AQA A-Level Chemistry Year 2 - Amazon Web Services



AQA A-Level Chemistry Year 2Student Book AnswersCHAPTER 1Assignment 1A1.a. ΔfH?[MgCl(s)] = 148 + 738 + 121 ? 349 ? 815 = ?157?kJ?mol?1b. ΔfH?[MgCl3(s)] = 148 + 738 + 1451 + 7733 + 121 ? 349 ? 5540 = +4302?kJ?mol?1A2.Its enthalpy of formation is highly endothermic.A3.Assignment 2A1.1000?gA2.98?gA3.98/18?molesA4.2.4 × 100 kJA5.(2.4 × 100)/(98/18) = 44.1?kJ?mol?1A6.373?KA7.(44.1 × 1000)/373 = 118?J?mol?1?K?1A8.For example, reduce energy loss by improving the kettle’s insulation; make sure that no water vapour condenses and returns to the kettle; try experiments with different quantities of water and compare results.Assignment 3A1.a.ΔfH? [KCl(s)] = (+89) + (+112) + (+419) + (?348) + (?711) = ?439 kJ mol?1b.ΔfH?[KCl2(s)] = +(+89) + (2 × +122) + (+419) + (+3051) + (2 × ?348) + (?2350) = +757 kJ mol?1Generally, a transfer of energy from the surroundings is needed to overcome attractive forces, and energy is transferred to the surroundings (released) when particles are attracted to each other and move towards each other.Atomisation is an endothermic change. A transfer of energy from the surroundings is needed.In metals, a transfer of energy from the surroundings is needed for atoms to break free from the lattice, where positive ions are held together by a sea of mobile, negative electrons.In covalent molecules, a transfer of energy from the surroundings is needed to form separate atoms by breaking covalent bonds.The ionisation enthalpy is endothermic. The negative electrons are attracted by the positive nucleus. A transfer of energy from the surroundings is needed to remove them.The electron affinity is exothermic for the first electron and endothermic for any further electrons. Electrons are attracted towards the positive nucleus, so energy is released. If more electrons are introduced then the electron–electron repulsion means that a transfer of energy from the surroundings is needed.The enthalpy of lattice formation is exothermic. There are attractive forces between the positive and negative ions, so energy is released. ΔfH for KCl is ?429?kJ?mol?1 (exothermic) and ΔfH for KCl2 is +757?kJ?mol?1 (endothermic); therefore, KCl is the more favoured in terms of enthalpy changes.A2. ΔlattH? (NH4Cl) = 705 kJ mol?1Therefore ΔsolH? = (+705) + (?307)+(?381) = 17 kJ mol?1A3.For a reaction to occur spontaneously, the Gibbs free energy must be zero or negative, and ΔG = ΔH ? T?ΔS. The process proceeds spontaneously because there is an increase in entropy as the ordered crystal arrangement changes to a disordered system in solution. Even though ΔH is positive, the increase in entropy (and, combined with the temperature, ?T?ΔS) is sufficiently negative to make ΔG negative.A4.ΔH = mc?ΔT = 200?g × 4.2?J?g?1?K?1 × 16.5?K = 13?860?J = 13.86?kJA5.The molar enthalpy change for ammonium nitrate is +26.5?kJ?mol?1, so the number of moles needed is 5 = 0.52 molPractice questions1aMg2+(g) + 2e? + 2Cl(g) (Note: this is the only answer for the top line)Mg2+(g) + 2e? + Cl2(g)Mg+(g) + e? + Cl2(g)Mg(g) + Cl2(g)1b642 + 150 + 736 + 2 × 121 = 2 × 364 + 2493 = = +1451 kJ mol?11cΔH = ?ΔH(lattice formation) + ΣΔH(hydration)= 2493 – 1920 – 2 × 364 = ?155 kJ mol?11d(i)Increase in disorder on dissolving or ΔS positiveΔG is negative or TΔS > ΔH1d(ii)Moles of NH4Cl = 2/53.5 = 0.0374Heat absorbed = 15 × 0.0374 = 0.561Q = m c ΔTΔT = Q/mc = (0.561×1000)/(50×4.2) = 2.6°CFinal temperature = 20 – 2.6 = 17.4°C2aΔH? = ΣΔH(formation products) - ΣΔH(formation reactants)= 3 × ?111 – (?1669) = +1336ΔS? = ΣS(products) - ΣS(reactants)= 2×28 + 3×198 – (51 + 3×6) = +581ΔG? = ΔH – TΔS= 1336 – (298×581)/1000 = 1163ΔG? is positive (or free energy (G) increases)2bWhen ΔG? = 0 or T = ΔH?/ΔS?= (1336×1000)/581 = 2299 K2cEnthalpy too high or reaction too slow2dMethod: electrolysisConditions: molten or high T or 500–1500°C or dissolved Cryolite3aParticles are in the maximum state of order (entropy is zero at 0 K by definition)13bIce melts (or water freezes)3cThe increase in disorder is bigger (at T2)3d(i)Moles of water = 1.53/18 = 0.085Heat change per mole = 3.49/0.085 = 41.1 kJ mol?13d(ii)ΔG = ΔH – TΔS3d(iii)ΔH = TΔS or ΔS = ΔH/TΔS = 41.1/373 = 0.110 kJ K?1 (mol?1 (or 110 (J K?1 (mol?1))4a1s2 2s2 2p6 3s2 3p64bS–(g)4cThe negative S– ion repels the electron being added4d(i)Enthalpy of atomisation of sulfur4d(ii)Second ionisation enthalpy of calcium4d(iii)Second electron affinity of sulfur4eElectron more strongly attracted nearer to the nucleus or attracted by Ca+ ion4f+178 + 279 + 590 +1145 – 200 + ΔH –3013 + 482 = 0ΔH = 5395aΔS = 238 + 189 – 214 – 3 × 131 = –180 J K–1 mol–1ΔG = ΔH – TΔS = –49 – 523 × (–180) 1000= +45.1 kJ mol–15bWhen ΔG = 0, ΔH = TΔSTherefore, T = ΔH/ΔS = –49 × 1000/–180 = 272 K5cDiagram of a molecule showing O–H bond and two lone pairs on each oxygenHydrogen bonding is a strong enough force to hold methanol molecules together in a liquid.CHAPTER 2Assignment 1A1.The temperature of the reaction mixture.A2.A pipette and safety filler. The size of the pipette depends on the volume required, but 10.0?cm3 is a reasonable choice. Smaller pipettes will be slightly less accurate; larger pipettes require a larger volume of reaction mixture to allow a sufficient sample to be taken.A3.Sodium hydroxide solution (concentration depends on initial concentration of ester).A4.Alkalis are corrosive. It is best to work with concentrations as low as the experiment allows. Eye and hand protection should be used and a safety filler used to fill the pipette. Should any alkali get onto a person’s skin, it should be washed off immediately with cold running water.A5. A6.a. At 0.200?mol?dm?3, rate = 2.3?mol?dm?3?s?1 (note: to calculate rate, time must be converted from minutes to seconds) At 0.150?mol?dm?3, rate = 1.9?mol?dm?3?s?1At 0.100?mol?dm?3, rate = 1.2?mol?dm?3?s?1At 0.050?mol?dm?3, rate = 0.7?mol?dm?3?s?1Because tangents are difficult to draw with accuracy, the rates are given to just 2 significant figures)b.Since the graph is a straight line, the order of the reaction with respect to methyl ethanoate is 1.c. The gradient of the rate against the ester concentration gives the rate constant k for the hydrolysis of methyl ethanoate as 4?s?1.Assignment 2A1.A light probe in, for example, a colorimeter can accurately record the moment at which the dark blue colour appears. Linking the probe to an electronic timer allows the time for the colour change to be measured accurately, avoiding human error when trying the judge the endpoint.A2.TemperatureA3.The final volume of reaction mixture in all experiments is 100?cm3, so the concentration of sodium thiosulfate is 0.005?mol?dm?3.Reaction 1: 2.00 × 10?4?mol?dm?3?s?1. Reaction 2: 1.61 × 10?4?mol?dm?3?s?1. Reaction 3: 1.22 × 10?4?mol?dm?3?s?1. Reaction 4: 0.81 × 10?4?mol?dm?3?s?1. Reaction 5: 0.48 × 10?4?mol?dm?3?s?1.A4.Reaction 1: 0.05?mol?dm?3. Reaction 2: 0.04?mol?dm?3. Reaction 3: 0.03?mol?dm?3. Reaction 4: 0.02?mol?dm?3. Reaction 5: 0.01?mol?dm?3.A5.A6.Straight-line graph, so first order with respect to hydrogen peroxide.Assignment 3A1.a. Changing the concentration of OH? has no effect on rate, so zero order with respect to OH?. Halving the concentration of C4H9Br halves the rate, so first order with respect to C4H9Br.b. Rate = k[C4H9Br]c. From experiment 1, k = 0.001/2.5 = 0.0004 (or 4 × 10?4)?s?1.d. SN1 since the rate-determining step involves just one species (C4H9Br)A2.a.b.[OH?] (mol?dm?3)Rate (mol?dm?3?s?1)0.2413.10.1952.40.1551.50.11810.0990.9Note: Values for rate are not accurate because of the difficulty in drawing tangents.c.d. Since the graph in c is not a straight line, it appears that both 1-bromobutane and sodium hydroxide affect the rate of reaction and so an SN2 mechanism is likely.Required practicalP1.a. Initial rate methodb. i. 10?cm3 pipette, ii. 25?cm3 pipette, iii. dropping pipette, iv. 15?cm3 pipette or, since these are rare, a 50?cm3 burette.c. It is simply an indicator to show the presence of iodine.d. Instead of making a judgement by eye, a light meter can measure the precise moment when a certain depth of darkness appears.P2.a. