A2 Chemistry Unit 4 Notes

March 26, 2018 | Author: Rebecca Siew | Category: Ph, Aldehyde, Acid, Chemical Equilibrium, Catalysis


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This is for the Unit 4 of Edexcel Chemistry A2 Level. Enjoy, and any feedback is very welcome.4.1 Rates of Reactions Reaction Rate = change in amount of reactants/products per unit time (units: mol dm-3s-1) Following a reaction;  gas volume produced (gas syringe)  mass lost (balance)  colour change (colorimeter)  clock reaction (sudden change at particular time means specific concentration of product has  been reached - the shorter the time taken, the faster the rate) electrical conductivity (number of ions will change as reaction occurs) Concentration-Time Graph Rate at any point can be found by drawing a tangent at that point on the graph and finding the gradient. Orders of Reaction The order of reaction = how the reactants concentration affects the rate INCREASE REACTANT – RATE STAYS THE SAME – ORDER OF 0 INCREASE REACTANT – RATE INCREASES BY 1 FACTOR – ORDER OF 1 INCREASE REACTANT – RATE INCREASES BY 2 FACTORS – ORDER OF 2 You can only find the order of a reaction *experimentally* – there is NO theoretical order system. Shapes of Rate-Concentration Graphs tell you the order. 6 4 2 0 time (s) [X] 6 4 2 0 time (s) [X] 6 4 2 0 time (s) [X] 5 4 3 2 1 0 [X] 6 4 rate 2 0 [X] rate 8 6 4 2 0 [X] rate ZERO ORDER FIRST ORDER SECOND ORDER *square brackets indicate concentration. For example [X] = concentration of X. Half-life = time taken for half the reactant to react If the half life is constant = first order If the half life is doubling = second order You can also calculate the half life using reaction rates. For example, if you’re given the rate constant (see below) and the order you can work out half life (you don’t need to know how, just to be aware of it) Rate Equations Rate equation = tell you how the rate is affected by the concentrations of reactants. E.G. Rate = k[A]m[B]n Where: m = order of A n = order of B n+m = overall order k = rate constant (always the same for a reaction at specific temp and pressure, increase temp = increase k = bigger value of k = faster reaction) 033 0. Then.2 0.8 1.g.4 0.4 0. Using data to deduce the order 1) The experiment: titrate sample solutions against sodium thiosulfate and starch to work out the concentration of the iodine. Lastly. Units of k can be found as you know concentration is moldm-3 and rate is moldm-3s-1 using a normal “cancelling” method. Repeat experiment.8 Here.4 0.002 0.002 0. first we changed the concentration of propanone for experiments 1.032 0.058 0.094 Change --Rate doubled Rate trebled No change No change Rate doubled Rate trebled 7 0.002 1.092 0. we changed concentration of H+ in experiments 6 and 7.4 0. Finding the gradient at time zero for each of these plots will give us the INITIAL rate of each.4 0.4 0.2 *Reaction rates won’t be exactly double or treble due to experimental errors etc. 2) From this table we can plot 7 Concentration-Time graphs. we changed the concentration of iodine in experiments 4 and 5.4 [iodine] 0.4 0.002 + [H ] 0. 2 and 3.002 0.004 0. Rate equation = k[propanone][H+] (because anything to the power of 0 is 1) How to calculate rate constant from the orders and rate? Rearrange to make k the subject and calculate. experiment 1 2 3 4 5 6 [propanone] 0.4 0.4 0. .062 0.006 0. 3) Compare the results e. changing only the concentration for ONE REACTANT at a time. Experiment 1 2 3 4 5 6 7 Change compared to experiment 1 --[propanone] doubled [propanone] trebled [iodine] doubled [iodine] trebled [H+] doubled [H+] doubled Rate of reaction 0.EXAMPLE Propanone + Iodine —> Iodopropanone + H+ + I(reaction occurs in acid) Info: First order with respect to propanone and H+ and zero order with respect to iodine Rate equation = k[propanone]1[H+]1[iodine]0 Simplify to.034 0. EXAMPLE: If the rate equation is: rate = k[X][Y] And the two different mechanisms are: 1) X + Y —> Z OR 2) X —> Y + Z From the rate equation. therefore. . we know that X and Y MUST be in the rate determining step. RATE = k*Cl•+*O3] Predicting Mechanisms: Once you know what the rate determining reactants are.4) Now we can work out the rate equation:    Rate is proportional to [propanone] so the reaction is of order 1 with respect to propanone. you can think about what reaction mechanism it follows. Chlorine free radicals in the ozone consist of 2 steps: Cl•(g) + O3(g) —> ClO•(g) + O2(g) ClO•(g) + O•(g) —> Cl•(g) + O2(g) slow rate determining step fast reaction Therefore. it’s mechanism 1 which is the right one. one molecule of X and 2 molecules of Y will be involved in the rate