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- 1. Teacher Guide
- 2. Philip Allan Updates, part of Hodder Education, an Hachette UK company, Market Place, Deddington, Oxfordshire OX15 0SE Orders Bookpoint Ltd, 130 Milton Park, Abingdon, Oxfordshire OX14 4SB tel: 01235 827720 fax: 01235 400454 e-mail: uk.orders@bookpoint.co.uk Lines are open 9.00 a.m.–5.00 p.m., Monday to Saturday, with a 24-hour message answering service. You can also order through the Philip Allan Updates website: www.philipallan.co.uk First published 2009 ISBN 978-0-340-95763-9 © 2009 Philip Allan Updates All rights reserved Printed by Marston Book Services Ltd, Didcot In all cases we have attempted to trace and credit copyright owners of material used. Copyright notice Any educational institution that has purchased one copy of this publication may make duplicate copies for use exclusively within that institution. Permission does not extend to reproduction, storage in a retrieval system, or transmittal, in any form or means, electronic, mechanical, photocopying, recording or otherwise, or duplicate copies for loaning, renting or selling to any other institution without the prior consent in writing of the publisher. Restrictions on the use of the CD-ROM All rights reserved. The CD-ROM must not be sold, rented, leased, sublicensed, lent, assigned or transferred, in whole or in part, to third parties. No part of the CD-ROM may be reformatted, adapted, varied or modified by the user other than specifically for teaching purposes. The CD-ROM may not be reproduced or transmitted in any form or by any means without the permission of the publisher, other than in the form of printed copies for teaching purposes within the purchasing institution. Environmental information The paper on which this title is printed is sourced from managed, sustainable forests. P01318© Philip Allan Updates Edexcel A2 Chemistry
- 3. Contents Introduction ..................................................................................................................................... 1 Unit 4 Rates, equilibria and further organic chemistry Chapter 1 Rates: how fast?...............................................................................................................5 Chapter 2 Entropy: how far? ..........................................................................................................14 Chapter 3 Equilibrium ...................................................................................................................20 Chapter 4 Applications of rate and equilibrium...............................................................................27 Chapter 5 Acid–base equilibria ......................................................................................................32 Chapter 6 Isomerism.....................................................................................................................43 Chapter 7 Carbonyl compounds ....................................................................................................47 Chapter 8 Carboxylic acids and their derivatives .............................................................................53 Chapter 9 Spectroscopy and chromatography................................................................................58 Practice Unit Test 4 ........................................................................................................62 Unit 5 Transition metals, arenes and organic nitrogen chemistry Chapter 10 Electrochemistry and redox equilibria ............................................................................70 Chapter 11 Transition metals and the d-block elements ....................................................................79 Chapter 12 Arenes and their derivatives...........................................................................................88 Chapter 13 Organic nitrogen compounds ........................................................................................96 Chapter 14 Organic analysis and synthesis .....................................................................................103 Practice Unit Test 5 ......................................................................................................113© Philip Allan Updates Edexcel A2 Chemistry iii
- 4. Introduction This Teacher Guide accompanies the Edexcel A2 Chemistry textbook (2nd edition), by George Facer, published by Philip Allan Updates. The answers to the end-of-chapter questions are not model answers. They are designed to help the student understand the answers. The examiner notes after each answer, indicated by the icon e, are additional points to help the student: Answers to the chapter summary worksheets (see the textbook online resource — www.hodderplus.co.uk/ philipallan) are given at the end of each chapter of this Teacher Guide. The answers to the practice unit tests are model answers that include all that examiners look for when awarding marks, together with some extra explanation. Each marking point is indicated by a . Relative atomic mass values are not given in the questions, because the student is expected to use the periodic table. All Edexcel’s theory exam papers have a periodic table printed on the back page. The textbook has an identical version inside the back cover. Note that the relative atomic masses in the periodic table are given to one decimal place. This guide includes a CD-ROM, designed so that answers to individual questions and the explanatory comments can be cut and pasted and either posted on the centre’s intranet or issued to students after they have done their homework. A-level examiners are instructed not to be too literal in following the mark scheme. The important point is that what has been written is correct and answers the question. There are several occasions in this Guide when different wording is used from that in the textbook, but either is equally acceptable. An example of this is: G textbook — the bombarding electrons in a mass spectrometer have a high kinetic energy G answer — the electrons are fast moving Some answers are expanded to show the logic of the calculation. For instance, in Question 8 of Chapter 3, the equilibrium moles of CH3OOCCOOCH3 is shown as 0.100 − 0.075 = 0.025, rather than just 0.025. Another example is the amount of salt in Question 19 of Chapter 5. Instead of just giving the numbers, this is written as: moles of salt = volume × concentration = 0.050 × 1.00 = 0.050 Specific terms used in exam questions State This means that no explanation is required and so none should be given. However, in this Guide an explanation is often given in order to help the student understand why the answer is correct. Name If the name of a substance is asked for, examiners will accept either the name or the formula, but if both are given, both must be correct. Identify The command word ‘identify’ is often used. The answer here can be a name or a formula. Give the formula If the formula is required, only the formula and not the name will score the mark.© Philip Allan Updates Edexcel A2 Chemistry 1
- 5. Introduction Molecular formula The added up formula, such as C2H4Cl2, is what is required. This is normally the type of answer to questions involving percentage composition of compounds. Give the formula (of an organic substance) An unambiguous formula must be given. Butanoic acid is CH3CH2CH2COOH or C2H5CH2COOH, not C3H7COOH, which could also be methylpropanoic acid. Ethanal is CH3CHO, not CH3COH; ethyl ethanoate is CH3COOC2H5 or CH3CO2C2H5, not CH3OCOC2H5 (this could be ethyl ethanoate or methyl propanoate). Give the structural formula The bonding in the functional group containing a π-bond must be clearly shown. For example, the structural formula of ethene should be written as H2C=CH2, that of ethanal as: H CH3 C O and that of ethanoic acid as OH CH3 C O Give the displayed or full structural formula All the atoms and the bonds between them should be shown. For example, propene is: H H H H C C C H H Observations The colour before as well as after the test must be given in the answer. For example, bromine water changes from brown to colourless, not ‘is decolorised’, and acidified potassium dichromate(VI) goes from orange to green, not ‘goes green’. Gas evolved is not an observation (it is a deduction). Bubbles or effervescence are observations. Hydrogen chloride is observed as misty fumes, not as a white smoke. Preferred colours Flame tests Cation Preferred colour Acceptable colours Lithium Red Carmine, deep red or crimson Sodium Yellow (not orange) Potassium Lilac Mauve or purple Calcium Red or brick red Orange-red Strontium Crimson Red Barium Pale green Apple green or yellowy-green© Philip Allan Updates Edexcel A2 Chemistry 2
