Edexcel A2 Chemistry by George Facer

Teacher Guide

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: [email protected] 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.

© Philip Allan Updates

Edexcel A2 Chemistry

P01318

Environmental information The paper on which this title is printed is sourced from managed, sustainable forests.

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

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

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 Lithium Sodium Potassium Calcium Strontium Barium

© Philip Allan Updates

Preferred colour Red Yellow (not orange) Lilac Red or brick red Crimson Pale green

Acceptable colours Carmine, deep red or crimson Mauve or purple Orange-red Red Apple green or yellowy-green

Edexcel A2 Chemistry

2

Introduction

Indicators (alkali in the burette, acid in the conical flask) Indicator Methyl orange Phenolphthalein

Starting colour Red Colourless

End-point colour Orange or peach Pink

Not Yellow or pink Red or purple

Indicators (acid in the burette, alkali in the conical flask) Indicator Methyl orange Phenolphthalein

Starting colour Yellow Red or purple

End-point colour Orange or peach Colourless

Not Red or pink 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

Unit

4

Rates, equilibria and further organic chemistry

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

2 a

Fraction of molecules with energy E

order to convert it from cm3 to dm3.

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

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 ratedetermining 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

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: k= e

rate 2.2 × 10–5 = = 1.1 × 10–3 mol–1 dm3 s–1 [C2H5I][KOH] 0.20 × 0.10

The units are given by: concentration × s–1 (concentration)2

= (concentration)–1 s–1 = mol–1 dm3 s–1

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: k=

rate 2.3 × 10–3 = = 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

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 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

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: k= e

1.2 × 10–5 = 6.3 × 10–3 mol–1 dm3 s–1 0.038 × 0.050

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 concentration 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 concentration of 2-bromopropane. If the first rate equation is correct, the initial slope will be twice as large (rate

10 a

[Cyclopropane]/mol dm–3

twice as fast). If the second rate equation is correct, the slope will be unchanged. 0.08 0.07 0.06 0.05 14 min

0.04 0.03

14 min

0.02

14 min

0.01 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

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

[A]/mol dm

12 a

–3

reaction is second order. 0.20

0.18 Tangent at 0.16 0.195 – 0.04 47

0.16

= 3.3 × 10–3

0.14

0.12

0.10

Tangent at 0.08 0.147 – 0.04 71

0.08

= 1.5 × 10–3

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 = = 0.45 ≈ –12 slope when [A] is 0.16 3.3 × 10–3 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

Chapter 1 Rates: how fast?

Ea2

Enthalpy

b

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 The units of k cannot be found from the calculation.

15 a

ln k –7.9 –6.8 –4.8 –2.9

1 — / –1 T K

0.00152 0.00147 0.00139 0.00130

ln k

e

0.0013 –2

0.0014

1 — / –1 T K

0.0015

–3 –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

Chapter 1 Rates: how fast?

b slope of the straight line =

−8 − (2.4) −5.6 = = −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 0.0083 120 0.0145 69 0.0278 36 14 0.0714

ln (1/time) Temperature/°C Temperature/K 1/temperature/K−1 17 290 0.00345 −4.79 25 298 0.00336 −4.23 35 308 0.00325 −3.58 50 323 0.00310 −2.64

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.

1 In time

c 0.0030 –2

0.0031

0.0032

0.0033

0.0034

1 /K–1 T 0.0035

–2.5

–3

–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

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 ΔT =

heat produced 702 J = = 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

Step 1

C2H 5 Slow

C H3C

C+

Cl

H3C

+ H

Cl –

H

C2H 5

Step 2

C+ H3C

C2H 5 OH – H

Fast C H3C

OH

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

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 polarisation 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

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 diatomic gaseous elements are based on –12 mol of the diatomic elements. This applies to elements such as hydrogen, nitrogen and chlorine. The value for nitrogen is given as 95.8, which is for –12 N2, so for N2 the value is 2 × 95.8 = 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

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) ∆Hlatt(LiF)

LiF(s) + aq

+

∆Hhyd(Li+)

∆Hsoln(LiF)

Li+(aq)

F–(g) ∆Hhyd(F–)

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

Chapter 2 Entropy: how far?

9 The Hess’s law diagram is: Ca2+(g) ∆Hlatt(CaCl2)

CaCl2(s) + aq

+

∆Hhyd(Ca2+)

∆Hsoln(CaCl2)

Ca2+(aq)

+

2Cl–(g) 2 × ∆Hhyd(Cl–)

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 ΔHhyd(Cl–(g)) = –12 × (–670) = –335 kJ mol–1 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

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

aqueous cations.

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

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.

13 a The reaction is going from 1–12 mol of gas to 1 mol of gas and so the entropy of the system decreases. 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 ΔSsurr =

– ΔH 57.4 57.4 = = = –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

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

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

Unit 4 Rates, equilibria and further organic chemistry

Chapter 3 Equilibrium 1 a Kc =

[CH3OH]eq [CO]eq[H2]2eq

b Kc =

[NO2]eq 1 [NO]eq[O2]2–eq

c Kc =

[NO]4eq[H2O]6eq [NH3]4eq[O2]5eq

d Kc =

[HBr]eq [C2H5Br]eq[ H2O]eq

e Kc =

[H3O+]eq[[Fe(H2O)5OH]2+]eq [[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 raised to the powers according to the stoichiometric numbers in the reaction equation. Thus, –12 O2(g) appears in equation b, so [O2] must be raised to the power –12 . 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: Kc =

[PCl5]eq [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.

3 a The fraction

[CO2][H2] 3/100 × 1/100 = = 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

Chapter 3 Equilibrium

b Kc = 0.20 = [H2]eq =

4 Kc =

0.20 × 0.035 × 0.0040 = 0.0028 mol dm–3 0.010

[SO2]2eq[O2]eq [SO3]2eq

= e

[CO2]eq[H2]eq 0.010 × [H2]eq = [CO]eq[H2O]eq 0.035 × 0.0040

(0.044/10)2 × (0.022/10) = 3.5 × 10–3 mol dm–3 at 700°C (0.035/10)2

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 d ratio =

moles ammonia reacted 0.04022 = = 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 G

The number of ammonia molecules must be a whole number, so 4.022 is converted to 4. 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.

6 a Kc = b

[CO]eq[H2]3eq [CH4]eq[H2O]eq CH4

Start moles 1.00 Change in moles –0.75 Equilibrium moles 0.25 Equilibrium 0.025 concentration

Kc = e

H2O 2.00 –0.75 1.25 0.125

CO 0 +0.75 0.75 0.075

3H2 0 +2.25 2.25 0.225

0.075 × (0.225)3 = 0.273 mol2 dm–6 at 1200 K 0.025 × 0.125

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

Chapter 3 Equilibrium

7 Start moles

2H2 0.300

CH3OH 0

–0.0300

–0.0600

+0.0300

Equilibrium moles

0.0700

0.240

0.0300

Equilibrium concentration

0.00700

0.0240

0.00300

Change in moles

Kc = e

CO 0.100

[CH3OH]eq 0.00300 = = 744 mol–2 dm6 at 400°C 0.00700 × (0.0240)2 [CO]eq[H2]2eq

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. Start moles Change in moles Equilibrium moles

CH3OOCCOOCH3 11.8/118 = 0.100 –0.075 0.100 – 0.075 = 0.025 0.025/0.015 = 1.67

Equilibrium concentration

Kc =

2H2O 5.40/18 = 0.300 2 × (–0.075) = –0.15 0.300 – 0.15 = 0.15

HOOCCOOH 0 +0.075 0.075

2CH3OH 0 2 × +0.075 = +0.15 0.15

0.15/0.015 = 10

0.075/0.015 = 5

0.15/0.015 = 10

[HOOCCOOH]eq[CH3OH]2eq 5 × 102 = [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.)

9 a Kp =

p(SO2)eq p(Cl2)eq p(SO2Cl2)eq

b Kp =

p(NO2)2eq p(NO)2eq p(O2)eq

c Kp =

p(SO3)2eq p(SO2)2eq p(O2)eq

e

Never use square brackets in a Kp expression.

© Philip Allan Updates

Edexcel A2 Chemistry

22

Chapter 3 Equilibrium

10 Start moles Change in moles Equilibrium moles Mole fraction Partial pressure/atm

Kp = e

N2O4 1 –0.15 1 – 0.15 = 0.85 0.85/1.15 = 0.739

2NO2 0 +0.30 0.30 0.30/1.15 = 0.261

0.739 × 1.2 = 0.887

0.261 × 1.2 = 0.313

Total moles

1.15

p(NO2)2eq (0.313)2 = = 0.11 atm p(N2O4)eq 0.887

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)

11 a Kp =

p(NH3)2eq p(N2)eq p(H2)3eq

b Start moles Change in moles

3H2

2NH3

1

3

0

–0.15

3 × (–0.15) = –0.45

2 × (+0.15) = +0.30

1 – 0.15 = 0.85

3 – 0.45 = 2.55

0.30

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

Equilibrium moles

Kp = e

N2

Total moles

3.70

p(NH3)2eq (2.43)2 = = 9.65 × 10–5 atm–2 3 p(N2)eq p(H2) eq 6.90 × (20.7)3

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 Start moles 1 1 Change in moles –x –x Equilibrium moles 1–x 1–x Equilibrium (1 – x)/50 (1 – x)/50 concentration

Kc = 49 =

2HI 0 +2x 2x 2x/50

[HI]2eq 4x2 = [H2]eq[I2]eq (1 – x)2

Taking the square root of both sides: 7=

2x 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

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 b Kp = e

p(CO)eq p(H2)eq p(H2O)eq

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.

14 a Kc = e

[H+]2eq[CrO42–]2eq [Cr2O72–]eq

In this reaction the water is a solvent, so its concentration is constant. Therefore, [H2O] is omitted from the expression for Kc.

b Kc = e

[CH3COOCH2CH2OOCCH3]eq[H2O]2eq [CH3COOH]2eq[HOCH2CH2OH]eq

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

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 Start moles

A 0.10

2B 0.20

C 0

3D 0

+0.080 +0.080

3 × +0.080 = +0.24 0.24

Equilibrium moles

–0 .8 × 0.10 = –0.080 2 × –0.080 = –0.16 0.10 – 0.080 = 0.020 0.20 – 0.16 = 0.040

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

2 × 0.0526 = 0.105

2 × 0.105 = 0.210

2 × 0.211 = 0.422

2 × 0.632 = 1.26

Change

Partial pressure/atm

Kp =

[C][D]3 0.422 atm × 1.263 atm3 = = 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

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

Unit 4 Rates, equilibria and further organic chemistry

Chapter 4 Applications of rate

and equilibrium 1

[SO32] (2.0/20)2 = = 2.0 × 104 mol–1 dm3 2 [SO2] [O2] (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

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

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. 11 a Kc =

p[NO]4eq[H2O]6eq [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

Chapter 4 Applications of rate and equilibrium

c Adding excess air would not alter the value of K (only temperature does), but it makes the denominator bigger, so the fraction no longer equals K. The position of equilibrium moves to the right, increasing the numerator until the fraction and K are once again equal. This means that the position of equilibrium shifts to the right (increasing the equilibrium yield of NO from ammonia).

d A catalyst has no effect on the value of the equilibrium constant or on the fraction. So K remains the same value and the position of the equilibrium also remains unchanged. e

Equilibrium is reached more quickly as the catalyst speeds up the forward and the back reactions.

12 a None. Adding argon has no effect on the value of the equilibrium constant. Only a change in temperature will alter the value of Kc. Kc =

[NH3]2eq [N2]eq [H2]3eq

When argon is added at constant volume, the value of the concentrations of all three species does not alter, as concentration = moles/volume. Thus, neither the value of K nor the value of the concentration term alters, so the system is still in equilibrium

b The equilibrium yield of ammonia would increase. As before, the value of Kc does not alter. However, adding argon at constant pressure means that the volume must increase. This will cause all the concentrations to decrease, but as the top line of the concentration term is (concentration)2 and the bottom line (concentration)4, the bottom line will decrease more. To get the concentration term back to equalling Kc, the position of equilibrium will have to shift to the right, making more ammonia.

© Philip Allan Updates

Edexcel A2 Chemistry

30

Chapter 4 Applications of rate and equilibrium

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 A Only temperature alters the value of K for a given reaction. Both temperature and pressure alter the position of equilibrium. The presence of a catalyst has no effect on either the value or position of equilibrium.

2 C Beware — this is a negative question. Options A and B are true statements, because this is an exothermic reaction. Option D is also true as there are fewer gas moles on the left. (Careful! The pressure was lowered.) The one false statement is C, which is the correct answer to this negative question.

3 B This question is about rate of reaction. As it is a homogeneous gas reaction, a high pressure and a high temperature will both increase the rate and so result in equilibrium being reached sooner.

4 D This reaction is exothermic, so low temperature will favour a high yield. This means that options A and B are incorrect. There are more gas moles on the left-hand side, so high pressure will drive the equilibrium to the right, increasing the yield. Therefore, option C is incorrect.

5 A The fact that a high pressure is not used tells you that the value of the equilibrium constant is large. This means that options B and C are wrong. The fact that the reaction works well at room temperature and pressure, shows that the reaction rate is fast under these conditions. Option D is a true statement because all the atoms in the reactants become product. However, this is irrelevant to the choice of conditions.

6 A This is another negative question. Read the whole question and put ‘true’ or ‘false’ next to each statement. Option A is false: it is the number of gas molecules that matters. However, before choosing A as your answer, it is safer to check the other options. Option B is true as there are fewer gas molecules on the left-hand side. Option C is true as it is an endothermic reaction and the rate of an endothermic reaction is always increased by an increase in temperature. Option D is also true — removing a product will drive the equilibrium to the right. So option A is the false statement and is the correct answer to this negative question.

7 B In an exothermic reaction, an increase in temperature always reduces the value of the equilibrium constant, so options C and D are incorrect. The logic in option A is wrong. The equilibrium position moves to the left because the value of K decreases, not the other way round. K depends on ΔStotal, which equals ΔSsystem − ΔH/T. The term −ΔH/T is a positive number and so, as T increases, it will get smaller, causing ΔStotal and K to get smaller.

8 B An increase in temperature of an endothermic reaction will cause −ΔH/T to become less negative. Thus option D is incorrect. As −ΔH/T becomes less negative, ΔStotal will become more positive. Therefore, option A is also wrong. At A-level it is assumed that ΔSsystem is not altered by temperature, so option C is incorrect. e

A quick way of getting the correct answer is to realise that for an endothermic reaction an increase in temperature will increase the value of K. This is because the value of ΔStotal must have increased (become more positive), so answer B is correct.

9 A An increase in temperature of an exothermic reaction causes the value of the equilibrium constant to decrease. Therefore, options C and D are incorrect. All reactions are speeded up by an increase in temperature, so equilibrium is reached sooner. This is the same as saying that the time to reach equilibrium will be less. Thus option B is incorrect.

© Philip Allan Updates

Edexcel A2 Chemistry

31

Unit 4 Rates, equilibria and further organic chemistry

Chapter 5 Acid–base equilibria 1 The equation is: HCl + H2O → H3O+ + Cl– The HCl donates a proton (H+ ion) to a water molecule, and so acts as a B–L acid. The solution is acidic because H3O+ ions are produced, and so [H3O+] > [OH–]. e

H3O+ ions are formed. H3O+ is the conjugate acid of the base H2O. Cl– ions are also formed. Cl– is the conjugate base of the acid HCl.

2 H2SO4 is acting as an acid and its conjugate base is the HSO4– ion. CH3COOH is acting as a base and its conjugate acid is CH3COOH2+.

base 2 CH3COOH

e

+

acid 2 CH3COOH2+

H2SO4 acid 1

+

HSO4− base 1

Use the same number for the acid and its conjugate base when identifying pairs.

3 a The conjugate base of HCN is CN–. b The conjugate base of NH3 is NH2–. c The conjugate base of HClO3 is ClO3–. d The conjugate base of OH– is O2–. e The conjugate base of [Fe(H2O)6]3+ is [Fe(H2O)5OH]2+. e

The conjugate base is formed by removing a proton from the acid. The conjugate base will always be more negative or less positive than its parent acid.

4 a The conjugate acid of NH3 is NH4+. b The conjugate acid of CH3NH2 is CH3NH3+. c The conjugate acid of OH– is H2O. d The conjugate acid of HNO3 is H2NO3+. e

The conjugate acid is formed by adding a proton to the base. The conjugate acid will always be more positive or less negative than its parent base.

5 Water ionises according to the equation: H2O  H+ + OH–

ΔH = +57 kJ mol–1

Kw = [H+] × [OH–] = 1.0 × 10–14 mol2 dm–6 at 25°C A solution is neutral when [H+] = [OH–] = At 25°C [H+] =

Kw.

Kw = 1.0 × 10–7, so neutral pH = –log (1.0 × 10–7) = 7.0.

The value of Kw depends on the temperature. As the reaction is endothermic left to right, Kw is greater than 1.0 × 10–14 mol2 dm–6 above 25°C. This means that pure water at the higher temperature has [H+] > 1.0 × 10–7, so its pH is less than 7.0.

© Philip Allan Updates

Edexcel A2 Chemistry

32

Chapter 5 Acid-base equibria

Below 25°C, Kw is less than 1.0 × 10–14 mol2 dm–6 and so pure water has [H+] < 1.0 × 10–7 and its pH is greater than 7.0. e

Remember that the higher the value of [H+], the lower the value of the pH.

6 pKw = 14 Kw = 10–pKw = 1.0 × 10–14 mol2 dm–6 Kw = [H+] × [OH–], so [OH–] = Kw/[H+] = 1.0 × 10–14/[H+]

a If [H+] = 1.0 × 10–2 mol dm–3, [OH–] = 1.0 × 10–14/1.0 × 10–2 = 1.0 × 10–12 mol dm–3. [OH–] < [H+], so the solution is acidic.

b If [H+] = 2.2 × 10–7 mol dm–3, [OH–] = 1.0 × 10–14/2.2 × 10–7 = 4.5 × 10–8 mol dm–3. [OH–] < [H+], so the solution is acidic.

c If [H+] = 3.3 × 10–10 mol dm–3, [OH–] = 1.0 × 10–14/3.3 × 10–10 = 3.0 × 10–5 moldm–3. [OH–] > [H+], so the solution is alkaline. e

The pH of the solution can be calculated from [H+]. This can be used to find the pOH and hence [OH–]. For example, for the solution in b: pH = –log (2.2 × 10–7) = 6.66 pOH = 14 – pH = 14 – 6.66 = 7.34 [OH–] = 10–7.34 = 4.6 × 10–8 mol dm–3 (If all the numbers are left in full on the calculator, the value obtained is 4.5 × 10–8 mol dm–3.)

7 a pH = –log [H+] = –log (4.4 × 10–5) = 4.36 b pH = –log (5.5 × 10–9) = 8.26 c pOH = –log [OH–] = –log (6.6 × 10–2) = 1.18 pH = 14 – pOH = 14 – 1.18 = 12.82

d pOH = –log (7.7 × 10–11) = 10.11 pH = 14 – 10.11 = 3.89 e

The alternative method for calculating the pH in c and d is: Kw = [H+] × [OH–] = 1.0 × 10–14 mol2 dm–6, so [H+] = 1.0 × 10–14/[OH–] For the solution in c: [H+] = 1.0 × 10–14/6.6 × 10–2 = 1.52 × 10–13, so pH = –log [H+] = 12.82

8 a [H+] = 10−pH = 10−1.33 = 0.0468 mol dm−3 b [H+] = 10−7.00 = 1.00 × 10−7 mol dm−3 c [H+] = 10−13.67 = 2.14 × 10−14 mol dm−3 9 a At pH = 7, [H+] = [OH–], so the ratio is 1:1. b At pH = 10, [H+] = 10–pH = 1.0 × 10–10 [OH–] = 1.0 × 10–14/1.0 × 10–10 = 1.0 × 10–4 mol dm–3 Ratio [H+]:[OH–] = 1.0 × 10–10/1.0 × 10–4 = 1:106 or 1 to a million

c At pH = 3, [H+] = 10–pH = 1.0 × 10–3

© Philip Allan Updates

Edexcel A2 Chemistry

33

Chapter 5 Acid-base equibria

[OH–] = 1.0 × 10–14/1.0 × 10–3 = 1.0 × 10–11 mol dm–3 Ratio [H+]:[OH–] = 1.0 × 10–3/1.0 × 10–11 = 1.0 × 108:1 or one hundred million to one e

In these calculations it is assumed that the temperature is 25°C. An alternative method involves using the expression pH + pOH = 14, in part b: If pH = 10, [H+] = 10–pH = 1.0 × 10–10 pOH = 14 – 10 = 4, so [OH–] = 1.0 × 10–4 mol dm–3

10 a HBr is a strong acid and so is totally ionised: [H+] = [HBr] = 0.200 mol dm–3 pH = –log 0.200 = 0.70

b LiOH is a strong base and so is totally ionised: [OH–] = [LiOH] = 0.200 mol dm–3 pOH = –log 0.200 = 0.70 pH = 14 – pOH = 14 – 0.70 = 13.30

c Sr(OH)2 is a strong base with two OH– ions per mole: [OH–] = 2 × [Sr(OH)2] = 2 × 0.0500 = 0.100 mol dm–3 pOH = –log 0.100 = 1.00 pH = 14 – pOH = 14 – 1.00 = 13.00 e

Sulfuric acid, H2SO4, is a strong acid in its first ionisation only. A 0.0500 mol dm–3 solution of the acid will have [H+] slightly greater than 0.0500 but not as high as 0.100 mol dm–3.

11 As it is a strong acid, [H+] = 2.0 mol dm−3, so pH = −log 2.0 = −0.30.

e

I

When diluted 10 times, [H+] = 0.20 mol dm−3, so pH = 0.70.

I

When diluted 100 times, [H+] = 0.020 mol dm−3, so pH = 1.70.

I

When diluted 1 000 000 times, [H+] = 0.0000020 mol dm−3, so pH = 5.70.

Note that pH values can be negative. This occurs when [H+] > 1 mol dm−3. They can also be greater than 14, if [OH−] > 1 mol dm−3. It is a fallacy to think that the pH scale goes from 0 to 14.

12 a HOCl  H+ + ClO– Ka =

[H+][ClO–] = 3.02 × 10–11 mol dm–3 [HOCl]

b It is assumed that [H+] = [ClO–]: Ka = [H+]2/[HOCl] [H+] =

Ka × [HOCl] =

3.02 × 10–11 × 0.213 = 2.54 × 10–6 mol dm–3

pH = –log (2.54 × 10–6) = 5.60 e

The assumption that [H+] = [ClO–] is not valid for calculations involving buffer solutions. However, in weak acid and in buffer solution calculations, it is assumed that a negligible amount of the weak acid ionises. Thus, [HOCl]eq = [HOCl]initial = 0.213 mol dm–3. Always check that your answer to the pH of a solution of an acid is less than 7.

© Philip Allan Updates

Edexcel A2 Chemistry

34

Chapter 5 Acid-base equibria

13 HNO2  H+ + NO2– Ka =

[H+][NO2–] [HNO2]

As with all weak acid calculations, it is assumed that [H+] = [NO2–]: [H+] = 10–pH = 10–2.02 = 9.55 × 10–3 mol dm–3 = [NO2–] The assumption can also be made that a negligible amount of the HNO2 ionises: [HNO2]eq = [HNO2]initial = 0.200 mol dm–3 Ka = e

(9.55 × 10–3)2 = 4.56 × 10–4 mol dm–3 0.200

The assumption that [HNO2]eq = [HNO2]initial need not be made here, as [NO2–] = 9.55 × 10–3 mol dm–3 and so 9.55 × 10–3 mol dm–3 of HNO2 must have ionised. Thus, [HNO2]eq = 0.200 – 9.55 × 10–3 = 0.190 mol dm–3. This gives a more accurate value of Ka = 4.80 × 10–4 mol dm–3.

14 C2H5COOH  H+ + C2H5COO– Ka =

[H+][C2H5COO–] = 1.35 × 10–5 mol dm–3 [C2H5COOH]

[H+] = [C2H5COO–] = 10–pH = 10–3.09 = 8.13 × 10–4 mol dm–3 [C2H5COOH] =

= e

[H+][C2H5COO–] Ka (8.13 × 10–4)2 = 0.0489 mol dm– 3 1.35 × 10–5

Check that your value of the concentration of a weak acid solution is somewhere between 2 and 0.01 mol dm–3. If it is not, then you probably forgot to square the numerator or failed to treat one of the powers of ten as a negative number.

15 CH2(OH)COOH  H+ + CH2(OH)COO– Ka =

[H+][CH2(OH)COO–] = 10–pKa = 10–3.83 = 1.48 × 10–4 mol dm–3 [CH2(OH)COOH]

[H+] = [CH2(OH)COO–] and [CH2(OH)COOH] = 1.05 mol dm–3 [H+]2 = Ka × [CH2(OH)COOH] [H+] =

Ka × [CH2(OH)COOH] =

4.36 × 10−4 × 1.05 = 0.0125 mol dm–3

pH = –log 0.0125 = 1.90

16 a Kb = e

[CH3NH3+][OH–] [CH3NH2]

The term [H2O] is left out because it is the solvent.

b [CH3NH3+] = [OH–] [OH–]2 = Kb × [CH3NH2] [OH–] =

Kb × [CH3NH2] =

4.36 × 10−4 × 0.200

= 9.34 × 10–3 mol dm–3 pOH = –log(9.34 × 10–3) = 2.03 pH = 14 – pOH = 14 – 2.03 = 11.97

© Philip Allan Updates

Edexcel A2 Chemistry

35

Chapter 5 Acid-base equibria

e

The calculation is similar to those for weak acids, with similar assumptions. The difference is that [OH–] is calculated first and used to evaluate the pH.

17 a A buffer solution is one in which the pH hardly alters when small amounts of acid or base are added. e

Do not say that the pH is constant. The definition of a buffer as ‘a solution that resists change in pH when small amounts of acid or base are added’ is also acceptable.

b Ka =

[H+][CH3COO–] [H+][salt] = = 1.8 × 10–5 mol dm–3 [CH3COOH] [acid]

molar mass of sodium ethanoate (CH3COONa) = 3.0 + (2 × 12.0) + (2 × 16.0) + 23.0 = 82.0 g mol–1 amount of sodium ethanoate = 5.65 g/82.0 g mol–1 = 0.0689 mol [salt] = 0.0689/0.100 = 0.689 mol dm–3 [H+] =

Ka × [acid] 1.8 × 10–5 × 1.00 = = 2.61 × 10–5 mol dm–3 [salt] 0.689

pH = –log [H+] = –log (2.61 × 10–5) = 4.58 e

The assumptions made for buffer solutions are: G

The salt is totally ionised, so [CH3COO–] = [salt].

G

Ionisation of the acid is low and is also suppressed by the anions from the salt, so [CH3COOH]eq = [acid]initial.

A useful way of checking your answer in buffer calculations is to compare the amounts of acid and salt in the buffer. If there are more moles of acid than salt, the pH of the buffer will be less (more acidic) than the pKa value of the weak acid. In this question, pKa = –log Ka = 4.74 and [acid] > [salt], so the calculated pH (4.58) should be less than 4.74. Similarly, if the pH is greater than the pKa of the weak acid, there will be more moles of salt than weak acid in the buffer.