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)b. Hydrogen – measure the volume produced.c. Temperature.d. Measure the volume of hydrogen produced, either using a gas syringe or by the downward displacement of water.e. Plot a graph of the volume of hydrogen produced against time.If the graph is a straight line, the reaction is zero order and the rate constant is given by its gradient.If the graph is a curve, measure gradients at various points to calculate the rate of reaction at different stages of the reaction. Plot rate against concentration, and if a straight-line graph is produced the reaction is first order.Practice questions1a Initial rate of reaction is proportional to [(CH3)3Br] 1b Initial rate of reaction is not affected by changes in [OH–].1c Rate =k × [CH33Br] so either by using a graph plot or by using values directly in equation:k= rate[CH33Br]=6.0 × 10-32.0 × 10-3=3.0 s–11d Rate =k × CH33Br=3.0 × 4.0 × 10-3=12 mol dm-3 s-1 2a Table of dataExperimentInitial [A] /mol dm–3Initial [B] /mol dm–3Initial rate of reaction /mol dm–3 s–110.0200.0201.2 × 10-420.0400.0409.6 × 10-430.0100.0402.4 × 10-440.0600.0308.1 × 10-450.0400.0357.2 × 10-42b k= rate[A][B]2=1.2 × 10-40.020 × (0.020)2=15.0 mol-2 dm6 s–12cOrder with respect to A = second, order with respect to B = first2dOrder with respect to A = zero, order with respect to B = zero3a Table of dataExperimentInitial [X] /mol dm–3Initial [Y] /mol dm–3Initial rate of reaction /mol dm–3 s–113.0 × 10-24.0 × 10-21.6 × 10-526.0 × 10-24.0 × 10-26.4 × 10-533.0 × 10-216.0 × 10-26.4 × 10-541.5 × 10-216.0 × 10-21.6 × 10-5 3b k= rate[Y][X]2=1.6 × 10-54.0 × 10-2 × (3.0 × 10-2)2=0.444 mol-2 dm6 s–13c Rate constant increased as temperature is increased.3d Rate constant unchanged as concentration of Y is increased at a fixed temperature.4a(i)Expt 2 2.68 ×10?4Expt 3 10.7(2) ×10?4Expt 4 2.08 ×10?34a(ii)= 186 mol?1dm3s?14bIncreases (exponentially): straight line, not a curve5a(i)k = 0 65/(0 15)(0 24)2 = 75.23 to 74.7 mol?2dm6s?15a(ii)0.081 (min sig. figs required)5b(i)25b(ii)06a(i)26a(ii)16a(iii)06bk = rate/[D]2[E] or8 36×10?4/(0.84)2(1.16)= 1.0(2) × 10?3 to 1.05 × 10?3mol?2dm6s?17ak = rate/[CH3CH2COOCH3][H+]or= 1?15×10?4/(0.150)(0.555)= 1.38×10?3 to 1.4×10?3mol?1dm3s?17bans = rate constant × ( ? × 0.150) × (? × 0.555)= rate constant × 0.02082.88 × 10?58aConsider experiments 1 and 2:[B constant], [A] increases × 3: rate increases by 32Therefore, 2nd order with respect to A Consider experiments 2 and 3:[A] increases × 2: rate should increase × 22 but only increases × 2Therefore, halving [B] halves rate and so 1st order with respect to B Rate equation: rate = k[A]2[B]8brate = k [C]2[D]; therefore, k = rate / [C]2[D]k = 7.2 × 10?4/ ((1.9 × 10?2)2(3.5 × 10?2)) = 57.0mol–2 dm6 s–18crate = 57.0 × (3.6 × 10–2)2 × 5.4 × 10–2 = 3.99 × 10–3 (mol dm–3 s–1)8dReaction occurs when molecules have E≥Ea Doubling T causes many more molecules to have this E Whereas doubling [E] only doubles the number with this E 8eEa = RT(lnA – lnk)/1000 Ea = 8.31 × 300 (23.97 – (–5.03))/1000 = 72.3 (kJ mol–1) CHAPTER 3Assignment 1A1.Only a small amount of product is expected, so the reaction can be considered as unlikely to occur.A2.At the higher temperature, the value of Kp will be higher and the yield of NO(g) will increase.A3.Endothermic because Kp increases with an increase in temperature, whereso the yield of NO (right-hand side of equation) increases. Le Chatelier’s principle predicts that if temperature is increased, a reaction will shift in the endothermic direction.A4.At midnight (the start of the graph), the concentration of NO is quite low. As traffic starts to build up, the concentration of NO increases. Following the production of NO, it is then converted to NO2 by oxidation by oxygen in the atmosphere. The NO is being removed as it is formed, so its concentration reaches a maximum value. Also, as the light intensity increases, the concentration of NO2 decreases as it undergoes a series of photochemical reactions (one of the products is ‘ground-level ozone’). As the level of traffic decreases later in the morning, the concentration of NO2 decreases. There is then a slight increase in the level of NO from increased traffic in the afternoon (but this is removed as it is formed, by the ground-level ozone present in the atmosphere from the previous reactions). From then the levels of NO and NO2 decrease.A5.Arguments supporting opinion in favour of LEZs: Fossil-fuel-powered vehicles produce pollution. The photograph shows a brown haze of nitrogen dioxide (which in turn produces low-level ozone). This can cause breathing problems. Arguments against LEZs: People may have to drive around cities to carry out their legitimate work, for example taxi drivers and delivery drivers.Practice questions1a1bKp=(C2H2)(H2)3(CH4)2Kpunits=kPa ×kPa ×kPa ×kPa kPa ×kPa =kPa2Total number of moles = 0.44 + 0.28 + 0.12 = 0.84 molesP(CH4)=200 × 0.440.84=104.76 kPaP(C2H2)=200 × 0.120.84=28.57 kPaP(H2)=200 × 0.280.84=66.67 kPaKp=(28.57)(66.67)3(104.76)2=771.45 kPa21cTo right or to product(s) or forwards; Increase1dTo left or to reagent or backwards; no effect2a(i)Increase because higher P gives lower yield or moves to left, equilibrium shifts to reduce P or equilibrium favours side with fewer moles2a(ii)Endothermic because increase T increases yield or moves to right, equilibrium shifts to reduce T or equilibrium favours endothermic direction2b(i)Moles of iodine = 0.023Moles of HI = 0.1722b(ii)2b(iii)= 0.0179 or 1.79 × 10?23a(i)Moles of C2F2 = 0.40Moles of HCl = 0.803a(ii)3a(iii)mol dm?33b(i)Increase3b(ii)Decrease3cAddition or radical4amol Cl2 = 1.2total mol = 3.84bmol fraction PCl5 = 1.4/3 8 = 0.368mol fraction Cl2 = 1.2/3 8 = 0.3164c(i)pp = mol fraction × total P4c(ii)min: pp PCl5 = 0.368 × 125 = 46(.0) 0.37 × 125 = 46.3max: pp Cl2 = 0.316 × 125 = 39.47 0.32 × 125 = 40.04d 4e(i)No effect4e(ii)Increase4f42 62 / 36.9 = 49.2kPaCHAPTER 4Assignment 1A1.CH3CH(OH)COOH(aq) + H2O(l) ? CH3CH(OH)COO?(aq) + H3O+(aq)A2.CH3CH(OH)COOH(aq) + OH?(aq) ? CH3CH(OH)COO?(aq) + H2O(l)A3.Ka = 1.38 × 10?4?mol?dm?3A4.Lactic acid in venous blood = 0.5 × 10?3?mol?dm?3 to 2.2 × 10?3?mol?dm?3[H+] range = to 2.63 × 10?4?mol?dm?3 to 5.51 × 10?4?mol?dm?3The pH range would be from 3.58 to 3.26.Lactic acid in arterial blood = 0.5 × 10?3?mol?dm?3 to 1.6 × 10?3?mol?dm?3[H+] range = to 2.63 × 10?4?mol?dm?3 to 4.70 × 10?4?mol?dm?3The pH range would be from 3.58 to 3.23.A5.a. The pH values are much lower than blood pH.b. Blood is buffered to resist changes in pH when acid is produced in blood.Assignment 2A1.Ideas that may be included: CO2 causes warming by absorbing infrared. CO2 is released into the atmosphere by burning, respiration and decomposition. The amount of CO2 released has increased rapidly in the last 200 years. CO2 dissolves in the oceans, sets up an equilibrium and lowers the pH. CO2 reacts and forms HCO3? and CO32?, which react to form shells of sea creatures and eventually rocks. Carbonates act as a buffer. The resulting change in pH and the ability of oceans to buffer this must be fully understood. The impact of climate change on weather and marine life is not fully understood.Required practicalP1.Apparatus and instruments for measurement have either a graduation mark or a scale. Regular checks are needed and, if necessary, adjustments are made using standard materials (ones whose properties are known to an appropriate level of accuracy). For a pH meter, the standards used are buffer solutions of known pH. The checks and adjustments are called calibration.P2.a. A pipette has a single graduation mark (though some graduated pipettes have more than one mark). It is used to measure the required volume of the weak acid into a beaker because this is a fixed volume that needs to be measured out accurately.b. A burette is used to measure quantities of the strong base because the volume needed to react completely with the acid is being measured. Since this is not known, a burette has many graduation marks and, therefore, allows the volume of solution dispensed from it to be measured accurately and to the appropriate precision.P3.A measured volume of the weak base is placed in a conical flask. A strong acid of known concentration is added from a burette, with the pH being recorded. The endpoint is identified from the change in the shape of the pH curve.We need the equation for the reaction between the weak base and the strong acid, as this gives the stoichiometry of the reaction.The number of moles of acid added to reach the endpoint is calculated from the volume of the acid and its concentration. The equation for the acid–base reaction enables the number of moles of base to be calculated. Now, both the volume and the number of moles are known and the concentration of the base can be calculated.Practice questions 1a(i)Kw = [H+][OH–]1a(ii)2.34 ×10?71a(iii)2.34 ×10?71a(iv)5.48 to 5.50 ×10?141b[H+] = 10?14 / 0.136= 7.35 ×10?14 OR pOH = 0.87 1pH = 13.132aKa = [H+] = √(1.35 ×10?5 × 0.169)pH = 2.822b(i)CH3CH2COOH + NaOH → CH3CH2COONa + H2Oor CH3CH2COOH + OH– → CH3CH2COO– + H2O 2b(ii)mol propanoic acid = 0.250 – 0.015 = 0.235mol propanoate ions = 0.190 + 0.015 = 0.2052b(iii)(= 1.548 ×10–5)4.813a Rearranging the dissociation constant equation gives[H+]2= Ka × CH3CH2COOH=1.35 × 10-5 × 0.550=7.43 × 10-6 mol2 dm-6So H+=0.002 72 mol dm-3 and pH=–log10H+=2.563b(i) Amount of propanoic acid = 0.0165 mol3b(ii) Amount of sodium hydroxide = 0.002 30 mol3b(iii) Amount of propanoic acid in buffer solution = 0.0165 – 0.00230 = 0.0142 mol3b(iv) pH of buffer -log10H+=-log101.35 × 10-5 × 0.01420.010 + 0.030=2.664a(i)Proton donor - alone4a(ii)Completely dissociated4b(i)7.05 ×10?3 × 103/50 = 0.1414b(ii)?log [H+] or log 1/[H+]4b(iii)0.854b(iv)pH = 1, [H+] = 0.10 (mol dm?3 )(7.05 ×10?3)/0.10vol = 7.05 ×10?2 dm3 or 70.5 cm34c(i)4c(ii)4d(i)[H+] = 1.66 ×10?4= 7.22 × 10?5pKa = 4.144d(ii)Effect = none/ negligible/v small decrease/v small changeSalt or Y? reacts with extra H+ orequilibrium shifts to LHS orH+ is removed as equilibrium shifts to LHSTherefore, [H+] or ratio [HY]/[Y-] or ratio [Y-]/[HY] remains almost constant5a?log [H+]4.57 × 10?35b(i) [5b(ii) = 1.39 × 10?4mol dm?35b(iii)pKa = 3.865c(i)100030/1000 × 0.480 = 0.0144 or 1.44 ×10?25c(ii)18/1000 × 0.350 = 0.0063 or 6.3 × 10?35c(iii)0.0144 × 2(0.0063) = 1.80 × 10?35c(iv)1.80 × 10-3 × 1000/48 = 0.03755c(v)10?14 / 0.0375 = 2.66 × 10?13or pOH = 1.426pH = 12.576a(i)[H+][OH?]1–log [H+]6a(ii)[H+] = [OH?]6a(iii)(2.0 × 10?3) × 0.5 = 1.0 × 10?36a(iv)pH = 10.406b(i)[H+] = √(1.35 × 10?5) × 0.125 ( = 1.30 × 10?3)pH = 2.896c(i)(50.0 × 10?3) × 0.125 = 