- 6. Introduction Indicators (alkali in the burette, acid in the conical flask) Indicator Starting colour End-point colour Not Methyl orange Red Orange or peach Yellow or pink Phenolphthalein Colourless Pink Red or purple Indicators (acid in the burette, alkali in the conical flask) Indicator Starting colour End-point colour Not Methyl orange Yellow Orange or peach Red or pink Phenolphthalein Red or purple Colourless Clear Significant figures If a question asks for appropriate or a specific number of significant figures, the answer must be given to the correct number of significant figures. Never round up to one figure, even if the answer is 5.0. The exception to this is when calculating the number of atoms or the number of molecules of water of crystallisation in a formula. In calculations from titration data, always give the answer to three or more significant figures. Signs Always give a + or a – sign in your answers to ΔH, ΔS and E questions and to those on oxidation numbers. Conditions You must state: G the temperature range necessary or whether the substance has to be heated under reflux G the name of the solvent G the name of any catalyst required© Philip Allan Updates Edexcel A2 Chemistry 3
- 7. Unit 4Rates, equilibriaand furtherorganic chemistry
- 8. Unit 4 Rates, equilibria and further organic chemistry Chapter 1 Rates: how fast? 1 The rate of reaction is the change in concentration divided by the time for that change. amount of hydrogen peroxide decreases by (1.46 – 1.32) × 10–3 mol = 1.4 × 10–4 mol change in concentration = 1.4 × 10–4 mol/0.050 dm3 = 2.8 × 10–3 mol dm–3 rate of reaction = 2.8 × 10–3 mol dm–3/45 s = 6.2 × 10–5 mol dm–3 s–1 e There are two common errors in this type of calculation. The first is to think that the rate equals the change in moles per unit time, rather than the change in concentration per unit time. The second is to fail to divide the volume by 1000 in order to convert it from cm3 to dm3. 2 a Fraction of molecules with energy E T2 T1 Kinetic energy, E e Make sure that the peak for the lower temperature curve (T2 in this question) is higher and to the left of that for the T1 curve. b The average kinetic energy of the molecules at the lower temperature (T2) is less than that at the higher temperature (T1). This has two effects on the rate of reaction. Fewer of the colliding molecules have a combined energy greater than or equal to the activation energy. Thus, a smaller proportion of the collisions results in reaction. This means that the rate of reaction is considerably reduced, even for a small lowering of temperature. The second effect is that the frequency of collision is also slightly reduced. This also lowers the rate of reaction. e Be careful. Most questions, but not all, ask for an explanation of the effect of an increase in temperature. Never state that there are fewer collisions with energy ≥ Ea or that there are fewer successful collisions. Over time, there will be the same number of successful collisions. It is the proportion of the colliding molecules with energy ≥ Ea that is smaller. c By far the more important factor is the reduction in the proportion of collisions that have energy ≥ Ea. As an approximate guide, a 10°C reduction in temperature will cause the rate of a reaction to halve. A decrease in temperature from 25°C to 15°C will cause the frequency of collision, and hence the rate, to decrease by less than 2%. d When the pressure of a gaseous mixture is increased, the molecules have the same average kinetic energy, but they are packed together more closely. This closer contact causes the frequency of collision to increase. As the distribution of energies has not been altered, the proportion of collision with energy ≥ Ea remains unaltered. An increase in pressure produces an increase in frequency of collision, with the same fraction of collisions resulting in reaction, so the rate of reaction increases. e Do not state that there are more collisions; it is the frequency (number of collisions per second) that has increased.© Philip Allan Updates Edexcel A2 Chemistry 5
- 9. Chapter 1 Rates: how fast? 3 The manganate(VII) ion is an oxidising agent. The ions are coloured. a I Known volumes of known concentrations of all the reagents are mixed and a stop-clock is started. I At a noted time, a 25 cm3 sample is removed with a pipette and put into a conical flask containing iced water, in order to quench (stop) the reaction. I The remaining unreacted potassium manganate(VII) in the pipetted sample is then titrated against standard iron(II) sulfate solution until the purple solution goes just colourless. I At later times, other samples are removed and treated as before. I The volume of the iron(II) sulfate solution is proportional to the amount of potassium manganate(VII) left in the reaction vessel. b 25 cm3 of a solution of potassium manganate(VII) is mixed with 50 cm3 of water and the intensity of its colour measured using a colorimeter. I 25 cm3 of the original potassium manganate(VII) solution is mixed with 25 cm3 of sulfuric acid and 25 cm3 of ethanedioic acid solution and a stop-clock is started. I The mixture is placed in the colorimeter and the intensity of the purple colour of the solution is measured at regular intervals. The intensity of the colour compared with the original intensity is a measure of the concentration of the manganate(VII) ions. e The potassium manganate(VII) solution is diluted so that its concentration is exactly the same as that in the initial reaction mixture. This allows a comparison to be made between the intensity of the colour and, therefore, the amount of manganate(VII) ions left as the reaction proceeds. c This increase in rate followed by a decrease is caused by one of the products acting as a catalyst. In this experiment the catalyst is the Mn2+ ion. This can be proved by repeating the experiment and adding some Mn2+ ions initially. The fact that the reaction becomes faster shows that the Mn2+ ion is a catalyst. e The temperature must be kept constant. Many exothermic reactions carried out in a beaker accelerate first before slowing down. This is because the temperature rises, causing the reaction to speed up before slowing down as the reactants are used up. 4 The rate equation is: rate = k[A][B][C]2 e The order with respect to a reagent is the power to which its concentration is raised in the rate equation. 5 The two most likely rate equations are: rate = k[A][B] and rate = k[A]2 e A possible mechanism for the first suggested rate equation would be: A + B → X in a slow, rate-determining step, then X + A → products A possible mechanism for the second suggested rate equation would be: 2A → X in a slow, rate-determining step, then X + B → products (where X is an intermediate in both mechanisms) The rate equation k[B]2 is possible but highly unlikely — two B species would have to react in the first and rate- determining step, and one B would have to be reformed in a later step. 6 a In experiments 1 and 2, [KOH] remains constant and [C2H5I] is doubled. The rate doubles (increases by a factor of 21), so the reaction is first order with respect to iodoethane. In experiments 1 and 3, [C2H5I] remains constant and [KOH] is doubled. The rate doubles (increases by a factor of 21), so the reaction is also first order with respect to potassium hydroxide.© Philip Allan Updates Edexcel A2 Chemistry 6