18 NaOH + CH3COOH → CH3COONa + H2O When the sodium hydroxide is added, the salt sodium ethanoate is formed. There is excess acid, so the mixture is a buffer solution. amount of salt = amount of NaOH = 2.00 mol dm–3 × 0.050 dm3 = 0.100 mol initial amount of acid = 1.00 mol dm–3 × 0.150 dm3 = 0.150 mol amount of acid reacted = amount of NaOH added = 0.100 mol amount of acid in excess = 0.150 – 0.100 = 0.050 mol total volume of solution = 50 cm3 + 150 cm3 = 200 cm3 = 0.200 dm3 [acid in excess] = 0.050/0.200 = 0.25 mol dm–3 [salt] = 0.100/0.200 = 0.50 mol dm–3 Ka = 1.8 × 10–5 mol dm–3 =

[H+] =

[H+][CH3COO–] [H+][salt] = [CH3COOH] [acid]

Ka × [CH3COOH] 1.8 × 10–5 × 0.25 = = 9.0 × 10–6 mol dm–3 [CH3COO–] 0.50

pH = –log [H+] = –log (9.0 × 10–6) = 5.05 e

You must not make the mistake of thinking of sodium hydroxide as the conjugate base in the buffer mixture. It reacts with some of the acid to form the anion, CH3COO–, which is the conjugate base of ethanoic acid. The second common error is to fail to divide the moles of each component by the total volume (which must be in dm3). This type of buffer calculation is the hardest type tested at A-level.

© Philip Allan Updates

Edexcel A2 Chemistry

36

Chapter 5 Acid-base equibria

19 C2H5COOH  H+ + C2H5COO– Ka = 10–pKa = 10–4.87 = 1.35 × 10–5 = =

[H+][C2H5COO–] [C2H5COOH] [H+] × moles of salt moles of acid

[H+] = 10–pH = 10–4.50 = 3.16 × 10–5 mol dm–3 moles of salt = volume × concentration = 0.050 × 1.00 = 0.050 moles of acid =

[H+] × moles of salt 3.16 × 10–5 × 0.050 = = 0.117 Ka 1.35 × 10–5

volume of acid = e

moles 0.117 = = 0.117 dm3 = 117 cm3 concentration 1.00

As the acid and the salt are in the same solution in the buffer, the ratio [salt]/[acid] is the same as the ratio (moles of salt)/(moles of acid).

20 CH3COOH  H+ + CH3COO– Ka = 1.80 × 10–5 =

[H+][CH3COO–] [CH3COOH]

[H+] = 10–pH = 10–5.00 = 1.00 × 10–5 mol dm–3 [CH3COO–] =

Ka × [CH3COOH] 1.80 × 10–5 × 1.25 = = 2.25 mol dm–3 [H+] 1.00 × 10–5

amount of CH3COO– ions in 100 cm3 = 0.225 mol As 1 mol calcium ethanoate, Ca(CH3COO–)2, contains 2 mol CH3COO– ions: moles of calcium ethanoate = 0.225/2 = 0.1125 molar mass of Ca(CH3COO)2 = 40.1 + (4 × 12.0) + (4 × 16.0) + 6 = 158.1 g mol–1 mass of calcium ethanoate = moles × molar mass = 0.1125 × 158.1 = 17.8 g e

In this example, [CH3COO–] = 2 × [salt].

21 CH3COOH  H+ + CH3COO– Ka = 1.80 × 10–5 =

[H+][CH3COO–] [H+][salt] = [CH3COOH] [acid]

[H+] = 10–pH = 10–4.44 = 3.63 × 10–5 mol dm–3 As both salt and acid are mixed in the same solution: [salt] moles of salt = [acid] moles of acid =

Ka 1.80 × 10–5 1 = = + 2 [H ] 3.63 × 10–5

This means that one-third of the acid must be neutralised, forming one-third salt and two-thirds unreacted acid. The initial volume of acid is 100 cm3 and both the acid and the sodium hydroxide are of the same concentration: –13 × 100 = 33.3 cm3 of sodium hydroxide must be added

© Philip Allan Updates

Edexcel A2 Chemistry

37

Chapter 5 Acid-base equibria

e

There are two important points in this question. The first is that the sodium hydroxide reacts with some of the ethanoic acid to form the salt, sodium ethanoate. The second point is that there must be twice as much acid left as salt formed. This means that one-third of the acid must react.

22 a CH3NH3+ + H2O  CH3NH2 + H3O+ As H3O+ ions are formed, the solution will be acidic (pH < 7). e

The equation is similar to that of the ammonium ion reacting with water.

b CN– + H2O  HCN + OH– e

As OH– ions are formed, the solution will be alkaline (pH > 7).

c CO32– + H2O  HCO3– + OH–, forming an alkaline solution. e

A further reaction can take place: HCO3– + H2O  H2CO3 + OH–, but this occurs to a very small extent because the OH– ions produced in the first reaction suppress this second reaction.

d I– + H2O → no reaction, and so the solution will be neutral (pH = 7 at 25°C). HI is a strong acid (stronger than HCl), and so I– is too weak a conjugate base to react with an acid as weak as water. e –

I ions will react with very strong acids, such as concentrated sulfuric acid, to form HI (which is subsequently oxidised to

iodine by concentrated sulfuric acid).

23 a

pH 12 Buffer 10 8 6 4 2 0 0

e

10

20

30 40 Volume of HCl/cm3

Note the rule of two to determine the pH from the strong-base value. G

Ammonia is a weak base, so the starting pH ≈ 11 (the strong-base value of 13 – 2).

G

The pH of the equivalence point is ≈ 5 (the strong-base value of 7 – 2).

G

The pH after an excess of strong acid has been added ≈ 1 (the strong-acid value).

Note also that as the concentration of the acid is twice that of the alkali, the equivalence point is when 20 cm3 of HCl has been added to 40 cm3 of NH3 solution.

The whole pH range of the indicator must lie on the vertical section of the graph. In this example, where a strong acid is being added to a weak base, the pH of the vertical section is from about pH 7 to pH 3. The ranges for methyl orange, bromophenol blue, bromocresol green and methyl red all lie fully within pH 7–3 and so would be good indicators for the titration. e

A common error is to state that the pKind value must be somewhere on the vertical section. The whole range (usually pKind ± 1) must lie within the vertical section.

© Philip Allan Updates

Edexcel A2 Chemistry

38

Chapter 5 Acid-base equibria

b A solution is a buffer when it contains significant amounts of both the acid and its conjugate base. Here that is when some, but not all, of the ammonia has been neutralised by the hydrochloric acid. The solution then contains both NH3 and NH4+. It is best as a buffer when half the ammonia has been neutralised, that is when 10 cm3 of hydrochloric acid have been added.

24 a pH

e

12 11 10 9 8 7 6 5 4 3 2 1 0

pH 11.0

pH 7.1

0

5

10

15

20

25 30 35 Volume NaOH/cm3

Note the initial rapid increase in pH followed by the flatter (buffered region), then the rapid rise around the equivalence point, flattening off to the pH value for strong alkali.

b The pKa equals the pH when half the acid has been neutralised. As the acid and the alkali are both 0.100 mol dm–3, half-neutralisation will occur when 10 cm3 of NaOH has been added. Thus, pKa = pH at 10 cm3 = 4.65. Ka = 10–pKa = 10–4.65 = 2.24 × 10–5 mol dm–3

c When the acid is fully neutralised, [salt] = 1–2 [acid]initial = 0.050 mol dm–3. Neutralisation is at the point halfway up the vertical section of the graph. The pH at this point is 9.05, so the pH of the salt solution is 9.05. e

The addition of sodium hydroxide increases the volume of the solution. At the equivalence point, the volume has gone from 20 cm3 to 40 cm3, so the concentration of the salt is 0.050 mol dm–3.

d The full colour change (the pH range) of the indicator must be within the vertical part of the titration curve. Here the vertical part goes from 7.1 to 11.0 and therefore thymol blue or phenolphthalein would be suitable indicators. e

Do learn to spell ‘phenolphthalein’ correctly. A missing ‘h’ or the ending ‘-ene’ is acceptable to examiners, but ‘phenyl-’ or ‘-palin’ is not.

25 HThy  H+ + Thy– yellow

blue

I

When acid is added, [H+] is increased. The equilibrium is shifted to the left and the indicator turns yellow.

I

When alkali is added, the OH– ions react with the H+ ions from the indicator and reduce the value of [H+]. This drives the equilibrium to the right and the colour changes from yellow to green and finally to blue.

26 a HBr and HCl are strong acids, which means that they are totally ionised. The neutralisation reaction for both is H+ + OH– → H2O, so the enthalpy of neutralisation is the same.

b The enthalpy change for neutralisation of a weak acid can be considered as the sum of two reactions: HA  H+ + A–

ΔH = x kJ mol–1

H+ + OH– → H2O

ΔH = –57 kJ mol–1

© Philip Allan Updates

Edexcel A2 Chemistry

39

Chapter 5 Acid-base equibria

As alkali is added, the OH– ions react with the H+ ions and so the first equilibrium is driven to the right. The overall ΔHneut = x + (–57) kJ mol–1 If the first reaction is endothermic and the value of x is small, the enthalpy of neutralisation of the weak acid is only a little less exothermic than that of a strong acid. This is true regardless of the extent of ionisation of the weak acid. What controls the difference between the two enthalpies of neutralisation is the value of x.

c Hydrocyanic acid, HCN, is a weak acid and the CN– ions are only weakly hydrated in solution. This means that the value of x is much more endothermic than the value for ethanoic acid and so the enthalpy of neutralisation is significantly less exothermic than –57 kJ mol–1. e

The value of ΔHneut for a weak acid depends mostly on how endothermic its enthalpy of ionisation is, rather than on the extent of ionisation.

27 a Chloric(V) acid, HClO3: O O

Cl

H

O

b Hydrocyanic acid, HCN: C

H

N

c Sulfurous acid, H2SO3: O O

S

H

O

H

28 The acid is slightly ionised: HOCl  H+ + ClO– As the degree of ionisation is very low, [ClO–] is also low. When OH– ions are added, they react with undissociated HOCl molecules: HOCl + OH– → ClO– + H2O However, as [ClO–] was initially very small, even small amounts of OH– will cause its value to increase significantly. Since [H+] =

Ka × [HOCl] [ClO–]

the value of [H+] and, hence, the pH will alter significantly. If H+ ions are added, the equilibrium is driven to the left, but as [ClO–] is small initially, its value decreases [HOCl] significantly and so the ratio and hence the value of [H+] will change markedly. [ClO–] e

For a solution to be a buffer, the concentrations of both the weak acid and its conjugate base must be large relative to any amounts of added H+ or OH–.

29

e

Try Googling ‘blood buffer solutions’.

© Philip Allan Updates

Edexcel A2 Chemistry

40

Chapter 5 Acid-base equibria

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 D A higher pH means that the solution is less acidic or more alkaline. The correct answer, therefore, is the solution with the lowest [H+]. The numbers in options A, B and C are all larger than 1.0 × 10−7, so they are all more acidic and have a lower pH than the solution in option D.

2 A A buffer solution must consist of an acid/base conjugate pair, i.e. a weak acid and its salt or a weak base and its salt. If the methanoic acid in option B is in excess, the solution will contain methanoic acid and sodium methanoate, as a result of some of the methanoic acid having been neutralised. So this is a buffer solution. Ammonia is a weak base and the ammonium ions in ammonium sulfate are its conjugate acid, so option C is also a buffer. Carbon dioxide reacts with water to form carbonic acid which is a weak acid and the hydrogencarbonate ion is its conjugate base, so option D is also a buffer solution. The answer to this negative question is option A, as this mixture is not a buffer. The solution will consist of either a strong acid and its salt or a strong base and its salt, depending on whether the hydrochloric acid or the sodium hydroxide is in excess.

3 B This is another negative question. It may be helpful to write ‘true’ or ‘false’ beside each option. Option A is a true statement — it is the definition of an acidic solution. Statement B is false. The pH of a neutral solution is only 7 at 25°C. It is always worth checking the other options. Option C is a true statement. A weaker acid will have a smaller value of Ka, say 10−6 compared with, say, 10−5 for the less weak acid. Thus pKa of the weaker acid is –log Ka = +6; –log Ka of the less weak acid is +5, which is smaller than the pKa of the weaker acid. Statement D is true, as all aqueous solutions contain both ions and, at 25°C, [H+] × [OH−] = 1.0 × 10−14.

4 C The pH of a buffer solution of a specified acid/base conjugate pair depends on the ratio of [acid] to [conjugate base]. This does not change on dilution. Option A is incorrect because the pH of an acid rises (because less acidic) on dilution. Option B is incorrect because diluting an alkali decreases its pH, as the solution becomes less alkaline. On dilution, the pH of a weak acid changes less than that of a strong acid, so option D is incorrect.

5 C The pH of the salt of a weak acid and a strong base will be around 9, and the vertical part of the pH curve will be about ±2 pH units from this value, i.e. from 7 to 11. This means that an indicator must change colour completely within the range 7 to 11, so its pKind value should be between 8 and 10. The only indicator in the list that satisfies this condition is thymol blue, option C.

6 B The conjugate acid of a base is obtained by adding an H+ ion. H3O+ is the conjugate acid of H2O, not of the OH− ion. O2− is the conjugate base of OH−, not its conjugate acid. NaOH contains OH− ions and is the base itself.

7 C Option A is incorrect as the pH does alter (but only slightly) when even small amounts of acid or base are added. Option B is incorrect because the pH will change significantly if large amounts of acid or base are added. At first sight, option D looks true. However, it is wrong, as a buffer could also be a mixture of a weak base and its salt.

8 B All aqueous solutions contain some OH− ions, so option A is incorrect. The calculation of the concentration of OH− ions from pH can be carried out in two ways: pOH = 14 − pH = 14 − 2 = 12 [OH−] = 10−pOH = 10−12 = 1.0 × 10−12 mol dm−3 or [H+] = 10−pH = 10−2 = 0.010 mol dm−3 [OH−] = Kw/[H+] = 1.0 × 10−14/0.010 = 1.0 × 10−12 mol dm−3 Option C is the value of [H+] not [OH−]. Option D is just plain wrong! © Philip Allan Updates

Edexcel A2 Chemistry

41

Chapter 5 Acid-base equibria

9 A In this reaction, the sulfuric acid has protonated the HNO3 molecule. This means that sulfuric acid is acting as an acid and nitric acid as a base. This can only happen if sulfuric acid is a stronger acid than nitric acid. Therefore, option B is wrong. Option C is a true statement about the two substances but the reaction does not show this, so it is not the answer to this question. Option D is incorrect because sulfuric acid is reacting as an acid.

10 D Ammonia is a weak base. Hydroiodic acid, HI, is an even stronger acid than hydrochloric acid, HCl. The pH of the salt of a weak base and a strong acid is less than 7. Option A is the pH of a salt of a weak acid and a strong base. Option B is incorrect as many salts have pH values greater than or less than 7. Option C is incorrect as hydroiodic acid is a strong acid.

11 A The amount of acid = mass/molar mass = 0.20/158.1 = 0.00127 mol. [acid] = moles/volume = 0.00127/0.100 = 0.0127 mol dm−3. As it is a strong acid, the pH = −log [acid]= −log 0.0127 = 1.90. In option B, the mass has been divided by 100. Here, two errors have been made — not converting mass to moles and using the volume in cm3 not dm3. In option C the number of moles has been calculated correctly but this has been used as the concentration. In option D, the correct number of moles has been divided by 100 cm3, rather than by 0.100 dm3.

© Philip Allan Updates

Edexcel A2 Chemistry

42

Unit 4 Rates, equilibria and further organic chemistry

Chapter 6 Isomerism 1 a The four structural isomers are:

Cl

Cl

H

H

C

C

C

H

H

H

H

H

1,1-dichloropropane

H

Cl

Cl

H

C

C

C

H

H

H

H

Cl

H

C

C

C

H

Cl

H

H

2,2-dichloropropane

H

H

1,2-dichloropropane

Cl

H

Cl

C

C

C

H

H

H

H

1,3-dichloropropane

b 1,2-dichloropropane contains an asymmetric carbon atom: four different groups are attached to the middle carbon (a –CH2Cl group, a Cl, an H and a –CH3 group). The two optical isomers are: CH 2 Cl

C H3 C e

CH 2 Cl

Cl

Cl

H

H

C CH 3

Make sure that you draw the two optical isomers as object and mirror image. Each structure should have one bond shown as a wedge and one as a dashed or dotted bond. The other two bonds must not be at 180° to each other.

2 The conditions for geometric isomerism are: I

a double bond (this restricts rotation)

I

two different groups on each of the carbon atoms of the double bond

1-chloropropene fulfils both of these conditions as one of the carbon atoms of the double bond has a –CH3 group and an H atom and the other has a Cl atom and an H atom. 3-chloropropene has two hydrogen atoms on one of the doubly bonded carbon atoms, so geometric isomerism cannot occur. e

A ring structure can also cause geometric isomerism because rotation is impossible without totally twisting the ring. The two chlorine atoms in 1,2-dichlorocyclohexane can be located above the ring (cis) or one above and one below (trans). A double bond consists of a σ-bond, with a head-on overlap of two atomic orbitals, and a π-bond, which is formed by a side-by-side overlap of two p-orbitals. Rotation about the double bond would involve breaking the π-bond. This process is energetically unfavourable at room temperature. Geometric isomers should be drawn with 120° bond angles around the carbon–carbon double bond.

© Philip Allan Updates

Edexcel A2 Chemistry

43

Chapter 6 Isomerism

3 The two geometric isomers are: +

Cl 180°

H3N

+

Cl 90°

H3N

NH3

Cr

Cl

Cr

H3N

H3N

NH3

NH3 NH3

Cl

trans-

cis-

Geometric isomerism can also occur with certain planar, as well as some octahedral, species. In the example above, the two chlorine atoms can be at 180° or at 90° to each other. Similarly, in the planar molecule [Pt(NH3)2Cl2], the two chlorine atoms can be at 180° or 90° to each other.

4 Geraniol fulfils both conditions for geometric isomerism: I

It has a double bond.

I

One of the carbon atoms in the double bond on the right of the molecule has a –C6H11 group and a –CH3 group attached and the other carbon atom has an H atom and a –CH2OH group attached. This means that the CH3 and the CH2OH can be on the same side of the planar double bond or on opposite sides.

e

Geraniol is a natural substance and exists as just one of the two geometric isomers. The other geometric isomer is also a natural compound — the terpene nerol.

5 a The name is E-1-chloro-1-fluoroprop-1-ene. The hydrogen atom has a priority of 1 and the –CH3 carbon has a priority of 6. The priority of fluorine is 9; that of chlorine is 17. Thus the two lowest priority groups are on opposite sides of the double bond, so the compound is an E-isomer. Both the fluorine and the chorine atoms are attached to the first carbon atom of the propene chain. The two halogens are listed alphabetically.

b Z-1-bromo-1-chloro-2-fluoroethene e

Bromine has a higher priority than chloride, and fluorine has a higher priority than hydrogen. Thus it is a Z-isomer. These are good examples of when the cis-/trans- method of naming geometric isomers cannot be used. The priority of an atom is its atomic number.

6

C2H5

H5C2 C

C OH

H3C

HO

H e

7

Make sure that you draw the two isomers as object and mirror image. CH2

H2 C C H3 C

e

CH3 H

CH

*CH

C CH2

CH3

CH2

The carbon atom marked * has four different groups attached — the C(CH3)=CH2 group, an H atom, the bottom part of the ring and the top part, which is different as it has a double bond.

8 a A chiral species is one that has no plane of symmetry. This means that the compound exists in two forms, one of which is a non-superimposable mirror image of the other.

© Philip Allan Updates

Edexcel A2 Chemistry

44

Chapter 6 Isomerism

b The two optical isomers of alanine are: COOH

HOOC

C H3 C e

H2N

NH2

H

H

C CH3

Alanine exists naturally as just one of these optical isomers — the one shown on the right. The other isomer is not a naturally occurring amino acid.

c If a beam of plane-polarised light is shone into a solution of the optically active substance, one isomer will rotate the plane of polarisation in one direction and the other isomer will rotate it in the opposite direction. A racemic (50/50) mixture of the two optical isomers will have no effect on the plane of polarisation. e

The key word is rotate. Do not say that the light is bent, refracted or twisted.

9 a There are two chiral centres in this molecule: COOH

COOH *C *C

HO H

COOH

OH which is often written as H

H

*C

OH

H

*C

OH

COOH

b The optical isomers associated with a chiral centre are usually labelled + (rotates the plane clockwise) and – (rotates it anticlockwise). Since there are two chiral centres, there are four combinations: + +, + –, – + and – –. However, + – is the same as – +, so there are only three isomers, not four.

10 The method is to shine plane-polarised (monochromatic) light through solutions of both optical isomers. One solution will rotate the plane of polarisation clockwise and the other will rotate it anticlockwise. e

Do not state that the isomers bend or refract the plane of polarisation, nor just that they rotate the light.

11

e

A Google search for cisplatin and anticancer drugs provides many useful sites. The history of how an ordinary experiment using platinum electrodes led to the discovery that cells were not dividing under these conditions is a story of how an unexpected result can be applied to a different field of science.

© Philip Allan Updates

Edexcel A2 Chemistry

45

Chapter 6 Isomerism

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 A The two stereoisomers are H3C

C2H5 C

C

C

H

H

H3C

H Z-pent-2-ene

C

H

C2H5 E-pent-2-ene

Pent-1-ene, CH2=CHCH2CH2CH3, and both 2-methylbut-1-ene, CH2=CH(CH3)CH2CH3, and 3-methylbut1-ene, CH2=CHCH(CH3)2, have two H atoms on the terminal carbon and so cannot exhibit geometric isomerism. The other structural isomer, 2-methylbut-2-ene, (CH3)2C=CHCH3, has two –CH3 groups on the left-hand carbon in the double bond and so also cannot exhibit stereoisomerism. There are no isomers with four different groups on the same carbon atom, so optical isomerism is not possible.

2 C The cis–trans method of naming geometric isomers cannot be used here because there are four different groups, two on each carbon in the double bond. The E–Z- system has to be used. Therefore, options A and B are incorrect. The priorities are Br > Cl > F > H, so the higher priority atom on the left-hand carbon is bromine and on the right-hand carbon is fluorine. These are on opposite sides of the double bond, so the prefix E- is correct. e

You could work this out by the atoms of the lowest priorities being on opposite sides, hence E-1,1-bromochloro-2fluoroethene.

3 D The SN1 mechanism involves a planar intermediate, so the NH3 nucleophile can attack from either above or below the plane, producing equal amounts of each optical isomer. This is called a racemic mixture and will not rotate the plane of polarisation of the plane-polarised light. Option C is incorrect because the product, CH3CH(NH2)C2H5 is chiral.

4 C In the product of this reaction, the Cl atom of chlorofluoromethylamine has been substituted by a second NH2 group. This product does not have four different groups on the same carbon atom, so it is not chiral. If the original compound had had an iodine atom rather than an NH2 group, the SN2 mechanism would cause an inversion of the chirality. This does not mean that the plane of plane-polarised light would be rotated anticlockwise. Since the compound is different, you cannot tell which way a particular enantiomer will rotate the light. The only certainty is that it will rotate it in the opposite direction to its enantiomer.

5 B There are two centres that cause stereoisomerism — the asymmetric double bond and the fourth carbon atom, which is chiral. Thus there are four stereoisomers, E+, E–, Z+ and Z–. Response A ignores the chirality. Responses C and D include structural isomers such as 5-hydroxypent-2-ene.

6 C The molecule has two chiral centres and so there will be more than two enantiomers. The two chiral centres each rotate the plane of plane-polarised light in either the + or the − direction. There are therefore four combinations: + +, − −, + − and − +. However, as the molecule is symmetrical, + − is the same as − +, so there are three, not four, optical isomers.

© Philip Allan Updates

Edexcel A2 Chemistry

46

Unit 4 Rates, equilibria and further organic chemistry

Chapter 7 Carbonyl compounds 1 Both compounds contain the C=O carbonyl group. Aldehydes have an H atom bonded to the carbonyl carbon atom, but ketones do not.

2 The ketone isomers are: H3C

C

CH3 H 3C

CH2CH2CH3

CH2

C

CH2

CH3

H3C

O

O

C

CH

O

CH3

3 The preparation of propanal from propan-1-ol is an oxidation reaction. The target product, propanal, is more easily oxidised than the starting reactant, propan-1-ol, so the aldehyde must be boiled off as it is made. The temperature must be high enough to boil off the propanal (i.e. above the boiling point of propanal) but not so high as to boil off any propan-1-ol (i.e. below the boiling point of propan-1-ol). This means that the reaction temperature must be kept between the two boiling points. e

Aldehydes can be oxidised further to carboxylic acids. This subsequent reaction does not affect the preparation of ketones from secondary alcohols, as ketones are not oxidised by acidified potassium dichromate.

4 Ethanal contains a δ– oxygen atom in the –CHO group. This can form hydrogen bonds with the δ+ hydrogen atoms in water. The remainder of the molecule is a CH3 group. Although CH3 is hydrophobic, the energy released by the formation of the hydrogen bond more than compensates for the lack of attraction between the non-polar hydrocarbon group and water molecules. Pentanal has a δ– oxygen atom, but it also has a bulky hydrophobic C4H9 group, and so insufficient energy is released from hydrogen bonding to make pentanal soluble in water. δ– O

H δ+

H3C

C

H

δ H

+

O δ– represents a hydrogen bond H δ+ O δ– H

5 All three substances have similar numbers of electrons in their molecules. Butanone, CH3COCH2CH3, has 40 electrons and butan-1-ol, CH3CH2CH2CH2OH, and pentane, C5H12, each have 42. This means that the intermolecular induced dipole (dispersion) forces are almost the same between the molecules in all three substances. Butan-1-ol has a δ+ hydrogen atom as well as a δ– oxygen and so it can also form hydrogen bonds between its molecules. The other compounds cannot form intermolecular hydrogen bonds as they do not have any δ+ hydrogen atoms. This means that the intermolecular forces are the strongest in butan-1-ol and it has the highest boiling temperature. Butanone is polar and so, unlike in the non-polar pentane, permanent dipole–dipole, as well as induced dipole, forces exist between molecules. This makes the total strength of intermolecular forces greater in butanone than in pentane.

© Philip Allan Updates

Edexcel A2 Chemistry

47

Chapter 7 Carbonyl compounds

H O

δ–

CH2(CH2)2CH3

δ+ H δ– O

CH3(CH2)2CH2

H3C

CH2CH3

δ+ C

Butanone

Hδ

+

Oδ

Butan-1-ol

O

–

δ–

CH2 CH3CH2CH2CH2CH3

(CH2)2

Pentane

CH3 e

Butan-1-ol: hydrogen bonds plus the weaker dispersion forces Butanone: dispersion forces and the very weak permanent dipole–dipole forces Pentane: only the dispersion forces A general rule for the relative strengths of intermolecular forces is: hydrogen bonds > dispersion forces > permanent dipole–dipole forces The exception is molecules such as I2 that have a very large number of electrons. The dispersion forces are so strong that iodine is a solid at room temperature, whereas the hydrogen-bonded water is a liquid.