determining step. Rate is proportional to [H+] so the reaction is of order 1 with respect to [H+]. Rate determining step = slowest step in a multi-step reaction (if a reactant appears in the rate equation it MUST be a rate determining step including catalysts which may appear in a rate equation) PREDICITIONS The order of a reaction with respect to a reactant shows the number of molecules that the reactant is involved in with regard to the rate-determining step. Here. EXAMPLE: rate = k[X][Y]2. Rate does not change/is independent of [iodine] so the reaction is of order 0 with respect to iodine. Cl• and O3 must be in the rate equation as they are the reactants from the slowest step. g. This mechanism occurs: *C-Br bond breaks heterolytically (unevenly)    Primary – react by SN2 where 2 molecules/ions are involved in the rate determining step Secondary – react by SN1 and SN2 Tertiary – react by SN1 where 1 molecule/ion is involved in the rate determining step You can see by the rate equation if there are 1 or 2 molecules in the rate determining step. The bond looks like this: Cδ+ — XδThus. In this case the Carbon (Cδ+) to Halogen (Xδ-) bond is POLAR as halogens are much more electronegative than the carbon so they draw in electrons making the Carbon slightly/delta positive. This is where a nucleophile (e. EXAMPLE: Rate = k[X][Y] = 2 molecules in rate determining step = SN2 = primary/secondary halogenoalkane OR Rate = k[X] = 1 molecule in rate determining step = SN1 = tertiary/secondary halogenoalkane . which in turn. the carbon can be easily attacked by a nucleophile who likes positive areas. Br δ-).Halogenoalkanes – Nucleophilic Substitution (SN) Halogenoalkanes can be hydrolysed by OH. :OH-) attacks a molecule and is swapped/substituted for one of the attached groups (e.ions by nucleophilic substitution.g. tells you if the mechanism is SN1 or SN2. .(aq) + I2 (aq) Rate of reaction is inversely proportional to the time taken for the solution to change colour i. EXAMPLE: Iodine clock reaction S2O82. we get. Therefore at high temperatures.31 JK-1mol-1) 1) As EA increases. Therefore large activation energy. just understand the relationship) EA = activation energy (J) R = gas constant (8.(aq) + 2I. k will get smaller.Activation Energy We can calculate the activation energy using the Arrhenius equation: Where. rate will be quicker – this makes sense too! If we “ln” both sides of Arrhenius’ equation. means a slow rate – this makes sense! 2) As T increases. increased rate = decreased time taken k α 1/t We can say that 1/t is the same as k (rate constant) and we can substitute 1/t instead of k in Arrhenius’ equation and find the gradient again to find a value for EA.(aq) —> 2SO42.e. a number) This looks a bit like: y = mx + c If we plot ln k (y) against 1/T (x). k increases. ln A is just a constant. Then R is just a number that we know (8. the gradient we produce will be –EA/R (m).31 JK-1mol-1) we can rearrange and find the activation energy. ln k = – EA/RT + ln A (don’t forget. k = rate constant T = temperature (K) A = another constant Some relationships to note: (you don’t have to learn this. preventing reaction speeding up) example: sulphur in the Haber process is a “poison”– BAD Solid catalysts provide a large surface area for the reaction to occur e.G. this is a homogenous catalysis.g. mesh/powder E.Catalysts Catalyst = increases rate of a reaction by providing an alternative reaction pathway with a LOWER activation energy (EA). They are easily separated from products – GOOD Can be poisoned (i. Disadv: High specificity to the reactions they catalyse. when enzymes catalyse reactions in your body. a substance clings to a catalyst stronger than the reactant would.e. thus reusable. also they are remade. A catalyst will be chemically unchanged at the end of a reaction. HETEROGENOUS CATALYSTS These are catalysts in different physical states to the reactants. all reactants are aqueous. Adv: Small amount needed to catalyse a lot of reactions. E. There are two types of catalysts: HOMOGENOUS CATALYSTS These are catalysts in the same state as the reactants. vanadium pentoxide in the contact process to make sulphuric acid .G. the reaction will be spontaneous. how many different ways particles can be arranged. .G.g. whether the reaction is exo/endothermic. therefore Some randomness. however some endothermic reactions are spontaneous too.from solid to liquid entropy has increased a bit . Sѳ (H2O(g)) = 189 JK-1mol-1 (see below) (see below) (see below) *Note that zero entropy will only occur in a perfectly ordered crystal Affecting Factors: 1. overall entropy has increased = SPONTANEOUS (also depends on ΔH – see below) . entropy.4 JK-1mol-1 e. reactants to products = 2:3) And so.from liquid to gas entropy has increased a lot 4. X -> 2Y 2 moles of Y produced from 1 mole of X therefore entropy has increased 3. More particles = More arrangements = More entropy.4. More quanta (packets of energy) = More ways to arrange particles = More entropy 2.g. Sѳ (H2O(l)) = 70 JK-1mol-1 e. EXAMPLE: A gas will want to escape its bottle because the room it’s in is much bigger and the particles can be arranged in lots of different ways. Systems are MORE energetically stable when disorder/entropy is HIGH. EXAMPLE: NaHCO3(s) 1 mole Solid + H+(aq) —> 1 mole aqueous ions Na+(aq) 1 mole aqueous ions + CO2(g) + 1 mole gas H2O(l) 1 mole liquid Here. entropy. e. If entropy is high enough. Increase in temperature = Increase in energy = More entropy E. SOLID LIQUID GAS No randomness.g. highest lowest entropy. This is to do with entropy. We expect exothermic reactions to be the spontaneous ones. some Most randomness. The units are JK-1mol-1. gas) and there are more moles (e. Sѳ. is the entropy of one mole of a substance under standard conditions of 298K and 1atm. E.g. the products have high entropy states (e.2 Entropy Entropy = a measure of how much disorder there is in a substance.g. Complicated/complex molecules = more entropy DEFINITIONS: Standard entropy of a substance.G. Sѳ (H2O(s)) = 7. 284.8) = .5 JK-1mol-1 [Note: ΔH = -315kJmol-1 is in KILOJOULES. the sum of the entropy changes of the system and the surroundings EXAMPLE: NH3(g) Info: Sѳ (NH3(g)) = 192.Sreactants = 94. ΔSsys = Entropy change of a system.5 + 1057 = + 772.3 JK-1mol-1 Sѳ (HCl(g)) = 186. therefore x1000] [Note: must include sign (and units) with final answer] .(-315000)/298 = + 1057 JK-1mol-1 3) Find total entropy ΔStotal = ΔSsys + ΔSsurr = -284.Sreactants ΔStotal = ΔSsys + ΔSsurr Where. the entropy change between the reactants and the products ΔSsurr = Entropy change of a surrounding ΔStotal = Total entropy change.6 JK-1mol-1 1) Find entropy of the system ΔSsys = Sproducts .8 JK-1mol-1 + HCl(g) —> NH4Cl(s) ΔH = -315kJmol-1 Sѳ (NH4Cl(s)) = 94.3 + 186.5 JK-1mol-1 2) Find entropy of surroundings = .6 – (192.LEARN THESE: ΔSsys = Sproducts . so no further enthalpy change occurs on further dilution.8H2O(s) + 2NH2Cl(s) —> BaCl2(s) + 10H2O(l) + 2NH3(g) When you add barium hydroxide to ammonium chloride:   Smell of ammonia gas Temperature drops below 0˚C 2) Cold pack – NH4NO3(s) and H2O(l) NH4NO3(s) —H2O(l)—> NH4+(aq) + NO3-(aq) When you dissolve ammonium nitrate crystals in water: Looking at the states in both these experiments. the reaction will simply not occur. diamond –> graphite The enthalpy change of hydration. NaCl(s) —> NaCl(aq) .G. if negative X ENDOTHERMIC experiments that are spontaneous: 1) Ba(OH)2(s) and NH2Cl(s) Ba(OH)2. E.G. however the rate of reaction at RTP is so slow because the activation energy needed for it to start is so high.When will a reaction be spontaneous?    Total entropy must increase + ΔS total = kinetically favourable (wants to react. E. we have an INCREASE in entropy (from solids to liquids/aqueous). Na+(g) —> Na+(aq) The standard lattice enthalpy. E. E. DEFINITIONS: Thermodynamic stability – where the ΔStotal is negative. limestone –> CaO + CO2 Kinetic inertness – when the ΔStotal of a reaction is positive. at RTP. not spontaneous) * You can predict ionic compound solubility using the same idea. a reaction can happen spontaneously. Na+(g) + Cl-(g)—> NaCl(s) The enthalpy change of solution. if ΔStotal is positive √. spontaneous) — ΔStotal = kinetically stable (will not react on its own. These reactions are spontaneous EVEN THOUGH the ΔSsurr is negative (because if ΔH is positive for endothermic reactions the equation of ΔSsurr means the overall ΔSsurr will be negative – see above equation) the ΔSsys is GREAT ENOUGH to overcome it. ΔHsol – the enthalpy change when 1 mole of solute is dissolved in sufficient solvent. meaning ΔStotal will be positive still. E.G. ΔHhyd – the enthalpy change when 1 mole of aqueous ions is formed from gaseous ions. ΔHѳlatt – the enthalpy change when 1 mole of a solid ionic compound is formed from gaseous ions under standard conditions (298K and 1atm).G.G. Finding the enthalpy of solution where we use a similar principle to Hess’ Law. Sodium’s ionic radius is bigger than magnesium’s (because Mg has one more proton which has a stronger positive nuclear attraction to its electrons – see unit 1/2) therefore magnesium will have a more negative lattice enthalpy/hydration enthalpy. NaCl has ΔHѳlatt = -780kJmol-1 whereas MgCl2 has ΔHѳlatt = -2526kJmol-1 because magnesium has a charge of 2+ which is greater