ENGINEERING CHEMISTRY

MINISTRY OF SCIENCE AND TECHNOLOGYDEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION E.Ch-01011 ENGINEERING CHEMISTRY A.G.T.I(First Year) (Textile & Metallurgy) PART ONE CHAPTER 1 1. Atoms and Atomic theory 1.1 Dalton's Atomic Theory 1. Each chemical element is composed of minute, indestructible particles called atoms. Atoms can neither be created nor destroyed during a chemical change. 2. All atoms of the elements are alike in mass (weight) and other properties, but the atoms of one element are different from those of all other elements. 3. In each of their compounds, different elements combine in a simple numerical ratio: for example, one atom of A to one of B (AB), or one atom of A to two of B (AB2), 0r....... 1.2 The Discovery of Electrons In passing electricity through evacuated glass tubes, Faraday discovered cathode rays, a type of radiation emitted by the negative terminal or cathode that cross the evacuated tube to the positive terminal or anode. cathode rays travel in straight lines and have properties that are independent of the cathode material (i.e., whether it is iron, platinum, etc.); cathode rays are invisible and are deflected by electric and magnetic fields in the manner expected for negatively charged particles. Thomson concluded that cathode rays are negatively charged fundamental particles of matter found in all atoms and become known as electrons. Metal shield with slid Cathod green beam of light anode (-) fluorescent screen High voltage source (+) 1.3 X-rays and Radioactivity • X-rays: X-ray is a form of high-energy electromagnetic radiation. • Radioactivity: Radioactivity is a natural phenomenon that is spontaneous emission of radiation by mineral sources. • Three types of radiation from radioactive materials: Alpha rays (∝): are particles carrying two fundamental units of positive charge and having the same mass as He atom. Beta rays (β): are negatively charged particles produced by changes occurring within the nuclei of radioactive atoms and have the same properties as electrons. Gamma rays (γ): penetrating power. 1. are electromagnetic radiation of extremely high 1.4 Rutherford's atomic model; (the nuclear atom) Most of the mass and all of the positive charge of an atom are centered in a very small region called the nucleus. The atom is mostly empty space. The magnitude of the positive charge is different for different atoms and is approximately one half the atomic weight of the element. There exist as many electrons outside the nucleus as there are units of positive charge on the nucleus. The atom as a whole is electrically neutral. Characteristic properties of fundamental particles of matter Particle Mass Charge 1.5 Isotopes Two or more atoms having the same atomic number (Z) but different mass numbers (A) are called isotopes. 20 For example, 10 Ne,1021Ne, 22Ne are the three neon isotopes. 10 The atomic mass Proton (p) 1 amu Positive (+1) Neutron (n) 1 amu Neutral Electron (e-) 1/1840 amu Negative (-1) 2. 3. 1.1441 u at.9889 x 12. The mass of one mole of atoms of a pure element in grams is numerically equal to the atomic weight of that element in amu. (1) isotope (2) (2) element isotope (1) The mass transfer of carbon shows that 98.00335 u.867 u + 0.867 u contribution to fraction of all x mass 13C atom 13 = 13 at.7 The concept of mole: (The mole) The mole is the amount of substance that consists of a number of chemical units equal to the number of atoms in exactly 12 units of pure carbon-12. mass by C atoms that are C = 0.892% of carbon atoms are carbon-12 with a mass of 12 u.011 u. weighted according to the naturally occurring abundances of the isotopes of the element. mass of fractional mass of at. It’s numeral value is 6. .00335 u = 0. 1.01108 x 13.. mass by C atoms that are C = 0. and 1.108% are carbon-13 with a mass of 13. Example: 1..1 Solution: contribution to fraction of all x mass 12C atom 12 = 12 at.1441 u = 12.6 The Avogadro’s num er (or) Avogadro’s constant The number of atoms in exactly 12 g of pure carbon-12 (12C) is called Avogadro’s number (or) constant. Calculate the atomic mass of carbon. mass fractional of an = Example:abundance of x isotope + abundance of x isotope + . exactly. mass of naturally occurring carbon = (contribution by 12C) + (contribution by13C) = 11.0000 u = 11.The atomic mass (weight) of an element is the average of the isotopic masses.02x1023. 88 g of it.For instance.105 cm3 ? The density of Pb solution = 11. .105 cm3 x = 6.34 g Pb 1 cm solution 3 atoms.981 x 1021 atoms 6.2 How many 24Mg atoms are present in 8.023x1023 atoms 1 mol Mg Example: 1. 2. g. Example: 1. Pb x 1 mol.1 Electromagnetic radiation: Electromagnetic radiation is a form of energy transmission through a vacuum (empty space) or a medium (such as glass) in which electric and magnetic fields are propagated as waves. A wave is a disturbance that transmits energy through a medium.023x1023 atoms 1 mol Pb = 3. Pb mol. you will have one mole (or) 6. Pb 11.023x1023 atoms of titanium.3 How many Pb atoms are present in a small piece of lead with a volume of 0. Pb 207 g. Structure of Atom 2.4645 x 1021 atoms of Pb.88 amu is taken and measure out 47.34 g/cm3. CHAPTER 2.27x10-3 mol Mg ? Solution: 24 Mg atoms = 8. (Pb = 207) Solution: Conv: Vol.27x10-3 mol Mg x = 4.Pb Pb atoms = 0. a pure sample of the metallic element titanium (Ti) shoes atomic weight is 47. Wavelength. Balmer's equation. λ. h.626 x 10-34 J s. The light is dispersed into a band or spectrum of colors. for the wavelengths of atomic spectra lines of hydrogen. 1 _ 1 n2 22 1 . Atomic spectra: The spectra produced by certain gaseous substances consist of only a limited number of colored lines with dark spaces between them.3 Quantum Theory Planck postulated that the energy of a quantum of electromagnetic radiation is proportional to the frequency of the radiation—the higher the frequency. is constant velocity of electromagnetic radiation. Frequency. the wavelength components are refracted differently.997925 x 108 ms-1 in a vacuum. The speed of light. is the distance between the tops of two successive crests (or the bottoms of two troughs). written in a form proposed by J. Planck's equation is E = h ν.2881 x 1015 s2.2 The Electromagnetic spectrum The visible spectrum: White light consists of a large number of light waves with different wavelengths. is the number of crests or troughs that pass through a given point per unit of time. These discontinuous spectra are called atomic or line spectra. ν. called Planck's constant. The proportionality constant. 2. the greater the energy. When a beam of white light is passed through a transparent medium. and the relationship between this speed and the frequency (ν) and wavelength of electromagnetic radiation (λ) is c = ν. is ν = 3. has a value of 6. c. λ 2. Rydberg. 196 x 10-19 J/photon x 6.4 x 10-9 m c 2.1 For radiation of wavelength 242.179 x 10-18 J. the longest wavelength that will bring about the photodissociation of O2.936 x 105 J/ mol 2.237 x 1015 s-1 = 8. In such transitions. The allowed orbits are those in which certain properties of the electron have unique values. 3. with the motion described by classical physics.196 x 10-19 J/photon (b) E = 8.Example:2. energy is emitted. with its value lowered to En = -RH 2 n RH is a numerical constant with a value of 2.626 x 10-34 J s/photon x 1. The electron has only a fixed set of allowed orbits. The electron energy becomes negative. 2.998 x 108 m s-1 = 1. in accordance with Planck's equation. fixed discrete quantities of energy (quanta) are involved. .4 nm = 242.998 x 108 m s-1 λ = 242. What is the energy of (a) one photon and (b) a mole of photons of this light? Solution: c = 2. E = hν. An electron can pass only from one allowed orbit to another.022 x 1023 photons/ mol = 4. called stationary states.4 nm. When the electron is attracted to the nucleus and confined to the orbit n.237 x 1015 s-1 E = h ν = 6.4 Bohr's atomic model for a hydrogen atom 1. The electron moves in circular orbits about the nucleus.4 x 10-9 m (a) ν= λ = 242. ν. and velocity.179 x 10-18 J = -1.00 x 107 m s-1 . this is not an allowed energy level for the hydrogen atom.109 x 10-31 kg. and Planck's constant.00 x 10-20 J 2. En = .RH / En n2 -2. ν = 3. Wave-Particle Duality According to de Broglie's hypothesis.9 = = n = 14.1.109 x 10-31 kg) (3.2 pm Heisenberg uncertainty principle Heisenberg uncertainty principle is that we cannot measure position and momentum with great precision simultaneously. h = 6.26 x 10-34 J s. p. Momentum is the product of mass. h.00 x 10-20 J? Solutoin: n2 = .26 x 10-34 J s = 6. λ= = h h Example: 2.2 Is it likely that there is an energy level for the hydrogen atom.Example: 2. There must always be 6.00 x 107 m s-1) = 2. the wavelength associated with a particle is related to the particle momentum. m. Solution: m = 9.179 x 1020 217.26 x 10-34 kg m2 –1 λ = (9.3 p m What is the wavelength associatedνwith electrons traveling at one-tenth the speed of light? Mass of an electron = 9.76 Since the value of n is not an integer.109 x 10-31 kg ν = 3.00 x 107 m s-1 h = 6.26 x 10-34 kg m2 –1 .42 x 10-11 m = 24. but not larger than n . -l + 2. including zero. have only a positive.5 Quantum numbers and electron orbitals • Four sets of quantum numbers: (i) Principal quantum number: The principal quantum number. l = 0. 3. ms. may be a negative or positive integer. l = 2.uncertainties in measurement such that the product of the uncertainty in position ∆x. ∆p 〈 h 4π 2. 3. is ∆x. 2.4 Can an orbital have the quantum numbers n = 2. -l + 1. n = 1. -----. 2. n-1 (iii) Magnetic quantum number: It. ml = 2? Solution: . the value of ms does not depend on any of the other quantum numbers. -----. may have a value of + 1/2 ( also denoted by the arrow ) or – 1/2 ( denoted by the arrow ) . nonzero integral value. + l (iv) Spin quantum number: The electron spin quantum number. ------. 2. 1. ml = -l.1 (where n is the principal quantum number). ----(ii) Orbital (angular-momentum) quantum number: The orbital quantum number. l. Example: 2. may be zero or a positive integer. ∆p. 0. and the uncertainty in momentum. n. 4. ml. 1. and ranging from –l to +1 (where l is the orbital quantum number). ml must be 0 and. ml may range from –1 to +1. (a) n = 4. ml cannot be 2. if l = 1. if n =2.5 Give the orbital designation specified by the quantum numbers. Princi pal Q. 0.No (ml) (2l+1) 0 0 +1. l = 1 (i) All orbitals with the same value of n are in the same principal electronic shell or principal level. that is due to the followings. l can only be 0 or 1.No ( l) (n-1) 0 0 1 Magnetic Q. (3) Thus. Allowed values and orbital designations of quantum numbers. and (ii) all orbitals with the same n and l values are in the same subshell or sublevel. if l =0. (2) If l can only be 0 or 1. Thus. -1 Orbital designation 1s 1s 2p 2p 2p 2 .No (n) 1 Orbital Q. Example: 2. ( b) When l =1 the orbitals are p. • Principal shells and Subshells (b) n = 3.No. l = 0 Solution: (a) When l =0 the orbitals are s. (1) The l quantum number cannot be greater than (n – 1) . so that is 4s orbital. so that is 4s orbital. 5p. 0.6 Electron configuration The electron configuration of an atom is a designation of how electrons are distributed among various orbitals. the p subshell consists of three orbitals with a capacity of six electrons. 5f.+1. 2.-2 3s 3p 3p 3p 3d 3d 3d 3d 3d Practice Example: 1. -1.3p. -1 +2. The order in which orbitals fill is: 1s. ( iii ) When orbitals of identical energy are available. the n subshell consists of one orbital with a capacity of two electrons. ml = 0. 6p. .2p. 0. what are the possible vales of l. 2. 7s. 5d. Because of this limit of two electrons per orbital. 6s. 4p. 2s. and so on. 6d. electrons initially occupy these orbitals singly. 3s. For an orbital with n = 3 and ml = -1. Write an orbital designation corresponding to the quantum number n =4. 4f. 5s.0 3 1 2 0 +1. 3d. 4d. 4s. • Rules for assigning electrons to orbitals ( I ) Electrons occupy orbitals in way that minimizes the energy of the atom. l = 2. 7p ( ii ) No two electrons in an atom may have all four quantum number alike: Pauli exclusion principle: Only two electrons may exist in the same orbital and these electrons must have opposing spins. 2s2 2p4 Electron confi: (spdf Notation) 1s 2s 2p Element: Orbital diagram: 9F 1s2. 2s2 2s5 1s 2s 2p 10 Ne 1s2.Hund's rule: An atom tends to have as many unpaired electrons as possible. 2s2 2s6 3s 11 Na [Ne] 3s1 [Ne] . 2s2 2p3 1s 2s 2p 8O 1s2. Element: 1H Electron confi: (spdf Notation) 1s1 Orbital diagram: 1s 2s 2 He 2s2 1s 2s 2p 6C 1s2. • The Aufbau Process: Hypothetically the electron configurations of the atoms can be build up by placing electrons in the lowest available orbitals until the total number of electrons added is equal to the atomic number. 2s2 2p2 Electron confi: (spdf Notation) 1s 2s 2p Element: Orbital diagram: 7N 1s2. b. Solution: a. 1s2. Superscript numerals (2 + 2 + 6 + 2 + 5) = atomic number 17. (a) Identify the element having the electron configuration. As (Z = 33): [Ar] 3d104s24p3 . 3s23p5 (b) Write out the electron configuration of Arsenic.6: Using spdf Notation for an Electron Configuration. 2s2 2s6. the element is chlorine.3s 18 Ar 3p [Ne] 3s23p6 [Ar] 3d14s2 [Ar] 3d34s2 [Ar] 3d 4s 10 5 1 [Ne] 3d 4s 21 Sc V Cr [Ar] 3d 4s 23 [Ar] 3d 4s 24 [Ar] 3d 4s 1 29 Cu [Ar] 3d 4s [Ar] 3d 4s 10 [Ar] 3d 4s 30 Zn 2 [Ar] Example 2. 1B. 7B. f-block: Inner transition elements: lanthanide and actinide series. The group number = the sum of the ns and np electrons. Main group element 1 H 3 Li 11 Na 19 K 37 Rb 55 Cs 87 Fr *Lanthanum series *Actinum series 58 Ce 90 Th 59 Pr 91 Pa 2A 4 Be 12 Mg 20 Ca 38 Sr 56 Ba 88 Ra 60 Nd 92 U d-block • S p-block 3A 4A 5A 6A 7A 6 C 14 Si 32 Gr 50 Sn 82 Pb 7 N 15 P 33 As 51 Sb 83 Bi 8 O 16 S 34 Se 52 Te 84 Po 9 F 17 Cl 35 Br 53 I 85 At 2 He 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn 5 B 8B 27 Co 45 Rh 77 Ir 109 67 Ho 99 Es 68 Er 100 Fm 69 Tm 101 Md 70 Yb 102 No 71 Lu 103 Lr 1B 2B 29 30 Cu Zn 47 Ag 79 Au 48 Cd 80 Hg 13 Al 31 Ga 49 In 81 Tl Transition element 3B 21 Se 39 Y 57* La 89** Ac 61 Pm 93 Np 4B 22 Ti 40 Zr 72 Hf 104 62 Sm 94 Pu 5B 23 V 41 Nb 73 Ta 105 63 Eu 95 Am 6B 24 Cr 42 Mo 74 W 106 64 Gd 96 Cm 7B 25 Mn 43 Te 75 Re 107 65 Tb 97 Bk 26 Fe 44 Ru 76 Os 108 66 Dy 98 Cf 28 Ni 46 Pd 78 Pt 3. d-block: Groups 3B. 4B. 6A. 5B. The group number = number of outermost s orbital e-s . and 2B: transition elements :ns. 6B. The group number = the sum of the ns and (n-1) d electrons. The periodic table and some atomic properties 3. 7A: ns2 npx.configuration. p-block: Groups 3A. according to the subshells involved in the Aufbau process. and 8A: noble gases: ns2 np6 as representative elements. having (n-2) f orbitals filled. 8B.2 Atomic properties The sizes of atoms and ions: .1 Modern periodic table Most modern periodic tables arrange the elements in 18 groups and the table can be divided into four blocks of elements. • • • s-block: Group 1A (alkali metals: ns1 electron configuration) and 2A (alkaline earth metals:ns2) as main group or representative elements . 4A.CHAPTER 3 3. Group number of 1B and 2B correspond to the number of outermost electrons. 5A. (n-1) d e. . thus the outer-shell electrons experience a roughly comparable force of attraction to the nucleus throughout a transition series. • Variaton of atomic radii within a transition series: Atomic radii do not change very much within a transition series. Unit for atomic radius: 1 A0(angstrom) = 10-10 m SI units: 1 nm (nanometer) = 103pm (picometer) = 1x10-9m • Variation of atomic radii within a group: Down a group. Because the more electronic shells in an atom. Due to the fact that additional electrons go into an inner electron shell where they participate in shielding outer-shell electrons from the nucleus. at the same time. Metallic radius: One half the distance between the nuclei of two atoms in contact in the crystalline solid metal. the core and outer-shell electrons are attracted more strongly to the nucleus resulting in an overall contraction of the size of the atom. • Variation of atomic radii within a period: From left to right across a period. the larger the atom will be. atomic radius decreases. Ionic radius: The distance between the nuclei of ions joined by an ionic bond. atomic radius increases. As the effective nuclear charge (Zeff ) increase. the number of electrons in the outer-shell tends to remain constant.• • • • Covalent radius: One half the distance between the nuclei of two identical atoms joined by a single covalent bond. • Ionic radius: (i) Cations are smaller than the parent atoms. K+ : [Ne] 3s23p6 . ∴Ba atom is the largest of the three. Solution: [Ar] 4s23d1 [Xe] 6s2 56Ba: 2 10 4 34Se: [Ar] 4s 3d . For isoelectronic cation. (ii) Anions are larger than the parent atoms. the nuclear charge remains constant. so the electrons are not held as tightly. When a nonmetal atom gains one or more electrons to form an anion. Ba. a. K+. S2Solution: a.Because there is an excess of nuclear charge over the number of electrons in the resulting cation when a metal atom loses one or more electrons to form a positive ion. 1. the more positive the ionic charge is. Group:2A) (Period:4. Cl-. Repulsion among electrons increase. 4p 21Sc: (Period:4. Example 3. Ca2+ b.1: Determine which is the largest atom: Sc. but Zeff is reduced because of the additional electron (s). Se.2: Compare the ionic sizes in each of the following pairs. Sc < Se Zeff : Atomic size: Sc > Se Ba is in the sixth period and so has more electronic shells than either Sc or Se. the smaller is the ionic radius. • Example 3. Group:3B) (Period:6. Group:6A) • In the same period 4. Sc is closer to the left than Se. Due to the following facts. 2. K+. Example 3. Thus.< S2 – The order of increasing size is Ca2+ < K+ < Ar < Cl. Cationic size: Ca2+ < K+ Because K+ has the higher nuclear charge than Ar. for isoelectronic cations.ion. 2. 