6 a CH3COCH3 + HCN → CH3C(OH)(CN)CH3 b The nucleophile that attacks the δ+ carbonyl carbon atom is a CN– ion. HCN is a very weak acid (Ka = 4 × 10–10 mol dm–3), so there are too few CN– ions in the solution for the reaction to happen at a noticeable rate. If a little alkali is added, some of the HCN molecules are deprotonated to form CN– ions and the reaction takes place. e

The second step of the reaction is when the –O– group in the intermediate removes a proton from an HCN molecule. This regenerates CN– ions, which act as a catalyst in this reaction.

7

C H3C

then

O–

H 3C

H3 C

C

O H3C

H– O–

H3C

H+

H 3C

OH C

C H3C

H

H

H3C

H

8 The answer is that the product CH3C(CN)(OH)CH3 is not chiral, as there are only three different groups on the central carbon atom.

9 The product is chiral, CH3C(CN)(OH)C2H5, but butanone is planar at the attack site. This means that the CN− nucleophile can attack from either above or below the plane. The result is an equal mixture of the two enantiomers. Therefore, there is no effect on the plane of polarisation of plane-polarised light. e

In this mechanism, a nucleophile attacks a planar starting material whereas in an SN1 mechanism the nucleophile attacks a planar intermediate.

© Philip Allan Updates

Edexcel A2 Chemistry

48

Chapter 7 Carbonyl compounds

10 a

H

H C

H3C

N

N

CH2

O2N

NO2

Observation: a yellow-orange precipitate

b CH3CH2COOH e

If you are asked to give the structural formula, you must draw the –COOH group in full: O CH2

H3C

C O

H

Observation: the orange potassium dichromate(VI) would turn to a green solution of Cr3+ ions e

You must give the colour before and after for observations such as this and the next two reactions. You do not need to identify the coloured product unless specifically asked.

c CH3CH2COO–Na+ e

The formula CH3CH2COONa is acceptable (unless you are asked for the structural formula), but drawing a covalent bond between the oxygen and the sodium is not: O CH3CH2C O

Na

Observation: warming the reactants causes the blue Fehling’s solution to turn to a red precipitate (of copper(I) oxide)

d CH3CH2COONH4 or CH3CH2COO–NH4+ Observation: on warming, a silver mirror would be seen on the side of the test tube

e No reaction Observation: the solution would stay colourless e

Ethanal is the only aldehyde that undergoes the iodoform reaction.

11 a Y is either an aldehyde or a ketone as it forms a yellow precipitate with 2,4-dinitrophenylhydrazine. However, it did not reduce Fehling’s solution, so it is not an aldehyde. Y is a ketone with four carbon atoms and so it is butanone. As X was oxidised to a ketone, it must be a secondary alcohol. It has to be unbranched because the oxidation product is butanone, so X is butan-2-ol. The structural formulae are: X CH3CH2CH(OH)CH3 Y

O H3C

CH2

C CH3

© Philip Allan Updates

Edexcel A2 Chemistry

49

Chapter 7 Carbonyl compounds

e

The formula for X does not need to be written out in full as the question asks for only the structural (not the full structural) formula and the compound does not have any double bonds. Y must be drawn so as to show the double bond between the carbon and the oxygen.

b X contains the CH3CH(OH) group and reacts to form a yellow precipitate of iodoform, CHI3. Y contains a CH3C=O group and also gives a yellow precipitate of iodoform.

c CH3CH2COCH3 + 3I2 + 4NaOH → CHI3 + 3NaI + CH3CH2COONa + 3H2O e

This is a difficult equation to derive and balance. If you remember that only half the iodine ends up as iodoform, this means that there must be 3NaI on the right. The organic product is sodium propanoate, so 3 + 1 = 4 NaOH are needed on the left. If you cannot remember anything about the equation, just write down CHI3 and the sodium salt of the acid with one fewer carbon atoms on the right and you will get 1 mark at least.

d Z is also oxidised to a carbonyl compound. Since this is then oxidised by Fehling’s solution, it must be an aldehyde and Z must be a primary alcohol: CH3CH2CH2CH2OH and (CH3)2CHCH2OH Butan-1-ol

Methyl propan-1-ol

12 The product of the reaction of CH2=CHCH2CHO with: a LiAlH4 followed by hydrolysis is CH2=CHCH2CH2OH b H2 and a platinum catalyst is CH3CH2CH2CH2OH e

Only the polar C=O group is reduced by lithium tetrahydridoaluminate(III). Both groups are reduced by hydrogen in the presence of a suitable catalyst.

13 Preparation of butanone from butan-2-ol: I

Butan-2-ol is added to a solution of potassium dichromate(VI) and sulfuric acid in a round-bottomed flask.

I

A reflux condenser is then fitted and the flask heated for 15 min, using a water bath or an electric heater.

I

The flask is allowed to cool and the apparatus reassembled for distillation.

I

The flask is heated again, using a water bath or electric heater, and the liquid that boils over between 78°C and 82°C is collected.

I

If very pure butanone is required, the distillate can be dried over anhydrous calcium chloride and redistilled, collecting the fraction that boils over between 78°C and 82°C. (a)

Water out

(b)

Thermometer Still head Water out Condenser

Butan-2-ol + potassium dichromate in dilute sulphuric acid

Water in Mixture of butanone and unreacted butan-2-ol Roundbottomed flask Electric heater Heat under reflux

© Philip Allan Updates

Round-bottomed flask Electric heater

Distillation with addition of reactant

Water in Open beaker

Edexcel A2 Chemistry

50

Chapter 7 Carbonyl compounds

14 a A nucleophile is a species that uses its lone pair of electrons to form a covalent bond with a δ+ atom in another molecule. Oxidation occurs when a species loses one or more electrons. It occurs when an atom in a species increases its oxidation number.

b (i) Ethanal and hydrogen cyanide react in a nucleophilic addition reaction. e

The lone pair of electrons in the CN– ion acts as a nucleophile and forms a bond with the δ+ carbon atom in the C=O group.

(ii) Ethanal and Fehling’s solution react in a redox reaction — ethanal is oxidised and the copper(II) complex in Fehling’s solution is reduced. e

Ethanal is oxidised to ethanoate ions and the copper(II) complex is reduced to copper(I) oxide.

(iii) Ethanal and 2,4-dintrophenylhydrazine react in an addition/elimination reaction which can also be called a condensation reaction, as a simple molecule, H2O, is lost from the two reactants. CH3CHO + H2NNHC6H3(NO2)2 → CH3CH=NNHC6H3(NO2)2 + H2O e

The route is: G

The lone pair of electrons on the N of the NH2 group attacks the carbon atom in the C=O group as the 2,4-dinitrophenylhydrazine adds on.

G

Water is then eliminated.

15 The substance is a ketone. The identity of this ketone can be found by recrystallising the orange precipitate, drying it and then measuring its melting temperature. This value is then compared with those in a data source of the melting temperatures of the 2,4-dinitrophenylhydrazine derivatives of ketones.

© Philip Allan Updates

Edexcel A2 Chemistry

51

Chapter 7 Carbonyl compounds

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 B The compound reacts with 2,4-dinitrophenylhydrazine. Therefore, it is a carbonyl compound. This excludes option D. As the compound gives a precipitate of iodoform, it must have either a CH3CO group or a CH3CH(OH) group, so it cannot be option C; it must be either A or B. Since the compound gives a silver mirror with Tollens’ reagent, it must be an aldehyde. This excludes option A, so the unknown must be B. e

One way of working out the answer is to put a  or  beside each formula: So: A B C D

   

   

   

2 C The first step is the attack by a lone pair of electrons belonging to the carbon atom of the CN− ion on the carbon of the C=O group.

3 C To form intermolecular hydrogen bonds in these compounds, there must be a δ+ hydrogen and a δ– oxygen. The hydrogen atom must be bonded to an oxygen atom for it to be sufficiently δ+. All the compounds have a δ– oxygen, but there is no H atom bonded to an oxygen in compound C. The other three compounds all contain –OH groups.

4 B Only propan-1-ol has intermolecular hydrogen bonding so it has the strongest intermolecular forces and therefore the highest boiling temperature. So the correct answer must be either option A or option B. Propanone is the only other compound to have dipole–dipole forces so it has the next highest boiling temperature. It also has more electrons than the remaining two compounds and so has stronger induced dipole (London or dispersion or van der Waals) forces. Propane has more electrons (26) than propene (24) and packs more closely, so it has slightly stronger induced dipole forces and, therefore, a slightly higher boiling temperature.

5 D The condensation reaction with 2,4-dintrophenylhydrazine does not produce a chiral molecule, so options A and C are incorrect. The product of the symmetrical ketone pentan-3-one and HCN is C2H5C(CN)(OH)C2H5, which is not chiral. The product of pentan-2-one is CH3C(CN)(OH)CH2C2H5, which has four different groups around the second carbon atom and so is chiral. As the starting material is planar at the attack site, both enantiomers are produced in equal amounts. This is called a racemic mixture.

6 B Lithium aluminium hydride only reduces polar double and triple bonds. Thus it will reduce the C≡N in option A, the C=O in options C and D, but not the C=C in option B. So B is the correct answer to this negative question.

© Philip Allan Updates

Edexcel A2 Chemistry

52

Unit 4 Rates, equilibria and further organic chemistry

Chapter 8 Carboxylic acids and

their derivatives 1 a (CH3)2C(OH)COOH is called 2-hydroxy-2-methylpropanoic acid. e

This compound has a carbon chain of three, so the stem name is prop-. The –OH group and one of the –CH3 groups are on the second carbon atom in the chain. It is sometimes easier to draw the structure out in order to see the carbon-chain length clearly.

b CH3CHClCOCl is called 2-chloropropanoyl chloride. c C2H5COOCH(CH3)2 is an ester and is called 2-propyl propanoate. 2 For an organic substance to be water-soluble it must either form ions in solution or be able to hydrogen-bond with water molecules. Propanoic acid has a δ+ hydrogen atom on the −OH group and two δ– oxygen atoms and so can hydrogen-bond with water. It is also slightly ionised. Propane is a non-polar molecule and has no δ+ hydrogen atoms. It cannot form hydrogen bonds and is therefore insoluble in water. H δ− O

δ+ H

O H

CH3CH2C δ− O

δ+

O

H

H e

Polarity does not cause solubility in water: ethers, esters and halogenoalkanes are all polar molecules but are insoluble in water.

3 The molar mass of CH3COOH is 60 g mol–1. To get a molar mass of 120 g mol–1, two molecules of the acid must join together to form a dimer. The OH hydrogen on one acid molecule forms a hydrogen bond with the C=O oxygen of another molecule and vice versa. δ− O H3C

δ+ H

O

C

C O

H δ+

CH3

O δ−

4 2-hydroxypropanoic acid has the formula CH3CH(OH)COOH. It has two −OH groups and so will react with two molecules of PCl5, with both −OH groups being replaced by Cl atoms. The product is CH3CHClCOCl, 2-chloropropanoyl chloride. e

There are usually clues in a question in which two functional groups on an organic molecule both react with a single reagent. Either the use of the word ‘excess’ for the amount of reagent or that 2 marks are available for the formula of the product.

5 a Lithium tetrahydridoaluminate(III) (also called lithium aluminium hydride) reduces polar double bonds but not C=C groups. Thus, it reacts with the aldehyde and the acid groups in CHOCH=CHCH(OH)COOH and reduces them to primary alcohol groups. The product is CH2(OH)CH=CHCH(OH)CH2OH.

b Catalytic hydrogen reduces C=C and aldehydes but not acid groups. The product in this reaction is CH2(OH)CH2CH2CH(OH)COOH.

© Philip Allan Updates

Edexcel A2 Chemistry

53

Chapter 8 Carboxylic acids and their derivatives

e

Lithium aluminium hydride reduces C=O in aldehydes, ketones and carboxylic acids, esters and C≡N groups. Hydrogen with a nickel or platinum catalyst reduces C=C, C=O in aldehydes and C≡N, but not carboxylic acids or their derivatives. Ketones are reduced very slowly by hydrogen.

6 a CH3CH2COOH + NaOH → CH3CH2COONa + H2O b 2CH3CH2COOH + Mg → (CH3CH2COO)2Mg + H2 c CH3CH2COOH + CH3OH  CH3CH2COOCH3 + H2O e

The reactions are: a acid + base → salt + water b acid + metal → salt + hydrogen c acid + alcohol  ester + water

Esterification reactions must be carried out using a small amount of concentrated sulfuric acid catalyst and the mixture has to be heated under reflux. e

Heating under reflux is necessary if: G

the reaction is slow and, therefore, needs heating for several minutes

G

the reactants are volatile and would boil off when heated

7 Aspirin is an ester. The –OH on the benzene ring in the starting compound reacts with the acid chloride. The formula of aspirin is: O C OH O O

C CH3

e

It is a common error to think that acid chlorines react with a COOH group.

8 A carboxylic acid can be prepared from an ester by hydrolysis. If an acid catalyst is used, the reaction is slow and reversible and so the yield is small. The best method is to hydrolyse the ester by heating it with aqueous alkali and then acidifying the solution to produce the carboxylic acid from its salt. The equations are: C6H5COOC2H5 + NaOH → C6H5COONa + C2H5OH C6H5COONa + HCl → C6H5COOH + NaCl The method is: I

Place the ester in a round-bottomed flask and add aqueous sodium hydroxide.

I

Heat the mixture under reflux for 30 minutes.

I

Allow to cool and add excess dilute hydrochloric (or sulfuric) acid.

I

Filter off the precipitate of benzoic acid.

I

Recrystallise the benzoic acid by dissolving the solid in the minimum of hot water.

I

Use a warm funnel to filter off any insoluble impurities; allow the solution to cool in a flask.

I

When the solution is cool, filter the benzoic acid by vacuum filtration. Finally, wash the solid with a little cold water and allow it to dry.

© Philip Allan Updates

Edexcel A2 Chemistry

54

Chapter 8 Carboxylic acids and their derivatives

e

The answer is in two distinct parts. First, you must describe how the benzoic acid is prepared from the ester. Second, you must describe how it is purified. You might be told about its solubility in water, which allows the purification to be done by recrystallisation from water. This works because benzoic acid is soluble in hot water and insoluble in cold water.

9 Add some solid phosphorus pentachloride to dry propanoic acid: C2H5COOH + PCl5 → C2H5COCl + POCl3 + HCl Any unreacted propanoyl chloride can be removed from the reaction mixture by distillation.

10 a C2H5COCl + 2NH3 → C2H5CONH2 + NH4Cl e

The organic product is an amide, propanamide. A common error is to give hydrogen chloride as the second product. However, this reacts with ammonia, so 2NH3 are needed on the left-hand side of the equation.

b C2H5COCl + H2O → C2H5COOH + HCl e

This hydrolysis reaction demonstrates why acid chlorides should be kept dry. When a bottle containing an acid chloride is opened in air, steamy fumes are observed as the water vapour in the air reacts with the acid chloride, producing gaseous hydrogen chloride.

c 2C2H5COCl + NH2CH2CH2OH → C2H5CONHCH2CH2OOCC2H5 + 2HCl e

The acid chloride reacts with the amine group to form a secondary amide and with the alcohol group to form an ester.

11 There are two types of transesterification reaction: I

the reaction of an ester with an acid to form a different ester and a different acid, for example: CH3COOC2H5 + C3H7COOH → C3H7COOC2H5 + CH3COOH ester of acid I

I

acid II

ester of acid II

acid I

the reaction of an ester with an alcohol to form a different ester and a different alcohol, for example: C3H7COOC2H5 + CH3OH → C3H7COOCH3 + C2H5OH ester of alcohol I

alcohol II

ester of alcohol II

alcohol I

In both types of reaction, a catalyst is needed. This is usually an acid, such as dilute sulfuric acid.

12 Both animal fats and vegetable oils are esters of propan-1,2,3-triol (glycerol). However, the acid groups in these esters differ. In animal fats, most of the fatty acids are saturated, for example stearic acid, C17H35COOH. In vegetable oils, some of the fatty acids are saturated, some are monounsaturated and some are polyunsaturated. Fish oils contain many polyunsaturated acid groups, such as omega-3 and omega-6 acids.

13 The reaction of an acid chloride with an alcohol is a non-reversible reaction with a high atom economy, as only one H atom and one Cl atom present in the starting materials are not found in the product. The reaction of a carboxylic acid with an alcohol is a slow reversible reaction that requires a concentrated sulfuric acid catalyst. The yield is therefore much lower than with the method using the acid chloride. The equations are: CH3COCl + C2H5OH → CH3COOC2H5 + HCl CH3COOH + C2H5OH  CH3COOC2H5 + H2O

14 The –OH group on one end of the molecule can form an ester with the –COOH group of another molecule and so on, forming a long-chain polyester: CH3 O

CH3

C

C

H

O

O

C

C

H

O

one repeat unit

© Philip Allan Updates

Edexcel A2 Chemistry

55

Chapter 8 Carboxylic acids and their derivatives

15 Vegetable oil is a mixture of esters in which many of the fatty acid groups are polyunsaturated. If some of these are replaced by saturated acids, such as stearic acid, the new ester will have a higher melting temperature. Such a transesterification reaction does not alter the stereochemistry of the remaining unsaturated acids. This means that the geometry around the C=C groups remains cis-. The partial hydrogenation reaction saturates some of these C=C groups, and the catalyst causes some of them to change their geometry to form trans- isomers. Trans-fats are more harmful than cis-fats because they increase the level of cholesterol in the blood.

16 This requires a transesterification reaction. The ester in the vegetable oil is reacted with a simple alcohol such as methanol or ethanol. The products are three smaller esters and propan-1,2,3-triol. The smaller esters are easier to volatilise and so are more suitable as a fuel. A typical equation is: CH 2OOCR CHOOCR′ + 3CH3OH

RCOOCH3 + R′COOCH3 + R′′COOCH3 + CH2OHCH(OH)CH2OH

CH2COOR′′

17 Both Jatropha trees and soya beans yield an oil that is suitable for conversion to biodiesel. The Jatropha tree will grow on marginal land — land that is not suitable for growing food crops. It will also grow on the edges of roads and railway lines, neither of which are used to grow food crops. To obtain a high yield, soya beans need good fertile soil. Using soya as a source of biofuels will reduce the food production of a country. If this happens on a large scale, the price of food will rise significantly and may result in starvation in poor countries.

© Philip Allan Updates

Edexcel A2 Chemistry

56

Chapter 8 Carboxylic acids and their derivatives

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 B The simplest way to answer this question is to realise that the substance is an ester of ethanol, so the answer must be either A or B. The acid residue is: CH3 H 3C

CH

O C O

The methyl, –CH3, group is on the second carbon atom counting from the carbon of the –COO group. Ethyl butanoate is CH3CH2CH2COOC2H5 so option A is incorrect. Options C and D are based on an incorrect alcohol and acid. 1-propyl ethanoate is the ester of ethanoic acid and propan-1-ol and has the formula CH3COOCH2CH2CH3, so option C is incorrect. 2-propyl ethanoate is the ester of ethanoic acid and propan2-ol, CH3COOCH(CH3)2, so option D is also incorrect.

2 C For these oxygen-containing compounds, there must be an –OH group for intermolecular hydrogen bonding. Alcohols, acids and water all contain an –OH group which results in a δ+ H atom and a δ− O atom. Ester C has two δ− oxygen atoms but no δ+ hydrogen, so it cannot hydrogen bond with other ester molecules. e

For negative questions such as 2 and 3, it is worth putting a tick or cross beside each option. You should get three ticks and one cross. If you get two ticks and two crosses then you should look again carefully at the options you marked with a cross.

3

B Under suitable conditions, an alcohol, such as methanol, will react with ethanoic acid to form an ester. Therefore, option A gets a tick. A reactive metal, such as zinc, will react with an acid to form a salt and hydrogen. So, option C gets a tick. Phosphorus pentachloride reacts with compounds containing an –OH group, so option D gets a tick. As options A, C and D all have ticks, none is the answer to this negative question. Hydrogen, in the presence of a platinum catalyst, will reduce non-polar C=C but not the C=O group in acids (or esters) and so will not react with ethanoic acid. Option B gets a cross and is the correct response to this negative question.

4 D The chlorine atom joins with one of the H atoms of the –NH2 group, forming an amide. Option A is totally wrong. The structure in B shows confusion with the reaction of an acid chloride with alcohols to form an ester. Option C is wrong because only one hydrogen atom is substituted in the reaction between a primary amine and an acid chloride. This shows confusion with the reaction of a primary amine with a halogenoalkane, where both hydrogen atoms can be replaced.

5 C The four esters are:

O C2H5C

O CH3C

OCH3

O HC

OC2H5

O HC

OCH2CH2CH3

OCH(CH3)2

There are three possible acids — propanoic, ethanoic and methanoic. The ester of methanoic acid can be formed from two isomeric alcohols, propan-1-ol and propan-2-ol. e

It is essential to sketch out the structures when answering a question about structural isomers.

6 A For a condensation polymer to form, each monomer must have two functional groups. 1,6-hexandioic acid has two –COOH groups and 1,3-propanediol has two –OH groups, so they will form a condensation polymer. 3-hydroxybutanoic acid has a –COOH group at one end of the molecule and an –OH group at the other, so it too can form a condensation polymer. But-2-ene, like all alkenes, can form an addition polymer. Therefore, options B, C and D can all form polymers. Methanol in option A has only one –OH group and so cannot form a condensation polymer. therefore, option A is the correct response to this negative question.

© Philip Allan Updates

Edexcel A2 Chemistry

57

Unit 4 Rates, equilibria and further organic chemistry

Chapter 9 Spectroscopy and

chromatography 1 a Vibration of the O–H bonds in the water molecules will increase. b The molecules will rotate faster, gaining rotational energy. On colliding with other water molecules, this energy will be converted to kinetic energy and the water will heat up.

2 The poisonous species is soluble Ba2+ ions. Both barium carbonate and barium sulfate are insoluble in water. The stomach contains hydrochloric acid, which reacts with barium carbonate to form a solution of barium chloride, containing Ba2+(aq) ions: BaCO3(s) + 2HCl(aq) → BaCl2(aq) + H2O(l) + CO2(g) Barium sulfate does not react with hydrochloric acid, so no soluble barium compound is formed: BaSO4(s) + HCl(aq) → no reaction

3 The first step (initiation) is the homolytic fission of a chlorine molecule into two chlorine radicals: hν Cl2 → 2Cl• The propagation steps are: (1)

the removal of a hydrogen atom by a chlorine radical: Cl• + CH4 → HCl + CH3•

(2)

the removal of a chlorine atom from a Cl2 molecule by the CH3• radical to form a second chlorine radical: CH3• + Cl2 → CH3Cl + Cl• The second chlorine radical then continues the chain reaction.

There are some chain-termination processes that could stop the chain reaction: Cl• + Cl• → Cl2 CH3• + Cl• → CH3Cl CH3• + CH3• → C2H6 e

The presence of traces of ethane is strong evidence for this mechanism.

4 Polar molecules absorb energy in the infrared region of the electromagnetic spectrum. It is because oxygen is more electronegative than hydrogen and because the water molecule is not linear that a water molecule is polar. The light coming from the sun contains higher frequencies than the infrared radiated by the Earth back towards space. The water vapour absorbs some of this infrared radiation, trapping energy in the atmosphere and so keeping the Earth warm. e

Polar water molecules also absorb in the microwave region, which is why aqueous liquids heat up in a microwave oven.

5 Spectrum A: the peak at wavenumber 1715 cm–1 is caused by a C=O bond and the peak at 1421 cm–1 by a C–C bond. Spectrum B: the peak at wavenumber 1716 cm–1 is caused by a C=O bond and the peak at 1416 cm–1 by a C–C bond, but the peak at 2986 cm–1 is caused by an O–H bond. Both propanal and propanoic acid have C=O and C–C bonds, but only propanoic acid has an O–H bond. Thus, spectrum B is that of propanoic acid and spectrum A that of propanal. e

The broadness of the peak at 3000 cm–1 is caused by the intermolecular hydrogen bonding of the –OH group on one molecule with the C=O group on another.

© Philip Allan Updates

Edexcel A2 Chemistry

58

Chapter 9 Spectroscopy and chromatography

6

The equation does not balance for charge — the charge on the left-hand side is 1+; the charge on the righthand side is 2+. This is impossible. The product of a molecular ion splitting is one positive fragment and one or more neutral fragments.

7 a The molar mass of propanal , CH3CH2CHO , is 58 g mol−1. The peak at m/e = 58 is caused by the molecular ion (CH3CH2CHO)+, which is formed by the process: CH3CH2CHO + e− → (CH3CH2CHO)+ + 2e− The peak at m/e = 29 is 29 units less than the molecular-ion peak. It is caused by either the loss of CHO resulting in the CH3CH2+ ion or the loss of CH3CH2 resulting in the CHO+ ion. CH3CH2CHO+ → CH3CH2+ + •CHO CH3CH2CHO+ → CHO+ + CH3CH2• The peak at m/e = 15 is due to the CH3+ ion, also formed by fragmentation of the molecular ion: CH3CH2CHO+ → CH3+ + •CH2CHO e

A common error is to omit the positive charge on the ions whose m/e values are detected.

b The peak at m/e = 59 is due to the presence of one atom of 13C in a few of the molecules of propanal. 8 The mass spectrum of molecular bromine is represented below:

155

160

165 m/e

There are two ways of obtaining a mass of 160: 79Br–81Br and 81Br–79Br. Therefore, the peak at m/e = 160 is twice the height of that at 158, caused by (79Br–79Br)+, and that at 162, caused by (81Br–81Br)+. e

The 1:2:1 ratio is similar to that obtained when tossing two coins at once. The chances of two heads, a head and a tail, and two tails are in the ratio 1:2:1. There are also two peaks at m/e = 79 and 81, caused by 79Br+ and 81Br+, formed from the decomposition of a molecular ion.

9 If a peak in an NMR spectrum is split into six, there must be five neighbouring hydrogen atoms. This can only be the case if there is a CH3 group on one side and a CH2 on the other. Butan-1-ol is CH3CH2CH2CH2OH. Therefore, the spectrum is consistent with that of butan-1-ol. e

Neither butan-2-ol, CH3CH(OH)CH2CH3, nor 2-methylpropan-1-ol, (CH3)2CHCH2OH, would show this splitting pattern.

10 The hydrogen nuclei in butanone are in three different chemical environments: I

CH3 hydrogen nuclei in the CH3CH2 group

I

the hydrogen nuclei in the CH3C=O group

I

the hydrogen nuclei in the CH2 group

Thus, there are three peaks in the NMR spectrum. The ratio of the peaks is 3:2:3 because there are three hydrogen nuclei in the CH3 of the CH3CH2 group, two in the CH2 group and three in the CH3C=O group. e

Examiners will accept statements such as ‘The hydrogen atoms are in different chemical environments’.