than sodium’s 1+ 2) Ionic radii. Apples In Ukraine .G. ΔH1 = ΔH2 + ΔH3 REMEMBER (for ΔHsol ): GASEOUS IONS DOWN. 1) Ionic charge.G. = larger charge = more exothermic lattice energy = MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION E. = smaller ionic radii = more exothermic lattice enthalpy = higher charge density = MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION E. AQUEOUS IONS UP Guns In Detroit.Factors affecting ΔHѳlatt AND ΔHhyd include. which increases chances of successful collision.3 Equilibria RECAP: (for exothermic reaction) LE CHATELIER – oppose the motion! Increase Temperature Where does equilibrium move and why? Toward reactants.0moldm-3 H2 = 0. Both these experiments are good economically 1) Contact process – making sulphuric acid 2SO2(g) + O2(g) 2SO3(g) USES = fertilisers. Move to the endothermic side. medicines. Initial concentration: Equilibrium concentration: H2(g) + I2(g) H2 = 1. 1:1 . Higher kinetic energy so more chance of successful collision LOW temp = high yield = but slow process. Dynamic Equilibrium – a reaction that occurs in both ways at the same time (conditions..228moldm-3 From this we can see that the ratio has remained the same.0moldm-3 I2 = 0. therefore less products. in a closed system at constant temperature) Many industrial reactions are reversible. we use this sign for equilibria: E. HIGH pressure = high yield = expensive! NO EFFECT ON EQUILIBRIUM POSITION (will affect rate) Increase Pressure Introduce Catalyst At equilibrium the amount of reactants and products is the SAME.e.4.G.. Toward side with less molecules of gas (only affects gases). dyes. i.228moldm-3 2HI(g) I2 = 1. batteries 2) Haber process – making ammonia N2(g) + 3H2(g) 2NH3(g) USES = fertilisers.G. Particles are pushed together. producing nitrogen-based compounds EXPERIMENT: Hydrogen-Iodine Reaction (REVERSIBLE) There is a relationship between the concentration of initial reactants/products and the equilibrium concentrations which are produced from them E. Kp / Kc What is Kp / Kc? K p / Kc is the ratio of product concentration to reactant concentration. If the equilibrium is HETEROGENOUS (where reactants/products are in different states) then you must LEAVE OUT any concentrations that are solid.1 moldm-3 of iron (II) sulfate solution 2) Leave mixture in stoppered flask at 298K. For Kp.G. *Note: we dont use square brackets for equilibrium partial pressures EXPERIMENT: Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) 1) Add 500cm3 of 0. If the equilibrium is HETEROGENOUS then you only take into account the gases. in the hydrogen-iodine reaction Kc will be. and is commonly known as the equilibrium constant.1moldm-3 silver nitrate solution to 500cm3 of 0. 4X 2Y + 3Z *We can calculate Kp using partial pressures (see below) As long as the equilibrium is HOMOGENOUS (all reactants/products in the same state) then we can use this general rule for finding Kc. For example. HOMOGENOUS equilibriums can be calculated using. *Note: products are 2 because in a balanced equation. it will reach equilibrium 3) Take samples and titrate . there is a 2 in front – see below E. 9kPa = = 1. Step 2) Find the mole fraction.75 moles of PCl3 and so (3 .0061 (0.0439) Ag(s) 0 solid Units Calculating partial pressures. what is the partial pressure of PCl5? Step 1) Find moles at equilibrium of all reactants and products.0439 Ag+(aq) 0.75) will leave us with the moles at equilibrium for PCl5 which is 1.05 – 0.05 0.66 . Minty Fruits Taste Minty EXAMPLE: When 3.75 total moles at equilibrium. Mole fraction of a gas in mixture = Step 3) Find partial pressure. therefore we must also have 1.75 = 4.05 0. the equilibrium mixture has 1.25+1.75+1. Partial pressure of gas = 714 x = 187.0439 (1:1 ratio) Fe3+(aq) 0 0.25 moles.1.0 moles of PCl5 is heated in a closed system.CALCULATION: Reactant/Product Initial concentration (moldm-3) Equilibrium concentration (from titre results) Equilibrium constant Fe2+(aq) 0. If total pressure of the mixture is 714kPa. We know 1. Adding these together we get 1.75 moles of Cl2.75 moles of Cl. ∆Stotal. the equilibrium constant. If. K. K = 10-10 = reaction will not occur K = 10-5 = mostly reactants K = 1 = balanced products and reactants K = 105 = mostly products K = 1010 = reaction complete .Equilibrium and Entropy are related ∆Stotal = R lnK When the total entropy. increases. will also increase. 