6.25 × 10?36c(ii)(6.25 × 10?3) – (1.0 × 10?3) = 5.25 × 10?36c(iii)Mol salt formed = 1.0 × 10?3pH = 4.157a(i)BCA7a(ii)Cresolphthalein or thymolphthalein7b(i)-log[H+]; 17b(ii)[H+] = 1.259 × 10?12 or pOH = 14 – pHor = 2.10= 7.94 × 10?37c(i)Ka = [H+]2/[CH3CH2COOH] or [H+]2/[HA] or [H+] = [A?][H+] = √1.35 × 10?5 × 0.117 = 1.257 × 10?3pH = 2.907c(ii)Ka = [H+] or pKa = pHpH = 4.878apH = ?log[H+][H+] = [A?][H+] = √ 1.74 × 10?5 × 0.15pH = 2.798b(i)Solution which resists change in pH /maintains pH despite the addition of (small amounts of) acid/base (or dilution)8b(ii)CH3COO? + H+ → CH3COOH8c(i)= 2.61 × 10?5pH = 4.588c(ii)Moles H+ added = 10 × 10?3 × 1.0 = 0.01 1Moles ethanoic acid after addition = 0.15 + 0.01 = 0.16 1Moles ethanoate ions after addition = 0.10 – 0.01 = 0.09pH = 4.519aConcentration of acid: m1v1 = m2v2Hence 25 × m1 = 18.2 × 0.150ormoles NaOH = 2.73 ×10?3m1 = 18.2 × 0.150 / 25 = 0.1099b(i)Ka = [H+][A?] / [HA]9b(ii)pKa = –log Ka9b(iii)[A?] = [HA]Hence Ka = [H+][A?] / [HA] = [H+]and –log Ka = –log [H+]9cRatio [A?] : [HA] remains constantHence as [H+] = Ka [HA] / [A?], [H+] remains constant10[H+] = √Ka[HA] = √(1.15×10?4 × 0.5)=7.58×10?3 mol dm?3pH = -log10[H+] (or log or lg)pH = 2.1211aBurette, because it can deliver variable volumes11bThe change in pH is gradual / not rapid at the end pointAn indicator would change colour over a range of volumes of sodium hydroxide11c[H+] = 10–pH = 1.58 × 10–12 Kw = [H+] [OH–]; therefore [OH–] = Kw / [H+]Therefore, [OH–] = 1 × 10–14/1.58 × 10–12 = 6.33 × 10–3 (mol dm–3)11dAt this point, [NH3] = [H+]Therefore, [H+] = 10–4.6 = 2.51 × 10–5Ka = (2.51 × 10–5)2/2 = 3.15 × 10–10 (mol dm–3)11eWhen [NH3] = [NH4+], Ka = [H]Therefore – log Ka = –log [H+]Therefore pH = – log10(3.15 × 10–10) = 9.5012a Ethanedioic acid dissociates more readily than ethanoic acid, partly because there are two –COOH groups per molecule. Electronegativity measures the tendency of an atom to attract a bonding pair of electrons. Oxygen atoms are more electronegative than carbon atoms. –COOH groups tend to attract electrons onto the oxygen atoms, whereas –CH3 groups tend to push electrons away. Therefore ethanedioic acid has a greater tendency to pull the electrons towards the oxygen atoms within the molecule, meaning that the first proton more readily dissociates than with ethanoic acid, so ethanedioic acid is a stronger acid than ethanoic acid.12bMoles of NaOH = Moles of HOOCCOO– formed = 6.00 × 10–2Moles of HOOCCOOH remaining = 1.00 × 10–1 – 6.00 × 10–2= 4.00 × 10–2Ka = [H+][A–]/[HA][H+] = Ka × [HA]/[A–][H+] = 5.89 × 10–2 × (4.00 × 10–2/V)/(6.00 × 10–2/V) = 3.927 × 10–2pH = –log10(3.927 ×10–2) = 1.406 = 1.41CHAPTER 5Assignment 1A1.A volumetric flask can measure a specified volume precisely.A2.The measured rotation of sample 4 is anomalous. The mass of sugar was not weighed accurately and was too high, the volume of water was not measured accurately and was too small, or the polarimeter was not calibrated properly.A3.0.5?g?dm?3A4.Average measured rotation = 3.33°Specific rotation = A4.d-(+)-sucroseA5.The measured rotations are small and close to the precision of the instrument, which gives high percentage errors from the equipment. Larger values of measured rotations would give lower percentage errors. Increasing the concentration of the sugar solution by increasing the mass of sucrose or decreasing the volume of water would increase the measured rotation.A6.Nine.A7.Specific rotation = 66.6° = Concentration c = 2.25?g?dm?3Assignment 2A1.?102°A2.It contains a 50?:?50 ratio of d-(+)-limonene and l-(-)-limoneneA3.PolarimeterA4.IR spectroscopy, 1H NMR, 13C NMR, mass spectroscopy, HPLCA5.The enantiomers have identical chemical and physical properties, including spectra.Assignment 3 A1.A2.The type of isomerism is optical isomerism.A3.It is a 50?:?50 mixture of d- and l-enantiomers.A4.Both contain a chiral centre. For both, the d-isomer is biologically active. They both have a benzene ring and a carboxylic acid functional group. Parts of the structures are identical, as shown in the blue boxes.A5.In both drugs, the d-isomer is the active compound. In ibuprofen, the l-isomer is not biologically active but it is not toxic, and so it can be sold as a racemic mixture of isomers. For naproxen, because the l-isomer is toxic, it is important to only sell the d-isomer. Naproxen will have to be prepared as a single enantiomer and so it will be more expensive to prepare than the racemate of ibuprofen.Practice questions1a4 (monochloro isomers) 1b23The major product exists as a pair of enantiomersThe third isomer is 1-bromobutane (minor product)Because it is obtained via primary carbocation4Compounds/molecules with same structural formula but with bonds/atoms/groups arranged differently in space or in 3D. Use plane-polarised light, which will be rotated by samples in opposite directions.5.B. Four different atoms or groups attached to a carbon atom6.A. CH3CH2CH(OH)CH37.C. Three8.B. Rotation of plane-polarised light9.B. One compound is optically inactive and the enantiomers of the other compound are present in a 50?:?50 ratio.CHAPTER 6Assignment 1A1.Nucleophilic additionA2.A3.A4.No, it is a racemate because the addition of the cyanide anion can occur equally from either side of the planar aldehyde to give a 50?:?50 mixture of enantiomers.A5.It will be faster if ionic KCN is used (a source of cyanide ion), as hydrogen cyanide is a very weak acid. The degree of ionisation in solution, producing hydrogen ions (H+) and the reactive nucleophilic cyanide ions (:CN?), is very small (Ka for HCN = 4.0 × 10?10?mol?dm?3).A5.Potassium cyanide is a highly toxic material that causes respiratory failure, convulsions and death if inhaled or absorbed through the skin. Disposable gloves should be used and all reactions undertaken in a fume cupboard, where vapours can be removed. As the reaction involves the addition of acid in a second step, then any unreacted potassium cyanide will react to give hydrogen cyanide gas, which is also very toxic and might be accidently inhaled.Assignment 2A1.A2. This is a reduction or a nucleophilic addition.A3.A4.A5.Required PracticalP1.Sample preparationFor each of the solids, add about 0.5?g to a clean test tube and dissolve it in approximately 5?cm3 of deionised water. If the sample does not fully dissolve, warm in a water bath until dissolved. Label each test tube with a permanent marker in order not to confuse the samples.Fehling’s solutionFor each sample to be tested, make up a fresh sample of Fehling’s solution by mixing about 1?cm3 of Fehling’s A and Fehling’s B solutions in a clean test tube with gentle swirling. When transferring the solutions, use a clean pipette for each solution. Check the Fehling’s test solution is blue in colour. Next add a few drops of the solution of one of the unknown samples to be tested, swirling the tube gently. Warm the reaction mixture in a water bath (at approximately 60?°C) for a few minutes. Note any colour change in your lab book. Repeat the whole process (making fresh Fehling’s test solutions each time) for the other four samples.pound testedOutcome of testOxidation state of copperNo reactionSolution remains blueCu(II)ReactionBrick-red precipitate formedCu(I)No reactionSolution remains blueCu(II)ReactionBrick-red precipitate formedCu(I)No reactionSolution remains blueCu(II)P3.You can heat the sample by placing the test tube in a bath of hot water.P4.a. You could use either Tollens’ reagent or Fehling’s solution. These distinguish between aldehydes and ketones.b. You would require two clean test tubes. For Fehling’s solution, you would require Fehling’s A and B solutions, three clean pipettes and a hot water bath at 60?°C. For Tollens’ reagent, you would require Tollens’ reagent, two pipettes and possibly a water bath at 60?°C (the reaction may go at room temperature).c. You would add a few drops of each material to the test solution (Fehling’s or Tollens) in a clean test tube. Then warm the tube and observe if any colour change occurs.d. In the Tollens test, the aldehyde would be oxidised to a carboxylic acid and the Ag(I) reduced to an Ag(0) mirror. This would be observed coating the test tube. The ketone would not undergo any reaction and the colourless test solution would remain. In the Fehling’s test, the aldehyde would be oxidised to the carboxylic acid and the blue Cu(II) complex reduced to an insoluble Cu(I) brick-red precipitate. The ketone would not undergo any reaction and the blue test solution would remain.P5.a. For every 1?mole of AgNO3 we require 2?moles of NH3.5?cm3 of 1?M AgNO3 solution = 0.005?moles. We therefore require 0.01?moles of NH3. This is 5?cm3 of a 2?M aqueous NH3 solution.b. None. The Tollens reaction occurs only with aldehydes. Propanone is a ketone.c. For every mole of aldehyde oxidised, 2?moles of Ag(0) are formed. Therefore we would obtain 0.1?mol of Ag(0). The relative atomic mass of silver is 107.9?g per mole, so we would expect 10.79?g of silver.Practice questions 1aCH3CH2COCH3 + 2[H] → CH3CH2CH(OH)CH31bThe carbonyl bond is planar in shape, so the [H] has an equal likelihood of attacking from either side of the molecule. This means that when the product is chiral a racemic mixture will be formed and so will not rotate the plane of polarised light.2b. 2-methylbutan-2-ol3aNucleophilic addition2-hydroxy-2-methylpentan(e)nitrile3bProduct