4. anionic size: Cl. having the electron configuration of Ar: 1s2 . the higher the charge is.3: Arrange the following species in order of increasing size: Ar. the larger the ion and so. 3. Cl-. I. Ca2+ Solution: 1. b. Mg (g) Mg +(g) + e- .Ca2+: [Ne] 3s23p6 Both are isoelectronic cations.2s22p6 . The five species are isoelectronic . Thus. For isoelectronic cations. [Ne] 3s23p6 Cl. the higher the charge is. for isoelectronic anions. S2-. the larger is the ion.< S2 – Ionization energy • The ionization energy. is the quantity of energy a gaseous atom must absorb so that an electron is stripped from the neutral atom. the smaller is the ion. S2. the higher the charge is. Ca2+ is smaller than K+ion. the smaller is the ion and thus. 5.: [Ne] 3s23p6 Both are isoelectronic anions. the higher the charge is.: S2.is larger than Cl. size: K+ < Ar Because Ar has higher nuclear charge than Cl--. size: Ar < ClFor isoelectronic anions.3s23p6 . 2A) 2 1 (Period. 15P 16S : : [Ne] 3s 3p (Period.3.• • • First ionization energy: the energy required to strip one electron from a neutral gaseous atom. Second ionization energy: the energy required to strip one electron from a gaseous ion with a charge of +1. • Variation of ionization energy within a group: Down a group. ionization energy decreases as atomic radii increases. Group. Variation of ionization energy within a period: From left to right across a period.3.3. • Example 3. ionization energy increase as atomic radii decreases. Because the larger effective nuclear charge makes removal of the electron more difficult.5A) (Period.4: Predict which member of each pair has the higher ionization energy. Group.6A) [Ne] . Group. Al or Mg b. Second ion energy: I2 is larger than the first I1: Mg (g) Mg+ (g) I1 I2 Mg +(g) + eMg+2 (g) + e- This is due to the fact that the attraction between the positive ion and the electron is stronger than between the neutral atom and the electron.3. : [Ne] 3s2 (Period. a.3A) 13Al : [Ne] 3s 3p More energy is required to strip an electron from the lower energy 3s orbital in Mg than from the 3p orbital in Al. the more easily it can be extracted. P of S Solution: a. ∴ Ionization energy: Mg > Al. Group. [exception] 12Mg 3s 3p b. Because the farther an electron is from the nucleus. A diamagnetic species is weakly repelled by a magnetic field. Ne (g) + e(1s22s22p6) Ne-(g) EA = + 29 kJ/mol (1s22s22p6. • For some atoms the gain of an electron required that energy be absorbed. The process is endothermic. a srong repulsion is felt and the energy of the system increase. energy is given off. O (g) O-(g) + e+ eO -(g) O2. is a measure of the energy change that occurs when a gaseous atom gains an electron. F(g) + eF-(g) EA = -322. When the second electron added is approaching a negative ion. This is the case for the noble gases.4 kJ/mol Magnetic properties • A diamagnetc atom or ion has paired electrons.4 kJ/mol EA2 =+141.I1 for S is slightly lower than for P because repulsion between the pair of electrons in the filled 3p orbital of an S atom makes it easier to remove one of those electrons than an unpaired electron from a half-filled 3p orbital of a P atom. 3s1) • For an element such as oxygen EA1 is negative and EA2 is positive. 3s 12 Mg [Ne] 3s 2 [Ne] . EA. the electron affinity is a negative quantity.2 kJ/mol When an F atom gains an electron. [exception] Electron affinity • Electron affinity. For example. and these individual magnetic effects cancel out. The process is exothermic. where the added electron enters the empty s orbital of the next electronic shell.(g) EA1 = -141. and the EA has a positive value. They must be paired . • Variation within a period A few properties vary regularly across a period in general. and Ar have poor thermal and electrical conductivities. And in general. The more unpaired electrons present. • 3s 11 Na [Ne] 3s1 [Ne] Sodium is diamagnetic. Na. it is needed for the relationship between melting point and intermolecular forces of attraction: the stronger these forces. The unpaired electrons induce a magnetic field that causes the atom or ion to be attracted into an external magnetic field. The nonmetals P. The ability to conduct heat and the ability to conduct electricity are two that do. Chemical properties of the elements Some variations within groups of elements and some variations across a period will be considered as follows. the higher the molecular mass. the stronger the attraction. Cl. S. corresponding to a single 3s unpaired electron outside the Ne core. To explain this trend. the higher the melting point. • Reducing abilities of group 1A and 2A metals . the stronger the intermolecular forces in a substances.Magnesium is diamagnetic. Mg. corresponding to two 3s electrons. For example. 3. and the individual magnetic effects do not cancel out. as are all the other electrons. A paramagnetic atom or ion has unpaired electrons.3 Other period properties of the elements Physical properties of the elements • Variation within a group The value of a property often changes uniformly down a group for compounds as well as for elements . Al have good thermal and electrical conductivity. Thus among the third period elements. Indeed. the better the nonmetal is as an oxidizing agent. K. The reducing agent makes possibly a reduction half-reaction and itself. In-group 1A metals. by losing electrons. two halogens. This is on account of their higher ionization energies compared with the other group 2A metals. by gaining electrons. electron affinity is related to the gain of electrons. We might expect an atom with a strong tendency to gain electrons (a large negative electron affinity) to take electrons away from atoms that lose electrons without much difficulty (low ionization energies). have lower ionization energy than does the next number of the fourth period. Mg and Be do not react with cold water as do the other alkaline earth metals. one in molecular form and the other in ionic form. exchange places. is oxidized. is reduced.The lower the ionization energy. reducing ability depends on the ionization energy. 2 Na (s) + Cl2 (g) 2 NaCl (s) (ii) In displacement reaction. An oxidizing agent makes possibly an oxidation half reaction and itself. The expectation is that K reacts more vigorously with water than does Ca. • Oxidizing abilities of the halogen elements The higher the electron affinity. (i) In these term. the better the metal is as a reducing agent and the more vigorous its reaction with water. for instance. it is understandable that active metals (group1A) form ionic compounds with active nonmetals (group 7A). Cl2(g) + 2I -(aq) • I2(aq) + 2 Cl-(aq) Acid-base nature of element oxides . Hence. Ca. 2K (s) + 2 H2O Ca (s) + 2 H2O 2 K+(aq) + 2 OH-(aq) + H2 (g) Ca2+(aq) + 2 OH-(aq) + H2 (g) With respect to group 2A. CO2 (g) + H2O H2CO3 (aq) (iii) It should be expected that the metal oxides at the left of the period to be basic and the nonmetal oxides at the right to be acidic. produce acidic solutions in water. react with water to produce basic metal hydroxides. called basic oxides. such as Li2O. Li2O (s) + H2O 2 Li -(aq) + 2 OH-(aq) (ii) Some nonmetal oxides. These nonmetal oxides are acidic oxides. .(i) Some metal oxides. like CO2 (g) . 1 Chemical reactions A chemical reaction is a process in which one set of substances called reactants is converted to a new set of substances called products. 1 mole of compound contains 6. For a compound. and catalyst State of matter: (s) = solid. 4 4.02x1023 molecules.02x1023 atoms /molecule For an element. Chemical reactions and Stoichiometric Calculation 4.CHAPTER. (aq) = aqueous (water) solution 4. (g) = gas. Sometimes products reacts to reform the original reactants that is called reversible reactions.02x1023 atoms.2 Stoichiometry It means the quantitative (mole) relationships among the reactants and products in a reaction. decomposition of silver oxide is illustrated as follow. (l) = liquid. 2 Ag2O(s) _4 Ag ( s) + O2 (g) stoichiometric coefficients Reaction conditions: Temperature. . N2 (g) + O2 (g) Stoichiometry: 1 mole 1 mole 2 NO(g) 2 moles Avagadro's constant: 6. pressure. 1 mole of an element contains 6. For example. 02x1023 atoms. stoichiometric factor based on N2 and H2 are 2 and 2/3 respectively. 1molar mass = 108 g Ag = 1 mol of Ag = 6. • 3 mole of H2 produce 2 moles of NH3 . = 40g NaOH = 1 mole.3 Chemical reactions in solutions: Solute + solvent solution . NaOH = 6. (3) . 4. *stoichiometric factor: (2) 1 molar mass. N2 (g) + 3 H2 (g) 2 NH3(g) • 1 mole of N2 produces 2 moles of NH3 .Stoichiometric calculations: Reactant (A) mass of substance A 1 Product (B) 2 mole of substance A mole of substance B 3 mass of substance B *molar mass: (1). Hence.02x1023 molecules. 4 Determining the limiting reagent Limiting reagent: The reagent or reactant that is completely consumed in a reaction is known as limiting reagent or limiting reactant. • • ni = n f Ci Vi = Cf Vf 1. Stoic. Molarity: Molarity = Amount of solutes (moles) Volume of solution (liters) Unit: mol /L n = C. For example. • The quantity of products formed depends on the quantity of limiting reagent. the amount of solute remains constant between the initial (i) and final (f) solution.The solvent is the component that is present in the greatest quantity in a solution and the solute is less quantity. Ratio: (Relationship) Reactants: 2 H2 (g) 2 mol [10mol] + O2 (g) + 1mol [7mol] 2 H2O(g) 2 mol Assuming that H2 is the limiting reagent: Required mol of O2 = 10 mol H2 x 1mol O2 2 mol H2 = 5 mol O2 .V Concentration: C = n (mol) V (liters) Solution dilution: When a solution is diluted. The solution is a homogeneous mixture of solvent and solute. [7 mol present – 5 mol consumed = 2 mol O2 in excess] ∴ O2 is in excess. Actual Yield The quantity of product that is actually produced is called the actual yield. Theoretical Yield The calculated quantity of product in a chemical reaction is called theoretical yield of a reaction. For example : 2 S (s) + 3 O2 (g) SO3 (g) + H2O (l) 2 SO3 (g) H2SO4 (l) Simultaneous reaction In simultaneous reaction. two or more substances react independently of each other in separate reactions occurring at the same time. For example: Mg(s) + 2 HCl(aq) Zn(s) + 2 HCl(aq) Mg Cl2 (aq) + H2 (g) Zn Cl2 (aq) + H2 (g) . Percent Yield Percent Yield = Actual yield Theoretical yield x 100 % Consecutive reaction Reaction that are carried out one after another in sequence to yield a final product are called consecutive reaction . so H2 is the limiting reagent. (aq) (CH3COOH) 4. Nonelectrolyte Strong electrolyte A nonelectrolyte is a substance that is not ionized and does not conduct electric current. HCl (aq) H+ (aq) + Cl.6 Predicting Precipitation Reactions A Set of Solubility Rules: . HC2H3O2 (aq) H+(aq) + C2H3O2.(aq) Weak electrolyte A weak electrolyte is partially ionized in water solution and the solution is only a fair conductor of electricity.5 Introduction to Reactions in Aqueous solutions. Solute + Nonelectrolyte Strong electrolyte Weak electrolyte Water itself is nonconductor Conductor or nonconductor Solvent Solution Characteristic classification of the solute.4. A strong electrolyte is completely ionized in water solution and the solution is a good conductor. acids HCl. H+(aq) + NO3-(aq) HNO3(aq) . of Ca2+.Salts SO42. bromides and.salts • • Ag+ and. S2. HCLO4. Pb2+ and.Salts Chlorides. Ba2+are slightly Soluble Molecular equation: Mg (OH)2 (s) + 2 HCl (aq) Net ionic equation: Mg (OH)2 (s) + H+ (aq) Mg2+(aq) + 2 H2O(l) Mg Cl2 (aq) + 2 H2O (l) 4. such as. (Hydronium ion. Hg2+2 Sulfates of Sr2+. I. and H2SO4. HBr. HNO3.and. Hg2+2 • Hydroxides and sulfides . • • Cl-. Br-and. Ba2+.7 Acid-Base Reaction An acid: A substance that provides hydrogen ions (H+) in solution. OH. H3O+) aqueous Strong Completely ionized in aqueous solutions. HI.Soluble salts • Group IA metal salts • NH4+ salts • NO3. iodides of Pb2+.Salts Insoluble salts (precipitate) • • CO32-. Sr2+. an acid and a base react to form water and aqueous solution of an ionic compound called a salt. eg.8 Oxidation-Reduction Reaction: • Oxidation: Oxidation is a process in which the oxidation state of element increases. bases Hydroxides of Gp. .g.IA and some Gp. e. NH4+ bases NH4+(aq) + OH. the Weak ionization is reversible. CH3COOH acids H+(aq) + C2H3O2.Partially ionized in aqueous solutions.IIA metals. Strong Completely ionized in aqueous solutions. NH3. HCOOH..(aq) HC2H3O2 (aq) A Base : A base is a substance capable of providing hydroxide ions (OH-) in aqueous solution. HCl(aq) + NaOH(aq) Na Cl(aq) + H2O(l) 4.e.(aq) NH3(aq) + H2O Neutralization: In a neutralization reaction... i. NaOH(aq) Na+(aq) + OH -(aq) Partially ionized in aqueous solutions. the Weak ionization is reversible. Verify: net charge (-6) net charge (-6) Practice Example: . SO32-(aq) + H2O MnO4-(aq) SO32-(aq) + H2O MnO4-(aq) + 8 H+(aq) SO42-(aq) + Mn2+(aq) + H2O(aq) SO42-(aq) Mn2+(aq) SO42-(aq) Mn2+(aq) + 4H2O SO42-(aq) + 2 H+(aq) Mn2+(aq) + 4H2O 3.• Reduction: Reduction is a process in which the oxidation state of element decreases. SO32-(aq)+ MnO4-(aq)+ H+ 1. SO32-(aq) MnO4-(aq) 2. • Balancing oxidation-reduction reaction: (Half-reaction method) Balancing the equation for a redox reaction in acid solution is as follow. SO32-(aq) + H2O SO42-(aq) + 2 H+(aq) + 2 eMnO4-(aq) + 8 H+(aq) + 5 eMn2+(aq) + 4H2O 4. 5 SO32-(aq)+ 5 H2O 5 SO42-(aq)+ 10 H+(aq)+ 10 e2MnO4-(aq)+ 16H+(aq)+ 10 e2Mn2+(aq) + 8H2O 5 SO32-(aq)+ 2MnO4-(aq)+ 6H+(aq) 5 SO42-(aq)+ 2Mn2+(aq) + 3H2O 5. Reducing agent. 1. Na makes Cl2 to be reduced to Cl.in NaCl and itself oxidized to Na+in NaCl. Cl2 is oxidizing agent. Cl2 makes Na to be oxidized to Na+ in NaCl and itself reduced to Cl-in NaCl. red: oxi: 0 -2 -1 2H 2 O 2 (aq) 2H 2 O + O 2 (g) 4. agent oxi.Balance the following equations for the redox reactions. in which the same substance is both oxidized and reduced. CrO42-(aq)+S2-(aq)+OH-(aq) 3.9 Disproportionation Reaction: Some oxidation-reduction reactions.10 Oxidizing and Reducing Agents: • • Oxidizing agent. H2O2 (aq) + Fe2+(aq)+ H+(aq) SO42-(aq)+Cr3+(aq)+H2O Cr(OH)3(s)+ S (s) + H2O S(s) + SO2 (g) + H2O H2O + Fe3+(aq) 4. S2O32-(aq) + H+(aq) 4. 4. Na is reducing agent. SO32-(aq)+ Cr2O72-(aq)+ H+(aq) 2. agent For the above reaction. reductant: The substance that causes some other substance to be reduced and so itself oxidized.11 Stoichiometric Calculations in Aqueous Solutions 2 NaCl (s) . oxidant: The substance that causes some other substance to be oxidized and so itself reduced. are called disproportionation reactions. 2 Na (s) + Cl2 (g) red. 808 x 10-3 mol HC2H3O2 Conc.808x10-3mol HC2H3O2 5. A 5 ml sample of a particular vinegar was titrated with 38.53 % HC2H3O2 x 1 ml vinegar x 100 % .= 3.808 x 10-3 mol OHmol HC2H3O2 = mol OH.HC2H3O2 1L 0. The legal minimum A/A content of vinegar is 4% by mass.05 gHC2H3O2 1 mol HC2H3O2 (ii) % HC2H3O2 = 0.01 g/mL) Solution: HC2H3O2(aq)+ OH-(aq) (i) ml NaOH L NaOH C2H3O2-(aq)+ H2O(l) mol NaOH conc.1mol NaOH 1L 1mol OH1mol NaOH mol OH- mol HC2H3O2 mol OH = 38.00x10 L -3 = 0. (i) Determine the concentration of HC2H3O2.08 ml x - 1000 ml x x = 3.HC2H3O2 = 3.08 ml of 0.1 M NaOH (aq).762 M C2H3O2 60.01 g i = 4.762mol HC2H3O2 1000 ml vinegar x 1. (ii) Does this sample exceed the minimum limit? (Density of vinegar = 1.Example: Vinegar is a dilute aqueous solution of acetic acid. ∴The vinegar sample does exceed the legal minimum limit.5% KOH.62 ml of an HCl (aq) solution for its titration.235 g sample of a solid that is 92. and there are two sources of OH-) . Practice Example: A 0. 7. What is the molarity of the HCl (aq)? (Hint: Write a net ionic equation.5% Ba (OH)2 by mass requires 45. • Barometric pressure To measure the atomospheric pressure.A.h. A force divided by the area over which the force is distributed F P= A SI unit: N/m2 (Pascal: 1P a = 1 N/m2) Kilopascal: kPa is more commonly used Liquid pressure The pressure of a liquid depends only on the height of column and the density of the liquid.1 What is the height of a column of water that exerts the same pressure as a column of mercury 76.CHAPTER 5. 5. The pressure exerted by a mercury column of exactly 760mm in height is called the standard atmosphere (atm).h.d = = = = = g. varies with atmospheric conditions and with altitude.1 The concept of pressure Pressure is defined as a force per unit area.0 cm (760mm) high? . P= F W g. The height of mercury in a barometer.80665 m s-2 h = Height of liquid. Example: 5.V.called the barometric pressure. The properties of gases 5.m g. 1 atm = 760 mmHg. barometer is used. d = density of liquid. 1 atm = 760 torr (Torricelli).d A A A A A g = Acceleration due togravity:9.d g. are these: the manometer is filled with liquid mercury (d = 131.∆P = 748. ∆P = negative value.0 cm x = 1. • Manometers One can compare the gas pressure and barometric pressure with a manometer. A different in height of the two arms signifies a difference between the gas pressure and barometric pressure. As the gas pressure is less than barometric pressure. and the difference in mercury levels is 8.03 x 10 3 cm = 10. gas pressure less than barometric pressure.00 g / cm 3 = g x 76. the lower the liquid density is the greater is the height of the liquid column.00 g / cm We see that the column height is inversely proportional to the liquid density: for a given liquid pressure.6g/cm3 Pressure of H2O column = g h H O d H O = g x h H O x 1.0 cm x 13.6 g / cm 3 h H 2O 13. Example: 5.6 mmHg.0 cm x 13.Solution: Pressure of Hg column = ghHg dHg = g x 76. the pressure we seek is Pgas = Pbar + ∆P.6 g/cm3).2 mmHg. ∆P = positive value. if the conditions. ∆P is negative.6 mmHg? Solution: Whether the situation corresponds to gas pressure less than barometric pressure.2 What is the gas pressure Pgas. If Pgas < Pbar . As long as the gas pressure being measured and the prevailing atmospheric (barometric) pressure are equal.6 mmHg = 739. the heights of the Hg column in the two arms of the manometer are equal. Pgas = Pbar .2 mmHg – 8. barometric pressure is 748. 