- 10. Chapter 1 Rates: how fast? e If you see the words ‘deduce’ or ‘justify your answer’, you must explain fully how you worked out the order for each reactant. You should make clear which two experimental rates you are comparing. You must also state that the concentration of one of the reactants does not alter. b The rate equation is: rate = k[C2H5I][KOH] e Do not forget to include the rate constant, k, in your answer and make sure that you write it as a lower-case k. An upper case K represents an equilibrium constant, not a rate constant. c Using the values from experiment 1: rate 2.2 × 10–5 k= = = 1.1 × 10–3 mol–1 dm3 s–1 [C2H5I][KOH] 0.20 × 0.10 e The units are given by: concentration × s–1 = (concentration)–1 s–1 = mol–1 dm3 s–1 (concentration)2 7 a In experiments 1 and 2, [A] and [C] remain constant and [B] is increased by a factor of 3. The rate increases by a factor of 31, so the reaction is first order with respect to B. In experiments 2 and 3, [B] and [C] remain constant and [A] is doubled. The rate doubles (increases by a factor of 21), so the reaction is first order with respect to A. e The ratio 1.4 × 10–2:6.9 × 10–3 ≈ 2:1. In experiments 1 and 4, [B] remains constant and [A] and [C] are both doubled. The rate doubles (increases by a factor of 21). However, as the reaction is first order with respect to A, a doubling of [A] will cause the rate to double, so increasing the value of [C] has no effect on the rate of reaction. This means that the reaction is zero order with respect to C. b The total order of reaction is 1 + 1 + 0 = 2. c The overall rate equation is: rate = k[A][B] d Using the values from experiment 1: rate 2.3 × 10–3 k= = = 2.3 × 10–3 mol–1 dm3 s–1 [A][B] 1.0 × 1.0 8 a The clock method for this reaction is as follows: I In one beaker place known volumes of sodium persulfate solution and sodium thiosulfate solution, plus a few drops of starch solution. I In another beaker place a known volume of potassium iodide solution. I Mix the two, start a clock and stir. Stop the clock when the solution turns blue. I Repeat the experiment with a different volume of sodium persulfate made up to the same total volume with water. I Repeat the experiment with a different volume of potassium iodide solution, again making up to the same total volume with water. I As the iodine is produced it reacts with the sodium thiosulfate, until all the sodium thiosulfate has been reacted. The next iodine produced turns the starch dark blue. volume of sodium thiosulfate I A measure of the rate is time taken b In experiments 1 and 2, [I–] remains constant and [S2O82–] is doubled. The rate doubles (increases by a factor of 21), so the reaction is first order with respect to persulfate, S2O82–, ions.© Philip Allan Updates Edexcel A2 Chemistry 7
- 11. Chapter 1 Rates: how fast? In experiments 2 and 3, [S2O82–] and [I–] are both doubled. The rate increases by a factor of 4. As the reaction is first order with respect to S2O82– ions, the rate doubles because of the doubling of [S2O82–]. Since the rate increases by 2 × 2, the reaction is also first order with respect to I– ions. c The rate equation is: rate = k[S2O82–][I–] Using the values from experiment 1, the rate constant is: 1.2 × 10–5 k= = 6.3 × 10–3 mol–1 dm3 s–1 0.038 × 0.050 e Only two significant figures are justified as the least accurate data are given to two significant figures. 9 a Since the slope of the tangent equals the rate of reaction, it can be seen that the rate halves as the concen- tration of OH– ions halves. This means that the reaction is first order (it has a total order of one). b The relative values of the slopes will be identical to those obtained by measuring the change in [OH–]. e Because the 2-bromopropane and the OH– ions react in a 1:1 ratio, the concentration of each reagent halves from 0.10 mol dm–3 to 0.050 mol dm–3. Thus, it is not possible to tell from these data whether the reaction is first order in OH– and zero order in 2-bromopropane or vice versa. c The rate equation is either: rate = k[OH–] or rate = k[CH3CHBrCH3] The experiment could be repeated using twice the initial concentration of OH– ions but an unaltered concen- tration of 2-bromopropane. If the first rate equation is correct, the initial slope will be twice as large (rate twice as fast). If the second rate equation is correct, the slope will be unchanged. 10 a 0.08 [Cyclopropane]/mol dm–3 0.07 0.06 0.05 14 min 0.04 0.03 14 min 0.02 0.01 14 min 0 0 10 20 30 40 50 Time/min b half-life starting from [cyclopropane] = 0.080 mol dm–3 = 14 min half-life starting from [cyclopropane] = 0.040 mol dm–3 = 14 min half-life starting from [cyclopropane] = 0.020 mol dm–3 = 14 min The half-lives are constant, so the reaction is first order.© Philip Allan Updates Edexcel A2 Chemistry 8
- 12. Chapter 1 Rates: how fast? c kt 1 = ln 2 – 2 average value of t 1 = 14 min – 2 Therefore, k = ln 2/t 1 = 0.693/14 = 0.050 min–1 – 2 e Make sure that you label the axes, using linear scales for concentration and for time. Then draw a smooth curve as closely through the points as possible. Mark on your graph how you measured the half-lives. The second half-life starts at the time of the point where [cyclopropane] = 0.040 mol dm–3, not from t = 0. 11 In reaction I, the half-lives are constant to within experimental error. This means that the reaction is first order. In the second reaction, the half-life (approximately) doubles as the concentration halves. This means that the reaction is second order. 12 a 0.20 –3 [A]/mol dm 0.18 Tangent at 0.16 0.195 – 0.04 0.16 = 3.3 × 10–3 47 0.14 0.12 0.10 Tangent at 0.08 0.147 – 0.04 = 1.5 × 10–3 71 0.08 0.06 0.04 0 10 20 30 40 50 60 70 80 Time/s b slope of the tangent where [A] is 0.16 mol dm–3 = 3.3 × 10–3 mol dm–3 s–1 slope of the tangent where [A] is 0.08 mol dm–3 = 1.5 × 10–3 mol dm–3 s–1 c The slope of the tangent at a given point on the graph equals the rate of reaction. slope when [A] is 0.08 1.5 × 10–3 1 = = 0.45 ≈ – slope when [A] is 0.16 3.3 × 10–3 2 Since the rate is approximately halved when [A] halves, the reaction is first order. e Be careful when working out the value of the slope. The graph drawn does not start at the origin, so the value of the tangent at [A] = 0.08 mol dm–3 is (0.147 – 0.04)/71, not 0.147/71. It is difficult to draw tangents accurately, but the ratio of rates will be a whole number ratio such as 1:1 or 1:2 or 1:4 — a high level of accuracy is not that important. You need only find out which of these three ratios is close to the ratio of the tangents’ slopes. 13 a The overall equation is: 2A + B → C + D© Philip Allan Updates Edexcel A2 Chemistry 9
- 13. Chapter 1 Rates: how fast? b Ea2 Enthalpy Ea1 2A + B A + Intermediate + C ∆H C+D e Note that the profile has two humps because an intermediate is formed. The energy level of the intermediate is below that of the reactants, but above that of the products, as both steps are exothermic. As step 2 is the rate-determining step, it is the slower of the two steps and its activation energy is greater. c The probable rate equation is: rate = k[A]2[B] The reaction is third order. e Because there are 2 mol of A up to and including the rate-determining step, the reaction is second order with respect to A. d If step 1 were rate determining, the rate equation would be: rate = k[A][B] The reaction would be second order. 14 The Arrhenius equation is: ln k = ln A – Ea/RT ln k = ln (1.2 × 1015) – 1.1 × 105/(8.31 × 303) = 34.72 – 43.69 = –8.97 k = e–8.97 = 1.28 × 10–4 e The units of k cannot be found from the calculation. 15 a 1 — /K–1 1 ln k — /K–1 T T 0.0013 0.0014 0.0015 –7.9 0.00152 –2 ln k –6.8 0.00147 –4.8 0.00139 –3 –2.9 0.00130 –4 –5 –5.6 –6 –7 –8 e Make sure that you label the axes, using linear scales for ln k and 1/T and that you draw the best-fit straight line. Remember that the values of ln k become more negative going down the y-axis.© Philip Allan Updates Edexcel A2 Chemistry 10
- 14. Chapter 1 Rates: how fast? −8 − (2.4) −5.6 b slope of the straight line = = = −2.29 × 104 0.001545 − 0.0013 0.000245 From the Arrhenius equation, slope = –Ea/R Hence Ea = –slope × R = –(–2.29 × 104) × 8.31 = +1.90 × 105 J mol–1 = +190 kJ mol–1 c T = 700 K, so 1/T = 1/700 = 1.43 × 10–3 K–1 ln k at this point = –5.6 (see graph) k = e–5.6 = 3.7 × 10–3 e The Arrhenius equation will always be given if it is needed to answer the question. 16 a Time/s 1/time/s−1 ln (1/time) Temperature/°C Temperature/K 1/temperature/K−1 120 0.0083 −4.79 17 290 0.00345 69 0.0145 −4.23 25 298 0.00336 36 0.0278 −3.58 35 308 0.00325 14 0.0714 −2.64 50 323 0.00310 b amount of CaCO3 (molar mass 100.1 g mol−1) = mass/molar mass = 0.20 g/100.1 g mol−1 = 0.0020 mol amount of acid needed to react with all the CaCO3 = 2 × 0.0020 = 0.0040 mol amount of acid taken = concentration × volume = 0.50 mol dm−3 × 0.050 dm3 = 0.025 mol % acid reacted = 0.0040 × 100/0.025 = 16% This is higher than ideal ( 10%) but makes the assumption reasonable. c 1 /K–1 T 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 –2 In time 1 –3 –2.5 –4 0.00040 –5 The gradient is −2.5/0.00040 = −6250. d gradient = −Ea/R Ea = −gradient × R = −(− 6250 ) × 8.31 = + 51 938 J mol−1 = +52 kJ mol−1 e This answer should be to 2 significant figures as the y-axis value when calculating the gradient is only accurate to 2 significant figures. e amount of CaCO3 = 0.0020 mol© Philip Allan Updates Edexcel A2 Chemistry 11