from Q is a racemic mixture/ equal amounts of enantiomersRacemic mixture is inactive or inactiveProduct from R is inactive (molecule) or has no chiral centre3aCH3CH2CHO + HCN → CH3CH2CH(OH)CN or C2H5CH(OH)CN2-Hydroxybutanenitrile OR 2-hydroxybutanonitrile 3bNucleophilic addition 3c(Propanone) slower or propanal faster Inductive effects of alkyl groups or C of C=O less δ+ in propanoneor alkyl groups in ketone hinder attack or easier to attack at end of chain4aNucleophilic addition4bPlanar C=O (bond/group) Attack (equally likely) from either side Product is a racemic mixture formed, or a 50:50 mixture of each enantiomer equally likely5Nucleophilic addition6aNucleophilic addition6b(i)6b(ii)(Plane) polarized light Rotated in opposite directions (equally) (only allow if M1 correct or close)6c2-Hydroxybutane(-1-)nitrile 6dWeak acid / (acid) only slightly / partially dissociated/ionised [CN–] very lowCHAPTER 7Assignment A1A1.A2.Olive oil contains two esters of oleic acid and one of palmitic acid, so we would need a ratio of glycerol?:?oleic acid?:?palmitic acid of 1?:?2?:?1.A3.If we mix the two carboxylic acids oleic acid and palmitic acid with glycerol, statistically it is possible to make a range of triglycerides:A4.Animal fat contains the saturated fatty acid stearic acid, while olive oil contains the unsaturated fatty acid oleic acid. Saturated fatty acids have higher melting points than unsaturated fatty acids because they have better packing in the solid state and greater van der Waals interactions. Consequently, animal fat has a higher melting point and is a solid at room temperature.A5.Trimyristin is made up of saturated fatty acids so will be a solid at room temperature.Assignment A2A1.A2.Moles of methyl benzoate = 4/136 = 0.03 moles.Moles of NaOH (20?cm3 of 2?mol?dm?3) = 0.04?moles.A3.The condenser was added so that the reaction mixture could boil without the loss of any of the reagents. Antibumping granules were added to ensure smooth boiling.A4.An excess of NaOH was used, so initially the reaction mixture will be basic and red litmus would turn blue.A5.In order to turn the carboxylate salt into the carboxylic acid. Blue litmus should turn red:A6.For efficient recrystallisation, a saturated solution of the compound in the hot solvent is required. When cooled, the compound will start to crystallise from the saturated solution as the solubility of the compound in the solvent decreases. If too much hot solvent is used initially, the compound might not crystallise out of solution, even at room temperature.A7.Impure compounds have wide melting point ranges which are lower than that of the pure compound. We can conclude that although the reaction worked, the recrystallisation did not provide us with a very pure product.Assignment A3A1.A2. Second generationA3. The amount of carbon (usually carbon dioxide) released is equal to the amount of carbon dioxide captured in its formation.A4. Particulates can cause asthma. Carbon monoxide is a poisonous gas which interferes with oxygen transport in the blood. Nitrogen oxides can convert oxygen to ground-level zone, which can affect breathing and the growth of plants. Greenhouse gases result in warming of the atmosphere and cause climate change. Sulfur oxides are acidic and can damage buildings and living organisms.A5. Land that could be used for growing food could be taken for biodiesel production, thus increasing world hunger.A6. Saturated fatty acid esters will have a high melting point because of the van der Waals interactions between the chains. Therefore the fuel could freeze in winter.A7.Assignment A4A1.Moles of salicylic acid used = 6.0/138 = 0.043 molesMoles of aspirin produced = 6.1/164 = 0.037 moles% yield = 0.037/0.043 × 100% = 86.0%A2.It is cheaper, reacts more slowly and does not evolve HCl.A3.% Atom economy = molecular mass of desired product/sum of molecular masses for all reactants × 100%.a. Molecule mass: ethanoyl chloride = 78.5, salicyclic acid = 138, aspirin = 164Atom economy = 164/216.5 = 75.7%b. Molecule mass: ethanoic anhydride = 102, salicyclic acid = 138, aspirin = 164Atom economy = 164/240 = 68.3%The use of ethanoyl chloride has greater atom economy.A4.Crush the sample into a powder using a spatula on a filter paper. Add a small amount of the sample to a melting point capillary by pushing the open end of the capillary into the powder. Tap the tube so that the powder settles to the bottom of the tube. Make sure that about 3–4?mm of the tube is filled with the sample.A5.The ethanol solution contained both the aspirin and any unreacted impurities, such as salicylic acid. Aspirin is not very soluble in water at room temperature, whereas salicylic acid (being a carboxylic acid) is more soluble. Upon cooling in water, the aspirin recrystallises and the salicylic acid remains in the water.Required Practical 10bP1.So that any pressure build-up can escape to the atmosphere.P2.If the apparatus was disassembled or the water supply to the condenser was shut off then the hot vapours in the flask would escape, which would be a safety hazard.P3.The reagents would evaporate and the reaction would fail.P4.Pour all the liquid back into the separating funnel, wait for the two layers to separate and start again.P5.a. The liberation of carbon dioxide, as well as the exothermic nature of the acid–base reaction causing some of the solvent to vaporise.b. To remove the acidic ethanoic acid from the organic layer by neutralising it with a base to provide the water-soluble sodium ethanoate salt.c. The ethyl ethanoate will be in the top (organic) layer.d. The anhydrous calcium chloride removes any residual water carried over from the separation using the separating funnel.P6. FiltrationP7. If the stopper is left in the top of the separating funnel and the tap opened, the liquid does not drain out easily. This is because as the liquid leaves the funnel, a partial vacuum is formed above the liquid, which stops the liquid draining.Required Practical 10aP1.If the impurities are soluble, the solid compound should be highly soluble in the solvent at high temperature but relatively insoluble at room temperature. The impurity should remain soluble at room temperature.If the impurities are insoluble, the solid compound should be highly soluble in the solvent at high temperature but relatively insoluble at room temperature. The impurity should remain insoluble at high temperature.P2.For efficient recrystallisation, a saturated solution of the compound in the hot solvent is required. When cooled, the compound will start to crystallise from the saturated solution as the solubility of the compound decreases. If too much hot solvent was used initially, the compound might not crystallise out of solution, even at room temperature.P3.a. 10?g of compound A in mixture, solubility at 90?°C = 20?g in 100?cm3 in water. So 10?g will be soluble in 50?cm3.b. At 20?°C the solubility of A is 0.2?g in 100?cm3, so in 50?cm3 only 0.1?g would remain in solution.c. The maximum recoverable amount is 9.9?g (or 99%). There is 1?g of B in the mixture. The solubility of B is 2.5?g per 100?cm3 of water, so in 50?cm3 a maximum of 1.25?g would remain in solution. As only 1?g of B is present, it will all remain in solution. Therefore recrystallised A will be pure.P4.To allow sufficient time for the heat to transfer from the heating solution through the capillary to the sample. If the sample is heated too quickly, the heating block will be at a higher temperature than the sample. In addition, slower heating allows a more accurate melting range to be reported.Practice questions 71(a)1(b)2a(Nucleophilic) addition-elimination2b(i)To ensure the hot solution would be saturated / crystals would form on cooling 2b(ii)Yield lower if warm / solubility higher if warm 2b(iii)Air passes through the sample not just round it 2b(iv)To wash away soluble impurities 2cWaterPress the sample of crystals between filter papers2dMr product = 135.0 3aPropan(e)-1,2,3-triol or 1,2,3- propan(e)triol3bSoaps3c(i)Biodiesel3c(ii)3c(iii)CH3(CH2)12COOCH3 + 21? O2 → 15CO2 + 15 H2O4(Nucleophilic) addition-eliminationN-ethylpropanamide5aCH3CH2CH2COOH + CH3CH2OH (or C2H5OH) → CH3CH2CH2COOCH2CH3 + H2OH2SO4 or HCl or H3PO45bH3CH2CH2CH2OH + (CH3CO)2O → CH3COOCH2CH2CH2CH3 + CH3COOH5c(Nucleophilic) addition-elimination5d6?Nucleophilic) addition-elimination N-propylethanamide 7a(i)3CH3OHHOCH2CH(OH)CH2OH7a(ii)C17H35COOCH3 + 27? or 55/2 O2 → 19CO2 + 19H2O8Electron pair donor or lone pair donor (Acid) anhydride9a(i)9a(ii)(Nucleophilic) addition-elimination or (nucleophilic) addition followed by elimination9a(iii)Any two from: ethanoic anhydride is ? less corrosive ? less vulnerable to hydrolysis ? less dangerous to use, ? less violent/exothermic/vigorous reaction OR more controllable rxn ? does not produce toxic/corrosive/harmful fumes (of HCl) OR does not produce HCl ? less volatile 9bCHAPTER 8Assignment 1A1.Accept 4-nitromethylbenzene, 1-methyl-4-nitrobenzene or 4-nitrotoluene.A2.Electrophilic substitution by nitronium ion (NO2+).A3. A4.Increases it.A5.Decreases it.Assignment 2A1. A2.The electrophile is NO2+:A3.As the acids are corrosive, rubber gloves must be worn, as well as a lab coat and safety spectacles. Because concentrated acids fume in air, the reaction should be carried out in a fume cupboard.A4.14?cm3 of liquids and 4.00?g of