© Philip Allan Updates

Edexcel A2 Chemistry

59

Chapter 9 Spectroscopy and chromatography

11 Propan-1-ol has the formula CH3CH2CH2OH and has hydrogen nuclei in four different chemical environments. Propan-2-ol has the formula CH3CH(OH)CH3 and has hydrogen nuclei in three different chemical environments. This means that propan-1-ol has four peaks in its NMR spectrum, whereas propan-2-ol has only three. e

The hydrogen nuclei in the two CH3 groups in propan-2-ol are in an identical environment because both are next to the same CH(OH) group.

12 The peak at δ ≈ 1 is due to the hydrogen nuclei in the CH3 group. It is split into three because of the two neighbouring hydrogen nuclei in the CH2 group. The peak at δ ≈ 3.6 is due to the hydrogen nuclei in the CH2 group. It is split into four by the neighbouring CH3 group (the rule of n + 1). The peak due to the hydrogen in the OH group is at δ ≈ 2.6. e

The position of the peak due to the hydrogen in OH varies considerably according to the conditions under which the spectrum is measured.

13 The three hydrogen nuclei in a CH3 group can have their spins aligned in four different ways — ↑↑↑, ↑↑↓, ↑↓↓ or ↓↓↓. This means that the neighbouring hydrogen atoms experience four slightly different magnetic fields. This causes the single peak of the neighbouring hydrogen atom observed in the low-resolution spectrum to split into four when examined with a high-resolution spectrometer. e

The split-peak heights will differ as there is only one way of getting ↑↑↑ or ↓↓↓, but three ways of getting two ↑ and one ↓ (↑↑↓, or ↑↓↑ or ↓↑↑) and three ways of getting two ↓ and one ↑ (↓↓↑or ↓↑↓ or ↑↓↓). The same statistics apply when tossing three coins together. The chances of three heads, two heads and a tail, two tails and a head, and three heads are in the ratio 1:3:3:1.

14 Deuterium, 2H, is an isotope of 1H, but because it has 1 proton and 1 neutron it is not NMR active. Thus the spectrum of C2H5OD does not have a peak at δ = 2.6 whereas that of ethanol does; this is due to the OH hydrogen nucleus.

15

e

Google ‘MRI’. Possible sites include netdoctor.co.uk, nhsdirect.nhs.uk, bupa.co.uk and Wikipedia. If the candidate has access to Encyclopaedia Brittannica, this would be a good source as well.

© Philip Allan Updates

Edexcel A2 Chemistry

60

Chapter 9 Spectroscopy and chromatography

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 C When microwaves are absorbed, their energy is transformed into the rotational energy in the molecule. It is IR radiation that causes an increase in bond stretching and bending and UV light that causes bonds to break, so options A, B and D are incorrect.

2 B Pentan-3-one is a symmetrical molecule, so the hydrogen atoms in the two CH2 groups are in identical environments, as are the hydrogen atoms in the two CH3 groups.

3 C The rule is that the peak is split into 1 more than the number of neighbouring hydrogen atoms. There are four neighbouring hydrogen atoms in CH3CH2CHO, so the CH2 peak will be split into five. e

A neighbouring hydrogen on an OH does not cause any splitting.

4 D The primary alcohol will be oxidised to either an aldehyde or to a carboxylic acid, both of which have a C=O group. Water contains O–H groups, so the aqueous solvent would interfere with any attempt to follow the change from alcohol to aldehyde using the O–H peak. Therefore, option A is incorrect. Both C–H and C–C are present in the alkyl group in both the reactant and the product. Therefore, neither of these could be used to follow the progress of the reaction, so options B and C are incorrect.

5 B Both CHO and C2H5 have an m/e value of 29, but only positive ions show up in a mass spectrometer, so option D is incorrect. Neutral radicals are not detected, so options A and C are also incorrect.

6 A The substances to be separated must be volatile so that they become gaseous when injected into the carrier gas. Amino acids and salts are not volatile and so GLC cannot be used to separate mixtures containing them so options C and D are incorrect. HPLC, not GLC, can be used to separate soluble solids, so option B is incorrect.

7 D The time taken for a particular component to pass through the HPLC column depends on the pressure, the length of the column and the nature of the eluent, so options A, B and C are not the answers to this negative question. The number of substances in the mixture being separated does not affect the retention time of any single constituent, so option D is the correct answer to this negative question.

© Philip Allan Updates

Edexcel A2 Chemistry

61

Unit 4 Rates, equilibria and further organic chemistry

Practice Unit Test 4 Section A 1 D All aqueous solutions, however acidic, contain some OH− ions so option A is incorrect. The concentration can be found in two ways: I

pOH = 14 − pH = 14 − 3 = 11 [OH−] = 10−pOH = 1 × 10−11 mol dm−3

or I

[H+] = 10−pH = 1 × 10−3 [OH−] = Kw/[H+] = 1 × 10−14/1 × 10−3 = 1 × 10−11 mol dm−3

Option B is the value of [H+] not [OH−]; option C is the value of [OH−] in a neutral solution at 25°C.

2

A To find the conjugate base of a species, a hydrogen ion has to be removed. Removing an H+ from OH− leaves O2−. Water, H2O, is the conjugate acid of OH−, not its conjugate base, so option B is wrong. H3O+ is the conjugate acid of H2O, so option C is incorrect as is option D — NaOH is simply a source of OH− ions.

3 C In an equilibrium reaction, the equilibrium constant only equals the concentration term when the system is at equilibrium. The only factor that alters the value of K is a change in temperature. As the reaction is endothermic, an increase in temperature will increase the value of K and so the system will react by producing more product. Thus option C is the correct answer. Carbon is a solid, so does not appear in the concentration term. So adding more carbon will not alter the equilibrium yield, so option A is wrong. An increase in pressure will drive the equilibrium to the side with fewer gas moles, which is to the left. Thus the equilibrium yield of hydrogen will get smaller, so option B is incorrect. A catalyst has no effect on the equilibrium yield, so option D is also wrong.

4 B ΔSsystem is made up of ΔS for the solid becoming more random and ΔS for the water becoming less random. The total is negative for anhydrous solids containing 2+ ions dissolving, so options C and D are incorrect. 1+ ions order the solvent water molecules slightly, but the bigger K+ ions do so less than Na+ ions, so ΔSsystem for KCl dissolving is more positive than ΔSsystem for NaCl dissolving.

5 D ΔHr° = ΣΔHf°products − ΣΔHf°reactants = 4 × −242 − ΔHf°(Fe3O4) = +150 ΔHf°(Fe3O4) = −968 − 150 = −1118 kJ mol−1 In option A the sign is wrong and incorrect stoichiometry has been used. In option B, the sign is correct but the wrong stoichiometry has been used. In option C the correct stoichiometry has been used but a mistake has been made in the algebra.

6 B ΔS°system = ΣS°products − ΣS°reactants = +151 3 × 27 + 4 × 190 − (S°(Fe3O4) + 4 × 131) = +151 S°(Fe3O4) = 81 + 760 − 524 – 151 = +166 J K−1 mol−1 In option A, the answer is absurd because the standard entropy of all substances is positive. In option C, the correct stoichiometry has not been used and it has been assumed that ΔSsystem = ΣSreactants − ΣSproducts. In option D, the correct stoichiometry has been used but an algebraic error has been made.

7 D Halving the concentration of [IO3−] and [I−] will cause the rate to decrease by a factor of 2 × 2 = 4 (as the reaction is first order in both). However, doubling [H+] will cause the rate to increase by a factor of 22 = 4 because the reaction is second order with respect to H+ ions. Thus the rate will be unaltered.

© Philip Allan Updates

Edexcel A2 Chemistry

62

Practice Unit Test 4

8 B The flask at the beginning contains a solution of a weak acid, which will have a pH ≈ 3. pH ≈1 is that of a strong acid, pH ≈ 11 is that of a weak base and pH ≈ 13 is that of a strong base.

9 C After 10 cm3 of the sodium hydroxide solution has been added the solution will be an equimolar solution of a weak acid and its salt. This means that the pH of the solution will equal pKa. Since Ka = 1 × 10−5, pKa = −log Ka = +5

10 C The indicator must change colour completely in the vertical range of the titration curve. This means that pKind ± 1 must lie within 9 ± 2, i.e. between 8 and 10. The only indicator that fits this is thymol blue. Methyl orange (A) and bromocresol blue (B) would have already changed colour by pH 6, which is well before the equivalence point. Alizarin yellow (D) does not start to change colour until pH = 11.5, which is well after the equivalence point.

11 C The expression for Kp =

p(CO)p(H2)3 p(CH4)p(H2O)

where the partial pressures are equilibrium values. This expression has units of atm4 on the numerator (top line) and atm2 on the denominator (bottom line), so the units of Kp are atm2.

12 B The total moles are 0.10 + 0.10 + 0.0040 + 0.0030 = 0.207, so the mole fraction of hydrogen is 0.0030/0.207. In options A and C, 3 × 0.0030 were used as the moles of hydrogen, the stoichiometry having been used mistakenly when working out the total moles. In options C and D, it was assumed that there were 3 × 0.0030 moles of hydrogen at equilibrium because the equation had 3H2.

13 A The calculation is mole fraction × 2 → partial pressure, then the values are substituted into the expression for Kp. Mole fraction

CH4 0.10/0.207 = 0.483

H2O 0.10/0.207 = 0.483

CO 0.0040/0.207 = 0.0193

3H2 0.0030/0.207 = 0.0145

Partial pressure/atm

0.483 × 2 = 0.966

0.483 × 2 = 0.966

0.0193 × 2 = 0.0386

0.0145 × 2 = 0.0290

Kp =

p(CO)p(H2)3 0.0386 × 0.02903 = = 1.0 × 10−6 p(CH4)p(H2O) 0.966 × 0.966

In option B, the error is using mole fractions, rather than partial pressures, in the expression for Kp. In option C the partial pressure of hydrogen has not been cubed; in option D the expression used was upside-down.

14 D The reaction is the iodoform reaction and the molecule must have either a CH3CO group, as in A, or a CH3CH(OH) group, as in B and C. Compound D does not have either of these groups and therefore will not produce iodoform. So, option D is the correct answer to this negative question.

15 A In these oxygen-containing substances, for intermolecular hydrogen bonding to take place, the molecule must have a δ+ hydrogen and a δ– oxygen. All four molecules have a δ– oxygen but propanone does not have a hydrogen atom joined to an oxygen atom. This means that it does not have a δ+ hydrogen atom and therefore will not form intermolecular hydrogen bonds. e

For negative questions such as 15 and 16, it is a good idea to put a cross or a tick beside each response.

16 B In propanone, the oxygen in the C=O group is very δ– and so forms hydrogen bonds with the δ+ hydrogen in water, so option A should have a tick. Carbon is more electronegative than hydrogen and so the oxygen in ethoxyethane is not sufficiently δ– to enable it to form hydrogen bonds with water molecules, so option B should have a cross. The acid (C) and the alcohol (D) both have an –OH group, which has a δ+ hydrogen that forms hydrogen bonds with the δ– oxygen in water; they also have a δ– oxygen that forms hydrogen bonds with the δ+ hydrogen in water. Therefore, options C and D should both have ticks. The only substance that does not form hydrogen bonds with water is ethoxyethane, which is, therefore, the correct answer to this negative question.

© Philip Allan Updates

Edexcel A2 Chemistry

63

Practice Unit Test 4

17 A Although the intermediate and hence the final product formed are chiral, the adduct is the racemic mixture. This is because the reaction site in ethanal is planar and so the nucleophile can attack from above or below the plane. Z-isomers refer to C=C compounds, so C is incorrect. The solvent does not alter this mechanism, so D is also incorrect.

18 C There are two stereoisomeric sites in this molecule. There are four groups attached to the second carbon atom, –CH3, –H, –OH and –CH=CHCH3, so this carbon atom is a chiral centre. There are two geometric isomers around C=C, producing a total of four stereoisomers: Z+, Z−, E+ and E−.

19 B The mass peak at m/e = 57 is due to the loss of 15 from the mass of the molecular ion, which is 72. This shows the presence of a CH3 group, so the correct answer cannot be option D. The IR peak at 1720 cm−1 indicates a C=O group, so options A and D are incorrect. The substance must be either B or C. Both these compounds have three peaks in their NMR spectra. However, the spectrum for compound B has these peaks in the ratio of 3:3:2, whereas the peaks in the spectrum of compound C are in the ratio 6:1:1. therefore, option B is the correct answer. e

It is sometimes helpful to put a tick or a cross for each piece of information for each compound. Thus: A B C D

   

   

   

20 A In order to form an ester with an alcohol, the isomer must be either an acid chloride or an acid. This rules out option C. Option D is not a carbonyl compound so will not react with 2,4-dinitrophenylhydrazine (and also it does not contain chlorine) so it cannot be the answer. Compound A is a ketone and B is an aldehyde, so both will give a precipitate with 2,4-dintrophenylhydrazine. However, the ketone A will not reduce acidified potassium dichromate(VI), whereas the aldehyde B will. Therefore, the correct answer is option A. e

As before: A B C D

   

   

   

Section B 21 a Kp = e

p(CH3OH) p(CO)p(H2)2



Do not use square brackets – they mean concentration, with units of mol dm−3.

b (i) The term heterogeneous means that the catalyst is in a different phase from the reactants  — in a gaseous reaction the catalyst is usually a solid.

(ii) Since the reaction proceeds via an alternative route with a lower activation energy , a greater number of molecules will have energy greater than or equal to this lower activation energy . This means that

a greater proportion of collisions will be successful  and so the reaction will be faster. e

This part of the question is marked *. This means that your ‘quality of written communication (QWC)’ is being tested, so make sure that you express yourself clearly and accurately. Spelling and grammar will not be judged, as long as the meaning is clear. In this answer, do not say that there will be more successful collisions — the number of collisions will be the same, but will take place over a much longer period without the catalyst. It is the proportion of successful collisions, not the number, which is increased by the use of a catalyst.

© Philip Allan Updates

Edexcel A2 Chemistry

64

Practice Unit Test 4

c (i) ΔSsystem = Sproducts − ΣSreactants = +238 − (198 + 2 × 131) = −222 J K−1 mol−1  (ii) ΔHr = ΔHproduct − ΔHreactant = −201 − (−111) = −90 kJ mol−1  ΔSsurr = −ΔHr/T = −(−90 000/523) = +172.1 J K−1 mol−1   e

Do not forget to convert ΔH from kJ to J as well as converting the temperature from °C to K

(iii) ΔStotal = ΔSsystem + ΔSsurr = −222 + 172.1 = −49.9 J K−1 mol−1  ΔStotal = R ln Kp  ln Kp = ΔStotal/R = −49.9/8.31 = −6.01 Kp = e−6.01 = 0.00246 atm−2  e

Remember that the expression is in logarithms to the base e not to the base 10, and that you must give a unit for the equilibrium constant.

(iv) Because ΔStotal is negative, the reaction is not thermodynamically feasible  at this temperature. e

Note that ‘not thermodynamically feasible’ means that the value of K is less than 1, so the yield of the reaction will be less than 50%.

(v) At 350°C, ΔSsurr = −(−90 000/623) = + 144 J K−1 mol−1 ΔStotal = −222 + 144 = −78 J K−1 mol−1  ln Kp =−78/8.31 = −9.39 Kp = e−9.39 = 8.39 × 10−5 atm−2  e

This is an exothermic reaction, so you should check that your value for Kp in (v) is smaller than the value at the lower temperature in (iii).

d As is shown in part c, the lower the temperature, the higher the yield, However, the rate will be slower .

Thus a compromise temperature  of 250°C is used. There are fewer gas moles on the right, so an increase

 , increasing the equilibrium yield of methanol. Even at 250°C, the value of Kp is quite small, so it is economic  to use a high pressure in order in pressure will drive the position of equilibrium to the right

to ensure a reasonable percentage conversion.

e Initial moles Change Equilibrium moles Mole fraction Partial pressure/atm

Kp =

= e

CO 2H2 1.00 2.00 1.00 – 0.25 = 0.75 2 × 0.75 = 1.50 0.25 2.00 – 1.50 = 0.50 0.25/1.50 = 0.167 0.50/1.50 = 0.333

CH3OH 0 0.75

0.75  0.75/1.50 = 0.50  0.167 × 100 = 16.7 0.333 × 100 = 33.3 0.50 × 100 = 50 

p(CH3OH) p(CO)p(H2) 2 50 atm = 0.0027  atm−2  16.7 atm × (33.3 atm)2

Be careful about the stoichiometry. If 0.75 mol of carbon monoxide reacts, then twice that amount of hydrogen reacts.

© Philip Allan Updates

Edexcel A2 Chemistry

65

Practice Unit Test 4

22 (a)(i)

NO2 H3C

H

✔ C

N

N

NO2

✔

H

(ii)

O H3C

✔

C –

O

(iii)

O CHI3

✔

H

✔

C O

e

–

The solutions in (ii) and (iii) are alkaline, so the carboxylate anion, rather than the acid, will be produced. If this error is made, it would only be penalised once.

(b)(i) [H+] = 10−pH = 1.0 × 10−5 mol dm−3 Ka =

[H+][CH3COO−] [CH3COOH]

[salt] =

=

[H+][salt] [acid]

 

Ka × [acid] 1.74 × 10−5 × 0.500 = = 0.87 mol dm−3 [H+] 1.0 × 10−5



moles of salt = concentration × volume = 0.87 mol dm−3 × 0.100 dm3 = 0.087 mol  mass of salt = molar mass × moles = 82.0 g mol−1 × 0.087 mol = 7.13 g 

(ii) The acid is a weak acid so it is only partially ionised, but the salt is totally ionised. When H+ ions are added they react with the CH3COO− ions CH3COO− + H+ → CH3COOH  Since the amounts of both acid and salt are large relative  to the small amount of H+ ions added, the

ratio of [acid] : [salt] does not alter significantly .

(iii) CH3COO− + H2O  CH3COOH + OH−  c (i) An example would be ethanoic acid reacting with an ester, in the presence of a catalyst, to form the ester of ethanoic acid. For example: CH3COOH + HCOOC2H5  CH3COOC2H5 + HCOOH 

(ii) Vegetable oils are esters of propane-1,2,3-triol and several different unsaturated fatty acids. In order to harden the oil, some of these unsaturated acids must be either converted into saturated acids or replaced by saturated acids . When the oil is reacted with a saturated acid such as stearic acid, C17H35COOH, transesterification takes place and some of the unsaturated acid residues are replaced : CH2OOCR CHOOCR′ + C17H35COOH CH2COOR′′

CH2OOCC17H35 RCOOH + CHOOCR CH2COOR′′

where R, R′ and R′′ are unsaturated fatty acid residues 

© Philip Allan Updates

Edexcel A2 Chemistry

66

Practice Unit Test 4

d

CH3 O

e

CH3

C

C

H

O

O

CH3

C

C

or

H

O

CH3

C

C

H

O

O

C

C

H

O

O

There is 1 mark are for one ester linkage shown in full and 1 mark for the rest of the double unit. Make sure that your double unit contains just the four oxygen atoms that are required for a polyester.

23 (a) The first half-life is 2.3 minutes . The second half-life is 2.3 minutes . As these are the same, the reaction is first order . e

The second of the two consecutive half-lives is the time taken for the concentration to fall from 0.10 mol dm−3 to 0.05 mol dm−3 (i.e. for it to halve for the second time) and is not the time for it to fall from 0.20 mol dm−3 to 0.05 mol dm−3 (i.e. not for its value to drop to one-quarter of the original), which would be 4.6 minutes.

(b)Doubling the concentration of hydroxide ions has no effect on the half-life, so the reaction is zero order with respect to hydroxide ions . The rate equation is: rate = k[2-bromopropane]  e

rate = k[2-bromopropane]1[OH−]0 is equally acceptable.

(c)

CH3

Step 1

C H3C

CH3

✔

C+

Br

H3C

✔

+

H

Br –

H

CH3

Step 2

C+ H3C

CH3

✔ OH

–

C

H

H3C

OH

H e

If a candidate states in part b that the rate equation is rate = k[2-bromopropane]1[OH−]1, then an SN2 mechanism would score full marks.

Section C 24 (a) e

This question requires candidates to use the Edexcel (or other) data booklet and to interpret the data in the question.

Carbon

% 77.9

Divide by Ar 77.9/12.0 = 6.49

Hydrogen

11.7

11.7/1.0 = 11.7

6.49/0.65 = 9.98 ≈ 10 11.7/0.65 = 18

Oxygen

10.4

10.4/16.0 = 0.65

0.65/0.65 = 1

Divide by smallest

}



The empirical formula is C10H18O, which is consistent with the molecular formula C10H18O .

(b)(i) The reaction is addition of Br2 to a C=C group. Since 2 mol Br2 reacts with 1 mol geraniol, there must be two C=C groups present .

(ii) Geraniol has an OH group, which must belong to an alcohol because there is only one oxygen atom in the molecule .

© Philip Allan Updates

Edexcel A2 Chemistry

67

Practice Unit Test 4

(iii) The peak at 1653 cm−1 is due to the alkene C=C stretching. The peak at 2985 cm−1 is due to an alkane C–H stretching. The peak at 3041 cm−1 is due to an alkene C–H stretching. The peak at 3512 cm−1 is due to an alcohol O–H stretching. e

All four correct earns 3 marks, any three correct earns 2 marks, two correct earns 1 mark. The value for the alkane C–H stretching is close enough to that in the data book. The more complex the molecule, the more the peak could differ from the value in the data book.

(c) (i) The molar mass of geraniol is 120.0 + 18.0 + 16.0 = 154.0 g mol−1. Compound X has a molar mass of 152 g mol−1 and so has two hydrogen atoms less than geraniol. This implies that the alcohol group in geraniol is oxidised to an aldehyde or ketone  group in compound X.

(ii) The precipitate confirms the presence of a carbonyl group  in compound X, thus X is either an aldehyde or a ketone.

(iii) A value of δ = 9.5 in the NMR spectrum, indicates that compound X is an aldehyde . (iv) Geraniol is a primary alcohol . (v) Geraniol must be a primary alcohol because it was oxidised to an aldehyde. The oxidation must not be carried out under reflux conditions because this would result in geraniol being oxidised to a carboxylic acid , rather than the aldehyde group in compound X.

(d)(i) CH2=C(CH3)CH=CH2 has hydrogen atoms in four different environments, so there will be four peaks in its NMR spectrum. These peaks will be in the ratio 2:3:1:2, which agrees with the data in the question

.

The other dialkene, CH2=CHCH2CH=CH2 is symmetrical around the central carbon atom.

Hydrogen atoms of both =CH2 groups are in the same environment, as are both =CH hydrogen atoms. The middle CH2 hydrogen atoms are in a third environment and so this dialkene will have only three

NMR peaks , which are in the ratio 4:2:2. e

You must say something about the NMR spectra of both dialkenes. It would be just enough to explain why one compound has four peaks in its spectrum and the other has three, but it would be safer to show the ratio of the peaks for the correct dialkene.

(ii) There is one hydrogen on the neighbouring carbon atom, so the peak will be split into two . (iii) One of the conditions for geometric isomerism is restricted rotation about a bond. Both dialkenes exhibit restricted rotation about their C=C groups . The second condition is that there must be two different groups on each carbon atom in the C=C group. Since both compounds have two hydrogen atoms on both terminal carbon atoms , geometric isomerism is impossible.

(e) The reaction mixture is dissolved in a suitable solvent and injected just above the top of the chromatography column. A suitable liquid eluent is then forced, under high pressure, through the column

.

Compound X separates from geraniol and other impurities because it has a different Rf value. This is because of the competition between being adsorbed onto the stationary phase and being dissolved in the eluent is different for different substances . Thus compound X will take a different time to pass through the column

 than the other substances.

© Philip Allan Updates

Edexcel A2 Chemistry

68

Unit

5

Transition metals, arenes and organic nitrogen chemistry

Unit 5 Transition metals, arenes and organic nitrogen chemistry

Chapter 10 Electrochemistry

and redox equilibria 1 a The oxidation number of oxygen in O2– is –2. b The oxidation number of oxygen in O22– is –1. c The oxidation number of oxygen in O2– is – 1–2 e

The sum of the oxidation numbers in an ion equals the charge on the ion. In O2–, the charge on the ion is 2–, so the oxidation number of oxygen is –2. In O22–, the sum of the oxidation numbers of the two oxygen atoms is –2, so each is –1. In O2–, the oxidation numbers add up to –1, so the oxidation number of each oxygen must be – –12 .

d The oxidation number of oxygen in OF2 is +2. e

The oxidation number of oxygen in compounds is always negative, except when bonded to fluorine, which is more electronegative and has an oxidation number –1 in all its compounds.

2 a The oxidation number of sulfur in H2S is –2. e

Hydrogen is always +1, except when combined with a metal, so 2H is +2 and the sulfur must be –2.

b The oxidation number of sulfur in SO42– is +6. e

The four oxygen atoms have a combined oxidation number of –8; the sulfate ion has a charge of 2–, so the sulfur must balance –6. Its oxidation number is +6.

c The oxidation number of sulfur in S2Cl2 is +1. e

Each chlorine is –1, so the two sulfur atoms are +2 in total, or +1 each.

d The oxidation number of sulfur in H2SO3 is +4. e

The sulfur in sulfuric acid, H2SO4, is +6. H2SO3 contains only three oxygens, so the sulfur has an oxidation number of +4.

3 a Cr2O72–(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l) e

As the Cr2O72– ions are reduced, the electrons must be on the left-hand side of the half-equation (OIL RIG: reduction is gain). The oxidation number of each chromium atom goes from +6 to +3, a decrease of 3. As there are two chromium atoms in the equation, the total change in oxidation number is +6, so there must be six electrons on the left. The number of H+ and H2O are obtained by traditional balancing (seven O atoms in Cr2O72–, so seven H2O required on the right).

b Cl2 + 2e– → 2Cl– e

–12 Cl2 + e– → Cl–

or

State symbols do not have to be used. The reaction in part a is in solution, so it is sensible, but not essential, to add state symbols. For part b, the conditions are not given in the question, so state symbols are not relevant.

c I– → –12 I2 + e– e

or

2I– → I2 + 2e–

As this is an oxidation, the electrons are on the right. Electrons are lost by the iodide ions (OIL RIG), so the equation could be written as 2I– – 2e– → I2.

d Fe2+ → Fe3+ + e– or Fe2+ – e– → Fe3+ e

Make sure that your ionic half-equations balance for charge and that the electrons are on the correct side — the left, if the equation is a reduction; the right if it is an oxidation.

© Philip Allan Updates

Edexcel A2 Chemistry

70

Chapter 10 Electrochemistry and redox equilibria

4 a Equation a in question 3 must be added to 6 × equation c. This puts both reactants on the left. The electrons cancel, as there are six on the right from equation a and six on the left from 6 × equation c. The overall equation is: Cr2O72–(aq) + 14H+(aq) + 6I–(aq) → 2Cr3+(aq) + 7H2O(l) + 3I2(s)

b Either equation b in question 3 must be halved or equation d doubled, giving either: 1 –2

Cl2 + Fe2+ → Cl– + Fe3+

or Cl2 + 2Fe2+ → 2Cl– + 2Fe3+ e

Make sure that both reactants are on the left-hand side and that the equations balance for charge. The equation Cl2 + Fe2+ → 2Cl– + Fe3+ is incorrect as it does not balance for charge. The left-hand side is 2+ but the right is only 1+.