4. You can work out the equilibrium constant in the same manner as we did before e. However. Equilibrium is set up with mostly reactants (to the left) Strong Acid Base Acid Base Weak Conjugate acid base pairs   HA and A. the equilibrium is very far left and so the equilibrium constant for this reaction is said to have a constant value. the Kc of water is 1. At 298K/1atm. it is always combined with H2O to form HYDROXONIUM IONS – H3O+ *NOTE: Acid-base equilibria involves the transfer of protons. The main ones to know are: Arrhenius definition – when acids/bases dissolve in water then completely/partially dissociate into charged particles (ions) Brønsted–Lowry definition – an acid is a proton donor and a base is a proton acceptor. *HCl has a pH of 0 = completely ionised Weak acids and bases only slightly ionise.g.7 Acid/base Equilibria From the timeline we can see the change in definition of acids through history.are conjugate pairs H2O and H3O+ are conjugate pairs WATER is special – it can behave as a base and an acid. example HCl(g) —> H+(aq) + Cl-(aq) NaOH(s) + H2O(l) —> Na+(aq) + OH-(aq) CH3COOH(aq) CH3COO-(aq) + H+(aq) NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) reason Strong acids and bases ionise almost completely in water. acids (proton donors) will never release a H+ on its own. either donated or accepted.0 x 10-14 mol2dm-6 (We often define this with its own notation – Kw) Kw = Kc x [H2O] = the ionic product of water = [H+][OH-] with UNITS: mol2dm-6 . e.09 x 10-4 pH = -log[5.3 (pH value is small – expected for a strong acid) 2) An acid has a pH of 2.is a measure of the hydrogen ion concentration pH = .29 .log[0.45.09 x 10-4] pH = 3. pH = . Ka = [H+]2/[CH3CH2COOH] [H+] = 5. There are some assumptions to make first: a) Only a tiny amount of product dissociates so initial concentration of reactant = equilibrium concentration of reactant b) All H+ ions come from the acid i.02 moldm-3 solution of propanoic acid (CH3CH2COOH).3 x 10-5 moldm-3. The Ka of propanoic acid is 1. concentration of product 1 = concentration of product 2 1) Calculate the hydrogen ion concentration and the pH of a 0.log[H+] pH = .05] = 1.55 x 10-3 moldm-3 *NOTE: H2SO4 dissociates to give 2[H+] and you will have to divide the final answer by 2 to find your hydrogen ion concentration CALCULATION: finding the pH of a weak acid Weak acids do not fully dissociate so it isn’t as straight forward as above.log [H+] CALCULATION: finding the pH of a strong acid 1) Calculate the pH of 0.05 moldm-3 of nitric acid.pH – “power of hydrogen” . Another constant called Ka is introduced. what is the hydrogen ion concentration? pH = .log[H+] [H+] = 10-pH = 3. [H+] = = = 1.ions and concentration of the base is the same.75 = -log[Ka] 1.g.98 x 10-3 pH = 2.log [Ka] 1) Calculate the pH of 0. drop by drop record amount of base needed to neutralise the acid Repeat for more accurate readings *end point: when the solution changes colour (also known as equivalence point – see below) INDICATORS Indicator Colour in pH when colour acid change Methyl orange red 3. pH = -log [1.g. Methanoic acid has a pKa of 3.1 moldm-3 of NaOH at 298K. HCl) by a factor of 10 increases the pH by 1 diluting a weak acid (e. Therefore.98 x 10-3] Lastly.   diluting a strong acid (e.ion fully dissociates per mole of base so the concentration of OH.78 x 10-4 [H+] = 2. 3.CALCULATION: finding the pH of a strong base One OH. you should be aware that.05moldm-3 of methanoic acid (HCOOH). we use our knowledge of (@ 298K). CH3COOH)by a factor of 10 increases the pH by 0.78 x 10-4 = [H+]2/[0. swirl conical flask for approximate end point* accurate titration.0 x 10-13 mol dm-3 Therefore.3-10 acid pH scale Colour in alkali yellow pink alkali . we need [H+].0 x 10-14 mol2dm-6 1) Find the pH of 0.0 (ph value is large – expected for a strong alkali) CALCULATION: finding the pKa pKa = .0 x 10-13] = 13.4 phenolphthalein colourless 8. Kw = 1.1-4.05] pH = -log[2.5 Ka = 1.53 1) 2) 3) 4) 5) add a measure of acid (with known concentration) to burette rough titration.75. However to work out pH from the formula. pH Curves Strong acid/Strong alkali Strong acid/Weak alkali Phenolphthalein indicator Methyl orange indicator Weak acid/Strong alkali Weak acid/Weak alkali Phenolphthalein indicator Equivalence point = where a tiny amount of alkali causes a sudden big change in pH. Thus. CALCULATION: finding the Ka of a weak acid using a pH curve The half equivalence point is the point where half the acid has been neutralised. At the half equivalence point [HA] = [A-] Therefore. a pH meter is the best thing to use to find the equivalence point as the colour change is gradual and unclear. where the acid is JUST neutralised. where half the volume of strong base has been added the weak acid before equivalence. For the last graph between a weak acid and weak alkali. then we can use Ka = 10-pKa . Equivalence point will vary depending on acid/alkali used. we can say that the