2.6 g / cm 3 = 76. 1. If Pgas>Pbar . .3 m 3 1.00 g / cm 3 2 2 2 g x h H 2O x 1. 2 The simple gas law • Boyle’s Law For a fixed amount of gas at a constant temperature.5 atm. and then it is connected to a 50. 1 or V P1V1 = P2 V2 Pα PV = a (cons tant ) Example: 5.3 The tank is first evacuated. .0 L x = 694 L P2 1.5 L is that of the cylinder. originally at 21. 50.0-L cylinder of compressed nitrogen gas. falls to 1. • Charle’s Law The volume of the fixed amount of gas at constant pressureis directly proportional to the Kelvin (absolute) temperature. What is the volume of the tank? Solution: P1V1 = P2 V2 V2 = V1 x P1 215 atm = 50. gas volume is inversely proportional to gas pressure. the volume of the gas is directly proportional to the amount of gas.5.15 STP condition: 0°C = 273. The gas pressure in the cylinder.15 K and 1 atm = 760 mmHg • Avogadro’s Law At a fixed temperature and pressure. The volume of the tank is 694 L – 50.55 atm after it is connected to the evacuated tank.0 L = 644 L. Vα T or V = bT (where b is constant) Relationship: T (K) = t (°C) + 273.5 atm Of this volume. 1636 L CO 2 1 mol 44g The Ideal gas equation From the gas law relationships.023 x 1023molecules Example: 5.193 mole 71g 1 atm 760 mmHg x 745 mmHg = 0.7 g = 0. Vα nT P and V=R nT P PV = nRT . 2. R = Ideal gas constant Example: 5.4 L at STP = 6.5 What is the volume occupied by 13.875 g 1 mol 22.Molar volume: 1 mol of gas = 22. • (?) g C3H8 = (?) L CO2 = 1 mol 42g x = 1.9803 atm T = 45°C + 273. Molar mass of Cl2 = 71 g/mol Solution: (?) n of Cl2 = P= 1mol x 13.318 K PV = nRT .7 g Cl2(g) at 45°C and 745 mmHg. 2.4 1.4 L 22. What is the volume of this gas at STP in Liters? Solution: 1. What is the mass of 1 L cyclopropane C3H6 at STP? 128 g pieces of solid CO2 (s) sublimates into CO2 (g) .4L 1 mol x x 128 g = 65. K x 318 K = P 0. when the O2 (g) is heated to 100°C. Released O2 (mole) = 0.4 L x 1L = 0.04464 mole At 100°C.0446 – 0.08205 L atm / mol.384 g O2 (g) O2 released = 1 mole 5.0446 mole x 273K 373 K = 0.0364 mole O2 (in flask) Thus.980263 atm = 5. V remains constant) T2 = 100°C + 273 = 373 K T1 = 0°C + 273 = 273 K.012 mole = 0. 32 g x 0. PV = nRT ∴ or R= P1V1 PV = 2 2 n 1T1 n 2 T2 PV nT n1T1 = n2 T2 (as P.03264 = 0. Suppose that.3 Applications of the ideal gas equation • Molar mass determination .V= nRT 0. and n2 = 0.6An oxygen gas is filled into 1L flask at STP. Example: 5. what mass of O2 must release from the flask? Solution: 1mole At STP. (?) mole O2 = 22. the pressure remains at 1 atm.1364 L.193 mole x 0.6 5. (?) mole O2.012 mole Molar mass of O2 = 32 g/mole. nxM n = xM V V m MP d= V = RT d= V= m Example: 5. that is n = m/M. substituting into the ideal gas equation.Since the number of moles of gas (n) is equal to the mass of gas (m) divided by the molar mass (M).8 Gases in chemical reactions What volume of N2(g) . d.987 atm? Solution: d= m MP 32.0 g NaN3 is decomposed? 2NaN3(g) ∆  → 2 Na(1) + 3 Na2(g) Solution: .987 atm = = 0. measured at 735 mmHg and 26C. PV = mRT M • Gas density To determine the density of gas.00 g / mol x 0.08206 Latm / mol.K x 298 K V RT = 1.29 g / L • Example: 5.7 What is the density of oxygen gas at 298°K and 0. is produced when 70. K T = 26OC + 273 = 299 K . in moles..08206 L atm/mol.1L Law of combining volume • Example:5.967 mmHg V=? n =1. 2ZnS(s) + 3 O2(g) ∆  → 2ZnO(s) + 2S02(g) = 0.967 atm.00 L O2(g) x • Mixture of gaes For the nonreactive gases.Example: 5.9 Zinc blend.62 mol R= 0.62 mol x 0.667 L S02 ? L S02(g) = 1. n = the total amount. of the gaseous mixture (n tot) . Roasting (strong heating) of ZnS is the first in the commercial production of zinc. is the most important zinc ore. ZnS.62 mol N2 P = 735 mmHg x 1atm/760 mmHg = 0.2NaN3(g) ? mol N2 = ∆  → 2 Na(1) + 3 Na2(g) = 1.10 .08206 Lmol-1K-1 x 299K 0.-::P = 1. = 41. 2ZnS(s) + 3 O2(g) ∆  → 2ZnO(s) + 2S02(g) What volume of SO(g) forms per liter of O2(g) consumed? Both gases are measured at 25 oC and 745 mmHg. 0 L • Daltn’s Law of partial pressures the partial pressures of the components of the mixture. nA /n tot .50 mol H2 + 1.4 atm 5.of the gas) The total pressure of a mixture of gases is the sum of • The mole fraction The mole fraction of a component.0L at 20°C? Solution: n. * Ptot = PA + PB + ----In case of the wet gas collected over water.75molx0.25 mol He = 1. *Ttot = Tbar = Pgas + H2O For the composition of gaseous mixtures in % by volume. n A PA VA = = n B PB VB Example: 5.0 g H2 x = 0.11 What are the partial pressure of Hz and He in the gaseous mixture described in example 5.(Vtot = total vol.75 mol mol gas P = nRT/V = 1.10? Solution: .00 g He when the mixture is confined to a volume of 5.tot = 1. in a mixture is the fraction of all molecules in the mixture contributed by that component. * Vtot = VA + VB + —.What is the pressure exerted by a mixture of 1.08206Lmol-1K-1x293K = 8.0 g H2 and 5. 4 atm and so.961 atm V = 81.12 Collecting a gas over a liquid (water).4 atm 1.961 atm x 0.2 ml of O2(g) is collected over water at 23°C and barometric pressure 7. In the following reaction 81.From example 5.51 mmHg. What must have been the mass of Ag2O(s) decomposed? (Vapor pressure of water at 23°C -= 21.75 mol Example: 5.49g Ag2O 1mol O 2 1 mol Ag 2 O .75 mol PHe = x Ptot = 1.2 ml = 0.0812 L -1 –1 0.50 mol x 8. = 751 mmHg – 21.1 mmHg.1 mmHg = 750 mmHg = (750/760) atm = 0. Ptot = 8.4 atm = 6 atm 1.10.) 2A2O(s) Solution: 4Ag(s) + O2(g) 1.0082 mol n = PV/RT = 0.08206 L atm mol K x 296K 2Ag2O(s) Ag2O(s) = 0.4 atm = 2.08206 Latm-1K-1 T = 23oC + 273 = 293 K n=? = 0.20 mol x 8.7g Ag 2 O x = 1.00321mol O2 x 4Ag(s) + O2(g) 2 mol Ag 2 O 231. PH = 2 x Ptot = 0.0812 L R = 0. 5.4 Kinecic molecular theory of gases 1. A gas is composed of a very large number of extremely small particles (molecules or. in some cases, atoms) m constant, random, straight-line motion. Great distances separate molecules of a gas. The gas is mostly empty space. (The molecules are treated as if they have mass but no volume, so called "point masses.") . Molecules colloid with one another and with the walls of their container. However, these collisions occur very rapidly and most of the time molecules are not colliding. There are assumed to be no forces between molecules except very briefly during collisions. That is, each molecule acts independently of all the others and is unaffectedly by their presence. Individual molecules may gain otiose energy as a result of collisions. In a collection of molecules at constant temperature, however, the total energy remains constant. This theory depends on the following facts. (i) The frequency of molecular collisions: The number of collision per second 2. 3. 4. 5. The higher this frequency, the greater the total force of these collisions. Collision frequency increases with the number of molecules per unit volume and with molecular speeds. (ii) The amount of translational kinetic energy: That is energy possessed by objects moving through space, represented as ek and has the value eK = 1/2 mu2 m = the mass of the molecule and u = speed of the molecules The faster molecules move, the greater are their translational kinetic energies and the greater is the force of their collisions. (iii) Molecules move in all direction and at different speeds: When all factors are properly considered, the result is this expression for the pressure of gas. The basic equation of the kinetic-molecular theory. P= 1N mu 2 3V N = the number of molecules present in the Volume V. m = mass of the molecules u2 = the average of the .squares of their speeds • Distribution of Molecular speed Root-mean-square speed, ursm thee square root of the average of the squares of the speeds of all the molecules in a sample. Urms = 3RT M (or) R = Gas constant, 8.3145 J mol-1 K-1 2 -2 -1 8.3145 Kg m s mol K-1 Example: 5.13 Which is The greater speed, that of a bullet from a high powered M-16 rifle (21~0 mi/h) or the root-mean-square speed of Hs molecules at 25C? Solution: 3RT M R = Gas constant, 8.3145 J mol-1 K-1 MH2 = 2.016 x 10-3' kg mol-1 Urms = Urms = 3 x 8.3145k gm 2 s -2 mol-1 K -1 x 298 K 2.016x10-3 Kg mol-l = 4.29 x 103 mi/h Comparing the two speeds, the root-mean-square speed of H2 molecules at 25C is greater than the speed of the high-powered rifle bullet. 5.5 Gas properties relating to the kinetic-molecular theory • . • • Diffusion: is the migration of molecules of different substances as a result of random molecular motion. Effusion: is the escape of gas molecules from their container through a tiny orifice or pinhole. The rates at which 'diffusion and effusion occur are directly proportional to molecular speeds. That is, molecules with high speeds diffuse and effuse faster than molecules with low speeds. Grahm 'a law • The rates of effusion (or diffusion) of two different gases are inversely proportional to the square root of their molar masses. The result shown in the following equation is a kinetic-theorv statement of Graham's law. 3RT/M A MA rate of efusion of A (u rms ) A = = = rate of effusion of B (u rms ) B 3RT/M B MB A ratio of effusion-rates, times, distances, .... is equal to the square root of a. ratio of molar masses. 1. molecular speeds 2. rates of effusion Ratio of 3. effusion times = ratio of two molar masses 4 .distance traveled by molecules 5. amount of gas effused Example: 5.14 A sample of Kr(g) escape through a tiny hole in 87.3 s, and unknown gas requires 42.9 s under identical conditions. What is the molar mass of the unknown gas? Solution: Again start with qualitative reasoning. Because the unknown gas effuses faster, it. must have a smaller molar mass than Kr. The ratio of molar masses in the setup below must be smaller than 1: Munk goes in the numerator. efusion time for unknown 42.9 s M unk = = = 0.491 efusion time for Kr 87.3 s M Kr Munk = (0.491)2 x MKr = (0.491)2 x 83.80 = 20.2 g/mol 5.5 Nonideal (real) gases Figure 5.1 The behavior of real gas compressibility factor as a function of pressure for three different gases at OoC. All gases behave ideally at sufficiently low pressures.2) b = the excluded volume per mole. the forces of the collisions of gas molecules with. The PV product is larger than predicted for an ideal gas and the compressibility factor is greater than 1. o Gases tend to behave nonideally at low temperatures and high pressures. where molecular motion slows down. but that deviations set in at increased pressures. Here is how we might explain nonideal gas behavior: Boyle's law predicts that at very high pressures a gas volume becomes extremely small and approaches zero. This cannot be. These forces become increasingly important at low temperatures.2 suggests that because of attractive forces among the molecules.2 Intermolecular attraction. forces of Figure 5. below I atm. . • The van der Waals equation [ P + (n2 a)/V2 ] (V – n b) = n RT V = the volume of n moles of gas. (n2a) /V2 is related to intermolecular . Intermolecular forces of attraction account for compressibility factors less than 1. the container walls is less than expected for an ideal gas. To summarize o Gases tend to behave ideally at high temperatures and low pressures. however. It is added to P because the measured pressure is lower than expected (recall Figure5. related to the volume of the gas molecules. because the molecules themselves occupy space and are practically incompressible. At very high pressures the compressibility factor is always greater than 1. Figure 5. We must also allow for the fact that mterrnolecular force exists in gases.forces of attraction. say. 49 L2 atm mol-2 b = 0.49 L2 atm n b = 1 mol x 0.0562)L 22 xL2 P = 11. varying with temperature and pressure. Example: 5. Solusion: Solve van der Waal"s equation for P.(V – n b) = free volume within the gas 'a'.9 atm .00 L at 273 K.08206 L atm mol-1K-1 n2a = (I mol)2 x 6. P= n RT n 2a − 2 V-nb V n = I mol V = 2 mol T = 273 K R = 0. "b" = specific value for particular gas.08206 L atm mol-1K -1 6.0562 L mol-1 = 0.62 atm = 9. (a = 6.49 L2 atm − (2 − 0.49 L2 atm mol-1 = 6.5 atm -1.13 Use the van der Waal's equation to calculate the pressure exerted by 1.0562 L mol--1).00 mol C12(g) when it is confined to a volume of 2.0562 L mol-1 P = Imol x 0. γ . such as within a drop of liquid. If cohesive forces are strong compared to adhesive forces. If adhesive forces are enough. is the energy or work required to increase the surface area of a liquid.CHAPTER 6 6. the interface between the water and the air above it. If the liquid in the tube is water. consisting of metallic bonds between Hg atoms. called a meniscus. is curved upward (concave). Cohesive forces are the intermolecular forces between like molecules. on the other hand. are strong. As the temperature—and hence the intensity of molecular motion— increases. When a drop of liquid spreads into a film across a surface. Water is drawn slightly up the walls by adhesive forces between water and the glass. The preoperties of lquids and solids 6. a drop maintains its shape. meaning that surface tension decreases with the increased temperature. Cohesive forces in mercury. Adheive forces are the intermolecular forces between unlike molecules. With liquid mercury the meniscus is curved downward (convex).1 Intermolecular forces and some properties of liquids ∗ Surface tension Surface tension. It has the units of energy per unit area. typically J/m2. molecules of a liquid and of the surface with which it is in contact. Whether a drop of liquid wets a surface or retains its spherical shape and stand on the surface depends on the strengths of two types of intermolecular forces. mercury does not wet glass. the energy requirement for spreading the drop into a film is met through the work done by the collapsing drop. . Less work is required to extend the surface of a liquid. intermolecular forces become less effective. we say that the liquid wets the surface. Liquid such as honey and heavy motor oil flow much more sluggishly.The effect of meniscus formation is greatly magnified in tubes of small diameter. ∆H vaporization = H vapor − H liquid Vaporization is an endothermic process. the term evaporation also used. Viscosity Viscosity refers to a liquid’s resistance to flow. They flow easily. ∆H vap is always positive. the greater is the viscosity. When a liquid flows. In this text enthalpies of vaporization are expressed in terms of ‘one mole of liquid vaporized. We say that they are viscous. Its magnitude depends on intermolecular forces of attraction and. in some cases. Vaporization of liquid The passage of molecules from the surface of a liquid into the gaseous or vapor state is called vaporization. • Enthalpy of vaporization: The enthalpy of vaporization is the quantity of heat that must be absorbed if a certain quantity of liquid is vaporized at a constant temperature. In the capillary action the water level inside the capillary is noticeably higher than outside. . Cohesive forces within the liquid create an ‘internal friction’ which reduce the rate of flow. The effect is weak in liquid of low viscosity such as ethyl alcohol and water. The stronger the intermolecular forces of attraction are. The tendency for a liquid to evaporate increases with increased temperature and decreased strength of intermolecular forces. called capillary tubes. on molecular sizes and shapes. one portion of the liquid moves with respect to neighboring portions. The weaker the intermolecular forces. ∆H condensatioon is always negative. 933 J of heat is required. that is the reverse of vaporization. What is the enthalpy of vaporization of acetone in kJ/mol? Solution: 1 kJ 1000 J = 1mol (CH 3 ) 2 CO 1. the more volatile the liquid (the higher its vapor pressure). Vapor pressure increases with temperature. • Vapor pressure: The pressure exerted by a vapor in dynamic equilibrium with its liquid is called vapor pressure.1 To vaporize 1.75 g (CH 3 ) 2 CO x 58. (CH3)2CO. and boiling is said to occur. ∆H condensatioon = H liquid − H vapor = − ∆H vap Condensation is an exothermic process.0 kJ/ mol (CH3)2CO • Condensation: Condensation is the passage of molecules from the gaseous state to the liquid state.08 g (CH 3 ) 2 CO 933 J x ∆H vapor = 31.75 g of acetone. In turn. . the conversion of a gas or vapor to a liquid is called condensation. Liquid with high vapor pressures are said to be volatile. • Boiling and the boiling point: The pressure exerted by escaping molecules equals that exerted by molecules of the atmosphere.Example: 6. and those with very low vapor pressures are nonvolatile. at 298 K. 0. The density of the liquid decreases.132 g H2O is produced and maintained at a temperature of 50. added pressure and lowering of temperature to a value below Tc are required.Tc. It is the highest temperature point on the vapor-pressure curve. Pc . this liquefaction can be accomplished just by applying sufficient pressure. that of the vapor increases. Tc and the pressure is the critical pressure. A gas can be liquefied only at temperatures below its critical temperature. If room temperature is below Tc. If room temperature is above Tc . The critical point refers to the temperature and pressure where a liquid and its vapor become identical. Will the water be present as vapor only or as liquid and vapor in equilibrium? Solution: . 2. The critical point represents the highest temperature at which the liquid can exist. the followings are observed. Example: 6.2 As a result of a chemical reaction. • The critical point: If a liquid in a glass tube is heated.0°C in a closed flask of 525 ml volume. and eventually the two densities become equal. The temperature at the critical point is the critical temperature. The surface tension of the liquid approaches zero.• Normal boiling point: The pressure at which the vapor pressure of a liquid is equal to standard atmospheric pressure (1 atm = 760 mmHg) is the normal boiling point. 