5

V Hydrogen gas at 1 atm

Salt bridge, KNO3(aq)

Platinum electrode MnO4–(aq), H+(aq), Mn2+(aq) all at 1.0 mol dm–3 e

Platinum electrode 1.0 mol dm–3 H+(aq)

The two half-equations, written as reduction half-equations, are: MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

E° = x

H+(aq) + e– → 1–2 H2(g)

E° = 0 V

The second equation must be reversed (and multiplied by 5) and then added to the first equation to give the overall equation, so Ecell = x – 0. The measured potential equals the standard reduction potential of MnO4– in acid solution. Note that all the ions in the manganate(VII) redox system must be present at a concentration of 1 mol dm–3.

6 a The reactants are C2O42− ions and MnO4− ions. These must be on the left-hand side. To get this the first equation has to be reversed. This changes the sign of E° from −0.49 V to +0.49 V. The standard cell potential is the sum of this and the E° for MnO4−. E° = +0.49 + 1.52 = +2.10 V This is a positive number so the reaction is feasible in terms of thermodynamics. e

If the overall equation were required, the first half-equation would have to be reversed and multiplied by 5 and then added to the second half-equation multiplied by 2. In this way the electrons cancel. However, do not multiply the E° values by 5 and 2.

b As the reaction is thermodynamically feasible and yet does not occur, the answer must lie in the kinetics. The activation energy for this reaction is too large for it to take place at 25°C. e

On warming, the reaction starts and then accelerates because it produces Mn2+, which is the catalyst. This is an example of autocatalysis.

7 Electrode potentials are written as reduction potentials with electrons on the left. For the standard hydrogen electrode, the half-equation is: H+(aq) + e– → 1–2 H2(g) © Philip Allan Updates

E° = 0 V

Edexcel A2 Chemistry

71

Chapter 10 Electrochemistry and redox equilibria

If the pressure of hydrogen is increased, the position of equilibrium is driven to the left. This makes the value of E negative. e

If the conditions become non-standard, and the position of equilibrium is driven to the left, the value of E becomes more negative or less positive. Similarly, if the equilibrium position is driven to the right, the value of E becomes more positive or less negative.

8 a The strongest oxidising agent in Table 10.1 (page 187 in the textbook) is fluorine, F2. e

Oxidising agents are reduced by gaining electrons and so appear on the left-hand side of reduction potential halfequations. The one with the most positive value of E° is the strongest oxidising agent.

b The weakest oxidising agent in the table is the lithium ion, Li+, as it has the most negative E° value. c The strongest reducing agent in the table is lithium. e

Reducing agents are oxidised and lose electrons, so they appear on the right-hand side of reduction half-equations. The species in the reduction half-equation with the most negative E° value is the best reducing agent. This is because the reverse reaction, when the species is oxidised, would be the most positive.

d The weakest reducing agent in the table is the fluoride ion, F–, because it occurs on the right-hand side of the most positive reduction half-equation.

9 a (i) The two reduction half-equations (see Table 10.1, page 187 in the textbook) are: Pb4+(aq) + 2e– → Pb2+(aq)

E° = +1.69 V

1 –2

E° = +1.36 V

Cl2 + e– → Cl–

In order to calculate E°cell for the reaction between chloride ions and lead(IV) ions, the second equation has to be reversed and added to the first equation. The E° value of the Cl2/Cl– equation has to be reversed in sign and then added to the E° value for the Pb4+/Pb2+ equation: E°cell = +1.69 + (–1.36) = +0.33 V Since the value of E°cell is positive, the reaction is feasible. e

Note that one of the reactants in the question appears on the right-hand side of one of the reduction half-equations — in this case, chloride ions. This is the equation that needs to be reversed to get the value of E°cell and the overall equation. When a half-equation is reversed, the sign of its E° value must be reversed.

(ii) The two reduction half-equations are: Sn4+(aq) + 2e– → Sn2+(aq)

E° = +0.15 V

Fe3+(aq) + e– → Fe2+(aq)

E° = +0.77 V

The reactants in the question are Sn4+ and Fe2+. Therefore, the second equation has to be reversed and the sign of its E° value changed: E°cell = +0.15 + (–0.77) = –0.62 V As the value of E°cell is negative, the reaction is not feasible.

(iii) The two reduction half-equations are: 1 –2 I2(s)

+ e– → I–(aq)

E° = +0.54 V

Fe3+(aq) + e– → Fe2+(aq)

E° = +0.77 V

The reactants are I– and Fe3+, so the first half-equation has to be reversed: E°cell = +0.77 + (–0.54) = +0.23 V As the value of E°cell is positive, the reaction is feasible.

© Philip Allan Updates

Edexcel A2 Chemistry

72

Chapter 10 Electrochemistry and redox equilibria

e

Do not state that the value of E°cell must be greater than a certain number for a reaction to be feasible. It needs only to be positive (> 0).

b To get the overall equation for part a (i), the second half-equation has to be doubled and reversed and then added to the first: Pb4+(aq) + 2e– → Pb2+(aq) 2Cl−(aq) → Cl2(g) + 2e– + 2Cl– → Pb2+(aq) + Cl2(g)

Pb4+(aq)

To get the overall equation for part a (iii), the first half-equation has to be doubled and reversed and then added to the second, which has to be doubled so that both half-equations contain two electrons: 2I–(aq) → I2(s) + 2e– 2Fe3+(aq) + 2e– → 2Fe2+(aq) 2I–(aq) + 2Fe3+(aq) → I2(s) + 2Fe2+(aq) e

Note that the number of electrons on both sides is equal, so they cancel when the equations are added. Remember that doubling a half-equation has no effect on its E° value.

10 The question is whether either of these oxidising agents will oxidise the chloride ions in the dilute hydrochloric acid. If this is the case, then the acid should not be used. To get the E°cell value, the E° value for –12 Cl2 + e−  Cl− must be reversed from +1.36 V to −1.36 V and then added, in turn, to the E° value of the oxidising agent. For potassium manganate(VII): E°cell = −1.36 + 1.52 = +0.16 V E°cell is positive and so manganate(VII) ions will oxidise chloride ions. Therefore, hydrochloric acid should not be used when manganate(VII) is to oxidise an organic substance. For potassium dichromate(VI): E°cell = −1.36 + 1.33 = −0.03 V E°cell is negative, so dichromate(VI) ions will not oxidise chloride ions under standard conditions. Therefore, when potassium dichromate(VI) is used as an oxidising agent in organic chemistry, it is safe to use dilute hydrochloric acid to acidify the solution.

11 a Disproportionation occurs when an element in a single species is simultaneously oxidised and reduced. e

There must be only one species containing the element on the left-hand side of the equation.

b There must be a starting oxidation number, a higher number (for the oxidised species) and a lower number (for the reduced species). This makes a total of three different oxidation states.

c The two reduction half-equations are: MnO42–(aq) + 2H2O(l) + 2e– → MnO2(s) + 4OH–(aq)

E° = +0.59 V

MnO4–(aq) + e– → MnO42–

E° = +0.56 V

The second equation has to be reversed, doubled and added to the first equation. This gives the following disproportionation equation: 3MnO42–(aq) + 2H2O(l) → 2MnO4–(aq) + MnO2(s) + 4OH–(aq) E°cell = +0.59 + (–0.56) = +0.03 V As E°cell is (just) positive, the disproportionation reaction is thermodynamically feasible.

© Philip Allan Updates

Edexcel A2 Chemistry

73

Chapter 10 Electrochemistry and redox equilibria

e

You have to find two reduction half-equations in Table 10.1 (page 187 of the textbook). One must contain MnO42– and MnO2 and the other MnO42– and MnO4–. Your disproportionation equation must have only one type of manganesecontaining species on the left-hand side, with an oxidation number between the other two values. That species must be MnO42–, in which Mn is +6 (i.e. between +7 and +4).

d If the pH of the solution is increased, then [OH–] increases. This drives the position of equilibrium of the first half-equation to the left, lowering its E value and making the disproportionation reaction unlikely. e

Remember that an increase in pH makes the solution less acidic and more alkaline.

12 In alkaline solution, the reduction half-equations are: 1 –2

O2 + H2O + 2e−  2OH−

Fe2+ + 2e−  Fe

E° = +0.40 V E° = +0.77 V

Reversing the second half-equation and adding the two gives: 1 –2 e

O2 + H2O + Fe  Fe2+ + 2OH−

E°cell = +0.40 + (−0.77) = −0.37 V

This reaction will not work under standard conditions (pH = 14), but will if the solution has a pH of 7.

In acid solution, the reduction half-equations are: O2 + 2H+ + 2e−  H2O

E° = +1.23 V

Fe2+ + 2e−  Fe

E° = +0.77 V

Reversing the second half-equation and adding the two gives: 1 –2 e

O2 + 2H+ + Fe  Fe2+ + H2O

E°cell = +1.23 + (−0.77) = 0.46 V

This reaction will work under standard conditions with pH = 0 as well as when the pH is 7.

13 Oxygen is reduced according to the half-equation: 1 –2

O2(g) + H2O(l) + 2e− → 2OH−(aq)

E° = +0.40 V

If the pH of the solution is increased, [OH–] is increased and the value of E° for the equation is made less positive. This lessens the likelihood of corrosion.

14 a mass of carbon dioxide produced = 1.51 − 0.93 = 0.58 g b amount (moles) of carbon dioxide = mass/molar mass = 0.58 g/44.0 g mol−1 = 0.01318 mol = moles of MCO3 molar mass of MCO3 = mass/moles = 1.51 g/0.01318 mol = 114.6 g mol−1

c relative atomic mass of M = 114.6 − (12.0 + 3 × 16.0) = 54.6 The metal is probably manganese, which has a relative atomic mass of 54.9.

d There were two weighings, each with a possible error of ± 0.01 g, so the error in the mass of MCO3 is ± 0.02. The percentage error is 0.02 × 100/1.51 = 1.32%

e

I

maximum mass = 1.53 g so, maximum value for the molar mass =1.53/0.01318 = 116.1 maximum relative atomic mass of M is 116.1 − 60.0 = 56.1

I

minimum mass = 1.49 g so, minimum value for the molar mass is 1.49/0.01318 = 113.1 minimum relative atomic mass of M = 113.1 – 60 = 53.1

e

Another way of calculating this is to say that the error in the molar mass is 1.32 × 114.6/100 = 1.5, so the maximum molar mass is 114.6 + 1.5 = 116.1 and the minimum is 114.6 – 1.5 = 113.1

© Philip Allan Updates

Edexcel A2 Chemistry

74

Chapter 10 Electrochemistry and redox equilibria

f These values show that M could be manganese (relative atomic mass = 54.9) or iron (relative atomic mass = 55.8)

g The reliability could be improved by using a balance that is more accurate, say to ±0.001 g. e

In the calculation, the error in weighing the crucible after heating has been ignored. This would give an error in the mass of carbon dioxide and hence in the moles of MCO3. As there are three weighings in this experiment, the total error is ±0.03 g or 2% in the molar mass of MCO3. This would give an error of ±2.3 in the relative atomic mass, which also results in uncertainty of the identity of M between manganese and iron.

15 a This question is answered by calculating the feasibility of, first the reduction of the ferrate(VI) ions to Fe3+, then the reduction of Fe3+ ions to Fe2+. In both cases the reducing agent is hydrogen peroxide. FeO42–(aq) + 8H+(aq) + 3e– → Fe3+(aq) + 4H2O(l)

E° = +2.20 V

H2O2(aq) → O2(g) + 2H+(aq) + 2e–

E° = –0.68 V

E°cell for the reduction to Fe3+ = +2.20 + (–0.68) = +1.52 V. This positive value means that the ferrate(VI) ions will be reduced. Fe3+(aq) + e– → Fe2+(aq)

E° = +0.77 V

H2O2(aq) → O2(g) + 2H+(aq) + 2e–

E° = –0.68 V

E°cell for the further reduction to Fe2+ ions = +0.77 + (–0.68) = +0.09 V, which is also positive. Therefore, the ferrate(VI) ions will be reduced from the +6 state via the +3 state to the +2 state.

b The oxidation state of the iron changes from +6 to +2, which is a change of four and requires the addition of four electrons. The hydrogen peroxide must provide the four electrons, so there must be 2 mol of H2O2 in the equation: FeO42–(aq) + 8H+(aq) + 2H2O2(aq) → Fe2+(aq) + 4H2O(l) + 2O2(g) + 4H+(aq) The 4H+(aq) on the right cancel with four of the 8H+(aq) on the left, giving: FeO42–(aq) + 4H+(aq) + 2H2O2(aq) → Fe2+(aq) + 4H2O(l) + 2O2(g) e

The number of electrons in a half-equation equals the change in oxidation number of the element being oxidised or reduced. The electrons must be cancelled out in the final overall redox equation.

16 The equations are: IO3− + 5I− + 6H+ → 3I2 + 3H2O I2 + 2S2O32− → 2I− + S4O62− amount of thiosulfate in titre = 0.104 mol dm−3 × 0.02320 dm3 = 0.002413 mol amount of iodine reacted with thiosulfate ions = 1–2 × 0.002413 = 0.001206 mol = amount iodine produced by reaction of iodate ions amount of iodate ions = 1–3 × amount of iodine produced = 1–3 × 0.001206 = 0.000402 mol concentration of potassium iodate = mol/volume = 0.000402 mol/0.02500 dm3 = 0.0161 mol dm−3

17 Sodium chlorate(I) is an oxidising agent. The method involves adding the sodium chlorate(I) to excess acidified potassium iodide and titrating the iodine produced with sodium thiosulfate solution. The details are: I

Use a pipette to add 25.0 cm3 of the sodium chlorate(I) solution to excess potassium iodide solution and add about 25 cm3 of dilute sulfuric acid.

I

Fill a burette with a standard solution of sodium thiosulfate.

I

Run in the sodium thiosulfate solution until the iodine colour fades to a pale straw colour.

© Philip Allan Updates

Edexcel A2 Chemistry

75

Chapter 10 Electrochemistry and redox equilibria

I

Add 10 drops of starch solution and add the sodium thiosulfate drop by drop until the blue colour disappears.

I

Repeat the titration until two consistent titres have been obtained. The reactions are: ClO– + 2I– + 2H+ → I2 + H2O + Cl– and I2 + 2S2O32– → 2I– + S4O62– If z mol of sodium thiosulfate is required, then –12 z mol of iodine is produced and –12 z mol of sodium chlorate(I) was present in the 25 cm3 of the solution. The concentration of the sodium chlorate(I) is 1–2 z/0.025 mol dm–3.

18 The advantages are: I

When the fuel is oxidised in the fuel cell, only water is produced.

I

Fuel cells are much more efficient than combustion engines.

The disadvantages are: I

It is difficult to store hydrogen in a motor vehicle because it cannot be liquefied except at extremely low temperatures.

I

Cylinders of pressurised hydrogen gas are very heavy.

I

It is extremely difficult and dangerous to refuel a vehicle with hydrogen gas.

I

The hydrogen has to be made either from methane and water or by electrolysis. The former releases carbon dioxide and the latter requires electricity, which has to be generated and which usually produces carbon dioxide.

19 The problems with potassium dichromate breathalysers are: I

They are not very accurate.

I

They cannot be reused.

The advantages of a fuel-cell breathalyser are: I

They are very accurate because the amount of electricity produced is directly proportional to the amount of ethanol in a measured amount of expired air.

I

The equipment can be reused, so only one breathalyser needs to be carried in the police car.

20 Water will also absorb infrared radiation due to O–H stretching and so the water vapour present in expired air would falsify the reading.

21 Iron(II) compounds are reducing agents, so they can be titrated against potassium manganate(VII) in acid solution. I

Weigh out a sample of hydrated iron(II) sulfate and place it in a beaker.

I

Add distilled water and dissolve the solid.

I

Transfer the solution and all washings into a 250 cm3 standard flask.

I

Make up to 250 cm3 with distilled water.

I

Thoroughly mix the contents of the flask.

I

Pipette a 25.0 cm3 portion of the solution into a conical flask and add about 25 cm3 of dilute sulfuric acid.

I

Fill a burette with standard potassium manganate(VII) solution.

I

Run in the potassium manganate(VII) solution from the burette until one drop turns the solution in the conical flask pink.

I

Repeat the titration until two consistent titres have been obtained.

© Philip Allan Updates

Edexcel A2 Chemistry

76

Chapter 10 Electrochemistry and redox equilibria

Outline of calculation: 5Fe2+(aq) + MnO4–(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) volume of titre → moles of KMnO4 → moles of Fe2+ in 25 cm3 → moles of iron(II) sulfate in 250 cm3 of solution molar mass of FeSO4.xH2O = mass weighed out/moles in 250 cm3 of solution Subtract molar mass of FeSO4 and divide the answer by 18 to get the value of x. e

The oxidation number of the manganese changes by 5 (from +7 to +2) and that of the iron by 1, so there must be 5Fe2+ ions per MnO4– ion to make the electrons cancel.

22

e

This is a complicated subject. An internet search engine such as Google will lead to many sources of information.

© Philip Allan Updates

Edexcel A2 Chemistry

77

Chapter 10 Electrochemistry and redox equilibria

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 B The four oxygen atoms have a combined oxidation number of −8. The ion is 2−, so the two carbon atoms have a combined oxidation number of +6 or +3 each.

2 C Any reaction that has an element on one side of the equation and its compound on the other side must be a redox reaction because the element has changed oxidation number. This means that options A and B are redox reactions. The vanadium in option C starts in the +5 state and ends in the +5 state, so this is not a redox reaction and is the correct answer to this negative question. Option D should be checked. The chlorine starts as +1 and ends as −1 in Cl− and +5 in ClO3−, so this is a redox reaction. e

Option D is also disproportionation, which is a special type of redox reaction.

3 A The reactants are Fe3+ and Sn2+. Both must be on the left-hand side of the final equation. The first equation has to be reversed, which means changing the sign of E° from +0.44 V to −0.44 V before adding it to the E° value of the second equation. This gives the standard cell potential as −0.44 + 0.77 V = +0.33 V which is option A. In option C, the sign of the first half-equation has not been reversed. In options B and D the E° value of the second half-equation was doubled because 2e− were needed. This is wrong. In option D the sign of the first half-equation was not reversed, which is another reason why option D is incorrect.

4 D Only a potential difference can be measured. The scale has hydrogen as zero, but this is not the reason why a reference electrode is used, so option A is not correct. The zinc becomes oxidised and is therefore the anode, so option B is incorrect. The zinc is oxidised but not because a reference electrode is used, so option C is wrong.

5 C A positive E°cell means that the reaction is thermodynamically feasible, but it will not happen if the activation energy is too high. The amount by which it is positive does not alter whether it is thermodynamically feasible, so option A is incorrect. If E°cell were negative, then ΔStotal would also be negative, so option B is incorrect. The salt bridge completes the circuit so that the reactants in one compartment can reduce the chemicals in the other by transfer of electrons through the wire. Thus option D is incorrect.

6 C A fuel must be flammable, so it is a necessary condition, rather than a disadvantage. Options A, B and D are all true statements and all are disadvantages. e

Although hydrogen has a high energy output per gram, this is not an advantage in a vehicle because it is overcome by the heaviness of the necessary container.

© Philip Allan Updates

Edexcel A2 Chemistry

78

Unit 5 Transition metals, arenes and organic nitrogen chemistry

Chapter 11 Transition metals

and the d-block elements 1 V: 1s2 2s2 2p6 3s2 3p6 3d3 4s2 e

This can be written 1s2 2s2 2p6 3s2 3p6 4s2 3d3, which is the order of the energy levels.

V3+: 1s2 2s2 2p6 3s2 3p6 3d2 4s0 e

The outer (4s) electrons (rather than the 3d-electrons) are removed first, as this produces a smaller cation, which is then more exothermically hydrated.

2 A half-full d5 configuration is more stable than a d4 configuration. This means that [Kr] 4d5 5s1 is more stable than [Kr] 4d4 5s2. e

Similarly, the half-full p3 is more stable than p2, but there is no difference in stability between s1 and s2. The stability difference is caused by a quantum-mechanical property called exchange energy, which is beyond the scope of an A-level text.

3 The number of protons in the nucleus increases on going from left to right in the periodic table. In the p-block, the number of inner screening electrons does not alter, so the force exerted by the positive nucleus on the outer electrons increases and the general trend in first ionisation energies is upwards. However, in moving across the d-block, the electrons are added to an inner shell. This means that the increase in the number of protons is compensated for by a corresponding rise in the number of inner screening electrons. The result is that the first ionisation energies hardly alter across the d-block. e

The first point to consider is the change in the number of protons in the nucleus and then any difference in the number of inner screening electrons. Other reasons for the differences in ionisation energies are the stability of a particular electron configuration (e.g. 3p3) and the radii of the atoms being compared. Remember that it is an outer electron that is removed in the first ionisation.

4 Both titanium and calcium are metals. The strength of a metallic bond is determined by the number of electrons delocalised in the solid metal and by the ionic radii of the positive ions formed. The two 4s-electrons of calcium are used in metallic bonding, but titanium has not only two 4s-electrons but also two 3d-electrons. The radius of a titanium ion is also smaller than that of a calcium ion. These two factors make the metallic bond much stronger in titanium than in calcium. This gives titanium a higher melting temperature than calcium. e

In order to melt a metal, the metallic lattice has to be broken down and the metal ions separated. The stronger the metallic bond, the harder it is to break down the lattice and, therefore, the melting temperature is higher. Do not say that the metallic bond is broken — it is still there, but the ions are further apart.

5 a The structural formula of the chromate(VI) ion, CrO42–, is: O– Cr

O

O

O–

Its dot-and-cross diagram is:

O* O Cr O O*

© Philip Allan Updates

2–

Edexcel A2 Chemistry

79

Chapter 11 Transition metals and the d-block elements

e

It is a good idea to first draw the structural formula of the ion or molecule showing all the bonds. Chromium has six valence electrons (two 4s and four 3d) and can form six covalent bonds. A single bond is a pair of electrons shared and a double bond is two pairs shared. If an atom gains a single electron, it has a single negative charge. If it loses one electron, it has a single positive charge. In this dot-and-cross diagram, the extra electrons on the two negative oxygen atoms are shown as *.

b The structural formula of the VO2+ ion is: +

V

O

O

Its dot-and-cross diagram is:

O V O e

+

Vanadium has five valence electrons, but one is lost in the formation of the positive ion, so it can form only four covalent bonds (two double bonds).

6 Titanium, calcium and zinc atoms have the following electron configurations: Ti: [Ar] 3d2 4s2 Ca: [Ar] 4s2 Zn: [Ar] 3d10 4s2 Titanium can lose both 4s-electrons to form a Ti2+ ion or both 4s- and one 3d-electron to form a Ti3+ ion. The extra energy required to remove the third electron, which is of almost identical energy to the 4s-electrons, is compensated for by the extra hydration energy produced when the aqua ion is formed. Calcium has no extra electron of similar energy and so does not form a Ca3+ ion. If zinc lost a 3d-electron, it would lose the considerable stability resulting from a full d-shell. This means that not enough energy is regained by hydration to make the formation of a Zn3+ ion energetically feasible. e

The extra energy required to remove another electron must be regained, either from the hydration enthalpy of the cation or from the lattice energy, depending on whether an aqueous ion or an ionic solid is formed. Remember that the values of both the hydration enthalpy and the lattice energy are greater if the ion is more highly charged.

7 a

I

Iron(III) oxide is a basic oxide and so reacts with acids to form a salt and water: Fe2O3 + 6HCl → 2FeCl3 + 3H2O

b

I

Iron(III) oxide is not amphoteric, so there is no reaction with sodium hydroxide.

I

Iron(VI) oxide is acidic. It is not amphoteric, so does not react with acids.

I

As it is acidic, iron(VI) oxide reacts with alkalis to form a salt with the iron in the anion: FeO3 + 2NaOH → Na2FeO4 + H2O

e

Each of these reactions can be written as an ionic equation: Fe2O3 + 6H+ → 2Fe3+ + 3H2O FeO3 + 2OH– → FeO42– + H2O These equations show clearly that the basic oxide forms the simple cation, Fe3+, and the acidic oxide forms an oxoanion, FeO42–, which has one more oxygen atom than the parent oxide. The latter reaction is typical of acidic oxides. For example, carbon dioxide reacts with bases to form the carbonate anion, CO32–; sulfur trioxide, SO3, forms the sulfate ion, SO42–. Remember that the oxides of transition metals in their lower valencies are basic and in their higher valencies are acidic.

8 The hexaaquachromium(III) ion is octahedral. Each bond angle between the central chromium ion and the oxygen atoms of the water molecules is 90°. The oxygen atoms are bonded to the chromium ion by dative bonds and to the hydrogen atoms by covalent bonds.

© Philip Allan Updates

Edexcel A2 Chemistry

80

Chapter 11 Transition metals and the d-block elements

H

3+

H O

3+

H2O H2O 90°

90°

OH2

OH2

H

O

H or

Cr

H2O

H 90°

O

H

Cr H

O

O

H

H2 O H

H O H

e

H

Make sure that the bonds to the chromium ion come from the oxygen atoms in the water. The complex ion has a charge of 3+, which must be included in the diagram that shows the shape. It is also a good idea to name the shape, in case your drawing is not clear.

9 a Chromium(II) chloride, CrCl2, and chromium(II) sulfate, CrSO4, are examples. b Chromium(III) chloride, CrCl3, and chromium(III) sulfate, Cr2(SO4)3, are examples. c Potassium dichromate(VI), K2Cr2O7, and potassium chromate(VI), K2CrO4, are examples. 10 a Copper(I) oxide, Cu2O, and copper(I) chloride, CuCl, are examples. b Copper(II) sulfate, CuSO4, is an obvious example. e

Copper(I) oxide is the red solid formed when an aldehyde reduces Fehling’s solution. Copper(I) chloride is a colourless, insoluble chloride.

11 a There are many examples of chromium(III) complexes, such as the hexaaqua species, [Cr(H2O)6]3+, and the soluble [Cr(OH)6]3–, which is produced when excess sodium hydroxide is added to a precipitate of chromium(III) hydroxide.

b The three copper(II) complexes that you should know are the hydrated copper(II) ion, [Cu(H2O)6]2+, the deep blue ammine [Cu(NH3)4(H2O)2]2+, produced when excess ammonia is added to a precipitate of copper(II) hydroxide, and yellow [CuCl4]2–, which is produced when concentrated hydrochloric acid is added to a copper(II) chloride solution. e

Be careful when working out the charge on a complex ion. Add together the charge on the metal ion (3+ for chromium(III)) and the total charge of the ligands (6– for the chromium–hydroxide complex). If a complex has six ligands (a coordination number of six), its shape is octahedral. Complex ions with four ligands will be either tetrahedral, for example [CuCl4]2–, or square planar, for example [Pt(NH3)2Cl2].

12 a A substance that is green absorbs the complementary colour red. b A substance that is red absorbs the complementary colour green. e

The ligands split the energy levels of the d-orbitals. When white light is shone into the solution, some of the light is absorbed and the energy is used to promote an electron from one of the lower split d-orbitals to a higher one. This is known as a d–d transition.