half equivalence point is also the pKa of the weak acid. no change in pH.to form CH3COOH so equilibrium shifts to left.6 moldm-3 sodium methanoate. Equation for Ka . so more H+ dissociate from NH4+ so equilibrium shifts to right. no change in pH.4 Food products – changes in pH occur due to fungi and bacteria Sodium citrate Citric acid citrate ions Or Phosphoric acid phosphate ions Or Benzoic acid benzoate ions Carbonic acid (H2CO3) H2CO3 H2O + CO2 Lungs . ADDING ALKALI: (small amount) [OH-] increases which combines with the H+ to form H2O which removes the H+ ions from solution. therefore mostly ammonium ions NH4+(aq) H+(aq) + NH3(aq) This only slightly dissociates. therefore mostly ethanoate ions CH3COOH(aq) CH3COO-(aq) + H+(aq) This only slightly dissociates. therefore mostly ammonium ADDING ACID: (small amount) [H+] increases which combines with the NH3 to form NH4 so equilibrium shifts to left. no change in pH. ADDING ALKALI: (small amount) [OH-] increases which combines with the H+ to form H2O which removes the H+ ions from solution.6 x 10-4 moldm-3. so more H+ dissociate from CH3COOH so equilibrium shifts to right. ACIDIC BUFFERS Weak acid + Salt CH3COO-Na+(aq)—> CH3COO-(aq) + Na+(aq) This fully dissociates. no change in pH.4 moldm-3 of methanoic acid and 0. What is the pH of the buffer? 3. Biological environments (don’t need to learn but be aware of) Example Cells – need constant pH for biochemical reactions to take place Buffer Controlled by the equilibrium between dihydrogen phosphate and hydrogen phosphate H2PO4H+ + HPO42Blood – need to be kept at pH 7. therefore mostly ethanoic acid ADDING ACID: (small amount) [H+] increases which combines with the CH3COO. For methanoic acid Ka= 1.Buffers RESIST changes in pH when small amounts of acid/alkali are added Doesn’t stop the pH from changing completely They only work when small amounts of acids/alkalis are added ALKALINE BUFFERS Weak base + Salt NH4Cl(aq)—> NH4+(aq) + Cl -(aq) This fully dissociates. levels of H2CO3 decrease and so equilibrium moves to the right H2CO3 H+ + HCO3Kidneys control this equilibrium CALCULATION: A buffer solution of 0.by breathing out CO2. Find pH 2. Rearranging for [H+] 1. e. They are optically active (they rotate plane-polarised light) – one will rotate clockwise and the other anticlockwise. SN1 means only 1 molecule will be involved in the rate determining step and SN2 means there are 2 molecules in the rate determining step . A racemic mixture contains equal quantities of each enantiomer of an optically active compound (rotates plane polarised light). the product will be a single enantiomer which will rotate the polarised light.(2 stages) *Remember: from rates of reaction. the product will be a racemic mixture of two optical isomers. groups are fixed in position. First the nucleophile attacks a carbon and then the electrons in the polar bond (Cδ+ — Xδ-) move heterolytically to the Xδ.8 Further Organic Chemistry Isomers Structural: compounds with the same molecular formulae but different structural formulae Stereoisomerism: Optical E/Z only occurs in double bonds where rotation is restricted i. If a reaction is SN1 and you start with one enantiomer.4. For example. nucleophillic substitution can occur in two different ways. Z-ISOMER CIS “same” side E-ISOMER TRANS “opposite sides” determined by how the heaviest molecules are distributed around the double bond. *known as enantiomers The chiral carbons has four different groups attached to it. they cannot be superimposed. Optical Activity can be used to work out a reaction mechanism.(1 stage) If a reaction is SN2 and you start with one enantiomer. The electrons move in the polar bond (Cδ+ — Xδ-) move heterolytically to the Xδ. mirror images of each other. Note: small ketones/Aldehydes will dissolve due to the polarity mentioned above. Ketone Orange No change Aldehyde Orange Silver mirror (Ag(s)) Aldehyde oxidised Fehlings/Benedicts solutions No change Brick red precipitate (Cu+ ions) Aldehyde oxidised Cu2+(aq) + e. however large ketones/Aldehydes will have very strong intermolecular forces and will not dissolve. Experiment must be conducted in fume cupboard.—> racemic mixture/two optical isomers. The electrons in the C=O bond move to the oxygen. Evidence of optical activity: carbonyl group is planar. For this reason. Ag(NH3)2+(aq) + e.4-dinitropheylhydrazine Colourless solution of silver nitrate dissolved in ammonia which gets reduced and changes colour. H+ from water/hydrogen cyanide bond to the oxygen forming OH. aldehydes and ketones have lower boiling points than alcohols (which can hydrogen bond) They can hydrogen bond with water due to their polar Cδ+=Oδ. Tests to identify TEST Bradys reagent Tollen’s reagent + heat (water bath – not flame as flammable!) Info 2. producing two different isomers.