solid are reacted in the conical flask, which must be stirred. As a conical flask should only be filled between a third and half full, either a 50?cm3 or 100?cm3 flask would be chosen.A5.Both the formation of the nitronium ion and the nitration reaction are exothermic. Cooling the reagents ensures that the reaction mixture does not boil and the reaction does not become too violent.A6.A Pasteur pipette.A7.a. It is important to prepare a hot saturated solution of the product to be recrystallised. This ensures the maximum yield when the solution cools to room temperature and the product crystallises out.b. You can heat the ethanol using a water bath.A8.To assess the purity of the recrystallised product.Assignment 3A1.A lone pair acceptorA2.A3.A4.Reduction reactionA5.Another reducing agent that can be used is nickel and hydrogen.A6.Number of moles of B = 1.34/134 = 0.1?mol. A 75% yield would give 0.075?mol of product C.0.075?mol C = 0.075 × 176 = 1.32?gA7.Ethanoyl chloride will react vigorously with water, producing ethanoic acid and hydrochloric acid gas. Precautions must be taken so that moisture is not allowed into the reaction flask. Clean, dry apparatus must be used.A8.It could be purified by distillationPractice questions 12aor C6H5NHCOCH3 + NO2+ → C6H4(NHCOCH3)NO2 + H+2bElectrophilic substitution 2cHydrolysis 2dSn/HCl3a(i)NO2+HNO3 + 2H2SO4 →?NO2+ + 2HSO4– + H3O+or HNO3 + H2SO4 →?NO2+ + HSO4– + H2O3a(ii)Electrophilic substitution3bH2/Ni or H2/Pt or Sn/HCl or Fe/HCl, dilute H2SO44aCH3CH2COCl or CH3CH2CClO or propanoyl chloride or (CH3CH2CO)2O or propanoic anhydrideAlCl3 or FeCl3CH3CH2COCl + AlCl3 → CH3CH2CO+ + AlCl4–4b4cTollens or ammoniacal silver nitrate 5aBenzene is more stable than cyclohexatrieneExpected ΔH? hydrogenation of C6H6 is 3(–120) = –360 kJ mol?1Actual ΔH? hydrogenation of benzene is 152 kJ mol?1 (less exothermic) or 152 kJ mol?1 different from expected because of delocalisation or electrons spread out or resonance 5bConc HNO3, Conc H2SO42 H2SO4 + HNO3 → 2 HSO4– + NO2+ + H3O+ or H2SO4 + HNO3 → HSO4– + NO2+ + H2Oor via two equations:H2SO4 + HNO3 → HSO4– + H2NO3+ H2NO3+ → NO2+ + H2O 6a(i)Conc HNO3Conc H2SO46a(ii)2 H2SO4 + HNO3 → 2 HSO4– + NO2+ + H3O+or H2SO4 + HNO3 → HSO4– + NO2+ + H2OVia two equations:H2SO4 + HNO3 → HSO4– + H2NO3+ H2NO3+ → NO2+ + H2O6a(iii)7aSn / HCl or Fe / HClEquation must use molecular formulaeC6H4N2O4 + 12 [H] → C6H8N2 + 4H2O7bH2 (Ni / Pt)CH2In benzene 120°In cyclohexane 109° 28′ or 109?°7c(i)Nucleophilic addition 7c(ii)Planar C=O (bond/group)Attack (equally likely) from either side About product: racemic mixture formed or 50:50 mixture or each enantiomer equally likely CHAPTER 9Assignment 1A1.Nucleophilic substitutionA2.Reduction reactionA3.Ethanoic anhydride and a base, or ethanoyl chloride and a base.A4.A5.The primary amine product is still nucleophilic and can react further, and mixtures of secondary and tertiary amines and quaternary ammonium salts are inevitably produced.A6.The overall yield of A → B → C → D is [0.91 × 0.87 × 0.96] × 100% = 76%.The overall yield of E → C → D is [0.65 × 0.96] × 100% = 62.4%.Despite the second route being shorter, the overall yield is lower so A → B → C → D would be favoured.This does not take into account the cost of the reagents and any cost of purification, and the safety considerations of using toxic KCN. When this is taken into account, the lower-yielding route might become more attractive.A7.Drug molecules must interact with their biological receptors. The binding sites have a defined three-dimensional shape. H-bonding, ionic bonding or van der Waals interactions between drugs and their receptors in the binding site are important for binding. The drug molecule must fit into the binding site and its functional groups must be displayed so that binding interactions are maximised. Molecule D is similar in size and shape to benzedrex and has a group capable of H-bonding in a similar position.Assignment 2A1. 0–5?°CA2 a. NaNO2 + HCl → HNO2 + NaClb. Nitrous acid and the diazonium salt are unstable and decompose above 5?°C.c. 4-MethylazobenzeneA3.The N=N double bond in the chromophore allows the delocalised electrons of one benzene ring to be extended into the second ring. The delocalised electrons in the chromophore can absorb certain wavelengths of visible light, and the light not absorbed gives rise to the observed colour.A4.The NH2 group can hydrogen-bond with the surface molecules of the fabric.A5.The CO2? group can make an ionic bond with the surface molecules of the fabric. Ionic bonds are normally much stronger than H-bonds.A6.Practice questions 132aDimethylamine2bNucleophilic substitution2cQuaternary ammonium salt(Cationic) surfactant / bactericide / detergent / fabric softener or conditioner / hair conditioner2d3a(i)Concentrated HNO3, concentrated H2SO4HNO3 + 2 H2SO4 → NO2+ + H3O+ + 2HSO4?or HNO3 + H2SO4 → NO2+ + H2O + HSO4?3a(ii)Electrophilic substitution3bSn or Fe / HCl or Ni / H23c(i)NH3Use an excess of ammonia3c(ii)Nucleophilic substitution3dLone pair on N less available, delocalised into the ring (Q of L)3e3f4a(Nucleophilic) addition-eliminationN-ethylpropanamide4bCH3CN or ethan(e)nitrile or ethanonitrileStep 1 Cl2uv or above 300 °CStep 2 KCNaqueous and alcoholic (both needed)Step 3 H2/Ni or LiAlH4 or Na/C2H5OH5aProton acceptor 5b(i)CH3CH2NH2 + H2O → CH3CH2NH3+ + OH– 5b(ii)Reaction/equilibrium lies to left or low [OH–] or little OH– formed or little ethylamine has reacted5cEthylamine Alkyl group is electron releasing/donatingor alkyl group has (positive) inductive effect that increases electron density on N(H2)or increased availability of lpor increases ability of lp (to accept H(+))5dCH3CH2NH3Cl5eExtra H+ reacts with ethylamine or OH– or CH3CH2NH2 + H+ → CH3CH2NH3+ or H+ + OH– → H2OEquilibrium shifts to RHS or ratio [CH3CH2NH3+]/[ CH3CH2NH2] remains almost constant6a(Nucleophilic) addition-elimination Propanamide 6bNucleophilic substitutionPropylamine or propan-1-amine or 1-aminopropane6cElectron-rich ring or benzene or pi cloud repels nucleophile/ammonia 7a?Nucleophilic) addition-elimination N-propylethanamide 7b(i)PrimarySecondaryTertiary7b(ii)Absorption at 3300–3500 (cm?1) in spectrum N―H (bond) (only) present in secondary amine or not present in tertiary amine orThis peak or N―H absorption (only) present in spectrum of secondary amine or not present in spectrum of tertiary amine7c(i)Route A: stage 1KCNAqueous or ethanolicRoute A IntermediateCH3CH2CN or propanenitrileName alone must be exactly correct to gain M1 but mark on if name closecorrect formula gains M1 (ignore name if close)contradiction of name and formula loses markRoute A: stage 2H2H loses M4 but mark onLiAlH4Ni or Pt or PdetherRoute BNH3Excess NH37c(ii)Route A disadvToxic /poisonous KCN or cyanide or CN? or HCNExpensive LiAlH4Ignore acidifiedOR lower yield because 2 stepsRoute B disadvFurther reaction/substitution likely8Lone pair on N labelled b more available / more able to be donated than lone pair on N labelled alp or electrons or electron density on N labelled a: delocalized into (benzene) ringlp or electrons or electron density on N labelled b: methyl/alkyl groups electron releasing or donating or (positive) inductive effect or push electrons or electron density 9aQuaternary (alkyl) ammonium salt / bromide CH3Br or bromomethane Excess (CH3Br or bromomethane) 9bNucleophilic substitution10a(i)Ammonia is a nucleophile Benzene repels nucleophiles10a(ii)H2/Ni or H2/Pt or Sn/HCl or Fe/HCl 10a(iii)Concentrated HNO3Concentrated H2SO4HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4– or using two equations:HNO3 + H2SO4 → H2NO3+ + HSO4– H2NO3+ → H2O + NO2+10a(iv)Electrophilic substitution CHAPTER 10Assignment 1A1.a. Corrosiveb. Corrosivec. Flammable, toxicA2.The reaction should be undertaken in a fume cupboard; lab coat, safety spectacles and disposable gloves should be worn.A3.a. 2.2?g of 1,6-diaminohexane (molecular weight = 116) = 2.2/116 = 0.02?moles, in 50?cm3 = 0.02/50 × 1000 = 0.4?M.b. 1.5?g of decandioyl dichloride (molecular weight = 239) = 1.5/239 = 0.006?moles in 50?cm3 = 0.006/50 × 1000 = 0.12?M.A4.Nylon 6,10A5.HClA6.1:1 ratioA7.During the condensation polymerisation, HCl is produced. The excess 1,6-diaminohexane reacts with the HCl, making an ammonium salt.A8.Unreacted monomers might be adhering to the fibre. As these are corrosive, they must be washed off the nylon before it is used or disposed of.Assignment 2A1.A2.A3.Hydrogen bondsA4.A5.Both types of polyamides are capable of H-bonding between chains. But the more rigid nature of the linking aromatic rings in Kevlar? and other aramids compared with the flexible alkyl groups in nylons makes aramids stronger and tougher.A6.Kevlar? because the 1,4 monomers cause the polymer chain to be less zigzag shaped, so the molecules lie closer together and the intermolecular forces can be more effective.Practice questions1Can be hydrolysed or can be reacted with/attacked by acid/base/nucleophiles/H2O/OH?2aAddition2b2cQ is biodegradablePolar C=O group or δ+ C in Q (but not in P) Therefore, can be attacked by nucleophiles (leading to breakdown) 3Kevlar is biodegradeable but polymerised alkenes are notKevlar has polar bonds / is a (poly) amide / has peptide linkCan be hydrolysed/attacked by nucleophiles/acids/bases/enzymesPolyalkenes are non polar /have non-polar bonds4Advantages:reduces landfillsaves