- 15. Chapter 1 Rates: how fast? heat produced = moles × ΔH = 0.0020 mol × 351 kJ mol−1 = 0.702 kJ = 702 J heat produced = mass × specific heat capacity × ΔT heat produced 702 J ΔT = = = 3.3°C mass × specific heat capacity 50 g × 4.2 J g−1 °C−1 17 a (i) Consider experiments 1 and 3: when [CH3CHClC2H5] was increased by a factor of 3 and the [OH−] kept constant, the rate increased by a factor of 4.1 × 10−4 / 1.4 × 10−4 = 2.9 ≈ 3, so the reaction is first order with respect to 2-chlorobutane. (ii) Consider experiments 1 and 2: when the concentrations of both reactants are doubled, the rate increases by a factor of 2.9 ×10−4/4 × 10−4 = 2.1 ≈ 2. This doubling of the rate would be caused by the doubling of the [CH3CHClC2H5] alone, so the doubling of the [OH−] has no effect. Therefore, the reaction is zero order with respect to OH−. b rate = k [CH3CHClC2H5] c k = rate/[CH3CHClC2H5] = 4.1 × 10−4 mol dm−3 s−1/0.30 mol dm−3 = 0.0014 s−1 d C2H 5 C2H 5 Step 1 Slow C C+ + Cl – Cl H H3C H3C H C2H 5 C2H 5 Step 2 Fast C+ OH – C H OH H3C H3C H e There would be no effect on the plane of polarisation of plane-polarised light even though the product is chiral. This is because the intermediate is planar (having three bond pairs and no lone pairs around the positive carbon). The OH− ion can attack this intermediate from either above or below, resulting in an equal amount of both enantiomers — a racemic mixture.© Philip Allan Updates Edexcel A2 Chemistry 12
- 16. Chapter 1 Rates: how fast? Summary worksheet (www.hodderplus.co.uk/philipallan) 1 A Option B is a measure of the average rate. Option C is incorrect as the faster the rate the shorter the time, and option D is an approximation of the rate that is only valid if the concentration of the reactant has not altered by more than 10%. 2 D An increase in temperature causes the molecules to have greater kinetic energy. Therefore, on collision, a greater proportion of molecules will have energy ≥ the activation energy. A change in temperature does not alter the activation energy (a catalyst does), so options B and C are incorrect. It is true that an increase in temperature increases the frequency of collision, but this is a minor factor compared with the proportion of collisions having the necessary energy. Option A is, therefore, a correct statement but does not answer this question. 3 D Neither propan-2-ol nor its oxidation product propanone is chiral, so a polarimeter would not be a suitable piece of apparatus for following the reaction. The potassium dichromate(VI) does change colour, so colorimetry (A) would work. The reaction could be stopped by quenching and the remaining dichromate(VI) ions could be titrated (B). IR spectroscopy (C) would show the steady increase of a peak around 1720 cm−1 due to the C=O in the ketone produced on oxidation of propan-2-ol. 4 B The rate equation shows that the reaction mechanism is SN1, so the first step is the heterolytic breaking of the C–Br bond to form a positive ion. Positive ions are cations, so the answer is option B, not C. If the mechanism were SN2, options A and D would both be true. 5 C The reaction site is planar and so the attacking CN− ions can approach from either above or below the plane producing both enantiomers in equal amounts. This racemic mixture has no effect on the plane of polari- sation of plane-polarised light. The product is chiral, so option A is incorrect. The intermediate is not planar as the carbon atom has four separate pairs of bonding electrons, so option B is incorrect. Option D is a true statement but is not an explanation of the absence of an effect of the product on plane-polarised light. 6 B For a third-order reaction, k = rate/concentration3. Rate has units of mol dm−3 s−1 and so the rate constant, k, has units of mol dm−3 s−1 = mol−2 dm6 s−1 mol3 dm−9 7 A A straight-line graph of concentration against time means that the reaction has a constant rate. This can only be the case if the reaction is zero order. A first-order reaction would give a curve, so B is wrong. The slope of the straight line depends on the other reactants’ orders, so C is wrong. D is just wrong. 8 D A reaction profile shows the energy levels of the reactants and products. The difference is the value of ΔH. It also shows the ‘hump’ or energy barrier. The height of the peak relative to the energy level of the reactants is the activation energy of the forward reaction. The height of the peak relative to the energy level of the products shows the activation energy of the reverse reaction. This means that options A, B and C are all shown by the diagram. The rate of reaction cannot be deduced from the diagram and so option D is the correct answer to this negative question. 9 B The rate will change (decrease) by a factor of 4 when the concentration is halved, but the half-life will only increase by a factor of 2. This means that option B, not C, is correct. The half-life is constant for a first-order reaction, so option A is incorrect. Half-lives apply to all orders, so option D is incorrect. 10 B The reaction is SN1 and the product is chiral, so a racemic mixture is formed. Thus options A and C are incorrect. The C–I bond is weaker than the C–F bond and so iodine is substituted rather than fluorine. Therefore, option D is incorrect.© Philip Allan Updates Edexcel A2 Chemistry 13
- 17. Unit 4 Rates, equilibria and further organic chemistry Chapter 2 Entropy: how far? 1 There are a number of possible answers to this question including: I the reaction of pure ethanoic acid with solid ammonium carbonate I the reaction of solid hydrated barium hydroxide with solid ammonium chloride I dissolving ammonium nitrate in water 2 The entropy of a perfect crystal of helium at absolute zero is zero, as is that of a perfect crystal of sodium chloride. This is because of the third law of thermodynamics: ‘All perfect crystals have zero entropy at absolute zero’. This applies to both elements and compounds. e This is not to be confused with enthalpy — the standard enthalpy of formation of an element is zero, but that of a compound is not. 3 a Both are solutions of molecular substances, so the solution of the substance with the more complex structure will have the greater entropy. Glucose has 24 atoms in each molecule whereas carbon dioxide has only three, so the glucose solution is more random and has greater entropy. e The comparison is with a solution of carbon dioxide, not with gaseous carbon dioxide. b As any substance is heated, its entropy increases. Aqueous sodium chloride at 50°C has a higher entropy than it does at 25°C. e This is shown by the Maxwell–Boltzmann distribution of energies of the particles; those at the higher temperature are more spread out than those at the lower temperature. 4 For a solution to ‘unmix’, the second law of thermodynamics would have to be broken. The total entropy of the dissolved copper sulfate is greater than that of separate solid copper sulfate and water. This is because the total entropy change is made up of the entropy change of the system (the difference between the entropies of the solution and of the solid and the water) and the entropy change of the surroundings (−ΔH/T) 5 a ΔSsystem = ΣSproducts − ΣSreactants = 2 × 192 − (192 + 3 × 131) = −201 J K−1 mol−1 ΔSsurr = −ΔH/T = − (−92 000/298) = +309 J K−1 mol−1 ΔStotal = ΔSsystem + ΔSsurr = −201 + 309 = +108 J K−1 mol−1 This is a positive number and so the reaction is thermodynamically feasible at 298 K. e Be careful when using the Edexcel or Nuffield data booklets. The values given for the standard entropies of all the 1 diatomic gaseous elements are based on – mol of the diatomic elements. This applies to elements such as hydrogen, 2 1 nitrogen and chlorine. The value for nitrogen is given as 95.8, which is for – N2, so for N2 the value is 2 × 95.8 = 2 192 J K−1 mol−1 (to 3 significant figures). b ΔSsystem = ΣSproducts − ΣSreactants = 40 + 214 − 93 = +161 J K−1 mol−1 ΔSsurr = −ΔH/T = −(+178 000/298) = −597 J K−1 mol−1 ΔStotal = ΔSsystem + ΔSsurr = +161 − 597 = −436 J K−1 mol−1 This is a negative number and so the reaction is not thermodynamically feasible at 298 K. e There are three important points here. G The values for the standard entropy of elements in some data books is per atom in the molecule. The data in Table 2.2 are the standard entropies per molecule, i.e. for 1 mol of nitrogen gas, N2, at a temperature of 298 K.© Philip Allan Updates Edexcel A2 Chemistry 14