1. The meniscus between the liquid and vapor becomes less distinct and eventually disappears. 02gH 2 Ox0.08206Latm mol K x 323.5 5.132g H 2 Ox P= P = 0.8 3.6 2. The water is present as liquid and vapor in equilibrium at 92.5 4.2K 0. some of the vapor must condense to liquid.5 mmHg).0 4. calculate the pressure that would exist if the water were present as vapor only.0 Water H2 O benzene C 6H 6 2.8 .2 3.4 2.0 6.First.6 3.5 6. P= nRT V −1 −1 18.370atm x The calculated pressure (281 mmHg) exceeds the vapor pressure (92.0 1/T x 103 3.525L 760 mmHg 1 atm = 281mmHg 1mol H 2 O 0.4 3.0 aniline C6H7N diethyl ether C4H10O lnP 5.0 2.2 2.5 mmHg. ∗ Clausis-Clapeyron equation: 7. 99) is marked by the arrow ( ). 1/T = 1/333 = 3.3 mmHg. P2 = 55. we need to have values for the two constants. For benzene at 60°C.0°C with the following data.1). To use equation (6. T2 = 40. The point corresponding to (3.2 K. ∆H vap = 44.2  = 5. 1.0. ‘A’ is related to the enthalpy of vaporization of the liquid: A = ∆H vap /R and it is customary to eliminate B by rewriting (6. Substitute the given values into the equation.2   313. T1 = 308.1) in a form called Clausius—Clapeyron equation. 1/T.0°C.00.Figure 6.00 x 10-3 .29 x 103 (0. ln P = ln 400 = 5. A and B.003193) .00.99.3 Calculate the vapor of water at 35. Example: 6. ln P2 ∆H vap  1 1  − = P1 R  T1 T2      3. 1/T x 103 = 3. which express the relationship of a straight line plotted for the liquids.0 x 10 3 J mol −1  1   1  −1 ln =  − K P1 8.3 mmHg 44. 55. 1 ln P = − A   + B T (6. the vapor pressure is 400 mmHg.2 Vapor pressure data plotted as in lnP vs. T = 60 + 273 = 333 K. 5.1) 2. A particular common form of a vapor-pressure equation is that shown below.0 kJ/mol. Solution: Let P1 be the unknown vapor pressure at 35.3145 J mol −1 K −1  308.0°C.003245 . The reverse process of sublimation.28 = 1. solids can also give off vapors. solids are generally not as volatile as liquids at a given temperature. Eventually a temperature is reached at which these vibrations disrupt the ordered crystalline structure.= 0. The reverse process of melting point. Melting is the transition of a solid to a liquid and occurs at the melting point.3 mmHg/1. the conversation of a liquid to a solid. The temperature at which the process of melting occurs is the melting point. or molecules vibrate more vigorously. The direct passage of molecules from the solid to the vapor state is called sublimation.28 Next. This process is called melting or fusion.32 = 41. although because of the stronger intermolecular forces present. The melting point of a solid and the freezing point of its liquid are identical. is called deposition. its atoms. At this temperature solid and liquid coexist in equilibrium. 55. atoms.32 P1 P1 = 55. is called freezing point. ions. or molecules can slip past one another. ∗ Sublimation Like liquids. Thus. . the passage of molecules from the vapor to the solid state.9 mmHg 6. the solid loses its definite shape and is converted to a liquid. determine that e0.2 Intermolecular forces and some properties of solids ∗ Melting point and heat of fusion As a crystalline solid is heated.32.3 mmHg = 1. Dispersion (London) forces are intermolecular forces associated with instantaneous and induced dipoles. The enthalpy of sublimation is the quantity of heat needed to convert a solid to vapor. δ+ δ− δ+ δ− δ+ δ− (a) (b) (c) . An induced dipole is an atom or molecule in which a separation of charge is produced by a neighboring dipole. The van der Waal’s equation is an equation of state for nonideal gases. collectively. These forces are the ones account for in the vander Waal’s equation for nonideal gases. a dynamic equilibrium exists between a solid and its vapor. It includes correction terms to account for intermolecular forces of attraction and for the volume occupied by the gas molecules themselves.When sublimation and deposition occur at equal rate.01 kJ/mol ∆H vap= 44. intermolecular forces of the London type and interactions between permanent dipoles. ∗ Instantaeous and induced dipoles An instantaneous dipole is an atom or molecule with a separation of charge produced by a momentary displacement of electrons from their normal distribution. H2O(s) H2O(l) H2O(s) H2O(l) H2O(g) H2O(g) ∆H fus = 6.3 The van der Waal’s forces van der Waal’s forces is a term used to describe.93 kJ/mol ∆H sub = ∆H vap + ∆H fus 6. The vapor exert a characteristic pressure called the sublimation pressure.92 kJ/mol ∆H sub = 55. Because dispersion forces become stronger as polarizability increases. These electrons are more easily displaced and the polarizability of the molecule increases. [ ]. δ+ δ− Induced dipole: The instantaneous dipole on the left induces a charge separation in the molecule on the right. (2) The strength of dispersion forces also depends on molecular shape. that is.Figure 6. mass (ii) α molecular shape Polarizability increases with increased number of electrons. The relations of the polarizability to the molecular mass and molecular shapes are addressed by the followings. melting point and boiling point of covalent substances generally increase with increasing molecular mass. are less firmly. • • Normal condition: A nonpolar molecule has a symmetrical charge distribution. the ease with which a dipole can be induced. H H H H C H H C H H C C H H C H H H H C H H C H H H C H C H H C H H . being farther from atomic nuclei. and the number of electrons increases with increased molecular mass.α mol. Intermolecular forces are stronger in pentane than in neopentane. For instant. of e-. Instantaneous condition: A displacement of the electronic charge produces an instantaneous dipole with a charge separation represented as [ ]. the elongated pentane molecule is more easily polarized than the compact neopentane molecule.3 Instantaneous and induced dipoles. The result is a dipole—dipole attraction. In large molecules some electrons. ∗ Mollecular shapes and Polarizability Polarizability describes the ease with which the electron cloud in an atom or molecule can be distorted in an electric field. Thus boiling point of Pentane is higher than that of neopentane. Polarizability: (i) α no. Dispersion (London) forces involve displacement of all the electrons in molecules. on the other hand. .∗ (a) Neopentane Bpt = 9. They are found only in substances with resultant dipole moments.mass 30 u Bpt 121. 2. 1. there is an electronegativity difference and the molecule has a slight dipole moment. N2 µ = 0 (nonpolar) Mol. Their existence adds to the effect of dispersion forces also present.39 K O2 µ=0 (nonpolar) mol. Hence. This additional ordering of molecules can cause a substance to persist as a solid or liquid at temperature higher than otherwise expected. consider the following. Consider N2. The strengths of these forces increase with increased molecular mass and also depend on molecular shapes. In NO. There are no electronegativity differences in N2 and O2 and both substances are nonpolar. Dispersion forces exist between all molecules.19 K Summery of van der Waal’s forces When assessing the important of van der Waal’s forces.mass 32 u Bpt 90. and NO.1°C Dipole-dipole attraction In a polar substance. NO has the highest boiling point of the three as shown below.153 D (polar) mol. molecules try to line up with the positive end of one dipole directed toward the negative ends of neighboring dipoles. Forces associated with permanent dipoles involve displacement of electron pairs in bonds rather than in molecules as a whole.34 K ∗ NO µ =0. O2.mass 28 u Bpt 77.5°C (b) Pentane Bpt = 36. or the organic solvent acetone (CH3)2CO? Solution: They have the same molecular mass (58u). and enthalpy of vaporization. HI as follows. C4H10. HBr. HBr.4 Which would you expect to have the higher boiling. It may be described how these statements relate to the data including a rough breakdown of van der Waal’s forces into dispersion forces and those due to dipoles for the selected substance: F2. Thus. Example: 6. (ii) Within the series HCl. but because HCl is polar it has significant larger ∆H vap and a higher boiling point than does F2. and HI. and ∆H vap and boiling points increase in the order: HCl < HBr < HI. Hence. ∆H vap and boiling points of the all substances are in increasing order of F2. the hydrocarbon fuel butane.3. The electronegativity difference between C and H is so small that H/C is generally nonpolar. HCl. dipole moment can produce significant differences in such properties as melting point. Acetone has a resultant dipole moment and is a polar molecule. When comparing substances of widely different molecular masses. HBr. boiling point. dispersion forces are usually more significant than dipole forces. 2. The acetone molecule has strong C –to.O bond moment . 4. HI. HCl. So we base our prediction on some other factors: 1. Practice example: . When comparing substances of roughly comparable molecular masses. (i) HCl and F2 have comparable molecular mases. the polar substance acetone has the higher boiling point. molecular mass increases sharply. At the melting point. This leaves the proton unshielded. When liquid water is heated above the melting point. only a fraction of the hydrogen bonds are broken. The molecules become even more closely packed. Hydrogen O O H bonding atom can occur only with H atoms because H all other atoms have inner-shell electrons to shield H H O H H their nuclei. Figure mentioned above shows how one water molecule is held to four neighbors in a tetrahedral arrangement by hydrogen bonds. water molecules are packed more tightly in liquid water than in ice. Cl2. a proton. One indication of this is the relatively low heat of fusion of ice (6.01 kJ/mol). and the density of the liquid water increases. with energies of the order of 15 to 40 kJ/mol. hydrogen bonds hold the water molecules in a rigid but rather open structure. As ice melts. In hydrogen bond formation. It is much less than we would expect if all the hydrogen bonds were to break during melting. In ice.4 Hydrogen bonding A hydrogen bond is formed when an H atom bonded to one highly electronegative atom—F. or N—is simultaneously attracted to a small highly electronegative atom of a neighboring molecule. H Ordinary water is certainly the most common O H H substance in which hydrogen bonding occurs. ClNO. Liquid water attains its maximum density at 3. hydrogen bonds continue to break.Arrange the following substances in the order in which you would expect their boiling points to increase: CCl4. This is why the liquid is more less than ice.98 C. O. 6. the highly electronegative atom to which an H atom is covalently bonded pulls an electron pair away from the H nucleus. Above this temperature the water behaves in a “normal” fashion: its density decreases with temperature. hydrogen bonding is unique O to certain hydrogen-containing compounds. N2. and it is attracted to a lone pair of electrons on a highly electronegative H atom of a neighboring molecule. Hydrogen bonds are rather strong intermolecular forces. Thus. . The p orbital is perpendicular to the plane. In these cases the entire crystal is held together by strong forces. Both diamond and silicon carbide (silicon atoms are substituted for onehalf the carbon atoms) are nonconductors of electricity and do not melt until very high temperatures are reached. atom is bonded to tetrahedral fashion. the entire crystal called a unit cell. The three sp2orbitals are directed in a plane at angles of 120°. Diamond: Each atom is bonded to four others. (a) A potion of Lewis (b) Crystal structure.4 (a)]. . a nonplanar hexagonal arrangement of carbon atoms is also seen [shown in Figure 6. C6H6. The best examples are described as follows. These orbitals are the same ones used for carbon atoms in benzene. When viewed from a particular direction. Figure 6. Each carbon four others in a The segment of shown here is (b) Graphite: In graphite the bonding involves the orbital set sp2+p.5 Chemical Bonds As Intermolecular Forces ∗ Network Covalent Solids Covalent bonds extend throughout a crystalline solid. directed above and below it. and diamond melts above 3500C.4 The diamond structure. the more dense water sinks to the bottom of the lake and the colder surface water freezes. The type of hybridization scheme that corresponds to four bonds directed from a central atom to the corners of a tetrahedron is sp3. When the water temperature falls below 4°C. The ice at the top of the lake then tends to insulate the water below the ice from further heat loss.This behavior explains why a freshwater lake freezes from the top down. (a) structure. SiC sublimes at 2700°C. 6. Ionic solid do not melt at ordinary temperature. Each carbon atom forms strong covalent bonds 142 pm with three neighboring carbon atoms in the same plane. but between layers it 335 pm is much weaker. Lattice energy is the quantity of energy released in the formation of one mole of a crystalline ionic solid from its separated gaseous ions. they migrate through the planes of carbon atoms when an electric field is applied. Diamond is not an electrical conductor because all of its valence electrons are localized or permanently fixed into single covalent bonds. ∗ Interionic Forces To predict properties of an ionic solid we face the lattice energy of the crystal. Figure 6. We can melt ionic solid by supplying enough thermal energy to disrupt the crystalline lattice. the following generalization works well.5. Bonding within layers is strong. The p electrons of the carbon atoms are delocalized. Because of weak bonding between layers. The higher the lattice energy is. which gives rise to layers of carbon atoms in a hexagonal arrangement. Because of the delocalized p electrons. . depending to some extent on the type of crystal structure. graphite conducts electricity.5 The graphite structure. either in dry form or suspended in oil. The C—C bond distance within a layer is 142 pm (compared to 139 pm in benzene).This type of bonding produces the crystal structure shown in Figure 6. between layer it is 335 pm. the layers can glide over one another rather easily. the higher is the melting point. As a result graphite is a good lubricant. The attractive force between a pair of oppositely charged ions increases with increased charged on the ions and with decreased ionic sizes. KI or CaO? Solution: . This idea is illustrated through Figure 6. the greater is the quantity of an ionic solid that can be dissolved in a given quantity of solvent..= 181 pm Mg2+ = 65 pm O2.6. the energy required to break up an ionic crystal results from the interaction of ions in the crystal with molecules of the solvent (e.is about seven times as great as below Na+ and Cl-.To dissolve an ionic compound. water molecules). the interionic attractive force between Mg2+ and O2. Which has the higher melting point.5 Predicting Physical Properties of Ionic Compounds.g. Example: 6.6 Interionic forces of attraction. Because of the higher charges on the ions and the closer proximity of their centers. Relative attractive force: Na+ Mg2+ ClO2- Radius: Na+ = 95 pm Cl.= 140 pm Radius sum = distance between center of ions: 276 pm 205 pm Figure 6. the lower the lattice energy is. In general. Unit cell is a small collection of atoms. In the simple cubic cell. Figure 6.centered cubic (bcc) crystal structure is one in which the unit cell has structure units at each corner and one in the center of the cube. three sets of parallel planes called a lattice. Also Ca2+ is smaller than K+ and O2. The space--filling models in the bottom row show contacts between spheres (atoms). (a) Simple cubic (b) (c) Body-centered cubic. ions.are more highly charged than K+ and I-. 6. thus CaO should have the higher melting point. A body. . A face-centered cubic (fcc) crystal structure is one in which the unit cell has structural units at the eight corners and in the center of each face of the unit cell.is smaller than I. This is called the cubic lattice.The lattice point of CaO is much larger than KI.Ca2+ and O2. molecules from which an entire crystal structure can be inferred. The planes are equidistant an mutually perpendicular (intersect at 90° angles).6 Unit cells in the cubic crystal system. Face-centered cubid.5 Crystal structures ∗ Crystal Lattice To describe the structures of crystal. we have to work with threedimensional patterns. In the top row only the centers of spheres (atoms) are shown at their respective positions in the unit cells. It is derived from the cubic closest packed arrangement of spheres. ∗ Closest packed structures A octahedral hole ral B hed B tra er e A er t lay cov les in ho hexagonal closest packed cov er hole octahe A dr s in laye al C rB B A tetrahedral hole cubic closest packed Figure 6.spheres come into contact along each edge. contact is along the diagonal of a face. In the faced-centered cubic (fcc) cell. The crystal on this type of packing is hcp. contact of the sphere is along the cube diagonal.7 Closest packed structures. is one of the two packed to space or voids structure based referred to as In the Figure a unit cell is highlighted in . In the body—centered cubic (bcc) cell. Hexagonal closest packed ways in which spheres can be minimize the amount of free among them. The corner atoms are shared among 8 adjoining unit cells. Zn Al. Ag Fe. Ti. The total number of atoms in a bcc unit cell is two [1 + (8 x 1 )]. Only the center atom belongs entirely to the bcc unit cell. Thus . Pb.1 Some features of close-packed structures in metals. the eight corner atoms collectively contribute the equivalent of one to the unit cell.atoms per unit cell 2 4 2 Examples Cd. The unit cell is not a tube. The atoms that are part of the cell are in solid color. W ∗ Crystal coordination number (CCN) and number of atoms per unit cell The crystal coordination number signifies the number of nearest neighboring atoms (or ions of opposite charge) to any given atom (or ion) in a crystal. Mg. Three adjoining unit cells are also shown.heavy black. Na. Crystal structure hcp fcc bcc Crystal coordination number 12 12 8 No. The united cell and broken-line regions together show the layering (ABC). 