13 In hydrated ions, the water molecules cause the d-orbitals to split into a lower-energy set of three and an upper set of two levels. The electron configurations of the two ions are: Ti4+: [Ar] 3d0 4s0 Ti3+: [Ar] 3d1 4s0

© Philip Allan Updates

Edexcel A2 Chemistry

81

Chapter 11 Transition metals and the d-block elements

The hydrated Ti4+ ion has no electrons in d-orbitals and so promotion of an electron from a lower to a higher level is impossible. The Ti3+ ion has one d-electron and can absorb visible light. This causes the d-electron to be promoted to one of the higher split 3d energy levels. The removal of some of the frequencies of visible light means that the ion is coloured. e

The colour of complex ions is due to the absorption of light energy as the electron is pushed up into a higher energy level. This is different from the process that gives rise to emission spectra in flame tests, in which heat pushes the electron up to an unstable higher level. Light is then given off as the electron falls back to its ground state.

14 When excess ammonia is added to hydrated copper(II) sulfate solution, ligand exchange takes place and four of the water molecules are replaced by ammonia ligands. The different ligands produce a new colour. Ammonia is a stronger ligand than water and the absorption moves from orange (complementary colour: blue) towards yellow (complementary colour: violet). e

In this question, the metal and its oxidation state stay the same. However, the ligands are different, hence the change in colour. A stronger ligand will move the absorption to a higher energy, which is in the general direction red → violet.

15 The colour of a complex ion depends on three factors:

e

I

the particular transition metal

I

its oxidation state

I

the nature of the ligands

This assumes that the transition metal ion has between one and nine d-electrons, so that d–d transitions can take place. If the ion has zero or ten electrons, its complex will be colourless.

16 There are no ligands in anhydrous copper(II) sulfate, so all five d-orbitals have the same energy and are not split. Copper(I) ions have the electronic configuration [Ar] 3d10 and so all the d-orbitals are full. The ligands split the energy levels of the five d-orbitals, but there is no empty or half-filled orbital into which an electron can be promoted. Thus the ion, even when complexed, cannot absorb radiation in the visible region. e

Ligands are needed to provide the electric field that splits the d-orbitals into three levels of lower energy and two of higher energy. Light energy can then promote an electron from a lower to a higher d-orbital as long as there is an empty or half-filled d-orbital. This causes the ion to be coloured.

17 a The green solution containing nickel(II) ions forms a green precipitate that does not dissolve on addition of excess sodium hydroxide. The equation is: [Ni(H2O)6]2+(aq) + 2OH–(aq) → [Ni(H2O)4(OH)2](s) + 2H2O(l) With ammonia solution, the green solution forms a green precipitate that dissolves in excess ammonia to form a pale blue solution: [Ni(H2O)6]2+(aq) + 2NH3(aq) → [ Ni(H2O)4(OH)2](s) + 2NH4+ [Ni(H2O)4(OH)2](s) + 4NH3(aq) → [Ni(NH3)4(H2O)2]2+(aq) + 2H2O(l) + 2OH–(aq)

b The blue solution containing copper(II) ions forms a blue precipitate that remains on addition of excess sodium hydroxide: [Cu(H2O)6]2+(aq) + 2OH–(aq) → [Cu(H2O)4(OH)2](s) + 2H2O(l) With ammonia, the pale blue solution forms a blue precipitate that dissolves, in a ligand-exchange reaction, to form the dark blue copper–ammonia complex: [Cu(H2O)6]2+(aq) + 2NH3(aq) → [Cu(H2O)4(OH)2](s) + 2NH4+ [Cu(H2O)4(OH)2](s) + 4NH3(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 2H2O(l) + 2OH–(aq)

© Philip Allan Updates

Edexcel A2 Chemistry

82

Chapter 11 Transition metals and the d-block elements

e

The formulae for the hydroxide precipitates can be written as Ni(OH)2 and Cu(OH)2. The reaction of transition metal ions with hydroxide ions and with ammonia to form the precipitate of metal hydroxide is called ‘deprotonation’.

18 The green precipitate obtained when alkali is added to a solution of iron(II) ions is iron(II) hydroxide — a deprotonation reaction. Addition of hydrogen peroxide to this precipitate oxidises it to iron(III) hydroxide, which is red-brown. e

There will also be some fizzing when the hydrogen peroxide is added, because it decomposes to water and oxygen gas. The iron(III) hydroxide acts as a catalyst for this decomposition.

19 Hydrated chromium(III) ions are deprotonated by water and by hydroxide ions. a In aqueous solution, the solvent water molecules partially deprotonate the hydrated ion: [Cr(H2O)6]3+(aq) + H2O(l)  [Cr(H2O)5(OH)]2+(aq) + H3O+(aq) e

The production of H3O+ ions makes the solution acidic.

b When some sodium hydroxide is added, ligand water molecules are successively deprotonated and a green precipitate of chromium(III) hydroxide is obtained: [Cr(H2O)6]3+(aq) + 3OH–(aq) → [Cr(H2O)3(OH)3](s) + 3H2O(l) With excess sodium hydroxide, further deprotonation takes place and a clear, green solution is obtained: [Cr(H2O)3(OH)3](s) + 3OH–(aq) → [Cr(OH)6]3–(aq) + 3H2O(l)

20 a The two reduction half-equations are: Fe3+ + e−  Fe2+

E° = +0.77 V

FeO4− + 8H+ + 3e−  Fe3+ + 4H2O

E° = +2.20 V

The value of E°cell is found by reversing the second equation (therefore changing the sign of the E° value) and then adding the two E° values. This is done in order to get the reactant, Fe3+ ions, on the left. E°cell = −(+2.20) + (+0.77) = −1.43V E°cell is negative, so the disproportionation reaction will not take place (under standard conditions).

b The two reduction half-equations are: [Fe(CN)6]3− + e−  [Fe(CN)6]4−

E° = +0.36 V

[Co(NH3)6]3+ + e−  [Co(NH3)6]2+

E° = +0.10 V

The reactants are [Fe(CN)6]3− and [Co(NH3)6]2+, so the second equation has to be reversed. The E° values are then added. E°cell = +0.36 + (−0.10) = +0.26 V E°cell is positive, so this redox reaction is thermodynamically feasible and will take place unless the activation energy is too high.

21 a amount of KMnO4 in titre = 0.0500 mol dm–3 × 0.0200 dm3 = 0.00100 mol b The oxidation number of manganese changes from +7 to +2, so each manganese gains five electrons. moles of electrons gained = 5 × 0.00100 = 0.00500 mol

c amount of V3+ ions = 0.100 mol dm–3 × 0.0250 dm3 = 0.00250 mol d 0.00250 mol of V3+ ions provides (to the MnO4– ions) 0.00500 mol of electrons, so each V3+ ion loses two electrons. As two electrons are lost, the oxidation number of vanadium increases by two from +3. Hence its oxidation state after reaction is +5.

© Philip Allan Updates

Edexcel A2 Chemistry

83

Chapter 11 Transition metals and the d-block elements

e

moles = concentration (in mol dm–3) × volume (in dm3), so the titre in cm3 must be divided by 1000 to change it into a volume in dm3. In acid solution, manganate(VII) ions are reduced to Mn2+ ions. Reduction is gain of electrons. The number gained by each manganese equals the change in its oxidation number (from +7 to +2 = 5). Oxidation is loss of electrons. The change in oxidation number equals the number of electrons lost by each vanadium ion.

22 a The reduction half-equations with their standard electrode potentials are: VO2+ + 2H+ + e– → VO2+ + H2O

E° = +1.00 V

VO2+ + 2H+ + e– → V3+ + H2O

E° = +0.34 V

S + 2H+ + 2e– → H2S

E° = +0.14 V

To check whether H2S will reduce VO2+ to VO2+, the sign of E° for the third equation must be reversed and added to that of the first: E°cell for the reduction of VO2+ to VO2+ = +1.00 + (–0.14) = +0.86 V This is positive, so the reduction is possible. To check whether the VO2+ ions can be reduced further, the sign of E° for the third equation has to be reversed and added to that of the second equation: E°cell for the reduction of VO2+ to V3+ = +0.34 + (–0.14) = +0.20 V, which is also positive, so this reduction is feasible. As both reductions are feasible, the VO2+ ions will be reduced to V3+ ions.

b Adding the first two equations to the reversed third equation gives the overall equation: VO2+ + 2H+ + H2S → V3+ + 2H2O + S e

These standard electrode potentials can be found in a data book or on pages 187 and 228 in Chapters 10 and 11 of the textbook.

Two electrons have to be added to the VO2+ ion to reduce it from the +5 to the +3 state. H2S has to lose two electrons for the sulfur to become oxidised from the –2 to the zero oxidation state, so there will be one of each species on the left-hand side of the equation.

23 a The standard reduction potentials are: V3+ + e– → V2+

E° = –0.26 V

VO2+ + 2H+ + e– → V3+ + H2O

E° = +0.34 V

VO2+ + 2H+ + e– → VO2+ + H2O

E° = +1.00 V

1 –2

E° = +1.23 V

O2 + 2H+ + 2e– → H2O

As the standard reduction potential of oxygen is more positive than all three potentials for vanadium going from +2 to +3, +3 to +4 and +4 to +5, oxygen should oxidise V2+ ions to VO2+ ions in acid solution.

b The change in oxidation number of the vanadium is from +2 to +5, a change of 3. Each oxygen atom in O2 changes by two, so there must be 2V2+ and 3 × 1–2 O2 in the overall equation, so that both change by a total of six. The overall reaction is obtained by adding the four equations: 2V2+ → 2V3++ 2e–

© Philip Allan Updates

Edexcel A2 Chemistry

84

Chapter 11 Transition metals and the d-block elements

2V3+ + 2H2O → 2VO2++ 4H+ + 2e– 2VO2+ + 2H2O → 2VO2+ + 4H+ + 2e– 2V2+ e

1 1–2 O2 + 6H+ + 6e– → 3H2O + 1 1–2 O2 + H2O → 2VO2+ + 2H+

Note that the 6H+ on the left cancel with six of the 8H+ on the right; three of the 4H2O on the left cancel with the 3H2O on the right; the 6e– on the left cancel with the 6e– on the right. E°cell for the oxidation of V2+ to V3+ is +1.23 + (+0.26) = +1.49 V E°cell for the oxidation of V3+ to VO2+ is +1.23 + (–0.34) = +0.89 V E°cell for the oxidation of VO2+ to VO2+ is +1.23 + (–1.00) = +0.23 V All the E°cell values are positive, so all the reactions will take place. This means that oxygen under standard conditions should oxidise V2+ to vanadium in the +5 state.

24 a The standard reduction potentials are: VO2+ + 2H+ + e– → V3+ + H2O

E° = +0.34 V

VO2+ + 2H+ + e– → VO2+ + H2O

E° = +1.00 V

1 –2

E° = +0.54 V

I2 + e– → I–

The E° value of iodine is more positive than that for the first equation but is less positive than that for the second equation. Thus, iodine will oxidise V3+ to VO2+, but no further.

b The equation is: V3+ + H2O + 1–2 I2 → VO2+ + 2H+ + I– e

Another way of answering this question is to work out the values of E°cell for the two possible oxidations. G

Oxidation of V3+ to VO2+: E°cell = +0.54 + (–0.34) = +0.20 V, which is positive and so the reaction is feasible.

G

Oxidation of VO2+ to VO2+: E°cell = +0.54 + (–1.00) = –0.46 V, which is negative and so the reaction will not take place.

25 The standard reduction potentials are: MnO2 + 4H+ + 2e– → Mn2+ + 2H2O

E° = +1.23 V

Cl2 + 2e– → 2Cl–

E° = +1.36 V

Br2 + 2e– → 2Br–

E° = +1.07 V

E°cell for the oxidation of chloride ions by MnO2 = +1.23 + (–1.36) = –0.13 V, which is negative and so the reaction will not take place under standard conditions. E°cell for the oxidation of bromide ions by MnO2 = +1.23 + (–1.07) = +0.16 V, which is positive and so the reaction is feasible. e

Do not state that a reaction will take place if the E°cell value is positive. The activation energy might be too high for the reaction to happen rapidly enough to be observed. The correct terminology is that the reaction is feasible or thermodynamically spontaneous.

26 The disproportionation reaction would be: 3Cr2+ → 2Cr3+ + Cr E°cell for this is given by: Cr2+ + 2e– → Cr

E° = –0.91 V

2Cr2+ → 2Cr3+ + 2e–

E° = –(–0.41) = +0.41 V

E°cell = –0.91 +0.41 = –0.50 V, which is negative, so disproportionation will not occur.

© Philip Allan Updates

Edexcel A2 Chemistry

85

Chapter 11 Transition metals and the d-block elements

27 a There are several possible answers to this question: I

Add a solution of the disodium salt of EDTA, followed by a few drops of alkali to form the EDTA4− ion. Then add to a solution of chromium(III) sulfate. Ligand exchange takes place: [Cr(H2O)6]3+ + EDTA4− → [Cr(EDTA)]− + 6H2O

I

Add a solution containing excess 1,2-diaminoethane (often written as ‘en’) to a solution of chromium(III) sulfate. Ligand exchange takes place: [Cr(H2O)6]3+ + 3en → [Cr(en)3]3+ + 6H2O

I

Add excess sodium hydroxide to a solution of chromium(III) sulfate. Complete deprotonation takes place and the complex ion chromium(III) [Cr(OH)6]3− is formed. [Cr(H2O)6]3+ + 6OH− → [Cr(OH)6]3− + 6H2O

b Chromate(III) ions are oxidised in alkaline solution by hydrogen peroxide. Excess aqueous sodium hydroxide is added to a solution of chromium(III) sulfate, followed by hydrogen peroxide solution. The mixture is then warmed on a water bath. The result is a yellow solution of sodium chromate(VI), containing CrO42− ions.

c Powdered zinc and concentrated hydrochloric acid are added to a solution of chromium(III) sulfate in a conical flask fitted with a Bunsen valve. The colour of the solution changes as chromium(II) ions are formed. The function of the Bunsen valve is to exclude air, which would oxidise the chromium(II) back to chromium(III).

28 a Ligand exchange is a reaction in which one type of ligand in a d-block metal complex ion is replaced by another, either totally or partially. An example is the addition of concentrated hydrochloric acid to a solution containing hydrated copper(II) ions: [Cu(H2O)6]2+ + 4Cl− → [CuCl4]2− + 6H2O

b A catalyst is a substance that speeds up a chemical reaction without itself being used up in the reaction. A heterogeneous catalyst is in a different phase from the reactants. An example is the solid iron catalyst used in the Haber process involving the reaction between nitrogen gas and hydrogen gas.

© Philip Allan Updates

Edexcel A2 Chemistry

86

Chapter 11 Transition metals and the d-block elements

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 B Ethene does not have a lone pair of electrons and so cannot be a ligand. Therefore, option B is the correct answer to this negative question. Options A, C and D can all act as ligands: F− has four lone pairs, the N of the –NH2 group and the O− of the –COO− group both have lone pairs that can form bonds with transition metal ions.

2 D Here, five ions are formed from one ion and six molecules. Since this is a reduction in particles from seven to five, the entropy will decrease. In options A and C more particles are produced, so there will be an increase in entropy. In option B a gas is produced, so this too will have a positive ΔSsystem.

3 B Mn2+ gives an off-white precipitate with alkali. This darkens on standing as it is oxidised slowly to MnO2. The cations in options A, C and D all give green precipitates with alkali. The addition of barium chloride showed that the substance is either a sulfate, a carbonate or a sulfite and the addition of hydrochloric acid proved it to be a sulfate.

4 D [CrCl4]− is tetrahedral, so option A is incorrect. The copper in [CuCl4]2− is in the +2 state, so option B is incorrect and the copper in [CuCl2]− is in the +1 state, so option C is also wrong. The complex ion in D has six ligands and so is octahedral, and the copper is in the +2 state. Therefore, option D is the correct answer.

5 A Option A is the electron configuration of chromium, which like copper breaks the 3dx 4s2 rule. Option B is zinc, which does not form an ion with an incomplete d-shell. Option C is scandium, which is always +3 in compounds and so also does not form an ion with an incomplete d-shell. The substance represented by the electron configuration in option D does not exist (it should be [Kr]4d75s2).

6 C Iron(II) sulfate gives a green precipitate of iron(II) hydroxide, which does not form a complex with ammonia, and so does not redissolve to form a solution. Silver(I) ions (option A) and copper(I) ions (option B) form colourless ammonia complexes. Nickel(II) (D) forms a blue complex. Therefore, options A, B and D all give hydroxide precipitates that redissolve in excess ammonia.

7 A Chromium(III), iron(III) and copper(II) compounds in solution give precipitates of hydroxides, but only chromium(III) hydroxide is amphoteric. Thus the precipitates remain with the iron(III) and copper(II) compounds, so B and C are incorrect. Copper(I) chloride, CuCl, is insoluble, so does not form a precipitate. Therefore, D is also incorrect.

8 B All transition metals, i.e. those in the series Ti to Cu have more than one oxidation state in their compounds. Not all transition metal ions are coloured — Ti4+ and Cu+ are colourless, so option A is incorrect. Copper metal has the configuration [Ar] 3d10 4s1. It is a transition metal and it has no unpaired d-electrons, so option C is incorrect. Option D is false because copper(III) does not exist.

© Philip Allan Updates

Edexcel A2 Chemistry

87

Unit 5 Transition metals, arenes and organic nitrogen chemistry

Chapter 12 Arenes and their

derivatives 1 a The Hess’s law cycle is: 6C(g)

+

∆H1 6C(s)

∆H3

6H(g) ∆H2

∆Hf

+ 3H2(g)

‘

’

ΔH1 = 6 × ΔHa(carbon) = 6 × +715 = +4290 kJ ΔH2 = 6 × ΔHa(hydrogen) = 6 × +218 = +1308 kJ ΔH3 = heat change on making 6 × C–H bonds, 3 × C–C bonds and 3 × C=C bonds = 6 × (–412) + 3 × (–348) + 3 × (–612) = –5352 kJ ΔHf(‘cyclohexatriene’(g)) = ΔH1 + ΔH2 + ΔH3 = 4290 + 1308 + (–5352) = +246 kJ mol–1 The enthalpy of formation of the theoretical gaseous ‘cyclohexatriene’ is +246 kJ mol–1. e

Remember that bond making is always exothermic (negative ΔH) and that the enthalpy of atomisation of hydrogen is per mole of atoms of hydrogen produced.

b The Hess’s law cycle is: 6C(s) + 3H2(g)

∆Hf(‘cyclohexatriene’(g))

‘

’ ∆Hstabilisation

∆Hf(benzene(g))

ΔHf(‘cyclohexatriene’(g)) + ΔHstabilisation = ΔHf(benzene(g)) ΔHstabilisation = ΔHf(benzene(g)) – ΔHf(‘cyclohexatriene’(g)) = +83 – (+246) = –163 kJ mol–1 The resonance stabilisation energy of benzene is 163 kJ mol–1. e

This means that benzene is 163 kJ more stable (at a lower energy) than the theoretical compound with localised double bonds. The value is slightly different from that obtained from hydrogenation data. This is because average bond enthalpies were used.

c The energy-level diagram is: Enthalpy

‘Cyclohexatriene’(g) –163 kJ +246 kJ Benzene(g) 6C(s) + 3H2(g)

© Philip Allan Updates

+83 kJ

Edexcel A2 Chemistry

88

Chapter 12 Arenes and their derivatives

2 The double bond between the carbon and the oxygen is delocalised between the two oxygen atoms in the ethanoate ion, thus stabilising it: –

O

O H3C

H3C

C

C

–

O e

O

It is the stabilisation energy due to the delocalisation that makes ethanoic acid an acid, as the anion is stabilised relative to the undissociated acid. The hydrogensulfate ion, HSO4–, has three resonance structures, stabilising it considerably and making sulfuric acid a strong acid.

3 The aromatic isomers of C6H4ClNO2 are: Cl

Cl

Cl

NO2

NO2 1-chloro-2-nitrobenzene

1-chloro-3-nitrobenzene NO2 1-chloro-4-nitrobenzene

e

The 1,6-structure is identical to the 1,2-isomer, as is the 1,5-structure to the 1,3-isomer. Cl

Cl

O2N

Cl

Cl

NO2 =

= O 2N

NO2

4 The aromatic isomers of C6H3Br2OH are: OH

OH Br

OH Br

Br

Br

Br

2,3-dibromophenol

2,5-dibromophenol

Br 2,4-dibromophenol

OH Br

OH

OH

Br

Br 2,6-dibromophenol

Br

Br 3,5-dibromophenol

Br 3,4-dibromophenol

© Philip Allan Updates

Edexcel A2 Chemistry

89

Chapter 12 Arenes and their derivatives

e

Put one bromine atom next to the OH (in the 2-position) and work out how many different places the other bromine can go (three more). Then put one bromine in the 3-position and see where the other bromine can go (one more). Putting a bromine in the 4-position would give one of the isomers that you have already drawn, so there are only six isomers. Note that 3,6-dibromophenol is the same as 2,5-dibromophenol.

5 a This is an addition reaction and so free-radical conditions are required. Mix liquid bromine with the nitrobenzene and heat under reflux in the presence of ultraviolet light. e

The reaction is slow and requires heat, so the reagents must be heated under reflux otherwise they would boil off.

b This is an electrophilic substitution reaction, so a catalyst of anhydrous aluminium chloride, iron(III) chloride or iron(III) bromide must be used under dry conditions. The mixture is heated under reflux. e

The iron(III) bromide is usually made in situ, by adding iron filings to the reaction mixture, which contains excess bromine. The iron reacts with some of the bromine: Fe + 1 –12 Br2 → FeBr3

6 a The equation for the formation of the NO2+ electrophile is: HNO3 + H2SO4 → NO2+ + H2O + HSO4– e

The first step is the protonation of the HNO3 molecule by the stronger acid, sulfuric acid. The water produced causes the ionisation of another sulfuric acid molecule.

b The equation for the production of the Br+ electrophile is: FeBr3 + Br2 → Br+ + FeBr4– e

If AlCl3 is used as the catalyst, the equation is: AlCl3 + Br2 → Br+ + AlCl3Br–

c The equation for the production of an alkyl electrophile, such as C2H5+, is: AlCl3 + C2H5Cl → C2H5+ + AlCl4–

7

Formation of electrophile O CH3C

+

+

AlCl3

CH3C

=O

+

–

AlCl4

Cl

H

Step 1 +

CH3C

=O

+

C

CH3

O CH3 Step 2

–

H

AlCl4

C

CH3

C O

+

+ HCl + AlCl3

O e

The stability of the benzene ring is regained in the second step. This is the driving force of the reaction.

© Philip Allan Updates

Edexcel A2 Chemistry

90

Chapter 12 Arenes and their derivatives

8 a

CH3

O

CH3

C CH3 + CH3COCl

+ HCl

Conditions: warm under reflux with anhydrous aluminium chloride as catalyst e

The electrophile will attach at either the 2- or the 4-position in this reaction and the subsequent reactions in this question.

b

CH 3

CH3 Br + Br2

+ HBr

Conditions: warm at 50°C under anhydrous conditions with some iron filings or with a catalyst of iron(III) chloride e

A mixture of 2-bromomethylbenzene and 4-bromomethylbenzene is produced.

c

CH3

CH3 NO2 + HNO3

+ H2O

Conditions: warm with a mixture of concentrated nitric and sulfuric acids e

The temperature of the mixture must not be allowed to go above 55°C or a second NO2 group will substitute into the ring.

9 a The electrons on the O– of the phenate ion are partially drawn into the ring and become delocalised with the π-electrons of the benzene ring. This stabilises the anion relative to the unionised phenol molecule, making phenol acidic. This delocalisation is not possible in the C2H5O– ion formed from ethanol. The ethoxide anion is not stabilised and the molecule is barely acidic.

b Both ethanol and phenol can form intermolecular hydrogen bonds. This is because they each have a δ+ hydrogen atom and a δ– oxygen atom. However, the dispersion (instantaneous induced dipole–induced dipole) forces are stronger between phenol molecules, because phenol has more electrons than ethanol. Thus, more energy is required to separate phenol molecules than ethanol molecules and phenol has the higher boiling temperature. e

The strength of covalent bonds is irrelevant to the boiling temperatures of these substances because no covalent bonds are broken on boiling. The strength of the dispersion forces (sometimes called van der Waals forces) depends mostly on the number of electrons and not on the masses of the molecules.

c Both substances can form hydrogen bonds with water molecules, which makes them soluble. However, the large, non-polar benzene ring is hydrophobic and reduces the solubility of phenol relative to that of the smaller ethanol molecule.

© Philip Allan Updates

Edexcel A2 Chemistry

91

Chapter 12 Arenes and their derivatives

e

Solubility in water is due mostly to hydrogen bonds between solute molecules and water molecules. Do not use statements such as ‘like dissolves like’ — they are meaningless. O – K+

10 a

e

Do not draw a covalent bond between the O and the K — the product, potassium phenate, is an ionic compound.

b

OH Cl

Cl

Cl e

A white precipitate would be observed. The reaction is of the same type as that between phenol and bromine in aqueous solution.

c

O C O

e

The product is the ester phenyl benzoate.

11 Phenol is a weak acid and dissociates according to the equation: C6H5OH  H+ + C6H5O– acid dissociation constant Ka =

[H+][C6H5O–] = 1.3 × 10–10 mol dm–3 [C6H5OH]

Ka = [H+]2/[C6H5OH] [H+]2 = Ka × [C6H5OH] [H+] =

Ka × [C6H5OH] =

1.3 × 10−10 × 0.20 =

2.6 × 10−11

= 5.10 × 10–6 mol dm–3 pH = –log [H+] = –log (5.10 × 10–6) = 5.29 e

The assumptions made in this calculation are that [H+] = [C6H5O–], that [C6H5OH]eq ≈ [C6H5OH]initial and that the temperature of the solution is 25°C.

© Philip Allan Updates

Edexcel A2 Chemistry

92

Chapter 12 Arenes and their derivatives

12 a The equation for the reaction of phenol with bromine is: OH

OH Br

Br

+ 3Br2

+ 3HBr

Br

The bromine reacts with phenol in a 3:1 ratio. molar mass of phenol = (6 × 12.0) + (6 × 1.0) + 16.0 = 94.0 g mol–1 amount of phenol = mass/molar mass = 1.23 g/94.0 g mol–1 = 0.0131 mol amount of bromine = 3 × 0.0131 = 0.0393 mol mass of bromine, Br2 = mol × molar mass = 0.0393 × (2 × 79.9) = 6.28 g e

if you keep all the figures on the calculator during the calculation (which is a better method), the answer is 6.27 g.

b theoretical amount (moles) of 2,4,6-tribromophenol produced = 0.0131 mol theoretical mass of 2,4,6-tribromophenol, C6H2Br3OH, produced = mol × molar mass = 0.0131 × [(6 × 12.0) + (3 × 1.0) + (3 × 79.9) + 16.0] = 0.0131 × 330.7 = 4.332 g percentage yield = mass produced × 100/theoretical mass = (4.25 × 100)/4.332 = 98.1% e

Another method is to calculate the moles of product formed (4.25/330.7 = 0.01285) and then to calculate the percentage yield as (moles produced × 100)/theoretical yield in moles. A common error is to think that the percentage yield is (mass of product × 100)/mass of reactant.