—> Ag(s) +2NH3 (aq) Blue solution of copper(II) ions dissolved in NaOH(aq) become Cu+ ions. This is what you expect if the CN can attack either side. Asymmetric (not symmetrical) ketone/aldehyde + CN.—> Cu+(aq) Iodine in alkali + heat Positive test = yellow precipitate (tests for CH3 on carbon If aldehyde positive = ethanal attached to oxygen) If ketone positive = one end is CH3 *NOTE: Brady’s: you can identify carbonyl compounds by the melting point of the orange precipitate against known values . acidified potassium cyanide is used to reduce the risk. Oxygen uses its lone pair to form hydrogen bonds with Hδ+ atoms on the water molecules. nucleophile can attack from either side. NOTE: HCN is a very toxic gas.bond. NUCLEOPHILIC ADDITION Hydrogen cyanide is a weak acid – it partially dissociates in water HCN H+ + CN- CN.is a nucleophile and attacks the slightly positive carbon atom and donates its electrons to it.Aldehydes and Ketones They do not hydrogen bond with themselves as they don’t have a polar Oδ-—Hδ+ bond. 017 moldm-3 Reduction.1 = 0. carboxylic acids are soluble. 1) CH3COOH —LiAlH4 (in dry ether)—> 2CH3OH 2) CH3COOH + PCl5 —> CH3COCl + POCl3 + HCl ethanoic acid ethyl chloride . Ketone —> Secondary alcohol Making a carboxylic acid: 1) Primary alcohol – oxidised – aldehyde – oxidised – carboxylic acid 2) Nitrile – hydrolysed (reflux with HCl then distil) – distilled product is carboxylic acid REACTIONS OF CARBOXYLIC ACIDS: Neutralisation.000417mol conc = mols ÷ vol 0. meaning the boiling point is also higher. carboxylic acids have very high boiling points.OXIDISING Aldehyde —> Carboxylic acid √ [heat with acidified potassium dichromate (VI) ions (oxidising agent)] colour change: ORANGE to GREEN Ketone —> Carboxylic acid X [acidified dichromate (VI) ions are not a strong enough oxidising agent] REDUCING: [LiAlH4 in dry ether] Aldehyde —> Primary alcohol Carboxylic Acids They hydrogen bond with themselves as they do have a polar Oδ-—Hδ+ bond.00125 ÷ 3 = 0. Therefore.000417 = 0.00125mols 3mol NaOH neutralised 1mol citric acid. They can hydrogen bond with water due to their polar Cδ+=Oδ. Dimer: when a molecule hydrogen bonds with just one other molecule.1 moldm-3 of NaOH exactly neutralises 25ml of orange juice.bond. What is the concentration of citric acid in the juice? 3NaOH + C6H8O7 —> Na3C6H5O7 + 3H2O 1) Find moles 2) Find ratio/moles 3) Find concentration mols = conc x vol = 0.G. increasing the size and intermolecular forces of the molecule. 12. Oxygen uses its lone pair to form hydrogen bonds with Hδ+ atoms on the water molecules. 1) CH3COOH + NaOH —> CH3COONa + H2O ethanoic acid sodium ethanoate *NOTE: CO2 causes effervescence 2) 2CH3COOH + Na2CO3 —> 2CH3COONa + H2O + CO2 E. For this reason. however as they get bigger they become less soluble as the intermolecular forces get too strong.0125 x 0.025 ÷ 0. 3:1 0.5 ml of 0. ethanol + methanoic acid will produce an ester call ethyl methanoate Acyl chlorides and Esters REACTIONS OF ACYL CHLORIDES WATER (produce carboxylic acid) . Use: making low fat spread from butter.e.Violent reaction @298K acyl chloride + OH —> COOCH + HCl AMMONIA (produce amide) . Base Hydrolysis Reflux an ester with DILUTE ALKALI (e.Violent reaction at 298K acyl chloride + NH3 —> CONH2 + HCl AMINE (produce Nsub-amide) Violent reaction at 298K acyl chloride + CNH2 —> CONH2C + HCl REACTIONS OF ESTERS Acid hydrolysis – adding water so that the ester splits into an acid and an alcohol (reverse of making ester) using reflux/heat/acid catalyst. biodiesel Problem: some trans isomers have been linked to various diseases Solution: to hydrogenation: trans-esterification. the alcohol that was added comes first i.g. and then mix with sodium hydrogen carbonate solution to remove any acid.Making an ester: Carboxylic acid + alcohol (heat/reflux/acid catalyst) ester It is a reversible reaction so in order to get the ester you must distil off the liquid at 80˚C. Then separate the top layer (ester) using a funnel. USES: ethyl ethanoate is used as a solvent in chromatography as well as pineapple flavouring. Ester + Alcohol —> New ester *NOTE: HCl gas is always given off (observation) Forming a polyester Dicarboxylic acid + Diol —> Polyester + Water . hydrolysing vegetable oils and animal fats (trimesters) and heating them with NaOH produces glycerol (tri-ol) and sodium salt (soap) that we use every day Trans-Esterification (TE) Hydrogenation: adding hydrogen to remove the double bonds. Naming. USES: making soaps.Vigorous reaction with cold