8 . Cu. Only one eighth of each corner atom should be thought of as belonging to a given unit cell. Cubic closest packed is one of the two ways in which spheres can be packed to minimize the amount of free space. Table 6. For the bcc structure: this is 8. K. For the hcp unit cell: CCN is 12. We also get 2 atoms per unit cell. X-rays interact with electrons in the atoms and the original beam is reradiated. diffracted (scattered) in all directions. The fcc unit cell contains four atoms. ∗ X-rays diffraction Cry Lead shield X-ray tube Photographic film Figure 6. The corner atoms account for ( 1 x 8) = 8 1 1 atom and those in the center of the faces for ( 2 x 6) = 3 atoms. When a beam of X-rays encounters atoms. detector d sinθ planes of atoms or ions .8 Diffraction of X-rays by a crystal. The scattered X-rays produce a visible pattern on a film. In the fcc unit cell: CNN is 12. The corner atoms account for ( 1 x 8) = 1 atom and the central atom 8 belongs entirely to the unit cell. Two rays in the monochromatic X-ray beam.or l2 + 2 (l)2 = 3 (l)2. diffracted to the parallel planes of atoms in a crystal. Example: 6. What is the the radius of an iron atom? l = 287 pm Figure 6. The additional distance is 2d sin θ. That is (l)2 + (l 2 )2 = ( )2.6 Using X-ray data to determine an atomic radius. At room temperature ion crystallizes in a bcc structure. l 3 Solution: 4r Pythagorean . labeled (a) and(b). (d = spacing between atomic planes).9 Determination of the atomic radius of iron.Figure 6. to Figure 6. o o o Wave (a) is reflected by one plane of atoms or ions in a crystal and wave (b) from the next plane below. Wave (b) travels a greater distance than wave (a). nλ = 2d sin θ.9 Determination of crystal structure by X-ray diffraction. that additional distance travel by wave (b) equal an integral (n) multiple of the wavelength of the X-rays (λ).9 is found to be 287 pm. The right triangle must conform to the formula: a2 + b2 = c2. one of eight is located at each of the eight corners of the cube and one at the center. Solution: 1 = 287 pm = 287 x 10-10 cm.8 Use data from Example 6. V = I3 = (2.6.85g 1 mol 2 atom 1 unit cell = x x 23 1mol 6. What is The volume of a unit cell? Solution: For the fee structure.7 Al crystallizes in a fcc structure. calculate the densitv of iron.1 pm.3639 x 10-23 cm3/ unit cell d= 55. 4r = l 2 r= 2 /4 x 143. The diagonal of a cube =l 3 The length of an edge.86 g/cm3 • Ionic crystal structure . l = 287 pm 4r = l 3 r= 3 x 287pm = 124 pm 4 Example: 6.8 x 10-10)3 cm3 = 2.628 x 107 (pm)3 Example: 6.Nine atoms are associated with a bcc unit cell. The 3 atoms along a cube diagonal are in contact.3639 x 10-23 cm3 = 7.1 pm V = r3 = 6. Given that the atomic radius of Al is 143.02 x 10 atms 1 unit cell 2. Total number of Cl. for Cs' and Ct are eight). 2. indicate the crystal coordination numbers of the ions. the total number of Na+ ions in the Unit cell is (12 x 1/4) (1 x 1) = 3 + 1 The unit cell has the equivalent of Na+ and Cl. by Translation in three dimensions generates the entire crystal. . The Na+ ion in the very center of the unit cell belongs entirely to that cell. (By contrast. There are 12 Na+ ions along the edges of the unit cell.jn the center of a face is shared by two unit cells. 3. corresponding to the formula NaCI.ion in a unit cell is (8 x 1/2) + (6 x 1/2) = 1 + 3 = 4.In defining a unit cell of an ionic crystal we must choose a unit cell that 1. the crystal coordination number of both ions are six.The relative size of cations and anions are important in establishing a particular packing arrangement. The ratio of the two is 4:4 = 1:1. and each edge is shared by tour unit cells. . Each Cr ion in a corner position is shared by eight unit cells. We must apportion the 27 ions in each unit cell. is consistent with the formula of the compound. Note: • • • • The Na ion in the center of 'unit cell is surrounded by six Cl. and each Cl.ions.ions. To establish the crystal coordination number in an ionic crystal count the number of nearest neighbor ions of opposite charge to any given ion in the crystal. 9 Relating ionic radius and the dimension of a unit cell of an ionic crystal. . The edge length = (rci) + ( rNa+) + ( rNa+) + (rci) = 2 ( rNa+) + 2 (rci) = [(2 x 95) + (2 x 181)] .= 552 pm 6.Figure 6.6 Lquid crystals Liquid crystal are a form of 'matter with some of the properties of a liquid and some of a crystalline solid. • • • Liquid crystals have the fluid properties of liquids and the optical properties of solids.10 (a)] two Cl. (a)The NaCI unit cell.ions are in contact with. (b)The CsCI unit cell Example: 6. Liquid crystals are observed more commonly in organic compounds that have cylindrically shaped (rod-like) molecules with masses of 200 to 500 u and lengths four to eight times their diameters. The ionic radius of Na+ and Cl. one Na+. Hence. Liquid crystals occur widely in living matter.in NaCI are 95 and 181 pm respectively. What is the length of the unit cell of Na Cl? Solution: along each edge of two unit cell [Figure 6.10. Liquid crystalline properties' have also been identified in various synthetic polymers. They are free to move in all directions but they can rotate only on their long axes. • Three forms of liquid crystals described as follows. Hardening of the arteries is caused by the deposition of liquid crystalline compounds of cholesterol. with the long axes of the molecules perpendicular to the plane of the layers. The molecular orientation in each layer is different from that in the layer immediately above and below it. . The cholesteric form is related to the nematic form.cell membranes and certain tissues have structures that can be described as liquid crystalline. In the nematic (meaning threadlike) form of the liquid crystalline state. The molecular motions possible here are translation within but not between layers and rotation about the long axis. the rodlike molecules 'are arranged in a parallel fashion. In the smectic (meaning greaselike) form. but molecules are stratified into: layers. rodlike molecules are arranged in layers. Na + Cl Na + Cl - Example: 7. this sharing of electrons is called a covalent bonds. a. MgCl2 Solution: a. 7. Ba + O Na + O - b.CHAPTER 7 7. H + Cl H Cl .1 Write Lewis structures for the following compounds. which attract each other through electrostatic forces called ionic bonds.2 Covalent bond Two or more pairs of electrons are shared between atoms. Basic concepts of Chemical Bonding 7. BaO b.1 Ionic bond Electrons are transferred from one atom to another to form positive and negative ions. double (B. the more tightly the atoms are held together. Hydrazine Dinitrogen difluoride Nitrogen H2N NH2 FN NF N N 145 pm 123 pm 109.O=1). Double bond between atoms is shorter than a single bond and a triple bond is the shortest.O=3). Multiple covalent bonds N N O O Bond order The term bond order describes whether a covalent is single (B. one with eight outer-shell electrons or an octet. triple (B. .Electrons are transferred or shared in a way that each atom acquires an especially stable electron configuration. The higher the bond order. the more electrons present.O=2).8 pm • Covalent radius The single bond radius is one-half the distance between the centers of identical atoms joined by a single bond. that is. Usually this is a noble-gas configuration. • Bond length sBond length is the distance between the centers of two atoms joined by a covalent bond. 2. H O Bond has the greater ionic character than H—Cl bond due to its greater ∆ EN value. . H H H polar Cl Nonpolar Electronegativity: EN The more metallic element has lower its EN. we expect the H—O bond to be the more polar bond. ENCl = 3. ∆ EN = 3.2.2 .2. and join the atoms in this structure by single covalent bonds. Which bond is more polar.2 = 1.4. H Cl or H O? b. 7.2 = 1.0 For the H—O bond. The more nonmetallic element has higher its EN.• Polar covalent bond A covalent bond in which electrons are not shared equally between two atoms is called polar covalent bond. For the H Cl bond. Example: 7.4 . ENH = 2. The strategy for writing a variety of Lewis structures may be described as follows. ∆ EN = 3. b. ii. Which bond has the greater ionic character? Solution: a. a.3 Writing Lewis structures i.2 Assessing electronegativity differences and polarity of bonds. ENO = 3. Because its ∆ EN is somewhat greater.2. Write the skeleton structure.2. Determine the total umber of valence electrons in the structure. With the valence electrons remaining. iv. Negative formal charges usually appear on the most electronegative atoms and positive formal charges on the least electronegative atoms. complete the octets of the central atom (s). v. so 6 v. we strive for a structure with no formal charges.iii. subtract two from the total number of valence electrons.3: Write the Lewis structure of HCN. Valence electrons 1 (no. if the central atom (s) lack an octet. b. H C N 4 v. first complete the octet of the terminal atoms.e-s = H (1) + C (4) + N (5) = 10 e-s. H C N (Incorrect) H C N H C N o • Formal charge: FC Formal charges are apparent charges associated with some atoms in a Lewis structure. generally achieving this object by noting that a. Bonding electrons) 2 – No.e-s remaining for the octet of N atom. we seek to keep formal charges to a minimum. FC = no. At this point. of v. but failing this. . to the extent possible. Lone pair electrons In writing Lewis structure. For each single bond thus formed. Example: 7. from multiple covalent bonds by converting lone-pair electrons from terminal atoms into bonding pairs.e-s used for 2 single bonds. Where formal charges are required. Then. these should be as small as possible. Solution: o o o Total no. NClO or ONCl A. SCS or CSS C. H H N O H (i) no formal charge H +1 -1 H N O H (ii) The more likely skeleton structure is (I). H2NOH or H2ONH Solution: B.4 Resonance The situation in which two or more plausible Lewis structure can be written but the “correct” can not be written at all is called resonance. B. A.4 Assign formal charges to the atoms in the following species and then select the more likely skeleton structure. The sum of the formal charges of the atoms in a Lewis structure must equal zero for a neutral molecule and must equal the ionic charge for a polyatomic ion.c. 7. C S (i) no formal charge S -2 +2 C S S (ii) The more likely skeleton structure is (I). -2 +2 N Cl O (i) O N Cl (ii) no formal charge The more likely skeleton structure is (I). . These rules on formal charges also lead to the result that the central atom in a structure is generally the atom with the lowest electronegativity. Example: 7. C. For NO molecule. Highly reactive molecular fragments with one or more unpaired electrons are called free radicals. for instance. (-1) (+1) (+1) (-1) (-1) F B F F F B F F F B F F F B F F (+1) The contributing structures with B to F double bonds have formal charges of –1 on B and +1 on F. Characteristic of BF3 . N O The present of unpaired electrons causes odd-electron species to be paramagnetic. this electron is put on the N atom to obtain a structure free of formal charges. the Lewis structure of the two free radicals are H H C H O H • Incomplete octet The representation of BF3 is a resonance hybrid based on four contributing structures. including one with an incomplete octet. Molecule with an even number of electrons are expected to have all electrons paired and to be diamagnetic. structure is its strong tendency to form a .5 Exceptions to the octet rule • Odd electron species If the number of valence electrons in Lewis structure is odd there must be an unpaired electron somewhere in the structure. contrary to the rules The Lewis structure with the incomplete octet has no formal charges. O S S O O O Resonance structure S O O Hybrid structure or true structure 7.The true structure is a resonance hybrid of plausible contributing structures. .coordinate covalent bond with a species capable of donating an electron pair to the B atom. Electron pairs assume orientations about an atom to minimize repulsions. whether they are in chemical bonds (bonding pair) or unshared (lone pairs). F F F F F B F F - - B F • Expanded octets Nonmetal of the second period have only s and p subshell. This can be seen in the formation of the BF 4 . The energy different between 2 p and 3 s is too great. F Cl P Cl octet F S F O F O S O expanded octet Cl F O F expanded octet 7.ion. The energy different between the np and nd is not that great and this open up the possibility of Lewis structures with more than eight electrons around the central atom. The central has only eight electrons an octet.6 The shape of molecules • Valence-shell electron-pair repulsion (VSEPR) theory Electron pairs repel one another. Nonmetal in the third period and beyond contain a d subshell. This. trigonal planar. 1. Establish the electron-pair geometry around the central atom: linear. 4.The valence-shell electron-pair repulsion theory focuses on pairs of electrons in the valence electronic shell. Draw a plausible Lewis structure of the species (molecule or polyatomic ion). Applying VSEPR theory Let us use this four step procedure for predicting the shapes of molecules. . trigonal. tetrahedral. 3. bipyramidal. or octahedral. in turn. 2. from data in table 4. Determine the number of electron-pair around the central atom and identify them as being either bonding pairs or lone pairs. Determine the molecular geometry from the positions around the central atom occupied by the other atomic nuclei. that is.1. results in particular geometrical shapes for molecules. Table 7.1 (Continued) .1 Molecular geometry as a function of e--pair geometry Table 7. s around the I Because these are exactly how we Lewis structure. There are six electron pair around the I atom. which has an expended octet. pairs. it does not matter show them around the I atom in the Cl I Cl Cl Cl . 2.1 leads to a square planar. We must place the four additional eatom.d Unit: D (Debye) -30 1 D = 3. 1. Solution: Apply the four steps outlined above.Example: 7. µ = δ.m) . To establish ionic charge of (–1) (1 x 7) + (4 x 7) + 1 = 36 To join 4Cl atoms to the central I atom and to provide octet for all the atoms we need 32 e-s.5 Using VSEPR theory to predict a geometrical shape. 4 bonding and 2 lone pairs. anion is of the Cl 4. 3. µ . The number of valence e-s is From I. The electron pair geometry. From Cl. The which according to Table molecular geometry that is • Dipole moment :µ that is. orientation of 6 electronoctahedral. Write the Lewis structure. the is Cl ICl4- I Cl Cl type AX4E2. two lone pairs. Dipole ment. Predict the molecular geometry of the polyatomic anionICl4-.34 x 10 Coloumb-meter (C. is the product of a charge (δ) and distance (d): bond length. 4. Which is it. XeF4. ICl. ICl and NO are diatomic molecules with an electronegativity difference between the bonded atoms. BF3.1. • Bond energy Energy is released when isolated atoms join to form a covalent bond. SO2. NO. SO2 is the bent molecule with an electronegativity difference between the S and O atoms. XeF4. Which of these molecules would you expect to be polar? Cl2.Example:7. Practice example: Only one of the following molecule is polar. BF3 is a symmetrical planar molecule (120° bond angles).. The B—F bond moments cancel each other. and why? SF6. BF3. Solution: Polar: ICl. NO.7 Determine the relationship between geometrical shapes and the resultant dipole moments of molecules. hence no electronegativity difference. Cl2 is a diatomic molecule of identical atoms. For BF3 and XeF4 refer to Table 4.C2H2. Energy is absorbed to break apart covalently bonded atoms. SO2. Nonpolar: Cl2. Endothermic (+) • Bond dissociation energy . XeF4 is a square planar molecule with the F atoms arranged symmetrically around the Xe atom. H2O2. Exothermic (-). The reaction of methane (CH4) and chlorine produces a mixture of products called chloromethanes. followed by the formation of one C—Cl bond and one H—Cl bond. D. CH3Cl. Triple bonds are strong still but they are not three times as larg.435.93 kJ Double bonds have higher bond energies than single bonds between the same atoms.339 kJ 1 mol H-Cl bonds = .7 Energy chages in the formation of ionic crystals . used un the preparation of silicones. One of these monochloromethane.93 kJ ∆ H = + 435. and one H—Cl bond.770 kJ Enthalpy of reaction: • Lattice energy ∆ Hrxm = 657 – 770 = .Bond dissociation energy.(SI) 2H (g) H2 (g) H2 (g) 2(H) (g) ∆ H = . ∆ H for net bond breakage: 1 mol C-H bonds = + 414 kJ 1 mol Cl-Cl bonds = + 243 kJ sum: = + 657 kJ ∆ H for net bond formation: 1 mol Cl-Cl bonds = . Calculate ∆ H for the reaction. • Enthalpy of reaction: ∆ H ∆ Hrxm = ∆ H (bond breakage) + ∆ H (bond formation) = ∑ BE ( reac tan ts ) − ∑ BE ( products) Example: 7. one C—Cl bond.113 kJ 7. is the quantity of energy required to break one of covalent bonds in a gaseous species. but they are not twice as large. The net change is the breaking of one C—H bond and one Cl—Cl bond.431 kJ sum: = .8 Calculating an enthalpy of reaction from bond energies. CH4 (g) + Cl2 (g) Solution: CH3Cl (g) + HCl (g) We would break four C—H bonds and one Cl—Cl bond and form three C—H bonds. Unit: kJ mol. 1. Na (g) 3. 1 Cl2 (g) 2 Na (g) ∆ H1 = ∆ Hsublimation = +108 kJ Na+(g)+e. come together to form 1 mol of a solid ionic compound. Na (s) 2.st ion.1 Enthaipy diagrtam for the formation of an ionic crystal. ∆ H1 + ∆ H2 + ∆ H3 + ∆ H4 + ∆ H5 + ∆ H6 = 0 Formation of an ionic crystal is represented through the first five steps. For the overall cycle .It is the energy given off when separated gaseous ion (positive and negative). ________________________________ Na+(g) + Cl (g) ________________________ ∆ H2 Na + (g) ∆ H3 ∆ H4 __________________ Na+(g) + Cl-(g) + 1 Cl2 (g) 2 1 Cl2 (g) 2 1 Cl 2 (g) 2 ∆ H6 ________________________ ∆ H1 Na (g) + ∆ H5 ________________________ Na (s) + __________________ NaCl (s) Figure 7. The six steps and their enthalpy changes for NaCl are written below. and the enthalpy changes are identified in terms of the property they represent.