- 18. Chapter 2 Entropy: how far? G As is usual in questions containing data about entropy and ΔH, make sure that both are in joules. This means that the ΔH value has to be multiplied by 1000. G Note that the first reaction is feasible at this temperature even though the entropy change for the system is negative. This is because the reaction is exothermic and the heat change outweighs the negative entropy change. The second reaction is not feasible even though the entropy change of the system is positive. The reaction is too endothermic for the reaction to be feasible at 298 K. 6 a ΔH° = ΣΔH°f (products) − ΣΔH°f (reactants) = −860 + 10 × −286 + 2 × −46 – (–3245 + 2 × −315) = +63 kJ mol−1 b ΔSsystem = ΔStotal − ΔSsurr = ΔStotal + ΔH/T = 150 + 63 000/298 = +361 J K−1 mol−1 = ΣSproducts − ΣSreactants = 130 + 700 + 386 – (x + 184) Entropy of hydrated barium hydroxide, x = 130 + 700 + 386 − 184 − 361 = +671 J K−1 mol−1 e Do not forget the stoichiometric numbers when calculating both ΔH and ΔSsystem. 7 ΔSsystem = 70 − (131 + 1⁄2 × 205) = −164 J K−1 mol−1 ΔSsurr = −ΔH/T = −(−286 000/298) = +960 J K−1 mol−1 ΔStotal = −164 + 960 = +796 J K−1 mol−1 which is positive, so the reactants are thermodynamically unstable relative to the products (reaction thermodynamically feasible). For M2+: charge/radius = 2/0.31 = 65. For Q3+: charge/radius = 3/0.095 = 32. M2+ has the larger charge density, so its hydration energy will be more exothermic. e Do not say that the reaction is thermodynamically unstable, but relate the stability of the reactants to the products. 8 The Hess’s law diagram is: Li+(g) + F–(g) ∆Hlatt(LiF) ∆Hhyd(Li+) ∆Hhyd(F–) ∆Hsoln(LiF) LiF(s) + aq Li+(aq) F–(aq) ΔHlatt(LiF(s)) + ΔHsoln(LiF(s)) = ΔHhyd(Li+(g)) + ΔHhyd(F–(g)) ΔHsoln(LiF(s)) = ΔHhyd(Li+(g)) + ΔHhyd(F–(g)) – ΔHlatt(LiF(s)) = –519 + (–506) – (–1022) = –3 kJ mol–1 The enthalpy change is so small that the solubility of lithium fluoride cannot be predicted. The solubility will be determined almost entirely by ΔSsystem. e You can work out the enthalpy of solution of an ionic solid using the formula: ΔHsoln = sum of hydration enthalpies of the ions – lattice energy Remember that if there are two anions in the formula, a value of 2 × ΔHhyd(anion) must be used. This is the case with hydroxides and halides of the group 2 metals.© Philip Allan Updates Edexcel A2 Chemistry 15
- 19. Chapter 2 Entropy: how far? 9 The Hess’s law diagram is: Ca2+(g) + 2Cl–(g) ∆Hlatt(CaCl2) ∆Hhyd(Ca2+) 2 × ∆Hhyd(Cl–) ∆Hsoln(CaCl2) CaCl2(s) + aq Ca2+(aq) + 2Cl–(aq) ΔHlatt(CaCl2(s)) + ΔHsoln(CaCl2(s)) = ΔHhyd(Ca2+(g)) + 2 × ΔHhyd(Cl–(g)) 2 × ΔHhyd(Cl–(g)) = ΔHlatt(CaCl2(s)) + ΔHsoln(CaCl2(s)) – ΔHhyd(Ca2+(g)) = –2237 + (–83) – (–1650) = –670 1 ΔHhyd(Cl–(g)) = – × (–670) = –335 kJ mol–1 2 e The calculation gives first a value of twice the hydration enthalpy of chloride ions, so you need to divide it by two. 10 The value of the hydration enthalpy depends on the charge and the ionic radius. A small radius results in a large hydration energy, as does a high charge. For M2+: charge/radius = 2/0.031 = 65. For Q3+: charge/radius = 3/0.095 = 32. M2+ has the larger charge density, so its hydration energy will be more exothermic. e M is beryllium in period 2 and Q is thallium in period 6. In most A-level questions, the more positive ion would also have the smaller radius, so that the effects of size and charge are in the same direction. 11 a Solubility is a balance between ΔSsystem and ΔSsurr. The value of ΔSsurr depends on the value of ΔH and the temperature. The compound that has the more positive, or less negative, ΔSsystem will be the more soluble. The compound that has the more exothermic ΔH and hence the more positive ΔSsurr, will be the more soluble. I The change in ΔSsystem going from MgF2 to CaF2 depends on the difference in the entropy values of the aqueous cations. I ΔSsystem becomes less negative by 83 J K−1 mol−1, tending to make CaF2 more soluble than MgF2. I ΔH becomes more endothermic by 31 000 J and so ΔSsurr becomes more negative by 31 000/298 = 104 J K−1 mol−1, tending to make CaF2 less soluble than MgF2. As the change in ΔSsurr is greater than the change in ΔSsystem, calcium fluoride will be less soluble than magnesium fluoride. e The value of S of the anion (the fluoride ion) is the same for both fluorides and so is not relevant to the explanation. b The change in ΔSsystem going from AgBr to AgI depends on the difference in the entropy values of the aqueous anions: ΔSsystem becomes more positive by 54 J K−1 mol−1, tending to make AgI more soluble than AgBr. ΔH becomes more endothermic by 27 000 J and so ΔSsurr becomes more negative by 27 000/298 = 91 J K−1 mol−1, tending to make AgI less soluble than AgBr. As the change in ΔSsurr is greater than the change in ΔSsystem, silver iodide is less soluble than silver bromide. e The value of S of the cation (the silver ion) is the same for both halides and so is not relevant to the explanation. e These calculations show that when comparing similar ionic compounds, it is the change in the value of the enthalpy of solution that is more significant than the change in entropy of the ions.© Philip Allan Updates Edexcel A2 Chemistry 16
- 20. Chapter 2 Entropy: how far? 12 a As a substance is heated, the motion of the particles becomes more random, so its entropy increases. Thus, liquid water at 25°C has a lower entropy than liquid water at 35°C. b The molecules in a liquid have short-range order, but those in the gas phase are randomly arranged. Therefore, liquid water at 100°C has lower entropy than gaseous water at the same temperature, i.e. there is an increase in entropy (disorder) when water at 100°C boils. 1 13 a The reaction is going from 1– mol of gas to 1 mol of gas and so the entropy of the system decreases. 2 b A solid (highly ordered) and ions in solution (fairly disordered) react to form a gas (highly disordered), ions in solution and a liquid (fairly disordered). Thus, there is an increase in entropy (ΔSsystem is positive). c An ordered solid forms another ordered solid plus a highly disordered gas. Therefore, there is an increase in entropy. d The reaction involves univalent ions dissolving, so the water is only slightly ordered. The solid is dispersed through the liquid resulting in a large increase in disorder (entropy). The second factor is larger than the first and so there is an increase in the entropy of the system. e In this example an anhydrous ionic compound containing 2+ ions is dissolving. This causes a large increase in order of the water, which may be larger than the increase in entropy of the solid becoming dispersed through the liquid. Unless the values are given, it is impossible to predict which factor will be dominant. e Do not make the mistake of thinking that all dissolving results in an increase in the entropy of the system. ΔHsolution for calcium sulfate is exothermic, but calcium sulfate is only slightly soluble. Thus ΔStotal must be slightly negative and, as ΔH is favourable, ΔSsystem must be unfavourable and therefore negative. 14 a ΔSsystem = ΔSproducts – ΔSreactants = 2 × (+256) – (+325) = +187 J K–1 mol–1 b ΔStotal = ΔSsystem + ΔSsurr – ΔH 57.4 57.4 ΔSsurr = = = = –0.160 kJ K–1 mol–1 T (273 + 85) 358 = –160 J K–1 mol–1 ΔStotal = +187 + (–160) = +27 J K–1 mol–1 e Beware of units. ΔSsystem is in joules, whereas ΔH, and hence ΔSsurr, are in kilojoules. c The total entropy change is positive, so the reaction is feasible (and will take place as long as the activation energy is not too high). d ΔStotal is positive at this temperature, so the reactants are thermodynamically unstable relative to the products. 15 a A change is favoured by exothermic (negative) ΔH and positive ΔSsystem. Change W is favourable on both counts and so is likely to take place. Change X has a favourable ΔH (negative) but an unfavourable ΔSsystem (negative). However: ΔSsurr = –ΔH/T = –(–170)/298 = +0.570 kJ K–1 mol–1 = +570 J K–1 mol–1 This outweighs the unfavourable ΔSsystem of –500 J K–1 mol–1, so the reaction is likely to take place. Change Y has an unfavourable ΔH (positive) and an unfavourable ΔSsystem (negative) and so will never take place.© Philip Allan Updates Edexcel A2 Chemistry 17