water acyl chloride + H2O —> COOH + HCl ALCOHOL (produce ester) . NaOH) producing a carboxylate ion (H3COO-) and an alcohol. sugars) rotate to line up with the field.Cl2 —UV—> 2Cl• . Cooking – Microwave oven Surgery – to kill cancer cells Chemical industry – heating Ultraviolet 400nm-10nm Initiating reactions Has enough energy to split molecules and produce free radicals Example: Initiating reactions such as substitution between halogen and alkane . Danger: n/a Mass Spectroscopy The base peak is the 100% relative abundance which is used to find the RFM M peak is caused by the whole molecular ion which breaks up into fragments of free radicals and positive ions. See below. fats. Some common RFM of fragment ions: CH3+ C2H6+ C3H7+ OH+ CHO+ COOH+ 15 29 43 17 29 45 .g.CF3Cl —UV—> CF3 + Cl• This initiation can cause one Cl• can cause the destruction of two O3 molecules and another Cl• Massive chain reaction. The other peaks are fragment ions of a broken ethanol molecule. but only the positive ion shows up on a mass spectrometer. food (also polar e.9 Spectroscopy and Chromatography EM Radiation: Wavelength: Why: How: Microwaves 1mm-1m Heating Radiation causes electric field. Dryer food with less water content will take longer to cook as water has polar Oδ-—Hδ+ bonds.4. Some protons are aligned in the direction of the magnetic field and others are opposing it. Those opposing it are at a higher energy level and can emit a radiowave to move to the lower energy level. this is because the magnetic field of neighbouring protons interact. Examples of different environments: 2 environments: 4 environments: Chemical shift – is the difference in absorption of a proton relative to TetraMethylSilane (Si(CH3)4). Those in the direction of the magnetic field are at a lower energy level and can absorb a radiowave and move to a higher radiowave. Where δ = 0 is the value of TMS.. Hydrogen is a single proton and so we can use proton NMR to find how many hydrogens there are and how they’re arranged. Normally protons are spinning randomly.. due to them being shielded by electrons experiencing the effects of the strong magnetic force instead. Protons in different environments absorb different amounts of energy. however when you apply a STRONG EXTERNAL MAGNETIC FIELD all the protons line up. In the graph opposite. Each peak = one environment. the peaks of an NMR usually split into smaller peaks. and applying a strong magnetic field will display accordingly. there are two environments (2 peaks) The less shielded a proton is. NMR measures the absorption of energy. the further left the shift will be. Spin-spin coupling – in high res. 2 splits [doublet] = 1 neighbouring proton (or hydrogen) 3 splits [triplet] = 2 neighbouring protons (or hydrogens) 4 splits [quartet] = 3 neighbouring protons (or hydrogens) .NMR Spectroscopy This gives you information about the structure using the idea that every atomic nucleus (with an odd number of protons/neutrons) has a weak magnetic field due to its nuclear spin. The peaks follow an n+1 rule whereby. Hydrogen nuclei in the water in patients body interacts with the radiowaves . checking purity in pharmaceutical industry ADV: non invasive.e. X-ray would be harmful Infrared Spectroscopy 1) IR beam goes through sample 2) IR energy is absorbed by the bonds.e. silica Tube is built into an oven Tube is not heated Sample injected and vaporised Sample forced through tube by high pressure Both rely on different amounts of the sample being moved from the top of the tube to the bottom known as the “retention time” ADV of HPLC over GC: HPLC can be used if sample is heat sensitive or has a high boiling point .g.Different frequencies of wave are absorbed by different densities of tissue . liquid/gas Stationary phase – where molecules can’t move i.A series of images can be produced by moving the beam to build a 3D image USES: cancer/bone and joint treatment. increasing their energy (vibrational) 3) Different bonds in different environments absorb different wavelengths 4) Any wavelengths that you need to know will be in the data book USES: in the chemical industry to determine the extent of a reaction by seeing what bonds are present Chromatography – good at separating and identifying things Mobile phase – where molecules can move i. solid Gas chromatography GC High performance liquid chromatography HPLC Stationary phase is a viscous liquid in a long Stationary phase is small particles of a solid coiled tube e.g. oil packed into a tube e.Patient is placed in a very large magnet and irradiated with radio waves .Magnetic Resonance . brains studies.
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