raw materials lower cost for recycling than making from scratch reduces CO2 emissions by not being incineratedDisadvantages:difficulty/cost of collecting/sorting/processing product not suitable for original purpose, easily contaminated 5a(i)5a(ii)In polyamides – H bondingIn polyalkenes – van der Waals forcesStronger forces (of attraction) in polyamides or H bonding is stronger6a(i)As a soap 6a(ii)Biodiesel or biofuel or fuel for cars/lorries 6a(iii)Cationic surfactant /detergent /fabric softener /germicide / shampoos /(hair) conditioners /spermicidal jelly6b(i)(Poly)ester Terylene or PET6b(ii)(Poly)amide Kevlar or nylons6b(iii)Hydrogen bonding in b(ii)Imfs in (b)(ii) are stronger or H bonding stronger than dipole-dipole/van der Waals/ dispersion/London forces in b(i)7aMethyl propanoate7b(i)Pentane-1,5-diol 7b(ii)7b(iii)(Base or alkaline) hydrolysisδ+ C in polyester reacts with OH– or hydroxide ion8a(i)Condensation 8a(ii)Propane-1,3-diol 8b(i)Addition 8b(ii)9Hydrogen bonding10d. NH2CH2COCl11b. Amide12c. There are no gas emissions from a landfill site.13a. Poly(ethene)CHAPTER 11Assignment 1A1.a. Negative electrodeb. Positive electrodeA2.No movement.A3.The structures are:Alanine: HOOCCH(CH3)NH2Aspartic acid: HOOCCH(CH2COOH)NH2Lysine: HOOCCH(CH2CH2CH2CH2 NH2)NH2A4.They are chemically and physically the same except that they cause plane polarised light to rotate in opposite directions.A5.Lysine has an extra basic NH2 group, causing it to be the most alkaline. Aspartic acid has an extra acid COOH group, causing it to be the most acidic.A6.abcA7.Perform an electrophoresis experiment at pH 6.1. Use a buffer to fix the pH.At pH 6.1, glycine will be in its zwitterionic form and will not move in an electric field.At pH 6.1 (above the isoelectric point), cysteine will be protonated on the nitrogen and move to the negative terminal.At pH 6.1 (below the isoelectric point), aspartic acid will be deprotonated at the carboxylic acid group and move to the positive terminal.Assignment 2A1.5?mol?dm?3 hydrochloric acid for 24 hours.A2.Glycine – no carbon atom is bonded to four different groups.A3.Dipeptides = A and C. Tripeptide = B.A4.A5.Assignment 3A1.The sequence of α-amino acids linked together in the protein chainA2.Yes. The α-helix is elastic and can be stretched but will hold its shape owing to the H-bonds. If it is stretched too far then the H-bonds will break.A3.α-Helix and β-sheetA4.The cysteine residues could covalently link adjacent α-helices of α-keratin together.A5.When hair is wetted the H-bonding of the α-helices is disrupted, meaning that the hair can be stretched.Assignment 4A1.A2. a. b. A3.H-bondA4.The active sites of enzymes are stereospecific and the two enantiomers interact with different enzymes, thereby producing different biological effects.A5.The structures of thalidomide and pomalidomide are very similar (differing only in one amine group). They both might be able fit into the active site of the target enzyme and so both inhibit it.A6.By forming a salt with a carboxylic acid active group in an enzymeA7.Owing to the similarity to thalidomide, the patient must not be pregnant and must agree to use contraception during the treatment.Assignment 5A1.AZT has an N3 group at the 3′-position on the sugar, while 2′-deoxythymidine-5′-monophosphate has a hydroxyl group.A2.It gets incorporated into the growing DNA strand as it binds to the complementary nucleotide during DNA replication.A3.The N3 group at the end of the growing chain cannot bond with another nucleotide, so the chain stops growing and the DNA strand cannot finish replicating.A4.In high doses AZT can stop human DNA replication as it can also be incorporated into human DNA, so it will be toxic. However, it is 100 times more toxic to the HIV virus than to human cells.Practice questions1aSecondary1bNitrogen and oxygen are very electronegativeTherefore, C=O and N–H are polar, which results in the formation of a hydrogen bond between O and H in which a lone pair of electrons on an oxygen atom is strongly attracted to the δ+H2aA nucleotide is the monomer that makes up DNA. Each nucleotide consists of three components linked together, a phosphate, a sugar and a base.2b2-deoxyribose does not have an alcohol group at the 2-position whereas ribose does.2cH-bonds between bases are being broken. A high temperature is required as there are a large number of H-bonds to break in the DNA strand.2dTCTCATGCAA3a2-Deoxyribose3bBase A:Top N–H forms hydrogen bonds to lone pair on O of guanine The lone pair of electrons on N bonds to H–N of guanine A lone pair of electrons on O bonds to lower H–N of guanine3cEither of the nitrogen atoms with a lone pair NOT involved in bonding to cytosine 3dUse in very small amounts / target the application to the tumour4a(i)4a(ii)4b4c5a5bIonic bonding in aminoethanoic acidStronger attractions than hydrogen bonding in hydroxyethanoic acid67(a) Glycine does not contain a chiral C atom i.e 1 with 4 different groups bonded to it and alanine and phenylalanine do.(b)(i)(ii)(c)(i)(ii)(d)8Facts (b) and (d) are correct9(a) A cation10Facts (b) and (d) are correctCHAPTER 12Assignment 1A1.A2.Concentrated nitric and concentrated sulfuric acid. Electrophilic substitution (nitration).A3.PhenylamineA4.N-methylphenylamineA5.CH3Cl in ethanol solvent; heatA6.Ethanoic anhydride. It is cheaper, reacts slowly and does not evolve HCl.A7.Only ibuprofen contains a carbon atom attached to four different groups and so exhibits optical isomerism.A8.Aluminium trichloride and 2-methylpropanoyl chloride.A9. A10.The electrophile is NO2+.Practice questions1aDiethylamine or ethyl ethanamine or ethyl aminoethane 1bThree valid routes: A, B, CRoute ARoute BRoute CFCH3CH2Br or CH3CH2ClC2H6CH3CH2OHGCH3CH2NH2 ethylamine OR ethanamine OR aminoethaneCH3CH2Br OR CH3CH2ClCH3CH2Br OR CH3CH2Cl1cRoute ARoute BRoute CStep 1Reagent(s)HBr OR HClH2/ Ni (Not NaBH4)H2O & H3PO4 OR H2O & H2SO4MechanismElectrophilic additionaddition (allow electrophilic OR catalytic but not nucleophilic) ignore hydrogenationElectrophilic additionStep 2Reagent(s)NH3Cl2 OR Br2HBr OR KBr & H2SO4 OR Pcl3 OR Pcl5 OR SOcl2MechanismNucleophilic substitution(free) radical substitutionNucleophilic substitutionStep 3Reagent(s)CH3CH2Br OR CH3CH2ClCH3CH2NH2 OR NH3 but penalise excess ammonia hereCH3CH2NH2 OR NH3 but penalise excess ammonia hereMechanismNucleophilic substitutionNucleophilic substitutionNucleophilic substitution1dTertiary amine or triethylamine or (CH3CH2)3NQuaternary ammonium salt or tetraethylammonium bromide or chloride or ion or (CH3CH2)4N+ (Br– or Cl–)Further substitution will take place or diethylamine is a better nucleophile than ethylamine2a [First part of question 2]Nucleophilic addition2b2-bromobutanenitrile2cReagent: ammonia, NH3Conditions: excess ammonia, heatMechanism: nucleophilic substitution2d(i)2d(ii)The electrostatic attraction between the oppositely charged parts of the ion accounts for the relatively high melting-point values of amino acids.2 [Second part of question 2]Compound LCompound MStep 1 NaBH4 or LiAlH4, (nucleophilic) addition Step 2 conc H2SO4 or conc H3PO4 or Al2O3, eliminationStep 3 HBr, electrophilic addition3a(i)3a(ii)3a(iii)3bCHAPTER 13Assignment 1A1.The sample contains paracetamol but it is impure; the impurity is a minor component of the mixture.A2.SilicaA3. a. Compound B (there are four peaks between 100 and 160?ppm)b. C=Oc. Compound B because it has an aromatic group, whereas A does not.A4.4A5.Z. This is the only structure that has two CH3 groups, which will give singlets. The triplet is from the CH2 next to the carbonyl group; it is coupled to the neighbouring CH2 group, which is at higher ppm (being next to the electronegative O atom).Required practicalP1.0.25P2.IbuprofenP3.D contains both paracetamol and ibuprofen. E contains a third painkiller that is not paracetamol or ibuprofen.P4. ParacetamolP5.I hour: mainly alcohol, a little aldehyde formed. 2 hours: mainly product aldehyde, a little alcohol remains.P6.After three hours, as pure aldehyde is produced.P7.Another product is forming, probably the carboxylic acid.Practice questions1a51ba, singlet QWCb, triplet QWC2a or Si(CH3)4Inert/non -toxic/volatile or low bp2b22c(i)a = quartet or 4b = triplet or 32c(ii)3230 – 3550 (cm?1)3a(i)3a(ii)3a(iii)CH2CH2 or two adjacent methylene groups3a(iv)or CH3COCH2CH2OCH33b(i)OH in acids or (carboxylic) acid present3b(ii)4IR:Absorption at 3360 cm–1 shows OH alcohol presentNMR:There are 4 peaks which indicates 4 different environments of hydrogen The integration ratio = 1.6 : 0.4 : 1.2 : 2.4 The simplest whole number ratio is 4 : 1 : 3 : 6 The singlet (integ 1) must be caused by H in OH alcohol The singlet (integ 3) must be due to a CH3 group with no adjacent H Quartet + triplet suggest CH2CH3 group Integration 4 and integration 6 indicates two equivalent CH2CH3 groups5aChromatography: GLC, TLC, GC, HPLC5b5320.0 or 322.05cUse of excess air/oxygen or high temperature (over 800 °C) or remove chlorine-containing compounds before incineration5d(i)Si(CH3)45d(ii)36a6b6cECH3CH2COOCH3FCH3COOCH2CH36d6e7aJ acid amide K secondary amine or amino7bδ = 3.1–3.9Doublet or duplet7c(i)Solvent must be proton-free or CHCl3 has protons or has H or gives a peak7c(ii)CDCl3 is polar or CCl4 is non-polar7d117e(i)Si(CH3)4 or SiC4H127e(ii)A single number or range within 21-257e(iii)CHAPTER 14Assignment 1A1.The first fuel cells powered a vehicle for only 30 seconds, but now with technological advances more than 300 miles is possible. The voltage of the fuel cell and the chemical reactions – which depend upon the scientific principles – are just the same.A2.No pollutants of carbon dioxide and nitrogen compounds are produced while the bus is using its fuel cell. Fuel cells are quiet.A3.Air is a mixture of nitrogen and oxygen; when the fuel burns at high temperatures, these two combine to form nitrogen oxides.A4.Fuel cells are nonpolluting when they are being used, and they have no moving parts, so they are reliable. But there needs to be a store of hydrogen, which is flammable and has to be pressurised.Assignment 2A1.a. Zn2+(aq) + 2e? ? Zn(s)b. Mn4+(aq) + 2e? ? Mn2+(aq)c. ?O2(g) + H+ + 2e? ? OH?