∆ H2 = 1.energy= +496 kJ Cl(g) ∆ H3= 1 2 Cl-Cl bond energy . A six step cycle for NaCl is illustrated here that begins with Na (s) and Cl 2 (g) and return to the starting point. NaCl(s) Cl. Na+(g)+ Cl-(g) 6. Cl(g)+ e5.(g) ∆ H4 = EA of Cl = -349 kJ NaCl(s) ∆ H5= lattice energy of NaCl=? Na(s)+ Cl2(g) 1 2 ∆ H6 = ∆ H°f [NaCl(s)] = +411 kJ ∆ H1 + ∆ H2 + ∆ H3 + ∆ H4 + ∆ H5 + ∆ H6 = 0 (108 + 496 + 122 – 349 + ∆ H5 + 411) kJ = 0 ∆ H5 = lattice energy = (-108 -496 -122 +349 -411) kJ = -788 kJ .= +122 kJ 4. Electrons at the surface of a metal are able to reradiate. if electrons from an external source enter a metal wire at one end. Energy band: A set of molecular orbitals with extremely small energy between each pair of successive levels. for instance. This of very closely spaced orbital energy levels is N an separation collection molecular called an . an ability to conduct electricity.8. Band theory • 1. the internal structure of the metal remains essentially unchanged. In this way. Thus metals absorb visible light. free electrons pass through the wire and leave the other end at the same rate. at the same frequency. such as a lustrous appearance. perhaps by hammering.2 Metallic bond The challenge to a bonding theory for metals is to explain how so much bonding can occur with so few electrons. electrical conductivity is explained. Also. the ions are Li+ and one electron per atom is contributed to the sea of electrons. Free electrons (those in the electron sea) are not limited in their ability to absorb photons of visible light as are electrons bound to an atom. Electrons in the sea are free (not attached to any particular ion) and they are mobile. no bonds are broken. they are opaque. and this explains the lustrous appearance of metals. and the sea of electrons rapidly adjusts to the new situation . In Li. the theory should account for certain properties that metals display to a far greater extent than nonmetals. • The electron sea model An oversimplified theory that can explain some of the properties of metals just cited pictures a solid metal as a network of positive ions immersed in a sea of electrons. light that strike the surface. If one layer of metal ions is forced across another. and an ease of deformation (metals are easily flattened into sheets and drawn into wires). • Explanation of the metallic properties by electron sea model Thus. The ease of deformation of metals can be explained in this way. This creates positive holes in the crystal. In consequence. Si. Electrons in the valence band may acquire enough energy (thermal energy) to jump to a level in the conduction band. Conduction band: The band of orbitals through which electrons can move is known as a conduction band. sunlight p-type silicon . the energy gap between the valence band and conduction band is fixed size. 3. Doping:Carefully adding impurities to the extrinsic semiconductor. Electrons in the crystal can move to filled positively charged holes and semiconductor is called a p-type semiconductor. If Si is doped with B atoms. • Semiconductors: Modern electronic components Light-emitting diodes: LED. called intrinsic semiconductors. which have only three valence electrons. This type is called n-type semiconductor. Ge. [Ne] 3s23p3: 5 valence electrons 2 2 14Si: [Ne] 3s 3p : 4 valence electrons 15P: If Si is doped with P. Solar cell In some of materials (Cds).energy band. there is a deficiency of one electron for every B atom. the valence band and conduction band are separated only by a small energy gap. Electrons from the valence band in the crystal move to form four bonds. the electrical conductivity of semiconductor increases with temperature. One electron promoted to the conduction band for every P atom. for example. 2s1 2p0 Valence band Conduction band In a semiconductor. 3Li: 1s2. 2. Valence band: The highest energy band of molecular orbitals in a metal that is filled with valence electrons is called the valence band. P atom enter the Si crystal structure by using four of their valence electrons to bond to Si atoms. an electric current. where they fill positive holes. and the process continues as long as light shines on the solar cell. e –flow Conduction electrons can easily cross the junction into the n-type semiconductor. creating holes in the valence band. electrons in the valence band can absorb energy and promoted to the conduction band. When the p-type semiconductor is struck by a beam of light. .A thin layer of a ptype semiconductor is in contact with an n-type n-type semiconductor in a region called the silicon junction. Further light absorption creates more conduction electrons and positive holes. Electrons can be carried by wire through an external load and eventually returned to the p-type semiconductor. This sets up a flow of electrons. A solvent: is the component that is present in the greatest quantity in a solution. Solid solution: Brass. Natural gas: (CH4 + C2H6+ several other gases). . A concentrated solution: has a relatively large quantity of dissolved solute. (Cu + Zn) Bronze. (Pb + Sn) Dental amalgam. Gaseous solution: Air: (N2 + O2 + several other gases). (Ag + Sn) Solid solution with a metal as a solvent is also called alloys.1 Solutions and Their Physical Properties Definition: A solution: is a homogeneous mixture of solvent and solute. A dilute solution has a relatively small quantity of dissolved solute. (Cu + Sn) Solder. A solute: is a component that is present in lesser quantity than the solvent in a solution. Liquid solution: Sea water: (H2O + NaCl + many other) Vinegar: (CH3COOH + H2O).CHAPTER 9 9. 9. Example: 9.0 g water = 10.1 What mass percent of a solution containing 8 g of NaOH in 70 g of water? Solution: Mass present = 8.9.0 g NaOH x100% 78.26 % • Mole fraction: XA = nA nA + nB = mole of A Total mole of componentin solution • Mole percent = XA x 100 % .2 Expression of concentration • Mass percent = mass of solute x 100 % mass of solution volume of solute x 100 % volumeof solution • Volume percent = • Mass/volume percent is used in medicine and pharmacy. .• Molarity: M = mole of solute volumeof solution [mol/L] mole of solute • Molality: m = [mol/kg-solvent] mass of solvent in kg Practice Example: 10 mL of ethanol (d = 0. What is the concentration of ethanol in this solution expressed as? (a) Vol% (b) mass % (c) mass/vol % (d) Ethanol (e) Mol% (f) molarity (g) molality • Solubilities of gases Solubility’s of gases decrease with increased temperature and increase as the gas pressure is increased.789 g/ml) is made up to 100 mL ethanol water solution by adding water. The resulting solution density is 0. .982 g/mL.HCl. • Henry’s law: The mass of a gas dissolved in a given mass of liquid is directly proportional to the pressure of the gas above the solution. Pgas (C = concentration) C ∴K= Pgas ∴ C C 1= 2 P P 1 2 Note: Gases that are highly soluble in water (NH3. SO2) do not obey Henry’s law. C = K. Solution: K = C/Pgas = 23. 2. The escaping tendency of the solvent molecules in a solution decreases as the concentration of the non-volatile solute increase. Colligative properties of solutions • Vapor pressure of solution: The vapor pressure of a solution is always lower than the vapor pressure of the solvent at the same temperature.000 atm = 4.54 mL N2/L.54) x 1.Example: 9. Posolvent = Vapor pressure of the pure solvent.0 mL N 2 /L PN 2 PN = (100/23. Find required gas pressure to increase the solubility of N2(g) to a value of 100 mL N2/L. This is because there are fewer solvent molecules per unit area on the surface of the solution. • Raoult’s Law (1887) of vapor pressure: The partial pressure exerted by solvent vapor above an ideal solution (PA) is the product of the mole fraction of solvent in the solution (XA) and the vapor pressure of the pure solvent at the given temperature ( PA0 ) . The greater the concentration of non-volatile solute in the solution. the lower the vapor pressure of the solvent. .3 1.54 mL N 2 /L 1. P solvent = Vapor pressure of solvent in the solution. ∆ Psolvent = P°solvent − Psolvent ∆ Psolvent = The lowering of vapor pressure. 3.00 atm PN = 2 100.1 The solubility of N2(g) at STP is 23.25 atm 2 9. 2 mmHg= 61.4 mmHg = 14.PA = XA PA0 The lowering of the vapor pressure in dilute solution of nonelectrolytes at constant temperature is proportional to the quantity of substance dissolved in a given weight of solvent.500 x 95.1 and 28. What are the partial pressures of the benzene and toluene above this solution? What is the total pressure? Solution: Benzene-toluene solution should be ideal.8 mmHg • Freezing point depression and boiling point elevation of nonelectrolyte solution .1 mmHg = 47. Example: 9.6 mmHg Ptol = Xtol P0tol = 0.2 The vapor pressure of pure benzene and toluene at 250C are 95.6 mmHg +14. Pbenz = Xbenz P0benz = 0.2 mmHg Ptot= Pbenz + Ptol =47. respectively.500. so we expect Raoult’s law to both solution components. A solution is prepared in which the mole fractions of benzene and toluene are both 0.500 x 28.4 mmHg. Example: 9.450°C = 0.= 0.Figure 9.86oC m-1) Solution: (a) Molality = ∆ Tf / Kf. m ∆ Tf = Kf.92 g H2O. what must be molar mass of nicotine? (Kf = 1. extracted from tobacco leaves.242 m 1.450oC? (b) If this solution is obtained by dissolving 1. ∆ Tb = Kb.921 g of nicotine in 48.86°C m −1 (b) M = molar mass of nicotine. m ∆Tb = Freezing point depression ∆ Tf = Boiling point elevation Kb and Kf = Proportionality const. (a) What is the molality of nicotine in an aqueous solution that started to freeze at –0. Therefore boiling point of the solvent is elevated. is a liquid completely miscible with water at temperatures below 60oC. So a higher temperature is required for the vapor pressure of the solution to become equal to atmospheric pressure.1 Vapor pressure lowering by a nonvolatile solute. . A solution that contains nonvolatile solute always has higher boiling point than pure solvent. The boiling point raised or the freezing point lowered is proportional to the molar concentration of solute for dilute solutions. Non-volatile solute lowers the vapor pressure of solvent.3 Nicotine. whose values depend only on the concentration of solute particles in solution and not on what he solute is.921 g/M = 0.Amount of solute in moles = 1. Note: These properties.04892 x 0. . are called colligative properties. If the pressure greater tan pressure is applied to the flow of water will be formed direction. This is the molecules be formed left osmosis a net reverse– condition right.242) mol Osmosis: spontaneous solvent semipermeabe of lower into a Osmosis is the process by which the molecules pass through a membrane from a solution concentration of solute solution of higher concentration of solute.04892kg water M= • 1. • Osmotic pressure: A pressure that would have to a solution to stop the passage from the pure solvent through semipermeable membrane into the solution.242 mol/kg water 0. be applied into of molecules a π= n RT = C.921 g = 162 g/mol (0.921 g/M Molality = 1.RT V n = The amount of solute in moles V = The volume of solution in liters n/V = C = Molarity of the solution. • Reverse osmosis: The net flow of water through the membrane will to right. Solution: MgCl2(aq) Mg 2+(aq) + 2Cl-(aq) Because three moles of ions are obtained per mole of formula units dissolved.86oC m-1 x 0. simply substitute i = 1.RT (i = Van’t Hoff factor) If these equations are used -(i) (ii) for non electrolytes. predict a value of i is suggested in the following example. we expect the value i = 3. For strong electrolytes.known as reverse osmosis. • Solution of electrolyte: ∆ Tb = i x Kb x m (m = molality of the solute) ∆ Tf = I x Kf x m π = i x C. ∆Tf = i x Kf x m ∆Tf = 3 x 1.00810C .3 Predict the freezing point of aqueous 0.00145 m = 0.00145 m MgCl2. Example:9. .(aq) an acid a base H2O water 10. A neutralization reaction involves the combination of hydrogen ions and hydroxide ions to form water.(aq) 2 NaOH  → Na+ (aq) + OH. Cl-.is the conjugate base. HCl Η Ο 2  → Η Ο H+ (aq) + Cl.1 Arrhenius Theory An acid is a compound which could produce (H+) in water solution and a base is a compound which could produce (OH-) in water solution . H2O is base.CHAPTER 10 Acid Base Theories: 10. HCL + H2O H 3O + + Cl- HCl is acid.2 Bronsted and Lowery An acid is a proton donor and a base is a proton acceptor. NH4+ is the conjugate acid of NH3. H3O+ is the conjugated acid of H2O. NH3 + H2O NH4+ + OH- NH3 is base.(aq) H+ (aq) + OH. HClO2 is an acid.+ H2O base (2) acid (1) c. OCl.is the conjugate base of H2O. H2O accepts the proton and becomes H3O+. H+. and ClO2. HClO2 + H2O OCl. Example: 10. Substances that can act either as an acid or a base. HClO2 + H2O acid (1) base (2) H3O+ + ClO2acid (2) base (1) b. b. Therefore.1 For each of the following identify the acids and bases in both the forward and reverse reactions. c. a. d. and H3O+is its conjugate acid. OCl.OH.is its conjugate base. H2O is a base.is a base and gains a proton from water. NH3 + H2PO4- HOCl + OHacid (2) base (1) NH4+ + HPO4acid (2) base (1) base (2) acid (1) .+ H2O NH3 + H2PO4 HCl + H2PO4- H3O+ + ClO2HOCl + OHNH4+ + HPO4H3PO4 + Cl- Solution: a. OH. they are said to be amphiprotic. HClO2 loses a proton.H2O is acid. to become ClO2-.is the conjugate base. another acts as a base and hydronium in (H3O+) and hydroxide ion (OH-) are formed. H2O + H2O H3O+ + OH- K= [H 3O + ][OH + ] [H2O] = const: [H 2O]2 K = [H3O+] [OH-] . or molecules)that is an electron-pair acceptor A Lewis base is a species that is an electron-pair donor.d.4 The ionic dissociation of water The self-ionization of water and the pH scale In the self-ionization of water.3 Lewis Theory: A Lewis acid is a species (an atom. for each H2O molecule that acts as an acid. HCl + H2PO4- H3PO4+ Clacid (2) base (1) acid (1) base (2) Reactions (a) and (b) show that H2O is amphiprotic Reactions (c) and (d) illustrate the same for H2PO4-. F F B F + H N H H F F B F H N H H Lewis acid Lewis base Salt 10. 10. This reaction is reversible. ion. student found the pH of (1) a rainwater sample to be 4. At 25oC. Kw = 5.28 =2. pH = .28. Kw = 9.35 = 4.log [OH-] Example: 10. Restate in terms of [H3O+]. (a) What is [H3O+] in the rainwater ? (b) What is [OH-] in the ammonia ? Solution: a.5 x 10-13 • PH and pOH PH is the negative of the logarithm of [H+]. The equilibrium constant for the self-ionization of water is the ion product of water and is symbolized as Kw.00 – 11.log [H3O+] pOH = .35 and (b) a sample of household ammonia to be 11. pH = -log [H3O+] or log [H3O+] = . Kw will have a different values.The very small electrical conductance of pure water corresponds to the very small [H+] and[OH-]. At 60oC. determine pOH with equation .2 In a laboratory experiment. Kw = [H3O+] [OH-] = 1 x 10-14.pH = .5 x 10-5 M b. Both ions have equal concentration. pOH = 14.35 [H3O+] = 10-4. By defination.4. At any other temperature.6 x 10-14 At 100oC. each being 1 x 10-7 mol dm-3. First.72 pOH = -log [OH-] .00 – pH = 14. HBr.015) [OH-] = 1 x 10-14 1. Ca (OH)2.0 x 10−14 [OH ] = = 0. an [OH-] in 0. Calculate [H3O+].72 [OH-] = 10-2.…. [H3O+] = 0.67 x10-12 = 6.7 x 10-13 M −2 1.. HI. H2SO4 Srong bases: LiOH.015 M HCl(aq).015 M Kw = [H3O+] [OH-] = 1 x 10-14 (0. Ba (OH)2. [Cl-]. • Strong acid and strong base: (Completely ionized) Strong acids: HCl.9 x 10-3 M. Hence.72 = 1. HClO4.3 Calculating ion concentration in an aqueous solution of a strong acid. KOH. HNO3.4 . Solution: We can assumed that HCl is completely ionized and is the sole source of H3O+ in solution. NaOH.015 M [Cl-] = [H3O+] = 0.log [OH-] = -pOH = .5x 10 - Example: 10. Example: 10.2. Solution: .log 0.100L = 0.09g Ca (OH) 2 0.0223 mol Ca (OH)2 2 mol OH x 1L 1 mol Ca (OH)2 = 0.00 – pOH = 14.0223 M Ca (OH)2 [OH-]= 0.0446 = 1.log Ka (Ka = weak acid ionization const:) pKb = .35 pH = 14.Ca (OH)2(s) is soluble in water only to the extent of 0.0 ml solution at 20oC ? What is the pH of saturated Ca (OH)2(aq) at 20oC ? Solution: 0.00 – 1.250 M aqueous solution of HC4H7O2 is found to have a pH of2.log Kb (Kb = weak base ionization const:) Esample:10.5 A 0.72.165g Ca (OH) 2 x Molarity = 1mol Ca (OH) 2 74.65 • Weak acid and weak base: pKa = .0446 MOHpOH = -log [OH-] = . determine Ka for butyric acid.35 = 12.165 g Ca (OH)2/100. At equilibrium. Solution: At equilibrium.250 – x) M H3O+ + C4H7O2--(+x M) + (+x M) xM xM --- .pH = .8.250 .72 = 1.5 x 10 − 5 = −3 0.9x10 ) Esample:10.6 Show by calculation that the pH of 0.(1.250 M.2.x = [HC4 H 7 O 2 ] 0. HC4H7O2 + H2O Initial con: Changes: equal conc: (-x M) 0. (0.250 – x) M H3O+ + C4H7O2--(+x M) + (+x M) xM xM --- But x is not unknown.100 M aqueous solution of HC4H7O2 should be about the value shown on the pH meter that is pH ≈ 2. and we can determine this from the pH.72 [H3O+] = 10-2.9x10 −3 ) = 1.9 x 10-3 = x Ka = [H 3O + ][C4 H 7 O 2 ] x. HC4H7O2 + H2O Initial con: Changes: equal conc: (-x M) 0.9x10 −3 ) (1. Log [H3O+] = .1250 − x (1.250 M. It is the [H3O+] in solution. (0. log [H3O+] = .3 x 10-3) = -(-2.100 − x We assumed that (0.log (1.5 Buffer solution Buffer solution is a solution whose pH values change only slightly upon the addition of small amounts of either an acid or a base.89) = 2.100. CH3COOH(aq) CH3COONa(aq) CH3COO-(aq) CH3COO-(aq) + + H+(aq) Na+(aq) .8 x 10 -5 [HC 2 H 3 O 2 ] 0. a mixture of ethanoic acid and sodium ethanoate. Common buffer solutions are(i) (ii) • a mixture of a weak acid and its conjugate base or a mixture of a weak base and its conjugate acid.100 x 1.100 – x) ≈ 0.x Ka = = = 1.8 x 10 -6 = 1.8 x 10-5 = 1. Acid condition of buffer solution: (pH = 4 to 7) For example. x2 = 0. H3PO4 H2PO4HPO42H+ + H2PO4H+ + HPO42H+ + PO43- 10.8 x 10-6 x = [H3O+] = 1.89 − • Polyprotic acids: Acid molecules have more than one hydrogen atom per molecule is known as polyprotic acid.