13 A nucleophile has a lone pair of electrons, which it uses to form a new covalent bond. In the reaction of phenol with an acid chloride, the lone pair on the oxygen of phenol attacks the δ+ carbon atom in the acid chloride, forming an ester and hydrogen chloride. An example is the reaction: C6H5OH + CH3COCl → CH3COOC6H5 + HCl

14

180° H

O C6H5

O

120°

C O

C H

C6H5

O

180° e

The hydrogen bond is due to the attraction between the lone pair of electrons on the δ– oxygen atom of the C=O group and the δ+ hydrogen of the –OH group in another benzoic acid molecule. This makes the OHO bond angle 180° (two pairs of electrons repelling) and the COH angle about 120° (one σ-pair, one lone pair and one pair in the hydrogen bond repelling).

15 If the temperature goes above about 65°C, a significant amount of 1,3-dinitrobenzene is formed. If the temperature falls much below this value, the nitration of benzene is too slow to be noticeable. e

Once some nitrobenzene is formed, it competes with unreacted benzene in the nitration reaction. As the benzene ring is slightly deactivated by the NO2 group, a higher temperature is needed to nitrate nitrobenzene than benzene, which is why a temperature below 65°C prevents formation of a significant amount of dinitrobenzene.

© Philip Allan Updates

Edexcel A2 Chemistry

93

Chapter 12 Arenes and their derivatives

16 Formation of electrophile HNO3

+

+

H2SO4

H2NO3

H2NO3+

H2O

+

–

+

HSO4

NO2+

CH3

CH3

CH3 H

Step 1 NO2+

NO2

+

and

+

NO2

H

CH3

CH3 HSO4–

H

Step 2

NO2 + H 2SO4

NO2

+

CH3

CH3 and

+ H 2SO4

NO2

H

–

HSO4

NO2

17 The deductions are: I

As it burns with a smoky flame, compound X is probably an arene.

I

The precipitate with Brady’s reagent means that it is either an aldehyde or a ketone.

I

The lack of a silver mirror with Tollens’ reagent means that it is a ketone and not an aldehyde.

I

The precipitate of iodoform indicates that it has a CH3C=O group.

I

The simplest compound that X could be is: O C CH3

18

e

Google ‘Lister’ and follow leads. Also Google ‘phenol’ and go to www.3dchem.com.

© Philip Allan Updates

Edexcel A2 Chemistry

94

Chapter 12 Arenes and their derivatives

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 C amount (moles) of nitrobenzene, C6H5NO2 = mass/molar mass = 12.3 g/123.0 g mol−1 = 0.100 mol amount (moles) of 1,3-dinitrobenzene actually produced = 0.100 × 44.4/100 = 0.0444 mol mass produced (yield) = 0.0444 mol × molar mass of C6H4(NO2)2 = 0.0444 × 168.0 = 7.46 g In options B and D C6H5(NO2)2 has been used as the formula of 1,3-dinitrobenzene, rather than C6H4(NO2)2. In options A and B it has been assumed that 0.1 mol of reactant would produce 0.1 mol of product, therefore ignoring the fact that the yield is only 44.4%.

2 C An addition reaction does occur between benzene and hydrogen under these conditions, so option A is incorrect. Option B is not true because benzene is stabilised by delocalisation of the π-electrons. Benzene is at a lower energy level than the theoretical cyclohexa-1,3,5-triene and so less, not more, energy is given out on hydrogenation. Therefore option C, and not D, is correct.

3 B The first step is addition of Br+. This is followed by loss of H+ resulting in overall substitution. The intermediate loses H+ so that it regains the resonance stabilisation energy. The reaction occurs at room temperature, so the activation energy must be reasonably low. Therefore option A is incorrect. Bromine is an electrophile under these conditions because, after reaction with the catalyst, it becomes Br+. Therefore, options C and D are incorrect.

4 D Step 1 is acylation, so B is not the correct response. Step 2 is nitration, so C is not the correct response. Step 3 is addition of hydrogen (also can be called reduction) and so A is not the correct response. None of the steps involves oxidation, so D is the correct response to this negative question.

5 B The first step in both acylation (Friedel-Crafts) and nitration is attack by an electrophile, so C and D are wrong. The overall reaction is substitution not addition, so A is also wrong.

6 B Lithium aluminium hydride is the reagent for reduction of polar π bonds and so is not the correct response. Aluminium oxide is never a catalyst in arene chemistry, so C is wrong. Dry aluminium chloride is the catalyst for Friedel-Crafts or halogenation and so D is also wrong.

7 B Any compound with a low hydrogen-to-carbon ratio is liable to burn with a smoky flame unless oxygen is in excess. Option A is wrong as arenes with one or two oxygen atoms, for example phenol, will burn with a smoky flame. Species other than arenes can form resonance hybrids. An example is the anion of a carboxylic acid, so option C is incorrect. The statement in D is irrelevant.

8 A The reason that phenol is so reactive towards electrophilic substitution is that the lone pz pair of electrons on the oxygen atom becomes incorporated into the delocalised ring. This makes the ring more susceptible to attack by electrophiles. Options B and C are true statements, but are irrelevant to this question. Option D is not true. The Br+ ion is a much stronger electrophile, but is not needed in the bromination of phenol because the ring is activated by the lone pair of electrons in the oxygen of the –OH group.

© Philip Allan Updates

Edexcel A2 Chemistry

95

Unit 5 Transition metals, arenes and organic nitrogen chemistry

Chapter 13 Organic nitrogen

compounds 1 Step 1: the ethene has to be converted to a halogenoalkane Mix ethene with hydrogen chloride gas at room temperature: CH2=CH2 + HCl → CH3CH2Cl Step 2: react the chloroethane with excess concentrated ammonia solution in a mixture of water and ethanol. Allow the mixture to stand at room temperature or heat it in a sealed tube: CH3CH2Cl + 2NH3 → CH3CH2NH2 + NH4Cl e

A good way to work out the answer to this type of question is to start at the end. Ask yourself how a primary amine can be prepared — there are three ways. Reduction of a nitrile would not work as the –CN group would add an extra carbon atom. Amides can be reduced to amines, but ethene cannot be converted in one step to ethanamide. The only method that would work is the reaction of a halogenoalkane with ammonia. Step 1 is then obvious.

2 a The primary amine with a branched chain is: NH2 CH3

CH

CH2

NH2

or

CH3

CH3

CH

CH3

CH3

b The symmetrical secondary amine with an unbranched chain is: CH3

CH2 H

N CH2

CH3

c The secondary amine with a branched chain is: CH3 CH H 3C

H

N H3C

d The tertiary amine is: CH3

CH2 N

CH3

CH3 e

A primary amine has the general formula RNH2, a secondary amine the formula RR′NH and a tertiary amine RR′R′′N, where R, R′ and R′′ are alkyl groups.

3 The nitrogen atom in methylamine has three bond pairs and one lone pair, causing the molecule to be pyramidal. The four electron pairs repel each other, but the repulsion between the lone pair and the bond pairs is stronger than that between the bond pairs. This means that the H–N–H bond angle is about 107°, which is smaller than the tetrahedral angle of 109.5°. e

Do not make the common error of stating that the bonds or the atoms repel each other. The theory of shapes is called the valence-shell electron-pair repulsion theory (VSEPR). Another error is failing to realise that there is a lone pair of electrons on the nitrogen atom.

© Philip Allan Updates

Edexcel A2 Chemistry

96

Chapter 13 Organic nitrogen compounds

4 The structural formula of ethanamide is: O C

CH3 120

NH2

The carbon atom in the functional group has one double bond and two single bonds (and no lone pairs). The π-bond does not affect the shape and the electron pairs in the three σ-bonds repel each other to a position of maximum separation. This gives a trigonal planar shape with an N–C–C bond angle of 120°. e

Remember that a double bond consists of a σ- and a π-bond in the same direction between two atoms. The shape is caused by the repulsion between the electrons in the σ-bonds.

5 Ethylamine can form hydrogen bonds with water molecules as it has a δ– nitrogen atom and δ+ hydrogen atoms. Therefore, it will dissolve in water. The δ– chlorine in chloroethane is too big to form a hydrogen bond with the δ+ hydrogen in water and ethane is non-polar. Thus, neither of these compounds dissolves in water. e

Solubility in water is caused either by ions becoming hydrated by water molecules or by hydrogen bonding between the solute and the water. Polarity in an organic molecule does not mean that it is soluble in polar solvents such as water.

6 Both ethanoic acid and propylamine can form intermolecular hydrogen bonds. Induced dipole (dispersion) forces also exist between their molecules, but these are much weaker than the hydrogen bonds. The oxygen atom in ethanoic acid is more δ– than the nitrogen atom in propylamine, so the O---H hydrogen bond is stronger than the N---H hydrogen bond. This means that more energy is required to separate the ethanoic acid molecules and so ethanoic acid has the higher boiling temperature. Butane has no δ+ hydrogen atoms (and no δ– F, N or O atoms) and so cannot form hydrogen bonds. Its only intermolecular interactions are the weaker dispersion forces and so butane has the lowest boiling temperature. e

Oxygen is more electronegative than nitrogen and so is more δ–.

7 As with question 1, it makes sense to work backwards, asking how the amine can be prepared. Amines can be prepared from halogenoalkanes, which can themselves be prepared from alcohols. Acids can be reduced to alcohols. Step 1: add lithium aluminium hydride in dry ether to the ethanoic acid and then hydrolyse with dilute sulfuric acid: CH3COOH → CH3CH2OH Step 2: add phosphorus pentachloride: CH3CH2OH → CH3CH2Cl Step 3: allow the chloroethane to stand with excess concentrated aqueous ethanolic ammonia: CH3CH2Cl → CH3CH2NH2 e

The critical step is deducing that a halogenoalkane reacts with ammonia to form a primary amine. If you are asked for an outline of a synthesis, you should give details of reagents, conditions (if any) and the formulae of the intermediates produced at each step. You do not need to give balanced equations.

8 a

Element

%

Divide by Ar

Divide by smallest

Carbon

39.3

39.3/12.0 = 3.275

3.275/1.64 = 2.0

Hydrogen

11.5

11.5/1.0 = 11.5

11.5/1.64 = 7.0

Oxygen

26.2

26.2/16.0 = 1.64

1.64/1.64 = 1

Nitrogen

23.0

23.0/14.0 = 1.64

1.64/1.64 = 1

The empirical formula is C2H7ON. e

Make sure that you divide the percentage by the atomic mass and not the atomic number.

© Philip Allan Updates

Edexcel A2 Chemistry

97

Chapter 13 Organic nitrogen compounds

b The molecular formula is also C2H7ON. Since the substance produces steamy fumes with PCl5, it must contain an –OH group. It cannot be a carboxylic acid as it has only one oxygen atom, so it must contain an alcohol group. Its structural formula is CH2OHCH2NH2 and its name is 2-hydroxyethylamine (or 2aminoethanol).

9 a NH2CH2CH2OH + HCl → [NH3CH2CH2OH]+Cl– e

The –NH2 group acts as a base and accepts an H+ ion from the acid. The reaction is similar to NH3 + HCl → NH4+Cl–. The reaction can also be written as an ionic equation: NH2CH2CH2OH + H+ → [NH3CH2CH2OH]+.

b NH2CH2CH2OH + 2CH3COCl → CH3CONHCH2CH2OOCCH3 + 2HCl e

The acid chloride reacts with the amine group to form a substituted amide and with the alcohol group to form an ester.

H3C N+

c NH2CH2CH2OH + CH3I → H

CH2CH2OH

I–

H e

Primary amines react with halogenoalkanes to form secondary (and tertiary) amines. Alcohols do not react, so only one CH3I is needed in the equation. The equation NH2CH2CH2OH + CH3I → CH3NHCH2CH2OH + HI would be acceptable, even though the secondary amine produced is a base and would react with the HI to form the salt, as shown above.

10 A base must have a lone pair of electrons that is used to form a bond with an H+ ion. The nitrogen atoms in ammonia and in methylamine each have a lone pair, but the methyl group ‘pushes’ electrons towards the nitrogen, which becomes more δ–. This makes the nitrogen in methylamine a stronger base, as it is more negative than the nitrogen in ammonia. Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom becomes partially delocalised with the electrons of the benzene ring. This lowers its ability to accept a proton. e

This is also the reason why the benzene ring in phenylamine is attacked readily by electrophiles.

11 Aminoethanoic acid is an amino acid. It has an acidic –COOH group and a basic –NH2 group. The –COOH group loses a proton and becomes a –COO– group. The –NH2 group accepts that proton and becomes an –NH3+ group. The molecule now has a full positive charge on one end of the molecule and a full negative charge on the other end. This species is called a zwitterion. Strong forces of attraction exist between zwitterions and so a large amount of energy is required to separate them, causing aminoethanoic acid to be a solid at room temperature. The positive charge on the nitrogen in the zwitterion causes strong attractions with the δ– oxygen atoms in water; the negative charge on the oxygen interacts similarly with the δ+ hydrogen atoms in water. This makes the amino acid water-soluble. e

You must make it clear that the forces of attraction that cause the amino acid to be a solid act between zwitterions.

12 When acid is added to ethylamine the volatile base is protonated to form a non-volatile ionic compound. C2H5NH2 + HCl → C2H5NH3+Cl− When alkali is added, the C2H5NH3+ ion is deprotoanted re-forming the volatile base ethylamine. Therefore, the characteristic smell returns. C2H5NH3+ + OH− → C2H5NH2 + H2O

13 The reagents are tin and concentrated hydrochloric acid.

© Philip Allan Updates

Edexcel A2 Chemistry

98

Chapter 13 Organic nitrogen compounds

14 a The –NH2 group in phenylamine contains two δ+ hydrogen atoms and a δ− nitrogen atom. This means that phenylamine can form intermolecular hydrogen bonds. Chlorobenzene does not have any δ+ hydrogen atoms or any small δ− atoms and so it cannot form intermolecular hydrogen bonds. The hydrogen bonds between phenylamine molecules are stronger than the dipole–dipole forces and the induced dipole forces present between molecules of each substance. Thus more energy is required to separate phenylamine molecules than is required to separate chlorobenzene molecules and so phenylamine has the higher boiling temperature.

b Although phenylamine can form some hydrogen bonds with water, the large hydrophobic benzene ring reduces its solubility. When acid is added, C6H5NH3+ ions are formed. These are hydrated by the water molecules , which results in the substance dissolving.

c Chlorobenzene cannot form hydrogen bonds with water as it has neither δ+ hydrogen atoms nor a δ− F, N or O atom. It also does not react with water, so despite being polar, it is insoluble. It does not dissolve in acid, because it is not a base, and so it does not react with H+ ions to form an ionic compound.

15 a C6H5NH2 + HCl → C6H5NH3+Cl− b C6H5NH2 + CH3COCl → C6H5NHCOCH3 + HCl e

Here phenylamine is reacting as a primary amine. In a it forms a salt and in b it forms a substituted amide that has the structural formula: O NH

C CH3

16 a If the temperature falls below 0°C, the reaction between nitrous acid and phenylamine is too slow. b If the temperature rises above 10°C, the diazonium ion will decompose as soon as it is formed. 17 a +

N

N

b +

N

N

–

OH + OH

+

N

e

N

OH + H2O

You must show the two nitrogen atoms separately with the correct bonds between them. Make sure that you put the positive charge on the appropriate nitrogen atom in the formula for the diazonium ion.

18 The oxygen atom in the C=O group is δ− and the two hydrogen atoms of the NH2 group are δ+. Because of this, ethanamide forms strong hydrogen bonds with water molecules and so is soluble.

19 nHOOC–(CH2)3–COOH + nH2NCH2CH2NH2

C O

(CH2)3

C

N

O

H

CH2

CH C 2

N H

+ (2n–1)H2O n

20 Terylene® is a polyamide and so has δ− oxygen, δ− nitrogen atoms atoms and δ+ hydrogen atoms spaced regularly along the polymer chain. These atoms form hydrogen bonds with other chains. On melting, these

© Philip Allan Updates

Edexcel A2 Chemistry

99

Chapter 13 Organic nitrogen compounds

hydrogen bonds have to be broken and because they are strong, a high temperature is needed to provide sufficient energy. Addition polymers such as poly(propene) do not form hydrogen bonds because they do not contain any δ+ hydrogen atoms or δ− F, N or O atoms. This means that their intermolecular forces (really forces between chains) are weaker.

21 Poly(ethenol) has an –OH group on every other carbon atom in the chain. This means that it can form many hydrogen bonds with water and so, despite the extremely large molecules, it is soluble.

22 a The general formula for a zwitterion of an amino acid is +NH3CHRCOO–. b The equation for the reaction of the zwitterion with acid is: +NH CHRCOO– 3

+ H+ → +NH3CHRCOOH

c For the reaction with base, the equation is: +NH CHRCOO– 3 e

+ OH– → NH2CHRCOO– + H2O

Remember that the positive charge is on the nitrogen atom. This charge must be shown there if the full structural formula is drawn: H

H H

N

+

O

C

C –

H

O

R

23 The two optical isomers of aminobutane-1,4-dioic acid are: HOOC

COOH

C HOOCH2C

C

H

H

CH2COOH

H2 N

NH2 L-aspartic acid

The left-hand compound is the natural isomer, aspartic acid, as it has the same configuration as L-alanine (textbook, p. 275).

24

H

H

H N H

C

C

N

C

H

O

H

CH3

or

C

N H

OH

H

H

H

O

C

C

N

C

CH3

O

H

H

O C OH

25 Uracil in mRNA forms two hydrogen bonds to adenine in DNA as shown below: Sugar O

N

H C

H

C

H N

C H

N C

H

C N

N C CH

O C H Uracil

© Philip Allan Updates

C N Adenine

N Sugar

Edexcel A2 Chemistry 100

Chapter 13 Organic nitrogen compounds

The information needed for questions 26–29 is not given in the A2 chemistry textbook. You are advised to do your own research to find answers to these questions. Possible sources are the internet, biology textbooks and other students who are studying biochemistry.

26 A DNA double helix has two strands. The sense strand has its bases in a particular order. The antisense strand (linked by hydrogen bonds between the two strands) has its bases in the complementary order (A opposite T and C opposite G). The double helix unwinds and a strand of messenger RNA (mRNA) is made with the order of bases exactly in the order of the sense strand of the DNA, except that uracil, U, replaces thymine, T. Threebase sections of the mRNA molecule act as a code for the insertion of a particular amino acid in the polypeptide being synthesised. Each three-letter section is called a codon. A codon for leucine is GAC in the antisense strand of the DNA and hence CUG in mRNA.

27 Chargaff’s rules are: I

the amount of adenine = amount of thymine (A = T)

I

the amount of cytosine = amount of guanine (C = G)

I

A+G=C+T

These rules apply to DNA but not to RNA.

28 Search for Chargaff’s rule, Crick or Watson on Google. Also don’t forget about the important contribution of Wilkins and Franklin.

29 Each human gene has a sequence of bases, repeated many times, that seem to have no particular function. These repeating sequences are called minisatellites. Each person, except identical twins, has unique repeats. One method of DNA fingerprinting is as follows: I

Some tissue is sampled — for example, blood, saliva, semen or even hair.

I

DNA is extracted from the cells in the sample.

I

The DNA is cut up into the minisatellite fragments by enzymes.

I

The fragments are then separated by electrophoresis — chromatography under an electric potential.

I

A DNA ‘probe’ is added — a fragment of DNA that has been labelled with a radioisotope. This sticks to certain fragments in the sample.

I

A photographic plate is placed over the electrophoresis plate and the X-rays from the radioactive isotope darken the photo plate, showing the position of each fragment.

I

Several different DNA probes are required to obtain a definite profile of a person’s DNA.

Your only defence against a DNA match is to say that ‘it must have been my identical twin’.

© Philip Allan Updates

Edexcel A2 Chemistry 101

Chapter 13 Organic nitrogen compounds

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 C An amine must have NH2, NH or N attached to one or more carbon atoms by single bonds. CH3NH2 (option A) is a primary amine, as is CH3COCH2NH2 (option D). (CH3)2NH (option B) is a secondary amine, but CH3CONH2 (option C) is an amide. Note that compound D is also a ketone. The only non-amine is the compound in option C, which is therefore the correct answer to this negative question.

2 B An alkali will dissolve in, or react with, water to produce OH− ions, so the answer is option B. The reaction in option A shows that 1-butylamine is a base; that in option C shows it to be a primary amine and that in option D shows that it can act as a ligand.

3 D Phenylamine is a base, so it reacts with acids not bases. Therefore, option A is incorrect. The temperature of 35°C is too high for a diazonium salt to be formed, so option B is wrong. The benzene ring in phenylamine is activated and so it reacts towards electrophiles in the same way as phenol. So the ‘standard’ benzene conditions (option C) are not required and it reacts with bromine water to form a white precipitate of the tri-substituted compound.

4 B The metal can be either tin or iron, but the acid must be concentrated, so option A is wrong. Potassium dichromate is an oxidising agent not a reducing agent, so option C is incorrect. The correct catalyst with hydrogen is either platinum or nickel, not copper. Therefore, option D is incorrect.

5 A There is an –OH group on every other carbon atom in the polymer chain and so it can form thousands of hydrogen bonds with water. It is this that makes it soluble. Polarity (B) is not a guide to solubility and it neither reacts (C) nor breaks down into its monomers in water (D).

6 C The monomer must have two functional groups that can react with each other to form a new covalent bond. The amino acid in option A has an –NH2 group and a –COOH groups that can form amide links. Option B has an alcohol and an acid group and can, therefore, form a polyester. Option D has an alcohol group and an acid chloride group, so it can also form a polyester. Since these three compounds can all form polymers, none is the correct answer. The compound in option C has an alcohol group and an amine group. These groups do not react, so this compound cannot form a polymer. Therefore, option C is the correct answer to this negative question.

7 B The reason for a high melting temperature is strong forces between the particles. Glycine exists as zwitterions that have strong ionic bonds with neighbouring zwitterions.

8 D Amino acids react with acids because of the amine group and with bases because of the carboxylic acid group, thus neither option A nor option B is correct. The acid group will form an ester, so option C is not the correct response. Bromine water does not react with either an –NH2 group or a –COOH group and so it will not react with amino acids. Therefore, option D is the correct response to this negative question.

9 A Separation by chromatography occurs because the different species have different Rf values. Some have different pH values that would result in different Rf values, but many have the same pH but different Rf values, so option B is incorrect. The reaction with ninhydrin (C) is how the separated amino acids are located, not how they are separated. Their molar mass (D) is irrelevant.

10 A There are five different hydrogen environments in the molecule — that in NH2, in CH, in CH2, in the alcoholic OH group and in the acidic COOH group.

© Philip Allan Updates

Edexcel A2 Chemistry 102

Unit 5 Transition metals, arenes and organic nitrogen chemistry

Chapter 14 Organic analysis

and synthesis 1 The first step is to hydrolyse the halogenoalkane by warming it with aqueous sodium hydroxide and a few drops of ethanol for several minutes. The C–Cl and the C–I groups are hydrolysed to Cl– and I– ions. The solution is cooled and excess nitric acid is added, followed by aqueous silver nitrate solution. Precipitates of AgCl and AgI will be obtained. The colour will be somewhere between white and pale yellow, depending on the relative amounts of chloride and iodide, so the colour does not prove which halogens are present. Dilute aqueous ammonia is then added. Some of the precipitate will dissolve. The solution is filtered. The filtrate is acidified with dilute nitric acid. If a white precipitate appears, the original organic compound contained chlorine. If the residue is pale yellow and insoluble in concentrated aqueous ammonia, the organic compound also contained iodine. e

After hydrolysis the solution must be made acid, otherwise the sodium hydroxide still present would form a black precipitate with the silver nitrate. After addition of silver nitrate the mixture must be filtered, because it is difficult to tell whether any of the precipitate dissolved in the dilute ammonia. Acidification of the filtrate reprecipitates the silver chloride, thus proving that the organic compound contained chlorine.

2 The presence of the C=C groups can be detected using bromine water. The brown bromine water will turn colourless. The presence of the –OH group can be detected by adding phosphorus pentachloride to the dry substance. Clouds of misty fumes (of hydrogen chloride) prove the presence of an –OH group. This result could be obtained with an acid or an alcohol, so the absence of a –COOH group has to be shown by adding a solution of sodium carbonate to the unknown. An acid would produce bubbles (of carbon dioxide), but an alcohol, such as linalool, would show no reaction. e

The colour before (brown) and afterwards (colourless) must be stated when giving the results of the bromine water test. The alcohol group could be detected by warming linalool with some ethanoic acid and a few drops of concentrated sulfuric acid, then pouring the mixture into a solution of sodium carbonate (to remove excess acids). An alcohol gives rise to the characteristic smell of an ester after this reaction.

3 a The decolorisation of bromine water shows that carvone must contain a C=C group. b The failure to change the colour of acidified potassium dichromate(VI) solution shows that carvone cannot be oxidised easily. Therefore, it is not a primary or a secondary alcohol, or an aldehyde.

c A yellow precipitate with Brady’s reagent shows that carvone is a ketone. (The test in part b has already shown that it cannot be an aldehyde.) e

In this example, the positive Brady’s test does not show that the unknown is either an aldehyde or ketone, since an aldehyde has been excluded by an earlier test.

4 a The production of steamy fumes means that compound Y contains an –OH group and so is either an acid, an alcohol or both.

b The fact that carbon dioxide is evolved on the addition of sodium carbonate means that Y is an acid. It could also contain an alcohol group since it has three oxygen atoms.

© Philip Allan Updates

Edexcel A2 Chemistry 103

Chapter 14 Organic analysis and synthesis

c The precipitate with 2,4-dinitrophenylhydrazine shows that Y is an aldehyde or a ketone. Since it has only three oxygen atoms and it also contains a carboxylic acid group, compound Y cannot be an alcohol.

d Compound Y reduces dichromate(VI) ions, so it is an aldehyde. (The formula of compound Y and the tests in parts b and c exclude the possibility that Y is a primary or secondary alcohol.)

e Compound Y does not contain a CH3C=O group. The structural formula of Y is: O

H C

CH2

C

O

O

H

5 a Either: warm a few drops of citral with some Fehling’s solution. The blue solution will form a red precipitate (of copper(I) oxide). Or: warm a few drops of citral with a solution of ammonia and silver nitrate. A silver mirror will form.

b Add some phosphorus pentachloride to a dry sample of geraniol. Steamy fumes (of hydrogen chloride) will be produced. Then add some geraniol to a solution of sodium carbonate. An absence of bubbles proves that geraniol is an alcohol, containing an –OH group, rather than a carboxylic acid, containing a –COOH group.

c Both compounds would turn a brown solution of bromine water colourless. d Citral has a major peak in its infrared spectrum at around 1700 cm–1 due to the C=O group but does not have a peak at around 3000 cm–1. Geraniol does not have a peak around 1700 cm–1 but has one in the 3000 cm–1 region, due to the OH group.

e The aldehyde group has to be reduced. Suitable reagents are: I

lithium tetrahydridoaluminate (lithium aluminium hydride), LiAlH4, in dry ether solution, followed by hydrolysis with aqueous acid

I e

sodium tetrahydridoborate (sodium borohydride), NaBH4, in aqueous solution

Hydrogen gas and a heated nickel catalyst must not be used, as this would also reduce the C=C groups.