(aq)Reduction is more likely because the equilibrium goes from left to right. Therefore, the cell potential will be less negative.A2.a.i. Zn(s) | Zn2+(aq) ||ii. Li+(aq) | Li(s) ||iii. Pt(s) | H2(g) | H+(aq) ||b. Under standard conditions this is the standard hydrogen electrode. Its potential is assigned a value of zero and all other standard electrode potentials are measured relative to this.c. Hydrogen pressure = 1?bar, temperature = 298?K, [H+] = 1 mol dm?3.A3.a. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)b. Pt(s) |H2(g) | H+(aq) || Zn2+(aq) | Zn(s)A4.a. The zinc/silver cell gives a larger e.m.f. (1.56?V) than the zinc/copper cell (1.10?V).b.c. ?0.76 + (+0.80)?V = 1.56?Vd. It has a higher cell potential. The cell reactants have a lower density.A5.a. Silver will be displaced. For zinc to displace silver from aqueous silver solution, the following two half-reactions must occur:Zn(s) ? Zn2+(aq) + 2e?, E? = ?(?0.76)?VAg+(aq) + e? ? Ag(s), E? = +0.80?VThe sum of these two half-reactions isZn(s) + Ag+(aq) ? Zn2+(aq) + Ag(s), E? = +1.56?VFor this reaction to be spontaneous, E? must be positive. Therefore zinc does displace silver from solution.b. No reaction. This is the reverse of the reaction in part a. E? is negative. For this reaction to be spontaneous E? must be positive, so no reaction occurs.c. No reaction. For iodine to displace bromine from bromide solution, the following two half-reactions must occur:I2(s) + 2e? ? 2l?(aq), E? = +0.54?V2Br?(aq) ? Br2(l) + 2e?, E? = ?1.07?VThe sum of these two half-reactions is I2(s) + 2Br?(aq) ? 2l?(aq) + Br2(l).For this reaction to be spontaneous, E? must be positive. But (?1.07?V) + (+0.54?V) = ?0.53?V. Therefore, iodine does not displace bromine from solution.d. This is the reverse of the reaction in part c. E? is positive. +0.54?V ? (?1.07?V) = +0.53?V. Therefore, bromine does displace iodine from iodide solution.Required practical 8P1.The metals are cleaned with abrasive paper to remove the layer of metal oxide and then with propanone to remove any grease coming from handing themP2.Stay away from flames and sources of heat, use protective glasses and gloves, perform the operation in a fume hoodP3.The salt bridge maintains the charge neutrality in the solutionsP4.Temperature needs to be 25 degrees Celsius, pressure 1 Bar (100 KPa)P5.Take a number of readings at different temperatures e.g. every 10 degrees from 10 to 90 and plot them against the e.m.f.P6.Assemble an electrochemical cell consisting of two different metals, A and B, as described in Required Practical 8. Test each metal of the four in turn against the others in order to determine the cell with the highest e.m.f. Alternatively, test each metal against a standard electrochemical cell, in order to determine their standard electrochemical potential. The two metals with the highest electrochemical potentials will form the cell with the highest e.m.f.Practice questions1a(i)None or no reactionE(Zn2+/Zn) more negative than E(Fe2+/Fe)1a(ii)Fe2+Cr3+E(Fe3+/Fe2+) more positive than E(Cr3+/Cr2+)1bE.m.f. = -0.41 – (-0.76) = 0.35Zn + 2Cr3+ → Zn2+ + 2Cr2+2a(i) Iron(II), Fe2+2a(ii) Oxygen difluoride, F2O2a(iii) Fe2+, Cl–2b(i) E.m.f. = E (right) – E (left) = 1.52 – 0.77 = 0.75 V2b(ii) Fe2+ Fe3+ + e–2b(iii) Decrease Fe3+ concentration to increase e.m.f. of cell. Equilibrium shifts in favour of more Fe3+, so electrode potential for Fe3+/Fe2+ becomes less positive.3aMost powerful reducing agent: Zn3b(i)Reducing species: Fe2+3b(ii)Oxidising species: Cl23c(i)Standard electrode potential 1.25 V3c(ii)Equation: Tl3+ + 2 Fe2+ → 2Fe3+ + Tl+4a(i)0.60 V4a(ii)H2O + H2SO3 → SO42? + 4H+ + 2e?4b(i)2IO3? + 2H+ + 5H2O2 → 5O2 + I2 + 6H2O4b(ii)The concentration of the ions change or are no longer standard or the e.m.f. is determined when no current flows4b(iii)Unchanged4b(iv)IncreasedEquilibrium IO3? /I2 displaced to the rightElectrons more readily accepted or more reduction occurs or electrode becomes more positiveCHAPTER 15Assignment 1A1.[Cu(H2O)6]2+ ? [Cu(H2O)5(OH)]+ + H+; [Fe(H2O)6]3+ + H2O ? [Fe(H2O)5(OH)]2+ + H3O+A2.These are hydrolysis reactions because the ion reacts with water and releases hydrogen ions, giving an acidic solution.A3.The Fe3+ ion has a higher ionic charge than Cu2+, so the attraction for the lone pairs on the water molecule is greater and this weakens the O–H bond more.A4.The equilibrium for Fe3+ lies further to the right, the solution is more acidic and therefore the pH is lower for Fe3+ than for Cu2+.A5.Any test from the following.Reaction with aqueous sodium hydroxide gives a blue precipitate of copper hydroxide:[Cu(H2O)6]2+ + 2OH? → [Cu(H2O)4(OH)2] + 2H2OReaction with aqueous ammonia gives a blue precipitate of copper hydroxide:[Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4+A6.Fe2+ ions. The green precipitate is caused by the reaction[Fe(H2O)6]2+ + 2OH? → [Fe(H2O)4(OH)2] + 2H2OThe iron(II) slowly oxidises in air to give a brown precipitate of iron(III) hydroxide.Assignment 2A1.The water is acidic and contains copper ions. In acid conditions the equilibrium favours [Cu(H2O)6]2+, which is the blue ion, and does not favour the insoluble hydroxo complex, which would make the water cloudy.A2. a. A green–blue precipitate of CuCO3b. [Cu(H2O)4(OH)2]c. Experiment 1:[Cu(H2O)6]2+ + CO32? → CuCO3 + 6H2OExperiment 2:[Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4+which gives a blue precipitate, and then reaction with more aqueous ammonia to give [Cu(NH3)4(H2O)2]d. In low concentrations it is an irritant, and at higher concentrations it is corrosive and toxic. Use with gloves in a fume cupboard.A3. There may be hidden pollution problems, and companies may be liable for compensation claims in the future.A4.Plants growing in different areas will be exposed to different levels of copper ions in the soil.A5. Plant A, because the average amount of copper extracted from the soil is greater. With Plant A, the average amount of copper is 498?mg per kg of leaves, while the value for plant B is 67.4?mg per kg of leaves.A6.The hydrolysis reactions of transition metal aqua complexes make the soil acidic. Different levels of metals will give rise to different levels of pH.A7.If the post was made from iron, then the hexaaquairon(III) ions in solution in the soil could undergo hydrolysis,[Fe(H2O)6]3+ + H2O → [Fe(H2O)5(OH)]2+ + H3O+so as to acidify the soil. This would produce blue flowers.A8.Hydrolysis of the aluminium aqua ion in soft water lowers the pH of the water. Copper is more soluble at low pH values: the equilibrium [Cu(H2O)6]2+ ? [Cu(H2O)5(OH)]+ + H+ lies further to the left. In tap water, the copper comes from the copper pipes.Required practicalP1.At low concentration it is an irritant; at a higher concentration (>2?M) it is corrosive.P2.Tap water will contain dissolved metal ions, which can affect the analysis.P3.A = FeSO4[Fe(H2O)6]2+ + 2OH? → [Fe(H2O)4(OH)2] + 2H2OB = CuCl2[Cu(H2O)6]2+ + 2OH? → [Cu(H2O)4(OH)2] + 2H2OC = Fe(NO3)3[Fe(H2O)6]3+ + 3OH? → [Fe(H2O)3(OH)3] + 3H2OP4.Experiment a suggests the salt contains sulfate ions:Ba2+(aq) + SO42?(aq) → BaSO4(s)Experiment b suggests the salt contains Cr(III) ions as CO2 is produced. Metal(III) complexes are more acidic than metal(II) complexes and enough acid is liberated to turn carbonate into CO2 and water:2H3O+ + CO32? → CO2 + 3H2OThe precipitate is [Cr(H2O)3(OH)3]:2[Cr(H2O)6]3+ + 3CO32? → 2[Cr(H2O)3(OH)3] + 3CO2 + 3H2OTherefore the compound is Cr2(SO4)3.Practice questions1a(i)MgO: ionicP4O10: covalent1a(ii)Electronegativity difference small or electronegativities similar or big difference in electronegativity leads to ionic bonding1bNa2O + H2O → 2Na+ + 2OH? (or 2NaOH)SO2 + H2O → H2SO31cMgO + 2HCl → MgCl2 + H2O (or MgO + 2H+ → Mg2+ + H2O)1dP4O10 + 12NaOH → 4Na3PO4 +6H2O (or P4O10 + 12OH? → 4PO43? +6H2O)2Na2O: vigorous or violent or exothermic reaction; or forms a colourless solutionpH of solution formed = 13 or 14Na2O + H2O → 2NaOHP4O10 or P2O5: vigorous or violent or exothermic reaction; or forms aColourless solution; 1pH of solution formed = 0 or 1P4O10 + 6H2O → 4H3PO43 From sodium on the left-hand side of the Periodic Table, to sulfur on the right, the trend of the pH in solutions of the oxides is to decrease. Oxides on the left of the Period are alkaline, oxides on the right are acidic.Sodium oxide is ionic and contains the oxide ion, O2–, that readily combines with hydrogen ions.Na2O + H2O 2NaOHMagnesium oxide is ionic but reacts less strongly with water than sodium oxide.Aluminium oxide is amphoteric, so can react as either an acid or a base. It