[H 3 O + ][C 2 H 3 O 2 ] x.3 x 10-3 M pH = . is converted to its conjugate acid.• Alkaline condition of buffer solution: (pH = 7 to 11) For example. CH3COO-(aq) + H3O+ CH3COOH(aq) + H2O This prevent a change in pH of the solution. A reaction occurs in which CH3COO. Imagine adding a small amount of a strong base to acidic buffer solution. CH3COOH(aq) + OHCH3COO-(aq) + H2O This prevent a change in pH of the solution.] Ka = [CHCOOH] [H3O+] = Ka [CH 3 COOH] [CHCOO . • An Equation for Buffer Solution: (The Henderson-Hasselbalch equation) Let us consider a mixture of a weak acid (such as CH3COOH) and its salt (CH3COONa). NH4OH(aq) NH4Cl(aq) NH4+(aq) + OH-(aq) NH4+(aq) + Cl-(aq) Now imagine adding a small amount of a strong acid to acidic buffer solution.] . we start with the familiar expression: CH3COOH + H2O H3O+ + CH3COO- [H 3 O + ][CH 3 COO .] [CH 3 COOH] [CHCOO .log [H3O+] = -log Ka – log . a mixture of ammonium hydroxide and ammonium chloride. A reaction occurs in which CH3COOH is converted to its conjugate base CH3COO-. 01 [H+] = 1.] pH = pKa + log [CHCOOH] [conjugate base] [acid] [conjugate acid] [base] pH = pKa + log pOH = pKb + log Example:10.7 Calculate the pH of a buffer solution containing 0.001/0. CH3COOH [CH 3COO.] [CH3COO-]equil = [CH3COO-]salt+ [CH3COO-]acid sodium ethanoate is completely dissociated and ethanoic acid is negligibly dissociated in the presence of common ion.1 = 5.8 x 10-5 x CH3COO. Solution: For the acid.01 mole of ethanoic acid (Ka = 1.8 .] [H + ] Ka = CH3COOH [H+] = 1.1 mole of sodium ethanoate per cm3.74 Esample:10.8 x 10-5 x 0.[CH 3COO.+ H+ [CH3COOH] [CH 3COO. therefore [CH3COO-]equl = 0.8 x 10-5) and 0.1 [CH3COOH] = 0. 5g NaC2H5O2 x − imol NaC 2 H 3O 2 82. H3O + C2H3O2+ Weak acid Salt: Changes: Equil conc: 0.5 g NaC2H3O2 in a sufficient volume of 0.8 x10−5 [HC2 H 3O 2 ] 0.311molC2H3O21mol NaC 2 H 5 O 2 [C2H3O-]=0.04g NaC 2 H 3O 2 x 1mol C 2 H 3 O 2 = 0.55 X = [H3O+] = − 0.662+x) M ≈ 0.311mol C2H3O2-/0.662+x) M 0.log (1.What is the pH of a buffer solution prepared by dissolving 25.80 .550 M HC2H3O2 to make 500 ml of the buffer? Solution: Amount of C2H5O2.=25.550 x 1.8 x 10-5 0.622) Ka = = = 1.622 [H 3O + ] [C2 H 3O 2 ] (x)(0.55 M --xM (0.622 M C2H3O2Equilibrium calculation: HC2H3O2 + H2O Initial conc.5 L = 0.55 and (0.log [H3O+] = .622 PH = .55 – x) M ≈ 0.6 x x10-5) = 4.55 – x) M -xM -+xM + x M (0.622 M Let us assume that x is very small so that (0. What is the molar solubility of PbI2 in 0.10.+ H+ (weak electrolyte) When HCl (strong electrolyte) is added to the above solution. H+ become common ion.6 The common ion effect The repression of the ionization of a weak electrolyte by the presence of a common ion from a strong electrolyte is called the common ion effect.7 x 10-5) 10. Therefore the concentration of CH3COOH will increase.7 The solubility product constant: Ksp An equilibrium expression between a slightly soluble ionic compound and its ion in a saturated solution is called solubility product. Practice Example: Determine [H+] and [CH3COO-] in a solution containing 0. More [H+] repress the equilibrium to the left (Le Chatelier’s principle).9 Calculating the solubility of a slightly soluble solute in the presence of a common ion. (Ka = 1. PbCl2(s) Ksp = [Pb2+] [Cl-]2 Pb+(aq) + 2Cl-(aq) Example:10. CH3COOH CH3COO.1M in both CH3COOH and HCl.10 M KI(aq) ? Solution: . 1 x 10-9 Let us assume that s is much smaller than 0.1 x 10 .0 ml of 0. Will a precipitate of lead iodide form ? (Assume 1 drop = 0..10 s (0. so that 0.20 M KI are added to 100.1 x 10-7 M.10 )2 = 7.9 Three drops of 0.010 M Pb (NO3)2. Example:10.1 x 10-9 7.10 s s 2s (0.05 ml 1L 0.]2 = (s) (0.20 mol KI x x 1 drop 1 000 ml 1L .10+2s) Iinitial conc. and s = molar solubility of PbI2 = 7.05 mL) PbI2(s) Pb2+(aq) + 2I-(aq) Ksp = 7.M Ksp = [Pb2+][ I.= 3 drops x 0.1 x 10-7) is much smaller than 0. M From PbCl2.10 + 2s)2 = 7.10.10 + 2s ≈ 0.7 M s= 2 (0.10 M.1 x 10 -9 = 7.10) Our assumption is well justified: s (7.Equilibrium: PbI2(s) Pb2+(aq) + 2I-(aq) 0.M Equil conc.1 x 10-7 Amount I. of Pb (NO3)2 = 0.010 M. . so [Pb2+] = 0. dynamic equilibrium is said to be established. As time passes.1000L - Conc. When concentration of CH3OH is sufficiently high.]2 = (1 x 10-1) (3 x 10-4)2 = 9 x 10-10 This value is smaller than Ksp (7. principle of chemical equilibrium 11. we conclude that the precipitate of PbI2 will not form. Eventually two opposite reactions proceed at equal rate.01M Ksp = [Pb2+][ I. the reverse reaction starts and speeds up to form CO + H2.1 Dynamic equilibrium A condition in which both reactions are still proceeding but the rates of forward and reverse reactions are equal and there is no net change in concentration with time. the forward reaction slows down because of the decreasing concentration of CO and H2. CO(g) + 2H2(g) CH3OHg) Initially only the forward reaction occurs.1 x 10-9). and there is no further change in the net reaction. CHAPTER 11 11.1 mol I= 3 x 10 5 M x 1 mol KI 3 x 10-5 mol I[I ]= = 3x10-4 M 0. 6x108 = 2.6 x 108 What is the value of KC at 298 K for the reaction. the equilibrium constant K ′ =1/KC C 2NH3(g) N2(g) + 3H2(g) K ′C = 1/3.8x10-9 for the following reaction.2 The equilibrium constant expression The ratio of molar concentration of products and reactant is known as an equilibrium constant. CO(g) K= • + 2H2(g) CH3OH (g) [CH 3 OH] [CO][H 2 ] 2 General expression for K aA +bB [C]c [ d ]d KC = [A]a [ b ]b cC + dD [ ] = molar concentration: mol/L Example:11. .1 The following KC value is given at 298 K for the synthesis of NH3(g) from its elements.11. N2(g) + 3H2(g) 2NH3(g) KC = 3. NH3(g) Solution: 1 2 N2(g) + 3 2 H2(g) KC = ? For the reverse reaction. we divide all coefficients by 2. PV = nRT. KC = KP . KC = 2.2 .3 Equilibria involving gases aA +bB [C]c [ d ]d KC = [A]a [ b ]b KC = cC + dD [PC ]c [ PD ]d KP = [PA ]a [ PB ]b equilibrium constant based on concentration in a gaseous mixture (mole/L) KP= equilibrium constant based on the partial pressures of gases in the gaseous mixture.3 x 10 −5 11. [D] = D RT RT RT RT c d P c P d P xP [ ][ ] KC = RT a RT b = Ca D b x RT (c+ d)-(a + b) P P PA x PB [ RT ] [ RT ] C D A B ∆n = (c + d) – (a + b) = the difference in the stoichiometric coefficients of gaseous mixture.NH3(g) 1 2 N2(g) + 3 2 H2(g) To base the equation on 1 mol NH3(g). (RT) ∆ n Example:11.8 x 10 −9 = 5.Thus. v RT P P P P [A] = A . and so [ ]= n P = . we have to take the square root of K/ C. According to the ideal gas law. [C] = C . [B] = B . 4 11.08206 L atm mol-1 K-1) Solution: 2SO2(g) + O2(g) ∆ ngas = 2 – (2 + 1) = -1. Changes in system volume or pressure have no effect in the following equilibrium. the system reacts in a way that partially offsets the change while reacting a new state of equilibrium. (R = 0.08206 x 1000)-1 = 3. More NH3 will be produced. Equilibrium shift to an increase volume or produce a large number of moles of gas. formation of SO3(g) . the partial pressure of each gas and total pressure drop. (RT) ∆ n = (2. a net reaction occurs in the direction producing a small number of moles of gases.For the reaction at 1000 K. (or) When any one of the factors affecting the equilibrium of a chemical system such as temperature. 2SO3(g) (at 1000 K) KC = KP .4 Altering equilibrium conditions (Le Chatelier’s principle ) When an equilibrium system is subjected to a change in temperature.8 x 102) (0. When the volume of an equilibrium mixture of gases is reduced. N2(g) + 3H2(g) 2NH3(g) When the gaseous mixture is transferred to the larger flask. • Effect of changes in pressure or volume on equilibrium An increase in external pressure causes a decrease in the reaction volume. calculate the value of KP. This means that NH3 decomposes back to N2 and H2. H2 + I2 2HI . the system reacts in such a way as to nullify the effect of change. pressure or concentration of a reacting species. pressure or concentration is changed. This means the reaction favors the formation of NH3. CO + H2O CO2 + H2 There is the same number of moles of reactant and product. There will be no effect of pressure on the equilibrium. Equilibrium is reached more rapidly. but the equilibrium amounts are unchanged by the catalyst. an equilibrium mixture would have a higher concentration of SO2 and O2at higher temperature.00 L .0277 M 3. 2SO2 + O2 2SO3 ∆Ho = -180 KJ The reaction is exothermic reaction. Solution: For both gases. • Effect of a catalyst on equilibrium A catalyst in a reaction mixture can speed up both the forward and reverse reactions.3 In the reaction: N2O4(g) 2NO2(g) at 25oC.64 g N2O4 and 1.00 L vessel are 7. conversion: g mol mol/L 7. Raising the temperature favors the endothermic reaction. Therefore.64 g N 2O 4 x [N2O4] = 1 mol N 2O 4 92. the quantities of two gases present in a 3.5 KJ Lowering the temperature causes a shift in the forward direction of exothermic reaction. Example:11.01g N 2O 4 = 0. • Effect of temperature on equilibrium Raising the temperature of an equilibrium mixture shifts the forward direction of the endothermic reaction.56 g NO2. the reverse reaction of the exothermic reaction (that is the decomposition of SO3). N2 + O2 2NO ∆H0 = + 180. what is the value of KC for the reaction. 01g NO 2 = 0.0113 M 3.0058 mol 0.00142 mol SO3 is found to be present at equilibrium.0113) 2 = = 4. amt: 0.00 mol + O2(g) +0.52 L vessel at 900 K.54 g NO 2 x [NO2] = 1 mol NO 2 46.9 x 10-3 M .0142 mol/1.0142mol 0.8 x 10-3 M [O2] equil = 0.52 L = 9.00 mol 2SO2(g) 0.0058 mol SO3 (–) sign signifies that this amount of SO3 is consumed in equilibrium.0029 mol/1.0029 mol Change: -0.4 When a 0.0277 Example:11.0200 mol 0.concentration: [SO3] equil = 0. The reaction: Initial amt: 2SO3(g) 0.34 x 10-3 M [SO2] equil = 0.61 x 10 -3 KC = [N 2 O 4 ] 0.52 L = 3.1. What is the value of KP for the decomposition of SO3(g) at 900 K? 2SO3(g) Solution: 2SO2(g) + O2(g) KP = ? The change in amount of SO3 = [SO3]equil – [SO3]initially = 0.00 L [NO 2 ]2 (0.0200 mol sample of SO3 is introduced into an evacuated 1.0200 mol = – 0.52 L = 1.0058 mol Eqil.0058 mol/1.0058 mol +0.0142 mol – 0.0029 mol Equi. 0. [SO 2 ]2 [O 2 ] [3.8 x 10 -3 ]2 [1.9 x 10 -3 ] KC = = = 3.1 x 10 -4 2 -3 2 [SO 3 ] [9.34 x 10 ] K = K . (RT) ∆ n = 3.1 x 10-4 (0.0821 x 900)(2 + 1) - 2 C P = 2.3 x 10-2 PART TWO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (First Year) Textile & Metallurgy .T.I.G.Ch 01011 ENGINEERING CHEMISTRY A.MINISTRY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION Sample Questions & Worked Out Examples For E. beta. gamma rays. : Gamma rays are electromagnetic radiation of extremely high penetrating power. What is the significan features of nuclear atom □ Most of the mass and all of the positive charge of an atom are centered in a very small region called the nucleus. Isotopes : Two or more atoms having the same atomic number but different mass numbers are called isotopes. . Define the terms: Radioactivity : Radioactivity is a natural phenomenon that is spontaneous emission of radiation by mineral sources. Alpha rays Beta rays Gamma rays 3. 2.Chapter ( 1 ) 1. In passing electricity through evacuated glass tubes. The atom as a whole is electrically neutral. □ The magnitude of the positive charge is different for different atoms and is approximately one half the atomic weight of the element. a type of radiation emitted by the negative terminal (cathode) that crossed the evacuated tube to the positive terminal (anode). : Beta rays are negatively charged particles produced by changes occurring within the nuclei of radioactive atoms and have the same properties as electrons. Cathode rays travel in straight line and have properties that are independent of the cathode material Cathode rays are deflected by electric and magnetic fields. The atom is mostly empty space. Cathode rays are negatively charged fundamental particles of matter found in all atoms and becomes known as electrons. □ There exist as many electrons outside the nucleus as there are units of positive charge on the nucleus. Write an account for cathode rays. Describe the characteristics of alpha. Atomic mass : fractional   mass of fractional   mass of Atomic mass of isotope (1) × abundance of  + isotope (2) × abundance of  =     an element isotope (1) isotope (2) A mole : A mole is the amount of substance that contains of a number of chemical units equal to the number of atoms in exactly 12 g of pure carbon-12. : Alpha rays are particles carrying two fundamental units of positive charge and having the same mass as helium atom. 4. 01601 – 7.4944% = ( 100 .01601 u.000 u.941 u) is much closer to (7) than to(6).022 × 10 23 atom K 0.02 × 10 23 12 C atoms 1 mol 12 C = 1.01513 and 7.941 u.5056 % The percent abundance of 40K atom is 0. (a) Lithium (7) is more abundant because the weighted average atomic mass (6. corresponding to 1.941 u = 6. The two isotopes of 6Li and 7Li have masses of 6.098 ) 371 × 10 –3 g K × × 1 mol K 39.0.8570 ×1017 atom 7. The average atomic mass of lithium is 6.01601 u] 6. in gram. of 7Li isotope 6.6605 × 10-24 g .01513.2 5. (b)The fractional abundance of 6Li isotope= (x) -----------------------------------7Li isotope= (1 – x ) fractional   mass of fractional   mass of Atomic mass of isotope (1) × abundance of  + isotope (2) × abundance of  =     an element isotope (1) isotope (2) Atomic mass of Li = [ (x) × 6.u ] + [ ( 1 – x ) × 7. K What is the mass. of 6Li isotope The % abundance.? 12 g 12 C 1 atom 12 C 1 mol 12 C (?) g = 1 u × × × 12 u 6. (a) Which of these two isotopes is more abundant? (b) Calculate the percent natural abundance of the two.012%.074944 ) × 100% = 92. = 0. How many 40K atom do you ingest by drinking one cup of milk containing 371 mg of K? (K = 39.074944 x 100% = 7.0983 g K (?) 40K atoms = 6.012 atom 40 K × 1 mol K 100 atom K 40 = 6.01513 x + 7.01601 x x = 0.074944 The % abundance. 1. 2. (n) n = 1.no. called Pauli Exclusion Principle : No two electrons in an atom may have all four quantum numbers alike.no. Total particles in the nucleus = 133 No: of neutron in the nucleus = 133 – 55 = 78 78 ( % )neutrons in the nucleus = × 100 % = 67% 133 **************** Chapter ( 2 ) 1.no. These discontinuous spectra are called atomic spectra Stationary States stationary states. How many quantum numbers are required to specify an electron in a single orbital and what are the values? The four sets of quantum numbers. 2. 3. (n) (ii) Orbital Q.-l 1 1 ms = + . express the percentage. Atomic Spectra : The spectra produced by certain gaseous substances consist of only a limited number of colored lines with dark spaces between them. 2.− 2 2 . Hund’s rule : When orbitals of identical energy are available. electrons initially occupy these orbitals singly. (m1) (iv) Spin Q. : The electron has only a fixed set of allowed orbits. For the nulide 133 Cs .3 8. (i) Principle Q. by number of the fundamental 55 particles in the nucleus that are neutrons. (l) (iii) Magnetic Q. ……. What is meant by the following terms: Electromagnetic radiation: Electromagnetic radiation is a form of energy transmission through a vacuum or medium in which electric and magnetic fields are propagated as wave.no.(n-1) (ml)= + l …0 …. …… n l = 0. = 6.60931 km ) 93 million miles = 93 × 106 miles. is designated (a) n = 2. c = 2. l = 1.67 × 10-24 g = 1.626 ×10-34 J s.988 × 10 8 m = 499. l=1 l=2 l=0 → 2p → 4d → 5s 4.02 × 1023. ( alpha particles = 24 He2+ ) α particle = He 2+ 2 4 molar mass of He = 4 g/mol . h = 6.5× 107 m s-1. n = 5. h = plank’s constant = 6. (h = plank’s constant = 6. (b) n = 4. Avogadro’s no. 1.998 ×10 8 ms −1 E= = λ 589 ×10 −9 m = 3.4 3.67 × 10-24 g.998 × 108 ms-1. Calculate the energy emitted by the light of a sodium vapor lamp which has wavelength of 589 nm. Calculate the wavelength of an alpha particles if its velocity is 1.3748 10-19 J / photon 5.9676 × 104 m s-1 v= 7. c = 2. ( 1 mile = 1. How long does it take light from the sun 93 million miles away to the earth? c = 2. l = 2.626 × 10-34 kg m2s-1.60931 km 10 3 m 1s (?) second = 93 × 106 miles × × × 1 mile 1 km 2.67 × 10 − 27 kg ×10 × 10 −12 m v = 3. What type of orbital (c) n = 5. n = 4. mass of proton = 1.998 × 108 ms-1) λ = 589 nm = 589 × 10-9 m hc 6.67 × 10-27 kg λ= h mv 6.2189 s 6. Calculate the velocity of a beam of photon moving with a wavelength of 10pm.626 ×10 −34 Js photon −1 × 2.998 × 108 ms-1.626 × 10 −34 kg m 2 s −1 h = mλ 1.626 × 10-34 kg m2 s-1 ) λ = 10 pm = 10 × 10-12 m m =1. l = 0 ? n = 2. 65 ×10 −15 m − 27 7 −1 mv 6. State whether they are paramagnetic or diamagnetic.626 ×10 −34 kg m 2 s −1 h = = 6.5 1 mol 4g × 1 mol 6.626 ×10 −34 Js × 2.5 × 10 ms = 6.6423 × 10-24 g = 6.6423 × 10 kg × 1.4118 × 10-7 m = 241.023 ×10 23 photons c E =h λ hc 6. The ionization energy of Na is 496 kJ mol-1.022 × 10 23 particles = 6.65 ×10-6 nm 8.3646 ×10 − 20 J = 2. 12Mg m l ml ms 1 + K = [ Ar ] 4s1 4 0 0 19 2 1 2 10 + 4p2 4 1 0 32 Ge = [ Ar ] 4s 3d 2 1 2 4 3 1 -1 − 2 16S = [ Ne ] 3s 3p 1 2 − 3 0 0 12Mg = [ Ne] 3s 2 ************* . 16S.6423 × 10-24 g / particles (?) g of one α particle = m = 6.998 × 10 8 ms −1 λ= = E 82. What is the wavelength of a photon that will cause this ionization? 496 ×10 3 J 1 mol × = 82.6423 × 10-27 kg λ= 6. Describe four sets of quantum numbers for a last added electron of the following elements.3646 ×10 − 20 J / photon 1 mol 6. 19K . 32Ge.18 nm E= 9. (a) Which one is better reducing agent. 4s1 1s2. 2p6. Al and Mg. Therefore Al is better reducing agent. 2. Mg or Al? (a) Al has lower ionization energy than Mg and thus is more easily oxidized. the higher the melting point. K Ca (a) 1s2. The attraction between the positive ion and the electron is stronger than between the neutral atom and the electron. ************ . 3s2. 