- 21. Chapter 2 Entropy: how far? Change Z has an unfavourable ΔH (+170 kJ mol–1), but a favourable ΔSsystem (+500 J K–1 mol–1): ΔSsurr = –ΔH/T = –(+170 000/298) kJ K–1 mol–1 = –570 J K–1 mol–1 ΔStotal = ΔSsystem + ΔSsurr = +500 + (–570) = –70 J K–1 mol–1 ΔStotal is negative and so the reaction will not happen at this temperature. e Note that the enthalpy changes are quoted in kJ, but the entropy values are given in J. You should not state that a reaction will happen, as the system might be kinetically stable because of a high activation energy. b Only change Z becomes more favourable as the temperature increases. The value of –ΔH/T becomes less negative as the temperature rises and eventually it is smaller than the positive ΔSsystem. ΔStotal becomes positive, making the reaction feasible. The reactants become thermodynamically unstable relative to the products above this higher temperature. e The temperature at which the reactants change from being thermodynamically stable to being thermodynamically unstable is when ΔStotal = 0. At this point ΔSsystem – ΔH/T = 0, or T = ΔH/ΔSsystem = 170 × 103/500 = 340 K = 67ºC 16 As this is an endothermic reaction, ΔSsurr is negative. However, the reaction is spontaneous and so ΔStotal must be positive. This can only be the case if ΔSsystem is positive (and more positive than 58 000/323 = 180 J K−1 mol−1). e An endothermic enthalpy of reaction suggests a lack of reaction unless it is outweighed by a positive change in the entropy of the reactants and products. 17 a Both ethanoic acid and methanol are liquids, but ethanoic acid is a more complex molecule and so its standard entropy value will be greater. b ΔSsystem = S(ethanoic acid) − (S(carbon monoxide) + S(methanol)) = 160 − (198 + 127) = −165 J K−1 mol−1 c ΔSsurr = −ΔH/T = −(−137 000/298) = +460 J K−1 mol−1 d ΔStotal = ΔSsystem + ΔSsurr = −165 + (+460) = +295 J K−1 mol−1 This is a positive number and so the reaction will be spontaneous at a temperature of 298 K. e As the reaction is thermodynamically spontaneous, the reactants are thermodynamically unstable with respect to the products. As the reaction does not take place at room temperature but is thermodynamically feasible, the activation energy must be high, making the reactants kinetically inert with respect to the products. 18 e Type Boltzmann into Google and you will get many references. Boltzmann’s tombstone is inscribed with S = k log W (but it should have been loge W or ln W, rather than the ambiguous log W).© Philip Allan Updates Edexcel A2 Chemistry 18
- 22. Chapter 2 Entropy: how far? Summary worksheet (www.hodderplus.co.uk/philipallan) 1 C A solid and a solution are reacting to produce a solution, a liquid and a gas. It is the large randomness of the gas produced that causes the entropy change of the system to be positive. Options A and D both have a gas reacting to form a solid; in option B a gas turns into a liquid. So options B, C and D will result in an increase in order and, therefore, a decrease in entropy of the system. 2 C ΔSsurr = −ΔH/T = −(−123 000/298) = +413 J K−1 mol−1 In options A and B, the temperature in °C has not been converted into degrees kelvin. In options D and B the sign is wrong in the expression for ΔSsurr. 3 B To be thermodynamically spontaneous, ΔStotal must be positive. This is helped by a negative ΔH and a positive ΔSsystem, as in option B. Option C has a negative ΔStotal at all temperatures. Option A will be spontaneous if ΔSsystem outweighs the unfavourable (endothermic) ΔH, which will only happen at high temperatures. Option D will be spontaneous if ΔH outweighs the unfavourable (negative) ΔSsystem, which it will only at low temperatures. 4 A An increase in temperature shifts the position of equilibrium to the right in endothermic reactions only. Thus options B and D must be incorrect. Option C cannot be correct because the reaction will not be spontaneous as both ΔH and ΔSsystem are unfavourable. 5 A ΔSsystem = Sproduct − ΣSreactants = 240 − (198 + 2 × 131) = −220 J K−1 mol−1 In options C and D the entropy of hydrogen has not been multiplied by 2, which ignores the stoichiometry of the reaction. Options B and D both used Sreactants – Sproducts. 6 D Gaseous water has a higher entropy than liquid water, so option B cannot be correct. Of the three gases, ethanol, C2H5OH, has the most complex formula and so has the highest entropy. 7 A The extent of ordering of the solvent around the ions depends on the charge density of the ions. All are chlorides, so it is the relative charge densities of the cations that matter. The cations are all 2+, but Mg2+ has the smallest radius and so surrounds itself with more layers of water molecules than the others. e In questions about a group in the periodic table, the answer will always be the element at the top or the bottom of the group. Thus options B and C need not be considered.© Philip Allan Updates Edexcel A2 Chemistry 19
- 23. Unit 4 Rates, equilibria and further organic chemistry Chapter 3 Equilibrium [CH3OH]eq 1 a Kc = [CO]eq[H2]2eq [NO2]eq b Kc = 1 [NO]eq[O2]2eq – [NO]4eq[H2O]6eq c Kc = [NH3]4eq[O2]5eq [HBr]eq d Kc = [C2H5Br]eq[ H2O]eq [H3O+]eq[[Fe(H2O)5OH]2+]eq e Kc = [[Fe(H2O)6]3+]eq e Remember to put the concentrations of the products on the top line of the Kc expression. All concentrations must be 1 raised to the powers according to the stoichiometric numbers in the reaction equation. Thus, – O2(g) appears in 2 1 equation b, so [O2] must be raised to the power –. 2 A reactant that is also the solvent is not included in the expression for Kc. Thus in d, [C2H5OH] is omitted from the top line and in e [H2O] is omitted from the bottom line. The subscript ‘eq’ is often omitted, but you must remember that Kc equals this fraction only when the system is in equilibrium. Thus, the [reactant] and the [product] terms must be equilibrium values. 2 For the first reaction: [PCl5]eq Kc = [PCl3]eq[Cl2]eq and for the second reaction: [PCl3]eq[Cl2]eq K′c = [PCl5]eq Since K′c = 1/Kc: K′c = 1/(40 mol–1 dm3) = 0.025 mol dm–3 e Remember that Kc has units, unless the sum of the powers in the numerator equals the sum of the powers in the denominator. [CO2][H2] 3/100 × 1/100 3 a The fraction = = 1.33 [CO][H2O] 1.5/100 × 1.5/100 This does not equal the value of Kc at 1100 K, which is 1.0, so the system is not in equilibrium. As the fraction is greater than Kc, the system will adjust to the left, reducing the amounts of CO2 and H2 and increasing the quantities of CO and H2O until the fraction equals 1.0. e Do not forget to divide the moles of the substances by the volume to get concentration values. In this case, the final answer is unaffected, as there are the same number of molecules on each side of the equation. However, you would lose a mark, unless you made it clear that the volume cancels.© Philip Allan Updates Edexcel A2 Chemistry 20
- 24. Chapter 3 Equilibrium [CO2]eq[H2]eq 0.010 × [H2]eq b Kc = 0.20 = = [CO]eq[H2O]eq 0.035 × 0.0040 0.20 × 0.035 × 0.0040 [H2]eq = = 0.0028 mol dm–3 0.010 [SO2]2eq[O2]eq 4 Kc = [SO3]2eq (0.044/10)2 × (0.022/10) = = 3.5 × 10–3 mol dm–3 at 700°C (0.035/10)2 e The answer should be reported to two significant figures, as the data were given to two significant figures. 5 a [NH3 in water] = 30 × [NH3 in ethoxyethane] = 30 × 0.00980 = 0.294 mol dm−3 b amount of ammonia in water layer = 0.294 mol dm−3 × 0.200 dm3 = 0.0588 mol amount of ammonia in ethoxyethane layer = 0.00980 mol dm−3 × 0.100 dm3 = 0.000980 mol total free ammonia = 0.0588 + 0.000980 = 0.05978 e Do not forget that the volume of the aqueous layer is 200 cm3 (0.200 dm3) as two aqueous portions each of volume 100 cm3 were mixed before the 100 cm3 of ethoxyethane were added. c amount of ammonia reacted = initial amount − amount free = 0.100 − 0.05978 = 0.04022 mol moles ammonia reacted 0.04022 d ratio = = = 4.022 ≈ 4 initial moles copper ions 0.0100 So the formula of the ion is [Cu(NH3)4]2+. e Note two things. G The number of ammonia molecules must be a whole number, so 4.022 is converted to 4. G The calculation is much easier to understand if what is being calculated at each stage is written down clearly. Do not just write down a series of numbers. If you got the answer 7, you should have realised that you had made an error. The error would have been in thinking that the aqueous layer was 100 cm3, rather than 200 cm3. [CO]eq[H2]3eq 6 a Kc = [CH4]eq[H2O]eq b CH4 H2O CO 3H2 Start moles 1.00 2.00 0 0 Change in moles –0.75 –0.75 +0.75 +2.25 Equilibrium moles 0.25 1.25 0.75 2.25 Equilibrium 0.025 0.125 0.075 0.225 concentration 0.075 × (0.225)3 Kc = = 0.273 mol2 dm–6 at 1200 K 0.025 × 0.125 e Note that the number of moles of hydrogen produced is three times the amount of methane reacted. The units of Kc can be calculated as (concentration)4 divided by (concentration)2, which equals (concentration)2, i.e. mol2 dm–6.© Philip Allan Updates Edexcel A2 Chemistry 21
