueous layer required 8.15 cm3 of 0. The two layers were then separated. in which case the value calculated is 77. Calculate the efficiency of ethanoic acid extraction (in terms of %) at 298 K by shaking 100 cm3 of a 0. B and C. i Calculate the distribution coefficient of I2 between water and CCl4 at 298 K. A.Phase Equilibria Unit 5 92 2A 2 c i 96 1A 1 e i ii 98 2A 4 b i ii 99 1A 2 c i ii iii Page 4 Past Paper Question 92 2A 2 c i 2c i The distribution coefficient of a compound X between two immiscible liquids D and E is 12.50 M aqueous solution of ethanoic acid with i 200 cm3 of 2-methylpropan-1-ol. No change ½ mark because distribution coefficient is a function of temperature only ½ mark (Also accept K will change because there is a change in ionic strength/activity (1/0)) 98 2A 4 b i ii 4b Even though the partition coefficient of ethanoic acid between water and 2-methylpropan-1-ol is 3.0050 50 water ratio of concentration of I2 : CCl = 12. What mass of X can be extracted form 50cm3 of a solution of E which initially contains 4g of X when shaken with 100cm3 of D? Let α g be the mass of X in 1 cm3 of D and β g be the mass of X in 1 cm3 of E Since distribution coefficient is 12.0 cm3 of the CCl4 layer required 12. 20. will this affect the value of the distribution 1 coefficient ? Briefly explain. ii two successive portions of 100 cm3 of 2-methylpropan-1-ol. 2methylpropan-1-ol is still used to extract ethanoic acid from aqueous solutions. They probably did not read the question carefully.25 cm3 of 0.25 × 0. 99 1A 2 c i iii 2c A mixture of amino acids. X is more soluble in 3 D.012934 = 0.84 g 3 marks C i Many candidates gave the mass of X remaining in E instead of the mass of X extracted from E.05 at 298 K.3 (77)) (Award ½ marks for a correct expression of Kd ) ii If the 200 cm3 of distilled water contain some dissolved KI.105 1 mark 4 20 = 0.105 M S2O32-(aq) for titration. Explain your answer. 2 8.013 1 mark (Accept distribution coefficient expressed as [I2(CCl4)]/[I2(aq)]. 50 cm3 of a solution of I2 in CCl4 were mixed and shaken with 200 cm3 of distilled water in a separating funnel until equilibrium was attained.0050 M S2O32-(aq). B and C by paper chromatography.Phase Equilibria Unit 5 Page 5 (× is the starting point of the mixture) i Briefly describe the principle underlying the separation of A. ii The amino acid spots are invisible to naked eyes. . Suggest how to make them visible. iii Calculate the Rf value for A.