4. (a) Electron configuration (b) Most common ionic charge (c) First ionization energy (d) Atomic radius. 3p6. Na (g) → Na+(g) + eSecond ionization energy is always greater than the first ionization energy. Si. CCl4. (b) Which one is better oxidizing agent. (b) Are these any atoms for which the second ionization energy is larger than the first? Explain. the stronger intermolecular forces. Cl or Br? (c) Cl has higher electron affinity than Br and thus is more easily reduced. because the electron must be removed from a positively charged ions. Therefore Cl is better oxidizing agent. CH4 (a) Na < Mg < Al < Si (b) CH4 < CCl4 < CBr4 The larger the molecular mass. 4s2 (b) +1 +2 (c) K < Ca (d) K > Ca Arrange the following elements in order of inereasing their melting point (a) Na. (a) Define the term ionization energy. Compare the elements K and Ca with respect to the following properties.6 Chapter ( 3 ) 1. 3p6. (a) (b) The ionization energy is the quantity of energy a gaseous atom must absorb so that an electron is stripped from the neutral atom. 3. (b) CBr4. 2p6. 2s2. 2s2. 3s2. Al = 27 ). The alloy has a densily of 2.73 g × 79.7 g Al 2. In simultaneous reactions two or more substances react independently of each other in separate reactions occurring at the same time. but none of the Cu. O = 16. Consecutive reaction Reactions that are carried out one after another in sequence to yield a final product are called consecutive reactions. Theoretical yield The calculated quantity of product in a chemical reaction is called theoretical yield of a reaction.7 Chapter ( 4 ) 1. If 470. N = 14. 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) 1 mol H 2 2 mol Al 27 g Al 100 g alloy 1cm 3 alloy (?) cm alloy = 1gH2 × × × × × 2g H 2 3 mol H 2 1 mol Al 93. An alloy used in fabricating aircraft structures consist of 93. reacts with HCl(oq) ( H = 1. 7% Al and 6.3% Cu by mass.Explain the terms: Limiting reactant. Weak acid Weak acid Acids that are incompletely (partially) ionized in aqueous solution are called weak acids. in gram will be obtained? 2NH3 (g) + CO2 (g) → CO(NH2)2 (aq) + H2O(1) (C = 12. The reactant that is completely consumed in a reaction is known as limiting reactant. 85 g cm-3. Oxidizing agents The substance that causes some other substance to be oxidized and itself reduced.4 % = 500 g of Urea 100 % . 3. Strong electrolyte Strong electrolyte is completely ionized in water solution. and the solution is a good electrical conductor. H =1) (?) g Urea = 470.3434 cm alloy. Simultaneous reaction. What volume of the Al-Cu alloy must be dissolved in an excess of HCl(aq) to produce 1 g H2? If one assumes that all the Al.85 g alloy 3 = 3.4 %.2 LNH3 × 1mol NH 3 1mol Urea 60 g Urea × × 22.73 g Urea (Theoretical yield) Percentage yield = Actual yield = Actual yield × 100 % Theoretical yield 629. 2. What mass of urea.2 L of NH3 gas at STP is used in this reaction and percent yield of urea is 79.4 L NH 3 2 mol NH 3 1mol Urea = 629. 0415 M solution to increase the concentration of exactly 0. Cl = 35.83 ) mL = 229.88 mL 0.5.0415 M C2 = 0.5 M H2SO4 solution from 6M H2SO4 ? V1 = ? C1 = 6M C1 V1 = C2 V2 V1 = 250 mL × 0. 6. What is percent yield.(aq) has a density of 1.25 M volume of water evaporated = V1 – V2 = ( 650 – 134. if the reaction of 30 g P4 and 90 g Cl2 produes 85. (a) What volume of solution must be evaporated from 650 mL of 0. AgNO3.5 M volume of water added = V2 – V1 = ( 250 .12 mL.0415 M V1 = = 134.2 M C1 V1 = C1 V1 650 mL × 0. (Cl = 35.5 M = 20.20.265 mol AgNO 3 1 mol HCl × × 1L AgNO 3( aq ) 1 mol AgNO 3 1 mL HCl ( aq ) 36. 34 % by mass HCl.8 4.17 mL.83 mL 6M V2 = 250 mL C2 = 0.775 mL HCl ( aq ) 1 mol HCl 34 g HCl 1.6 g PCl3 ? ( P =31.025 g/mL.2 M ? V1 = 650 mL V2 = ? C1 = 0.5 g HCl 100 g HCl ( aq ) × × = 2.5 ) P4 (1) + 6Cl2 (g) → 4 PCl3 (1) .88 ) mL = 515. What volume (mL) of this solution would be required to reaSct with 100mL of 0.265 M. H =1) HCl(aq) + AgNO3(aq) → AgCl(s) + HNO3(aq) (?) mL HCl(aq) = 100 × 10-3 L AgNO3(aq) × 0.025 g HCl ( aq ) 5. (b) What volume of water (mL) must be added to prepare 250 mL of 0. (?) g of PCl3 = 0.8450 mol PCl3 × % Yield = 138 g PCl 3 = 116.9 Assuming that P4 is the limiting reagent.+ ClO-3 + 3 H2O 8. (?) mol of PCl 3 = 30 g P4 × 1 mol P4 4 mol PCl 3 × 124 g P 4 1 mol P4 = 0.8450 mol ) of PCl3 is formed.6 g = × 100% 116.61 g = 73. Cl2 is completely consumed. (a) MgCl2 + NaOH → ? Mg2+(aq) + OH-(aq) → Mg(OH)2(s) (b) CuSO4 + Na2CO3→ ? . ∴ Cl2 is limiting reagent.8450 mol PCl3 When the smaller amount ( 0.4 % 7. (?) mol of PCl 3 = 90 g Cl 2 × 1 mol Cl 2 4 mol PCl 3 × 71 g Cl 2 6 mol Cl 2 = 0.61 g 1 mol PCl 3 actual yield × 100 % theoretical yield 85.→ Cl. Write net ionic equation for the following reactions.+ + Mn2+ + H2O → 5 VO+2 + Mn2+ (in acid solution) + 2 H+ ClO-3 + H2O (in base solution) 3 Cl2 + 6 OH .9676 mol PCl3 Assuming that Cl2 is the limiting reagent.→ 5 Cl. (a) VO2+ + MnO4-1 → VO+2 ( ans: ) 5VO2+ + MnO4-1 + H2O (b) Cl2 ( ans: ) + OH . Na2C2O4 requires 25. in gram. What mass of Na2C2O4.8 mL of 0. of a saturated solution of sodium oxalate. □ A gas is composed of a very large number of extremely small particles (molecules or atoms) in constant. □ Molecules collide with one another and with the walls of their container. however.0214 mol KMnO 4 5 mol Na 2 C 2 O 4 × 1000 mLKMnO 4 2 mol KMnO 4 in 50 mL 134 g Na 2 C 2 O 4 = 0. straight-line motion.+ MnO 4 + H + → Mn 2+ + H 2 O + CO 2 (not balance) 4 − 5 C 2 O 2. The titration of 50 mL. would be present in 1 L of this saturated solution? ( C = 12. Deseribe any the assumptions of Kinetic molecular Theory of gases.10 Cu2+(aq) + CO32-(aq) → CuCO3(s) (c) CaCO3 + HCOOH → ? CaCO3 (s) + 2 HCOOH(aq)→ 2 HCOO-(aq) + Ca2+ (aq) + H2O(l) + CO2(g) 9. the total energy remains constant. O= 16. .0214 M. random. In a collision of molecules at constant temperature.1849 g 1 mol Na 2 C 2 O 4 (?) g /L Na2C2O4 = 0.+ 2 MnO 4 + 16 H + → 2 Mn 2+ + 8 H 2 O + 10 CO 2 4 5 mol (?) g Na2C2O4 = × 2 mol 25. □ Individual molecules may gain or lose energy as a result of collisions. The gas is mostly empty space.8 mL KMnO 4 × 0.1849 g 1000 mL × = 3. 6992 g / L 50 mL 1L Chapter ( 5 ) 1. KMnO4. Na = 23 ) − C 2 O 2. □ There are assumed to be no forces between molecules except very briefly during colliding. □ Molecules of a gas are separated by great distances. What mass of He(g) is present in the mixture? (He = 4. H = 1) □ Let mol of He (initial) = x mol (?) mol H2 ( initial ) = 4 g H2 x 1mol H 2 = 2 mol H 2 2 g H2 1mol H 2 = 5 mol H 2 2 g H2 (?) mol H2 ( added ) = 10 g H2 x n1 = ( x + 2 ) mol n2 = ( x + 7 ) mol V1 = V V2 = 2 V By Avogadro’s theory (V ∝ n ) V1 V2 = n1 n 2 ( x + 2) 1 V = ( x+7) 2V x = 3 mol (He) (?) g He = 3 mol He x 4 g He = 12 g He 1 mo He 4. □ Effusion is the escape of gas molecules from their container through a tiny hole.446 L when collected over hexane at 25°C and 738.6 mmHg pressure. A 1. If 10 g of H2 is added to the mixture while conditions are maintained at STP. Determine the vapour pressure of hexane at 25°C. A mixture of 4 g of H2 and unknown quantity of He(g) is maintain at STP.072 g sample of He(g) is found to occupy a volume of 8. 3. in mm Hg. ( He = 4. □ Explain the following terms: (a) diffusion (b) effusion Diffusion is the migration of molecules of different substances as a result of random molecular motion. T = 25°C + 273 = 298 K M = mRT PV . The volume of the gas is double. R = 08205 L atom mol-1 K-1 ).11 2. 3145 kg m 2 s -2 mol -1 K -1 × 308 K = 417.08205 L atm mol −1 K −1 × 308 K = = 44.7758 atm PHe = MV 4 g mol −1 × 8.9882 atm × 1 L PV = RT 08205 L atom mol -1 K -1 × 298 K = 0.6 ) mmHg = 149 mmHg 5.75 g 3 mol N 2 1 mol Na N 3 6.072 g × 08205 L atom mol -1 K -1 × 298 K = = 0.446 L = 589.13 g mol −1 PV 0.2 L of a gas sample measured at 35°C and 754 mm Hg is 2.6 mmHg Ptotal = PHe + Phexane Phexane = ( 738. mRT 2.13 g mol −1 U rms = .3145 kg m2 s-2 mol-1 K-1 . What mass of sodium azide must be dccomposed to produce 1 L of N2(g) at 25°C and 751 mmHg? 2NaN3 (s) → 2Na (s) + 3N2 (g) 1 atm = 0.079 g.24 ms −1 44.9921 atm 760 mm Hg R = 0. T = 35 °C+ 273 = 308 K P = 754 mm Hg × M= 1 atm = 0.6 .0404 mol N 2 × 2 mol NaN 3 65 g NaN 3 × = 1.079 g × 0.9921 atm × 1.589.12 mRT 1.0404 mol N2 (?) Na N 3 = 0.08205 L atom mol-1 K-1 ). The mass of 1. What is the root mean square velocity of this gas ? (R = 8.2 L 3 RT M = 3 × 8.9882 atm 760 mm Hg T = 25C + 273 = 298 K P = 751 mm Hg x PV = nRT n = 0. 6824 a = 40. Calculate the pressure exerted by 1 mol of CO2 (g) confined to a volume of 855 cm3 at 30°C. H = 1 . ( a = 3.8 atm (c) The lower pressure calculated by vender Waal's equation is the result of the intermolecular attraction between the real gas molecules.0 atm V ( vander Waal' s equation ) ( b = 0.5 ) Molar mass of NH3 = 17 g / mol Molar mass of HCl = 36. ****** Chapter ( 6 ) .5 g mol −1 = 0. 5 L2 atom mol-2 ) (a) PV = nRT P= nRT = 29.13 7. Use (a) ideal gas equation and (b) the vander waals equation (c) compare the results and explain. distance from HCl = a cm Distance from NH3 M NH3 dis tan ce from HCl = dis tan ce from NH 3 M HCl a = (100 − a ) 17 g mol −1 36. How many centimeters from HCl end of the tube will white smoke first form ? ( N = 14. Cl = 35.56 cm 8.5 g / mol = (100-a ) cm Let.0427 L mol-1 ) n2 (b) (P + a ) ( V − nb) = n R T V2 n R T n2 a P = − V − n b V2 P = 25. Suppose we put a cotton plug saturated with HCl solution into one end of 1 meter tube and simultaneously insert a plug saturated with aqueous NH3 into the other end. Which one of the following substance is a liquid at room temperature? Give explanation. Calculate (a) ∆Hvap and (b)normal boiling point. CH3NH2. What are the differences between cohesive forces and adhesive forces? 6. Ar. in mmHg? ( Br = 79. CO2 ( Tc = 304.7°C. enthalpy of vaporization. What is meant by the terms instantaneous dipole and induced dipole? Which of the following molecules have the higher boiling point? Explain (i) He (or) Ne (ii) n-pentane (or) neo-pentane. What type of intermolecular attractions are present in each of the following molecules? NH3. 7. 8 K ). SO2 ( Tc = 431K ) 4. Cyclohexanol used in the manufacture of nylon. (i) CH4 (ii) N2 (or) (or) CH3OH NH3 (or) (or) (or) (or) CHCl3 Hl CH4 C6H4(OH)2 12. Equilibrium is established between Br2 (1) and Br2 (g) at 25C.486 g. (a) Explain the meaning of localized and delocalized. N2. (b) Which electrons in graphite are delocalized. has a vapor pressure of 10 mmHg at 56°C and 100 mmHg at 103. 11. Describe briefly the meaning of: surface tension. normal boiling point. A 250 cm3 sample of the vapor weights 0. All electrons in diamond are localized. CH3OH. What is the vapor pressure of bromine at 25°C. Draw the following cubic system and describe the no. the critical point. 2. Which of the following molecules can be liquefied at room temperature ( about 25°C ) ? O2 ( Tc = 154. (Give a brief reason) (i) CH3OH (ii) l2 (iii)NH3 (iv) C6H6 unit cell. ( R = 8.3145 J mol-1 K-1 ) 5.9 ) 3. CH3OH. 8. CO 10. but that in graphite certain electrons are delocalized. Which member of each pair is more soluble in CCl4.14 1. 9.2 K ). LiBr. C3H8. HBr. Zn. of particles (atoms) per . (a) What is meant by the term crystal coordination number. (a) Simple cubic (b) Body – centered cubic (c) Face-centered cubic 14. Cu. Calculate the radius of Ba atom.533 nm on the edge. What is the distance between the plane atoms diffracting this X.15 (iii)CaCl2 (iv) CH3CH2OH (or) (or) CCl4 CH3OCH3 13. KH.5 g/cm3 and crystallized in n BCC. Draw the following cubic system and describe the no. (b) Describe the coordination number of following. What is the radius of K atom. of particles (atoms) per unit cell.534 g / cm3. Polar covalent bond. Chapter ( 7 ) 1. Mg. H2S.154nm from a nickel crystal occurs at an angle of 12°40’. Cr2O3 2. SiCl4. Na 19. 3. (a) Define.346 nm on an edge. Explain the structure of ice crystal lattice and why water is denser than ice. The length of unit all measures 0.ray beam ? 17. The first order reflection of X-rays of wavelength 0. Density of lithium is 0. Which of the following compounds are mostly ionic and which are mostly covalent. . Write an account on liquid crystal. K crystallize in a BCC unit cell measuring Ba has a density of 0. Covalent radius. 15. ( Ba = 137 ) 16. 20. What is the cubic crystal system occupied by lithium? ( Li = 7 ) 18. 16 (b) Indicate which of the following bond have the greatest bond length. And shortest bond length. O2, N2, Br2, BrCl 3. Explain why the incomplete octed of BF3 is found generally. 4. Alghough PCl5 is a known compound, NCl5 is not. What reason can you offer for the failure to observe NCl5 ? 5. What is meant by the terms. (i) Valence electron (ii) Resonance Draw the Lewis structure for N2F2, HNO3, SOCl2, F3SN, I3 -, CS2, PF5, XeF4, C2H4 Explain the electron pair repulsion theory with suitable example. Predict the shape of the following molecules and ions. NH-2, BH3, H2O+, IF5, POCl3, SiO4 Calculate the ionic character of H2O. It’s dipolement is 1.84 D and bond Discuss the nature of covalent bond in HF molecule by using experimental Percent ionic character of HF molecule is 43%. It’s dipolement is µ = 1.9 D. Estimate the C-H bond energy using a bond energy of 436 kJ mol-1 for H2 and the reaction scheme outlined below. C(g) + 2H2(g) → CH4(g) ∆ Hrxn = -792 kJ 6. 7. 8. 9. 10. length is 0.095 nm. 1 D = 3.34 × 10-30 °C.m. ( full ionic charge = 1.6× 10-19C) data. ( µ = 1.91 D, d = 0.092nm ) 11. 12. 13. Enthapy of formation of H2O2 is- 136 kJ mol-1, Estimate the O to O single bond energy. Bond energies for H-H ⇒ 436 kJ mol-1 O=O ⇒ 498 kJ mol-1 H-O ⇒ 464 kJ mol-1 14. Calculate the enthalpy of formation of methanol (CH3OH) → CH3OH (g) + H2 (g) ∆ Hrxn- ? CH4 (g) + H2O (g) Bond energies C – H 414 kJ mol-1 O– H 464 kJ mol-1 H – H 436 kJ mol-1 C – O 360 kJ mol-1 15. Explain the terms. (i) valence band (ii) conduction band 17 16. How does the electron sea model Explain. ? (i) The deformation of metals? (ii) Electrical conductivity of metals? (iii) Lustrous appearance of metals? Chapter 8 What mass of H2SO4 is contatned in 25g of 15% by mass solution? What is the molality of 20% by mass of H2SO4? 14.8 M aqueous ammonia solution has a density of 0.898 g/mL. What is XNH3 is this solution. ( N = 14, H = 1 ) 4. What is the molarity of 10 % by mass H2SO4 solution having a density of 1.07 g/ mL? 5. What mass of water must be added to 1 kg of 1.12 m CH3OH (aq:) to reduce the malality to 1.00 m CH3OH(aq:). 6. Define: Henry’s Law At 0°C and an O2 pressure of 1.00 atm, the aqueous solubility of O2(g) is 48.9 mL O2 per liter. What is the molarity of O2 is a saturated water solution when the O2 is under its nonmal partial pressure in air, 0.2095 atm? 7. Calculate the mole fraction of solute in the following aqueous solution. (C = 12, H = 1, O = 16, N = 14) (a) 13.2 % C2H5 OH by mass (b) 0.512 m urea CO (NH2)2 1. 2. 3. 8. Calculate the vapor pressure of solution at 25°C containing 26.9 g of Urea CO(NH2)2 is 712 g H2O. (Vapor pressure of water at 25°C is 23.8 mmHg) (N =4, H = 1, O = 16, C = 12) 9. An aqueous solution that is 0.205 m urea is formed to boil at 100.025°C. (Kg = 0.512°C m4). Is the actual atmospheric pressure above or below 760 mmHg? 10. 1.2 g sample of an unknown covalent compound is dissolved in 50 g of benzene and the solution freezes at 4.92°C. Determine molar mass of the compound. ( Kf = 5.12°C) 13. An osmotic pressure of 3.56 mmHg is measured for 0.288 g of protein in 25 mL of solution at 25°C. What is the molar mass of the protein? Chapter 9 18 1. Identify acid or base for each of the species in the following forward and reverse reactions. (a) H2O + OCl(b) H2PO − + NH3 4 ⇋ ⇋ HOCl + OHNH + + HPO 2− 4 4 2. If 535 ml of gaseous HCl, at 26.5ºC and 747 mmHg is dissolved in water to prepare 625 mL of solution. What is the pH of this solution. (R = 0.0812 L atm mol-1 K-1) 3. Calcualte the pH of KOH solution that is 3% by mass KOH and has a density of 1.0242 g mL-1. ( K = 39, O = 16, H = 1 ) 4. Calculate the pH of a buffer solution containing. (a) 0.1 M. methanoic acid and 0.01 M sodium methanoate. ( Ka = 1.8 × 10-5 ) (b) 0.15. acetic acid and 0.5 M CH3COONa. (Ka = 1.7 × 1 - 5) What mass of CH3COONa must be dissolved in 0.3L of 0.25 M, CH3COOH to produce a solution pH = 5.1 . (Ka = 1.74 x 10-5 ) ( C = 12, H = 1, O = 16, Na = 23 ) 6. Cocaine (C2H22O4N ) is an alkaloid, that is soluble in water to the extent of 0.17 g / 100 mL. pH of this solution is 10.08. What is the value of Kb for cocaine. 7. The solubiliry of certain metallic hydroxide M(OH)3 is 1.0 x 10-5 mol L-1 Calculate the Ksp of M (OH)3. 8. The solubility of silver carbonate at 25ºC is 1.16×10-1 M. Calculate the solubility product of that compound. 9. Lead (II) iodide saturated solution, its Ksp is 7.1 × 10-9 at 25ºC. Calculate the solubility of Pbl2 ? 5. Chapter 10 1. 0.001 mole of sample of S2 (g) dissociate in 0.5 L flask at 1000K.At equilibrium, 1 × 10-11 mole S (g) is present. What is Kc and Kp? S2 (g) ⇋ 2S(g) 2. What is the percent dissociation of N2O4(g) f 1.5 mole of N2O4(g) is introduced into a evacurated 2.5 L vessel at 25C ? 19 N2O4(g) ⇋ 3. 2NO2(g) Kc = 4.61× 10-3. ⇋ 2C(g) proceed to the The endothermic reaction A (g) + B (g) equilibrium condition at 200°C. Which of the following statements are true? (a) If the mixture is transferred to a reaction vessel of twice the volume. The concentration of reactants and products will remain unchanged. (b) Addition of B will result in the formation of a greater amount of C(g). (c) Lowering the reaction temperature to 100C will result in the formation of a greater amount of C (g). 4. ! g {C;5 os omtrpdiced omtp a 250 mL tlask and it is heated to 300C, where ⇋ PCl3(g) + Cl2(g) the following equilibrium establish. PCls(g) The quantity of Cl2 present at equilibrium is found to be 0.25g. What is the value of Kc for the dissociation reaction at 300C? (P = 31, Cl = 35.5) 5. Describe the relationship between Kc and Kp. 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