- 25. Chapter 3 Equilibrium 7 CO 2H2 CH3OH Start moles 0.100 0.300 0 Change in moles –0.0300 –0.0600 +0.0300 Equilibrium moles 0.0700 0.240 0.0300 Equilibrium 0.00700 0.0240 0.00300 concentration [CH3OH]eq 0.00300 Kc = = = 744 mol–2 dm6 at 400°C [CO]eq[H2]2eq 0.00700 × (0.0240)2 e To work out the amount of CO that reacted, find 30% of 0.100 (= 0.0300). The stoichiometry of the equation is such that 2 × 0.0300 mol of H2 reacted and 0.0300 mol of CH3OH were formed. Do not forget to divide the equilibrium moles by the volume to get the concentration. The units of Kc can be worked out from (concentration) divided by (concentration)3, which equals (concentration)–2, i.e. mol–2 dm6. 8 The molar mass of dimethyl ethanedioate is 118 g mol–1 and that of water is 18 g mol–1. CH3OOCCOOCH3 2H2O HOOCCOOH 2CH3OH Start moles 11.8/118 = 0.100 5.40/18 = 0.300 0 0 Change in moles –0.075 2 × (–0.075) = –0.15 +0.075 2 × +0.075 = +0.15 Equilibrium moles 0.100 – 0.075 = 0.025 0.300 – 0.15 = 0.15 0.075 0.15 Equilibrium concentration 0.025/0.015 = 1.67 0.15/0.015 = 10 0.075/0.015 = 5 0.15/0.015 = 10 [HOOCCOOH]eq[CH3OH]2eq 5 × 102 Kc = = [CH3OOCCOOCH3]eq[H2O]2eq 1.67 × 102 = 3.00 (no units) e Marks are awarded for: G calculating the two Mr values (1 mark) G using these values and the masses to calculate the starting moles (1 mark) G calculating the equilibrium moles (1 mark) G dividing by the volume to get equilibrium concentrations (1 mark) G substituting these values into the expression for Kc (1 mark) G evaluating the answer and working out the units of Kc (1 mark) Each mark is for the process involved and so depends on your answer to the previous point. If you make an error in calculating the molar mass, then your maximum mark would be 5 out of 6. The volume must be in dm3 and you must state that there are no units for Kc in this reaction. (Units are (concentration)3 divided by (concentration)3, so no units.) p(SO2)eq p(Cl2)eq 9 a Kp = p(SO2Cl2)eq p(NO2)2eq b Kp = p(NO)2eq p(O2)eq p(SO3)2eq c Kp = p(SO2)2eq p(O2)eq e Never use square brackets in a Kp expression.© Philip Allan Updates Edexcel A2 Chemistry 22
- 26. Chapter 3 Equilibrium 10 N2O4 2NO2 Total moles Start moles 1 0 Change in moles –0.15 +0.30 Equilibrium moles 1 – 0.15 = 0.85 0.30 1.15 Mole fraction 0.85/1.15 = 0.739 0.30/1.15 = 0.261 Partial pressure/atm 0.739 × 1.2 = 0.887 0.261 × 1.2 = 0.313 p(NO2)2eq (0.313)2 Kp = = = 0.11 atm p(N2O4)eq 0.887 e If you are not given the starting amount, assume that you have 1 mol. Be careful of the reaction stoichiometry — in this reaction, 1 mol of N2O4 forms 2 mol of NO2, so 0.15 mol forms 0.30 mol. Marks are awarded for: G the expression for Kp (1 mark) G the moles at equilibrium (1 mark) G dividing by the total number of moles to get the mole fraction of each component (1 mark) G multiplying by the pressure to get the partial pressure of each component (1 mark) G substituting these values into the expression for Kp (1 mark) G evaluating the answer and stating the units (1 mark) p(NH3)2eq 11 a Kp = p(N2)eq p(H2)3eq b N2 3H2 2NH3 Total moles Start moles 1 3 0 Change in moles –0.15 3 × (–0.15) = –0.45 2 × (+0.15) = +0.30 Equilibrium moles 1 – 0.15 = 0.85 3 – 0.45 = 2.55 0.30 3.70 Mole fraction 0.85/3.70 = 0.230 2.55/3.70 = 0.689 0.30/3.70 = 0.0811 Partial pressure 0.230 × 30 = 6.90 0.689 × 30 = 20.7 0.0811 × 30 = 2.43 p(NH3)2eq (2.43)2 Kp = = = 9.65 × 10–5 atm–2 3 p(N2)eq p(H2) eq 6.90 × (20.7)3 e As with all quantitative equilibrium questions, the stoichiometry of the reaction must be used to find the amounts reacted and produced. The units of Kp are atm2 divided by atm4, which gives atm–2. 12 Let the amount of hydrogen that reacts equal x: H2 I2 2HI Start moles 1 1 0 Change in moles –x –x +2x Equilibrium moles 1–x 1–x 2x Equilibrium (1 – x)/50 (1 – x)/50 2x/50 concentration [HI]2eq 4x2 Kc = 49 = = [H2]eq[I2]eq (1 – x)2 Taking the square root of both sides: 2x 7= 1–x 7 – 7x = 2x 9x = 7 x = 7/9 = 0.778, so 78% of the hydrogen reacts© Philip Allan Updates Edexcel A2 Chemistry 23
- 27. Chapter 3 Equilibrium e You are not expected to be able to solve quadratic equations of the form ax2 + bx + c = 0, but you are expected to take the square root of both sides of the equation if necessary. 13 a Kp = p(CO2)eq p(CO)eq p(H2)eq b Kp = p(H2O)eq e In heterogeneous equilibria, the expression for Kp does not include solids. The same is true for Kc expressions, in which [solid] and [solvent] are omitted. [H+]2eq[CrO42–]2eq 14 a Kc = [Cr2O72–]eq e In this reaction the water is a solvent, so its concentration is constant. Therefore, [H2O] is omitted from the expression for Kc. [CH3COOCH2CH2OOCCH3]eq[H2O]2eq b Kc = [CH3COOH]2eq[HOCH2CH2OH]eq e In this reaction the water is a reactant and not the solvent as well, so it must be included in the expression for Kc. 15 Kp = 1.48 atm = p(CO2)eq, so the partial pressure of carbon dioxide at equilibrium is 1.48 atm. e Do not include terms for the partial pressures of solids, as they are meaningless.© Philip Allan Updates Edexcel A2 Chemistry 24
- 28. Chapter 3 Equilibrium Summary worksheet (www.hodderplus.co.uk/philipallan) 1 C Square brackets must not be used in Kp expressions, so option A is incorrect. Water is only omitted from the expression if it is the solvent as well as the reactant, so option B is incorrect. In option D, the expression is upside-down — the reactant partial pressures are on the top of the expression, rather than on the bottom. 2 B In option A [CH3OH] and [H2O] have not been not squared, which ignores the stoichiometry of the equation. In options C and D, water has not been included in the expression. It should have been included because here water is a reactant and not the solvent. 3 D Solids are not included in Kp expressions so option A is incorrect. As there are 4CO and 4CO2 in the equation, the partial pressures of both must be raised to the fourth power, so options B and C are incorrect. Another reason why option C is wrong is that it has the reactant on the top of the expression not the bottom. 4 A Only temperature alters the value of the equilibrium constant. Both pressure and temperature may alter the position of equilibrium. A catalyst alters neither. 5 C The value of the fraction [C][D]/[A][B] on mixing equals 1 × 2/1 × 1 = 2 and so does not equal Kc. This means that the system is not in equilibrium and will react to make the fraction equal to 3, so the fraction has to get bigger. Thus the position of equilibrium moves to the right, increasing the values of [C] and [D] and decreasing the values of [A] and [B] until the fraction equals the value of the equilibrium constant. Option A is incorrect because the system is not in equilibrium as 2 does not equal 3! Option B is incorrect because the value of Kc is only altered by temperature. If the position moved to the left, the fraction would get smaller and become even less than 3, so option D is also wrong. 6 B The relationship between ΔStotal and K is ΔStotal = R ln K, where ΔStotal = ΔSsystem − ΔH/T. So ΔStotal = 150 −(+40 000/298) = +15.77 K = eΔS/R = e15.77/8.31 = 6.67 In option A, a calculator error has been made: K = ln ΔS/R was used, rather than K= e ΔS/R. Option C is incorrect because K = 10ΔS/R was used. In option D, ΔStotal = ΔSsystem + ΔH/T was used, rather than ΔSsystem − ΔH/T. 7 A A 2B C 3D Start moles 0.10 0.20 0 0 Change –0 .8 × 0.10 = –0.080 2 × –0.080 = –0.16 +0.080 3 × +0.080 = +0.24 Equilibrium moles 0.10 – 0.080 = 0.020 0.20 – 0.16 = 0.040 +0.080 0.24 Mole fraction 0.020/0.38 = 0.0526 0.040/0.38 = 0.105 0.080/0.38 = 0.211 0.24/0.38 = 0.632 Partial pressure/atm 2 × 0.0526 = 0.105 2 × 0.105 = 0.210 2 × 0.211 = 0.422 2 × 0.632 = 1.26 [C][D]3 0.422 atm × 1.263 atm3 Kp = = = 182 atm [A][B]2 0.105 atm × 0.2102 atm2 In option B an incorrect value of 0.08 moles was used for substance D. In option C, 0.080 was used as the change in moles of substance B whereas the change equals 2 × 0.080. Both these mistakes are made in option D. 8 A The value at 200°C need not be calculated, but there is a clue. The reaction is exothermic and so the value of K will decrease as the temperature rises. This means that options B and D are incorrect. The correct answer is calculated as follows: ΔStotal at 573 K = ΔSsystem − ΔH/T = −100 − (−75 000/573) = +30.9 J K−1 mol−1 Note that the enthalpy value was given in kJ and so has to be multiplied by 1000 to convert it to J and the temperature must be converted from °C into kelvin (300°C = 300 + 273 = 573 K).© Philip Allan Updates Edexcel A2 Chemistry 25
- 29. Chapter 3 Equilibrium In options B and C, the ΔH value has been divided by the difference in temperature (100°C), rather than the actual temperature (573 K). The expression ΔSsystem + ΔH/T was used in the incorrect options C and D. Another reason for option D being incorrect is that 75 kJ mol−1 were not converted to 75 000 J mol−1 at 573 K.© Philip Allan Updates Edexcel A2 Chemistry 26