is ionic but the oxide ions are too strongly held in the lattice to react with water.Silicon dioxide has a giant covalent structure which is difficult to break apart, and so does not react with water.Oxides of phosphorus are covalent and exist in different forms, they form weak acids in solution. For example P4O10 + 6H2O 4H3PO4Oxides of sulfur are covalent and exist in different forms: sulfur dioxide and sulfur trioxide. Sulfur dioxide forms a weak acid but sulfur trioxide forms a strong acidSO3 + H2O H2SO44a(i) Na2O + H2O 2NaOHpH 144a(ii) SO2 + H2O H2SO3But the main species present in solution is hydrated sulfur dioxide SO2.xH2OpH approximately 14b Generally, ionic bonded oxides are basic, so pH is high. Covalent bonded oxides are acidic, so pH is low.5aMgO (is a white solid that) forms a suspension (or slightly soluble)MgO + H2O → Mg(OH)2 (or → Mg2+ + 2OH?)pH is 8 to 10SO2 dissolves (or forms (colourless) solution)SO2 + H2O → H2SO3 (or → H+ + HSO3?) (or → 2H+ + SO32?)pH is 1 to 45bAl(OH)3 + OH? → Al(OH)4 (or forms Al(OH)63?)Al(OH)3 + 3H+ + 3H2O → Al(H2O)63+ (or forms [Al(H2O)5(OH)]2+, Al3+, AlCl36(i)Green solution (not blue-green or grey-green)[Cr(OH)6]3? (or Cr(H2O)(OH)5]2? or Cr(H2O)2(OH)4]?)6(ii)Green precipitateBubbles (or gas or fizzing or effervescence)Cr(H2O)3(OH)3 (or Cr(OH)3)7aThe number of protons increases (across the period) / nuclear charge increases Therefore, the attraction between the nucleus and electrons increases 7bS8 molecules are bigger than P4 molecules Therefore, van der Waals / dispersion / London forces between molecules are stronger in sulfur7cSodium oxide contains O2– ions These O2– ions react with water forming OH– ions7dP4O10 + 12OH– → 4PO43– + 6H2O8aAn electron pair on the ligand Is donated from the ligand to the central metal ion 8bBlue precipitate Dissolves to give a dark blue solution [Cu(H2O)6]2+ + 2NH3 → Cu(H2O)4(OH)2 + 2NH4+Cu(H2O)4(OH)2 + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2OH– + 2H2O9aMgO is ionicMelt itMolten oxide conducts electricity9bMacromolecular Covalent bonding Water cannot (supply enough energy to) break the covalent bonds / lattice9c(Phosphorus pentoxide’s melting point is) lowerMolecular with covalent bondingWeak / easily broken / not much energy to break intermolecular forces or weak vdW / dipole-dipole forces of attraction between molecules9dReagent (water or acid) Equation e.g. MgO + 2HCl → MgCl2 + H2O9e12NaOH + P4O10 4Na3PO4 + 6H2O10aYellow / purple (solution) Brown precipitate / solid [Fe(H2O)6]3+ + 3OH– → Fe(H2O)3(OH)3 + 3H2O10bBlue (solution) Dark / deep blue solution [Cu(H2O)6]2+ + 4NH3 → [Cu(H2O)2(NH3)4]2+ + 4H2O10cColourless (solution) White precipitate / solid Bubbles / effervescence / gas evolved / given off 2[Al(H2O)6]3+ + 3CO32– → 2Al(H2O)3(OH)3 + 3CO2 + 3H2O11Moles NaOH = 0.0212 × 0.5 = 0.0106Moles of H3PO4 = 1/3 moles of NaOH (= 0.00353)Moles of P in 25000 l = 0.00353 × 106 = 3.53× 103Moles of P4O10 = 3.53 × 103/4Mass of P4O10 = 3.53 × 103/4 × 284 = 0.251 × 106 g = 251 kgCHAPTER 16Assignment 1A1.Since the concentration is only given to one significant figure, precise measurements are not needed. Calculate (a) the mass of 1?mol of CuCl2.2H2O(s) and (b) the number of moles to be dissolved in 10?cm3 to make 0.1?mol?dm?3 of solution. Weigh out the calculated quantity and dissolve it in 10?cm3 of deionised water. Note: this is the preparation of a reagent solution, not a standard solution for volumetric analysis.A2.Both are corrosive and can cause burns. In both case the bottles should be stoppered when not in use. They should be stored in separate places.A3.Octahedral aqua complexA4.a. Water ligands are displaced by chloride ion ligands. The green colour is caused by a mixture of blue [Cu(H2O)6]2+ and yellow CuCl42? complex ions, and when there is little or no [Cu(H2O)6]2+ remaining the solution is yellow.b. Similar to a. except that [Co(H2O)6]2+ is pink and CoCl42? is blue.c. Successive replacement of water ligands in [Cu(H2O)6]2+ by ammonia to give a mixture of ammine complex ions; the most abundant one in concentrated ammonia solution is [Cu(NH3)5H2O]2+.d. Successive replacement of water ligands in [Co(H2O)6]2+ by ammonia to give [Co(NH3)6]2+.Assignment 2A1.60?s, c = 0.049?mol?dm?3; 120?s, c = 0.101?mol?dm?3; 180?s, c = 0.169?mol?dm?3; 240?s, c = 0.223?mol?dm?3; 300?s, c = 0.261?mol?dm?3A2.A3.This is given by the slope of the graph, 9.1 × 10?4?mol?dm?3?s?1.A4.0.22 = 9.1 × 10?4?t; t = 242?sA5.kr = 0.44/9.1 × 10?4 = 483?s?1Assignment 3A1.MnO4?(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)A2.a. Tablet 1b titre 17.1?cm3, therefore number of moles of MnO4? = (0.02 × 17.1)/1000 = 3.42 × 10?4b. 0.00171?mol of Fe2+c. Mass = 56 × 0.00171 = 95.76?mgd. 95.76/845 = 0.113, or 11.3% w/wA3.The expression is percentage of iron = titre (cm3) × 0.56/mass of the tablet (g).Tablet 1a: 17.4 × 0.56/0.850 = 11.5%Tablet 2a: 16.5 × 0.56/0.845 = 10.9%Tablet 2b: 16.9 × 0.56/0.855 = 11.1%A4.Only batch 1A5.a. ±0.135%b. The same apparatus, equipment and method are used for all tablets.A6.By taking more sample tablets from a batch and calculating the mean.Assignment 4A1.The unused feedstocks can be recycled to save costs and are cracked because the amounts in the feedstocks will not necessarily match the demand for the products.A2.A build-up of carbon will reduce the surface area of the catalyst available for reaction. Burning off the carbon will regenerate the catalyst.A3.To expose the maximum amount of catalyst to the reactants.A4.Catalytic converters remove hydrocarbons, carbon monoxide, sulfur oxides and nitrogen oxides. They do not remove carbon dioxide, a greenhouse gas.A5.The use of nonrenewable resources, dust hazards, health risks to miners and damage to the landscape due to mining.A6.A catalyst in a different phase from that of the reactants is considered heterogeneous.A7.To provide a greater surface area for the reactants.A8.The presence of double bonds increases the distance between adjacent molecules and this reduces the van der Waals forces of attraction between the molecules, reducing the melting point.A9.Animal fats contain a greater proportion of saturated fats, which, it is claimed, contribute to heart disease.Practice questions1a(i)SO2 + V2O5 → SO3 + V2O4V2O4 + 1/2O2 → V2O5V(IV) or 4 and V(V) or 51a(ii)MnO4– + 8H+ + 4Mn2+ → 5Mn3+ + 4H2O2Mn3+ + C2O42– → 2Mn2+ + 2CO2Mn(III) or 3 and Mn(II) or 21b[Co(NH3)6]2+ formedComplex easy to oxidiseH2O2 (or air or oxygen)1cMoles of dichromate = (29.2/1000)×0.04 = 0.001168Moles of Q2+ = (25/1000)×0.140 = 0.00350Each mole of dichromate needs 6 electrons or half equation with 6 e?Moles of electrons = 6×0.001168 = 0.007008 or moles Q2+:moles dichromate = 3:1Moles of electrons per mole of Q = 0.007008/0.0035 = 2.002 = 2Q(IV) or Q4+2a(i)Fe + 2HCl → FeCl2 + H2or Fe + 2H+ → Fe2+ + H22a(ii)PV = nRT, n = PV/RT= 4.53 × 10?3 mol2a(iii)Moles of iron = 4.53 × 10?3 mol (or = 4.25 × 10?3)Mass of iron = 4.53 × 10?3 × 55.8 = 0.253 g2a(iv)0.253 × 100/0.263 = 96.1%2b(i)Fe2+ → Fe3+ + e?Cr2O72? + 14H+ + 6e? → 2Cr3+ + 7H2OCr2O72? + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+2b(ii)Moles of dichromate = moles Fe2+/6= 4.53 × 10?3/6 = 7.55 × 10?4Volume of dichromate = moles/concentration= (7.55 × 10–4 × 1000)/0.0200V = 37.75 cm32b(iii)KMnO4 will also oxidise (or react with) Cl– (or chloride or HCl)3aInitially slow because reaction is between two negative ions (or between two negative reactants or two negative species) which repel each other.Then Mn2+ (or Mn3+) ions are formed acting as an autocatalyst (or Mn2+ ions formed in the reaction act as a catalyst)2MnO4? + 16H+ + 5C2O42? → 2Mn2+ + 8H2O + 4CO2MnO4? + 4Mn2+ + 8H+ → 5Mn3+ + 4H2OC2O42? + 2Mn3+ → 2Mn2+ + 2CO23bActive sites are where reactants are adsorbed onto a catalyst surface (or bind or react on a catalyst surface).Number of active sites increases if the surface area is increased (or catalyst is spread thinly or on honeycomb or powdered or decreased particle size.Active sites blocked by another species or poison (or species adsorbed more strongly or species adsorbed irreversibly or species not desorbed).4aEffect on reaction rate: catalyst provides an alternative reaction route with a lower Ea, more molecules able to react or rate increasedEquilibrium: forward and backward rates changes by the same amount ; hence concentration of reactants and products constant or yield unchanged4bHeterogeneous: catalyst in a different phase or state to that of the reactantsActive site: place where reactants adsorbed or attached or bond; where reaction occursReasons: large surface area; reduce cost or amount of catalystCatalyst poison: lead adsorbed; lead not desorbed or site blocked4cReaction slow as both ions negatively charged or ions repel2Fe2+ + S2O82? → 2Fe3+ + 2SO42–2Fe3+ + 2I? → 2Fe2+ + I25a3d75b[Co(H2O)6]2+Pink5c(i)[Co(NH3)6]2+Pale brown or straw5c(ii)[Co(H2O)6]2+ + 6NH3 → [Co(NH3)6]2+ + 6H2O5d[Co(NH3)6]3+An oxidising agent6aIronHeterogeneous; catalyst in a different phase from that of the reactantsPoison; a sulphur compoundPoison strongly adsorbed onto active sites / blockedPoison not desorbed or reactants not adsorbed or catalyst surface area reduced6bPale green solutionGreen precipitate formedInsoluble in excess ammoniaEquation:e.g. [Fe(H2O)6]2+ + 2NH3 → [Fe(H2O)4(OH)2] + 2NH4+ ................
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