- 30. Unit 4 Rates, equilibria and further organic chemistry Chapter 4 Applications of rate and equilibrium [SO32] (2.0/20)2 1 = = 2.0 × 104 mol–1 dm3 [SO2]2[O ] 2 (0.20/20)2 × (0.10/20) The value of the fraction does not equal Kc at 425°C (1.7 × 106 mol–1 dm3), so the system is not in equilibrium. As the fraction is smaller than Kc, the system will try to react to increase the fraction until it reaches the equilibrium value of 1.7 × 106 mol–1 dm3. To do this, the reaction will move from left to right when the catalyst is added. e The purpose of the catalyst is to make the reaction take place at 425°C. Without the catalyst, both the forward and the back reactions would be so slow that the amounts of the three gases would not change. 2 a As this reaction is endothermic (ΔSsurr negative) and is spontaneous, ΔSsystem must be positive. e The value of ΔSsystem must outweigh that of the negative ΔSsurr. This answer could have been predicted because 1 mol of solid turns to 2 mol of gas. b Kp = p(NH3) × p(HCl) c ΔSsurr = −ΔH/T, and, as the reaction is endothermic, this will be negative. At the lower temperature of 298 K its value will be more negative. This will make ΔStotal less positive, reducing the value of ln K and hence of K because ΔStotal = R ln K. e This is predicted by Le Chatelier’s principle. d ΔStotal = R ln K = 8.31 × ln 50 = 8.31 × 3.91 = +32.5 J K−1 mol−1 ΔSsurr = −ΔH/T = −(+176 000/700) = −251.4 J K−1 mol−1 ΔSsystem = ΔStotal − ΔSsurr = +32.5 −(−251.4) = +284 J K−1 mol−1 3 As the reaction is exothermic, ΔSsurr is positive and will decrease as the temperature is increased. Thus a higher temperature is not used because that would make the value of Kp even smaller. A lower temperature would result in a larger value of Kp, and thus a higher theoretical yield, but the rate would be much slower. This is because fewer of the molecules will have enough energy to overcome the activation energy barrier and so a smaller proportion of the collisions would result in reaction. Thus a catalyst is used with a compromise temperature so that a reasonable yield is achieved quickly. A pressure of 300 atm causes the position of equilibrium to be driven to the side with fewer gas moles, thus increasing the equilibrium yield. This is expensive. However, it is necessary because the value of the equilibrium constant is so small that without the high pressure the yield would be uneconomic. 4 a Kp = p(CO2) b The units are atm. c ΔStotal = ΔSsystem + ΔSsurr = ΔSsystem −ΔH/T (i) At 700 K, ΔStotal = +158 − (+177 000/700) = −95 J K−1 mol−1 ΔStotal = R ln K so, K= eΔS/R = e−95/8.31 = 1.14 ×10−5 atm© Philip Allan Updates Edexcel A2 Chemistry 27
- 31. Chapter 4 Applications of rate and equilibrium (ii) At 1200 K, ΔStotal = +158 – (+177 000/1200) = +10.5 J K−1 mol−1 ΔStotal = R ln K so, K= eΔS/R = e10.5/8.31 = 3.54 atm d Endothermic reactions becomes increasingly feasible as the temperature is increased. This reaction will change from being unfavourable to being favourable when ΔStotal = zero. Thus ΔSsystem − ΔH/T = +158 − (+177 000/T) = 0 T = 177 000/158 = 1120 K e At a temperature of 1120 K, the value of Kp = 1 atm. Above this temperature its value is 1 atm and so p(CO2) 1 atm; below this temperature Kp 1 and p(CO2) 1 atm 5 There are two reasons. First, the unreacted nitrogen and hydrogen are recycled repeatedly. This means that the conversion is almost 100%. Second all the atoms in the reactants are converted to ammonia. There is no other product. 6 a The reaction is exothermic. Therefore, the temperature rises as the gases pass through the catalyst. This lowers the value of the equilibrium constant and hence lowers the yield. When the gases are cooled and enter the second catalyst bed, more reaction takes place but the temperature hardly rises. Thus the yield is close to that expected from the initial temperature. b The concentrated sulfuric acid absorbs almost all the sulfur trioxide. This causes the fraction [SO3]2/[SO2]2[O2] to fall to almost zero. The system is now not at equilibrium and when passed through a third bed of catalyst, more sulfur dioxide and oxygen react until the fraction once again equals the value of the equilibrium constant. This results in a much greater total conversion of the sulfur dioxide into sulfur trioxide. 7 A reaction that is exothermic but has a negative ΔSsystem becomes less feasible as the temperature is increased. The change from feasibility to unfeasibility is when ΔStotal becomes zero. This is when ΔSsystem = ΔH/T. In this example it is when −860 = −250000/T, which is when T = 250000/860 = 291K. At this temperature the reaction is likely to be very slow and so the process is unlikely to be both kinetically and thermodynamically feasible. 8 a (i) There will be no change in the value of the equilibrium constant as it is only altered by a change in temperature. The concentration term is: [CuCl4]2− [Cu(H2O)62+][Cl−]4 and gets smaller as the concentration of Cl− ions is increased. e This results in the position of equilibrium shifting to the right until the fraction once again equals Kc. (ii) Because the reaction is exothermic, the value of the equilibrium constant decreases as the temperature is increased. This will cause the position of equilibrium to shift to the left, thus reducing the value of the concentration term until it equals the new K value. e The system is no longer at equilibrium and will shift left until the fraction is reduced so that it equals the new value of K. (iii) The addition of silver ions results in the formation of a precipitate of silver chloride. This will not affect the value of the equilibrium constant, but the value of the concentration term will increase. e The system is no longer at equilibrium and will shift to the left until the fraction is reduced so that it equals the original value of Kc.© Philip Allan Updates Edexcel A2 Chemistry 28
- 32. Chapter 4 Applications of rate and equilibrium b The solution will become yellow. e As was shown in answer a (i), the value of K does not alter but the fraction gets smaller. The hydrated copper(II) ions then react with the chloride ions to form more copper(II) chloride complex, increasing the value of the fraction until it once again equals the value of Kc. 9 Doubling the partial pressures of either gas has no effect on the value of the equilibrium constant but will increase the bottom line of the partial pressure term: p(NH3)2 p(N2) p(H2)3 This makes the fraction smaller and so the system will react to produce more ammonia until the fraction once again equals the value of Kp. Doubling the partial pressure of hydrogen will cause the value of the bottom line to increase by a factor of 23 = 8, whereas doubling the partial pressure of nitrogen will only double the value of the bottom line. 10 a Kp = p(CH4)eq. If the pressure is decreased, the value of Kp does not alter, but the partial pressure of methane decreases. Thus, more methane has to be produced so that the two quantities become equal again. This means that the position of equilibrium shifts to the right. e Altering pressure has no effect on the value of Kp, but it does alter the value of p(CH4), so the two are not equal and the system is no longer in equilibrium. The system then reacts until the new value of p(CH4) equals Kp once more. b As the reaction is endothermic from left to right, an increase in temperature causes the value of Kp to rise. The value of the partial pressure of the methane must also rise, so that the two quantities remain equal. This means that the position of equilibrium moves to the right with an increase in temperature. e Altering the temperature alters the value of Kp, so the system is no longer in equilibrium. The system must react until the value of p(CH4) reaches the new value of Kp. c Kp = p(CH4)eq and since p(CH4)eq = 0.86 atm, the value of Kp = 0.86 atm. p[NO]4eq[H2O]6eq 11 a Kc = [NH3]4eq[O2]5eq As the reaction is exothermic, an increase in temperature causes the value of the equilibrium constant, K, to decrease. The value of the fraction must also alter, to equal the new value of K. This is achieved by the equilibrium shifting from right to left. This reduces the value of the top line (the numerator) and increases the value of the bottom line (the denominator). So the position of equilibrium shifts to the left (the equilibrium yield decreases). e The reasoning is as follows: change in T → change in the value of K → change in position of equilibrium b An increase in pressure has no effect on the value of K, but it does alter the value of the fraction. As there are 10 gas moles on the right and only 9 on the left, the increase in the numerator is greater than that in the denominator. This means that the fraction becomes bigger than K, so the equilibrium must move to the left (the side with fewer gas molecules). The position of equilibrium shifts to the left (the equilibrium yield decreases). e The reasoning is as follows: increase in P → no change in K but change in the value of the fraction → fraction alters to regain equality with K In the industrial process, a pressure of between 2 atm and 4 atm is employed. A lower pressure cannot be used because the gases have to be forced through the industrial plant.© Philip Allan Updates Edexcel A2 Chemistry 29

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