6 a

Element Carbon Hydrogen Oxygen

% 55.8 7.00 37.2

Divide by Ar 55.8/12.0 = 4.65 7.00/1.0 = 7.00 37.2/16.0 = 2.325

Divide by smallest 4.65/2.325 = 2.0 7.00/ 2.325 = 3.0 2.325/2.325 = 1.0

The empirical formula is C2H3O.

b empirical mass of C2H3O = (2 × 12.0) + 3.0 + 16.0 = 43.0 relative molecular mass = largest m/e value = 86, which is twice the empirical mass The molecular formula is C4H6O2. I

Z does not change the colour of bromine water, so it does not contain any C=C bonds.

I

Z does not give steamy fumes with phosphorus pentachloride, so it does not contain an –OH group — it is neither an acid nor an alcohol.

I

Z gives a yellow precipitate with 2,4-dinitrophenylhydrazine (Brady’s reagent), so it could be an aldehyde or a ketone or both (since there are two oxygen atoms in the molecule).

I

A silver mirror forms when Z is warmed with Tollens’ reagent, so it is an aldehyde.

I

Z gives a precipitate of iodoform and, since it is not an alcohol (nor ethanal), it must be a methyl ketone, which has a CH3C=O group.

I

Z is neither an alcohol nor a carboxylic acid. There are two oxygen atoms in a molecule of Z: one is in an aldehyde group (positive Tollens’ test) and the other is in a methyl ketone (iodoform test).

© Philip Allan Updates

Edexcel A2 Chemistry 104

Chapter 14 Organic analysis and synthesis

The structural formula of Z is: H3C

H C

CH2

C

O e

O

The catch here is to think that Z is not a ketone because it gives a positive Tollens’ test. Note that there are two oxygen atoms in the molecule and that the compound is neither an acid nor an alcohol. As one oxygen atom is in an aldehyde group, the other could be in a ketone group or a second aldehyde group.

7 An ester is hydrolysed, after acidification, to an alcohol and a carboxylic acid. Compound F gives off carbon dioxide when added to a solution of sodium hydrogencarbonate, so F is the acid. Its empirical formula is given by: Element Carbon Hydrogen Oxygen

% 54.5 9.1 36.4

Divide by Ar 54.5/12.0 = 4.54 9.1/1.0 = 9.1 36.4/16.0 = 2.275

Divide by smallest 4.54/2.275 = 2.0 9.1/2.275 = 4.0 2.275/2.275 = 1.0

The empirical formula of F is C2H4O. However, it is a carboxylic acid and so must have at least two oxygen atoms. The most probable molecular formula is C4H8O2, which can exist as two isomers CH3CH2CH2COOH and (CH3)2CHCOOH. The first, unbranched, acid has hydrogen nuclei in four different chemical environments; the second isomer has only three environments. As there are three peaks in the NMR spectrum, the structural formula of F is: CH3 H3C

C

O C OH

H

Compound G is the alcohol. Since the ester has the formula C7H14O2 and the acid is C4H8O2, the alcohol must contain three carbon atoms. Propan-1-ol does not give a positive iodoform test, whereas propan-2-ol does. Compound G does not give a precipitate in the iodoform test, so G must be propan-1-ol. The structural formula of the ester E is: CH3 H 3C

C

O C O

H

CH2CH2CH3

The equation for the reaction of ester E with sodium hydroxide is: (CH3)2CHCOOCH2CH2CH3 + NaOH → (CH3)2CHCOONa + CH3CH2CH2OH

8

e

I

Dissolve the reaction product (which is an amide) in the minimum of boiling water.

I

Filter the hot solution into a conical flask using a heated funnel.

I

Allow the filtrate to cool.

I

Filter off the resulting solid under reduced pressure.

I

Wash the solid with a little cold water and allow it to dry at room temperature.

An alternative solvent, such as ethanol, could be used.

9 a The reagents for the oxidation of cyclohexanol to cyclohexanone are dilute sulfuric acid and a solution of potassium dichromate(VI). The conditions are to heat the mixture under reflux.

© Philip Allan Updates

Edexcel A2 Chemistry 105

Chapter 14 Organic analysis and synthesis

b As there is so little impurity, simple distillation is sufficient, collecting the distillate that comes over between 155°C and 157°C. e

The neck of the distillation flask acts as a mini-fractionating column, so the cyclohexanone is separated from the small amount of the impurity — cyclohexanol.

c As the amount of impurity is much greater, fractional distillation is necessary. The cyclohexanone distils off first, leaving pure cyclohexanol in the flask. As before, collect the fraction that boils between 155°C and 157°C.

10 a The reaction mixture is placed in a separating funnel. The two organic substances are less dense than water and form a single layer that floats on the aqueous layer. The water layer is run off, leaving the organic layer containing the cyclohexanone and the cyclohexanol.

b A solid drying agent, such as anhydrous calcium chloride, should be added to the organic fraction. c The organic liquid is dry when it is totally clear and not cloudy. 11 The molar mass of compound Z has the same value as the m/e of the molecular ion. There are two ways of finding the molecular formula: Method 1: mass of carbon in 11.9 g of CO2 = 11.9 × 12.0/44.0 = 3.245 g % carbon = 3.245 × 100/4.85 = 66.9% mass of hydrogen in 4.85 g of H2O = 4.85 × 2.0/18.0 = 0.5389 g % hydrogen = 0.5389 × 100/4.85 = 11.1% % oxygen = 100 − (66.9 + 11.1) = 22.0% Element Carbon Hydrogen Oxygen

% 66.9 11.1 22

Divide by Ar 66.9/12.0 = 5.575 11.1/1.0 = 11.1 22/16.0 = 1.375

Divide by smallest 5.575 /1.375 = 4.05 ≈ 4 11.1/1.375 = 8.07 ≈ 8 1.375/1.375 = 1

The empirical formula is C4H8O. The formula mass is 72, which is the same as the molar mass of Z. The molecular formula of Z is C4H8O. Method 2: moles of Z = 4.85/72.0 = 0.06736 moles of carbon = moles of CO2 = 11.9/44.0 = 0.2705 which is 4 × moles of Z moles of hydrogen = 2 × moles of H2O = 2 × 4.85/18.0 = 0.5389 which is 8 × moles of Z mass of oxygen = 72.0 − (4 × 12.0 + 8 × 1.0) = 16.0 which is 1 atom per molecule The molecular formula of Z is C4H8O. I

The precipitate with Brady’s reagent: indicates that Z has a C=O group. Therefore, Z is either an aldehyde

I

The lack of a precipitate of iodoform indicates that Z does not have a CH3CO group and so it is not butan-

or a ketone. The only ketone with the formula C4H8O is butan-2-one, CH3COCH2CH3. 2-one. It must, therefore, be an aldehyde — either 2-methylpropanal or butanal.

© Philip Allan Updates

Edexcel A2 Chemistry 106

Chapter 14 Organic analysis and synthesis

I

2-methylpropanal, (CH3)2CHCHO, has hydrogen atoms in three environments. Its NMR spectrum has three

I

Butanal, CH3CH2CH2CHO, has hydrogen atoms in four different environments. Its NMR spectrum with four

peaks of heights in the ratio 6:1:1. Thus compound Z is not 2-methylpropanal. peaks of heights in the ratio 3:2:2:1. Thus compound Z is butanal, and has the structural formula: O H3C

CH2

CH2

C H

e

When writing a structural formula it is advisable to draw out all the π-bonds.

12 The product is both a ketone and a primary amine. Amines can be produced from halogenoalkanes by reaction with ammonia. Ketones can be produced by the oxidation of secondary alcohols. Alkenes react by addition. The reaction of bromine water with an alkene adds an –OH and a Br, so this is the first step.

a Step 1: react but-2-ene with bromine water: CH3CH=CHCH3 → CH3CH(OH)CHBrCH3 Step 2: heat under reflux with acidified potassium dichromate(VII) CH3CH(OH)CHBrCH3 → CH3COCHBrCH3 Step 3: react the product with excess concentrated ammonia CH3COCHBrCH3 → CH3COCH(NH2)CH3 e

The question asks for reagents, conditions and intermediate compounds, so equations are not needed.

b The product is chiral, but the reaction is not stereospecific. Therefore, both enantiomers will be produced in equal quantities. The plane of polarisation of plane-polarised light will not be rotated by the resultant racemic mixture.

13 e

A structural formula should show all the double bonds. It is not necessary to write out a displayed formula.

a

Cl O H3C

C

C

O

H

C Cl

e

PCl5 reacts with –OH groups. The molecule has two –OH groups in the alcohol and the acid.

b

O H3C

C

CH(OH)

C O– Na+

O e

Sodium hydrogencarbonate neutralises the –COOH group. The product could be written as –ONa but not –O–Na because the latter implies a covalent bond.

c

O CHI3 and (Na+) –O

C

CH(OH)

C O–(Na+)

O e

This is the iodoform reaction with a methyl ketone. Since the reaction is carried out under alkaline conditions, the –COOH group will be deprotonated.

© Philip Allan Updates

Edexcel A2 Chemistry 107

Chapter 14 Organic analysis and synthesis

d

H3C C

N

NH

NO2

CH(OH) O

C

NO2 OH

e

This is a reaction of a carbonyl compound. The rest of the molecule is unchanged.

e H3C e

CH(OH)

CH(OH)

CH2OH

Lithium aluminium hydride reduces a carbonyl group to CH(OH) and an acid group to CH2OH. The result is butane-1,2,3triol.

14 Acid chlorides are made from acids by adding phosphorus pentachloride, so the first step must be the oxidation of the ethanol to ethanoic acid. Step 1

Reagents: an aqueous solution of potassium dichromate(VI) in sulfuric acid Conditions: heat under reflux Intermediate: ethanoic acid

Step 2

Reagent: phosphorus pentachloride Product: ethanoyl chloride

PCl5 Cr2O72–/H+(aq) CH3CH2OH —————— → CH3COOH → CH3COCl e

This problem is best solved working from the end — how is ethanoyl chloride made?

15 Based on the information given, the following conclusions can be drawn: I

Brady’s test: the compound does not contain a carbonyl group, so it must be either an acid, an ester or have two –OH groups.

I I

Bromine water test: the compound has at least one C=C group. Reaction with PCl5: as two moles of hydrogen chloride are produced per mol of the C4H8O2, there must be two –OH groups in the molecule.

I

Three NMR peaks mean that the hydrogen atoms are in three different environments. A peak at m/e = 57 in its mass spectrum shows a loss of 31 from the molecular ion peak at 88, which shows that the compound must have at least one –CH2OH group.

Possible structures containing a –CH2OH group are: I

Based on but-1-ene carbon chain — for example, CH(OH)=CHCH2CH2OH. This is not a symmetrical molecule, so the hydrogen atoms in the different groups are in different environments. This will give rise to six peaks in the NMR spectrum. The same is true whatever the relative positions of the two –OH groups are in the chain.

I

Based on but-2-ene chain: CH2OHC(OH)=CHCH3 has hydrogen in five environments. CH3C(OH)=CHCH2OH has hydrogen in five environments. CH2OHCH=CHCH2OH has hydrogen atoms in three environments as the molecule is symmetrical.

I

Based on methylpropene chain: CH3(CH2OH)C=CHOH has hydrogen in five environments. (CH2OH)2C=CH2 has hydrogen in three environments.

The unknown could be either CH2OHCH=CHCH2OH or (CH2OH)2C=CH2.

© Philip Allan Updates

Edexcel A2 Chemistry 108

Chapter 14 Organic analysis and synthesis

e

Whether the three peaks were split would distinguish between the two possibilities. The NMR spectrum of 1,4dihydroxybut-2-ene would have two of its peaks split in two; none of the peaks in the spectrum of the other molecule would be split. (Remember that the OH hydrogen atom does not cause splitting.) The statement ‘a formula’ implies that there could be more than one possible formula for the answer.

16 Carbon-containing side chains can be put on a benzene ring using the Friedel–Crafts reaction. In this case, the intermediate must be C6H5COCH3, which can then be reduced to C6H5CH(OH)CH3. Step 1 Reagent: ethanoyl chloride Conditions: anhydrous aluminium chloride Intermediate: phenylethanone, C6H5COCH3 Step 2

Reagent: lithium aluminium hydride (lithium tetrahydridoaluminate), LiAlH4 Conditions: dry ether solution followed by hydrolysis with dilute acid Product: 1-phenylethanol, C6H5CH(OH)CH3

CH3COCl 1 LiAlH4/2 H+(aq) C6H6 ———— → C6H5COCH3 ——————— → C6H5CH(OH)CH3

17 a The only difference between the two methods is that concentrated sulfuric acid is also an oxidising agent. This means that it will oxidise some of the hydrogen bromide to bromine and be reduced to sulfur dioxide. Both of these substances are gaseous under the experimental conditions and are poisonous, so the second method is less hazardous and hence poses less risk.

b The precautions (in addition to eye protection and a lab coat) that should be taken are: I

Wear gloves (as phosphoric acid is corrosive).

I

Carry out the preparation in a fume cupboard (as hydrogen bromide vapour is an irritant).

I

Heat the mixture using a water bath or an electric heater (as the two organic substances are flammable)

18 a Two –OH groups can be added on to the C=C by reaction with aqueous potassium manganate(VII). These –OH groups can then be oxidised to the desired product. Step 1 Reagent: aqueous potassium manganate(VII) Conditions: neutral or alkaline solution Intermediate: C6H5CH(OH)CH(OH)CH3 Step 2

Reagent: a solution of potassium dichromate and sulfuric acid Conditions: heat under reflux Product: C6H5COCOCH3

MnO4–(aq) Cr2O72–/H+(aq) C6H5CH=CHCH3 ————— → C6H5CH(OH)CH(OH)CH3 —————— → C6H5COCOCH3

b The products of the iodoform reaction are CHI3 and C6H5COCOO–. e

On acidification the ketoacid, C6H5COCOOH, is formed.

19 The theoretical yield is for 1 mol of benzene to form 1 mol of phenylamine, regardless of the number of steps. molar mass of benzene = (6 × 12.0) + 6 × 1.0 = 78.0 g mol–1 amount (moles) of benzene = mass/molar mass = 12.6/78.0 = 0.1615 mol = theoretical amount (moles) of phenylamine molar mass of phenylamine, C6H5NH2 = (6 × 12.0) + 7 × 1.0 + 14.0 = 93.0 g mol–1 theoretical mass of phenylamine = molar mass × moles = 93.0 × 0.1615 = 15.02 g actual mass of phenylamine = 6.75 g % yield = actual mass × 100/theoretical mass = (6.75 × 100)/15.02 = 44.9% © Philip Allan Updates

Edexcel A2 Chemistry 109

Chapter 14 Organic analysis and synthesis

e

You have to compare the actual mass with the theoretical mass of product or the actual moles with the theoretical moles of product in order to get the percentage yield. The percentage yield is not (mass of product × 100)/mass of reactant (which would give 6.75 × 100/12.6 = 53.6%). This is a common error.

20 The third step is wrong. Phosphorus pentachloride reacts with the OH in both the alcohol and the acid groups. The product of step 3 is (CH3)2CClCOCl, not (CH3)2C(OH)COCl. When ammonia is added in step 4, both the C–Cl chlorine and the COCl chlorine atoms are replaced. The final product would be (CH3)2CNH2CONH2, not (CH3)2C(OH)CONH2.

21 The first step is the addition of a cyanide ion to the carbon atom of the >C=O group. This carbon atom has three separate pairs of bonding electrons and so the reaction site is planar. The cyanide ion can attack from above or below the plane. The probabilities are the same, so both enantiomers will be produced in equal amounts — a racemic mixture.

22 a As the reaction is first order in 1-fluoro-1-iodoethane and in cyanide ions, it must be an SN2 reaction. This is a stereospecific mechanism in which the incoming nucleophile bonds opposite to the outgoing halide. The result is a single optical isomer with an inverted configuration. This means that if the cyanide ion reacts with the R-enantiomer, the S-enantiomer is produced.

b The bond that is broken is the weaker bond. Iodine is a much larger atom than fluorine, so the C–I bond is much weaker than the C–F bond. (The bond enthalpies are 484 kJ mol−1 for C–F and 238 kJ mol−1 for C–I.) Therefore, the cyanide ion replaces the iodine, not the fluorine.

23 The number is 32. Let a bead be represented by B and the two hydroxyacid chlorides by R and R': I I

First, B–R and B–R' are formed separately. These are mixed, divided into two and then reacted with B–R and B–R', which gives four products: B–R–R, B–R–R' and B–R'–R and B–R'–R'.

I

These are mixed, divided into four and then reacted again, which gives eight products: B–R–R–R, B–R–R–R', B–R–R'–R, B–R–R'–R', B–R'–R–R, B–R'–R–R', B–R'–R'–R and B–R'–R'–R'.

I

When the process is repeated, there are 16 products.

I

After the final repeat, there are 32 products.

The reasoning is that each polyester gives rise to two different polyesters on subsequent reaction.

© Philip Allan Updates

Edexcel A2 Chemistry 110

Chapter 14 Organic analysis and synthesis

Summary worksheet (www.hodderplus.co.uk/philipallan) 1 C

Element Carbon Hydrogen Oxygen

% 52.2 13 34.8

Divide by Ar 52.2/12.0 = 4.35 13.0/1.0 = 13 34.8/16.0 = 2.175

Divide by the smallest 4.35/2.175 = 2 13.0/2.175 = 5.98 ≈ 6 2.175/2.175 = 1

The error made in option A was to use the atomic number of oxygen, rather than the atomic mass. In option B the atomic numbers of both carbon and oxygen were used. In option D the atomic number of carbon was used, together with, correctly, the atomic mass for oxygen.

2 A The precipitate with 2,4-dinitrophenylhydrazine indicates that the unknown is a carbonyl compound. The precipitate with Fehling’s solution shows that it is an aldehyde, so it must be either option A or option B. The hydrogen atoms in the two CH3 groups in option A are in identical environments, so its NMR spectrum will have three peaks; that of option B will have four peaks, so B is incorrect. e

Option C would have three peaks in its NMR spectrum but would not react with Fehling’s solution. Option D would not react with either 2,4-dinitrophenylhydrazine or Fehling’s solution and its NMR spectrum would have five peaks.

3 B The –NH2 group is basic and would react with hydrochloric acid. Lithium aluminium hydride reduces polar π-bonds and so will not react with this compound. Therefore, option B is the correct response to this negative question. e

It is always sensible to check the other options. Bromine water reacts with >C=C< and ethanoic acid reacts with the alcohol group, so neither are correct. You should mark your question paper as: A reacts; B does not react; C reacts; D reacts or A ; B ; C ; D 

4 B This is best answered by putting ticks or crosses against each step for each response. A B C D

step 1  step 2  step 3  step 1  step 2  step 3  step 1  step 2  step 3  step 1  step 2  step 3 

Step 1: the formation of an ester from an acid requires the addition of an alcohol. Acid chlorides react with alcohols to form an ester. Step 2: the acids must be concentrated (unless the benzene ring is activated, as in phenol). Step 3: the hydrochloric acid must be concentrated.

5 D After adding the minimum of hot solvent, the mixture must be filtered while hot to remove any insoluble impurities. Therefore, options A and B are incorrect. Option C would not give any product, because the product is dissolved in the filtrate and is not the residue.

6 C Amines form complex ions with calcium ions and with transition metal ions, so options A and B are incorrect. Phenylamine is a base and would react with sulfuric acid, so option D is wrong. Potassium hydroxide is a suitable drying agent for basic substances.

7 D The best way to answer this question is to write either ‘necessary’ or ‘not necessary’ beside each response. Thus: I

A necessary (flammable substances used)

I

B necessary (sulfuric acid reacts exothermically with alcohols and carboxylic acids)

I

C necessary (sulfuric acid is corrosive)

I

D not necessary (as no irritant or toxic volatile substances are being used or formed)

© Philip Allan Updates

Edexcel A2 Chemistry 111

Chapter 14 Organic analysis and synthesis

8 A The best way to answer this question is to write ‘single ‘or ‘mixture’ next to each response. Thus: I

A (racemic) mixture

I

B single (with inversion)

I

C single (cis-addition, resulting in a single optical isomer)

I

D single (trans-addition, resulting in a single geometric isomer)

9 C The sodium hydrogencarbonate would react with any dissolved benzoic acid and all the hydrochloric acid catalyst to form non-volatile salts. The volatile methanol can then be removed by distillation.

© Philip Allan Updates

Edexcel A2 Chemistry 112

Unit 5 Transition metals, arenes and organic nitrogen chemistry

Practice Unit Test 5 Section A 1 C The oxidation number of oxygen is −2 (except in peroxides, superoxides and when bonded to fluorine). The oxidation numbers of vanadium and oxygen add up to +2 (the charge on the ion), so the oxidation number of the vanadium must be +4 because +4 − 2 = +2

2 A If an element appears on one side of an equation and a compound of that element on the other, then the reaction is redox. This means that options B, C and D are redox reactions, so option A is the correct answer to this negative question. This can be checked by working out the oxidation number of chromium. It is +6 in CrO42− and also +6 in Cr2O72−.

3 B The criterion for thermodynamic feasibility is that E°cell is positive. The larger the value, the larger is the equilibrium constant, K, for the reaction and so the more complete the reaction. It is a common misconception that E°cell must be bigger than a value such as 0.3 V or 0.03 V for a reaction to occur.

4 B Fe3+ and H2S are the reactants, so the S/H2S half-equation must be reversed. This means that the sign of its E° has to be reversed before it is added to the E° for the Fe3+/ Fe2+ half-equation. In option A, the wrong half-equation has been reversed. In option C, neither half-equation has been reversed. In option D the value of E° for the Fe3+/Fe2+ half-equation has been doubled because 2 electrons are needed when adding the reversed first half-equation to the second. Doubling a half-equation does not alter its E° value.

5 A One mole of iodate(V) produces three moles of iodine. Each mole of iodine requires two moles of thiosulfate ions, so the ratio of moles of IO3− to S2O32− is 1:6.

6 B Option A is a true statement — the more stable complex will always be formed under standard conditions. Option B is false. Seven particles are formed from two particles and, as they are all in solution, there will be an increase in ΔSsystem. It is advisable to check the remaining choices to ensure that your choice of B as the only false statement is correct. Option C is a true statement. The ammonia complex has six ligands and the EDTA complex has one ligand with six coordination sites, so both are octahedral. Option D is also true because a change in either the ligand or the oxidation state of the transition metal ion will cause a colour change. Here, it is the ligand that has changed.

7 D The energy levels of the 3d- and 4s-electrons are similar, so there is a steady increase in ionisation energy as these seven electrons are removed. The 3p-electrons have much lower energy and the jump occurs when one of these is removed, i.e. when the eighth electron is removed.

8 C There is no ligand in anhydrous copper(II) sulfate (option A), so there is no splitting of the d-orbitals. Therefore, there is no colour. In option B, the copper in [CuCl2]− is in the +1 state and has a d10 configuration. Splitting occurs but there is nowhere for a lower level d-electron to go, so the ion is colourless. The copper in [CuCl4]2− is in the +2 state and has a d9 configuration. There are ligands and there is (one) space in one of the higher energy split levels split levels, so the ion is coloured. In option D, the titanium compound is in the +4 state. There are no d-electrons to promote, so the ion is colourless.

9 B For a complex ion to be planar, there must be only four bonded pairs of electrons, so option C, which has three bidentate ligands, and option D, which has six monodentate ligands, cannot be correct. At A-level you should know that, that platinum(II) complexes are planar (B) and that chromium(III) complexes with four large ligands (A) are tetrahedral.

© Philip Allan Updates

Edexcel A2 Chemistry 113

Practice Unit Test 5

10 A Adding H+ ions increases their concentration and drives the equilibrium to the right. This will increase the value of E making it more positive. Adding OH− ions removes H+ ions and drives the equilibrium to the left, making E less positive. Note that adding H+ ions will make their concentration > 1, so the electrode potential is no longer a standard electrode potential.

11 C A delocalised double bond is longer than a localised double bond. Both are shorter than a single bond. Ethane is singly bonded and has the longest C–C bond, so option A is incorrect. Benzene has a delocalised π-system, so the C-to-C bond length is longer than that in ethene, which has a localised π-bond.

12 A Phenol has an activated benzene ring because the lone pair of pz-electrons in the oxygen atom overlaps, to some extent, with the delocalised π-system of the benzene ring. This means that it (and phenylamine) will react with bromine water and that three bromine atoms will substitute into the ring in the 2, 4 and 6 positions.

13 A Phenylethanone is one of the products, but not the only product, so option A is not true and is the correct answer to this negative question. The Freidel–Crafts reaction is an example of a substitution reaction and so there must be two products. In this example, the second product is hydrogen chloride. You would be wise to check that the other statements are true. Arenes react with ethanoyl chloride in an electrophilic substitution reaction, so option B is true. Aluminium chloride is the catalyst for Freidel–Crafts reactions and arenes react with electrophiles to form positively charged intermediates, so options C and D are also correct.

14 D The best way to answer this type of question is to put a tick or a cross for each reaction alongside the four responses. Smoky flame A B C D eG

   

Brady’s reagent Tollens’ reagent Solubility in NaOH

   

   

   

All aromatic compounds burn with a smoky flame, so all the choices have a tick.

G

All carbonyl compounds give a precipitate with Brady’s reagent, so all options have a tick.

G

Only aldehydes give a silver mirror with Tollens’ reagent, so only options C and D have ticks.

G

Phenols are sufficiently acidic to react with alkalis to give a soluble ionic compound, so only option D has a tick.

Option D is the only compound to have four ticks, so D is the correct answer.

15 A Phenol has an –OH group, as do alcohols. However, unlike an alcohol, phenol does not react with a carboxylic acid to form an ester. Its benzene ring is activated and it reacts with chlorine water (B) in the same way that it reacts with bromine water. The –OH group is sufficiently acidic for it to react with sodium (C) and with sodium hydroxide (D).

16 B Phenylamine is a base and so it reacts with acids (A), but not with alkalis (B). Therefore, option B is the correct response. As usual, the other options should be checked. Amines react with both acid chlorides (C) and with halogenoalkanes (D). This confirms that option B is the correct answer to this negative question.

17 D The double bonds are delocalised in benzene, so option A is incorrect. It is tempting to give option B as the right answer. However, in the reaction with hydrogen, energy is released not absorbed. Benzene reacts by electrophilic substitution but also burns in air and can undergo addition reactions with hydrogen, so option C is also incorrect.

18 C Ethenol does not exist, so option A is incorrect. (This polymer is made by first polymerising vinyl acetate and then carrying out a transesterification reaction with methanol). It is not a natural polymer, therefore it is not biodegradable, and so option B is incorrect.

© Philip Allan Updates

Edexcel A2 Chemistry 114

Practice Unit Test 5

It is not rigid (D) because of the difficulty of forming hydrogen bonds — the O of the –OH groups is sterically hindered from approach by the H of the –OH groups in other chains. The large number of –OH groups in the chain allow it to form thousands of hydrogen bonds with water, making it water-soluble. Therefore, option C is the correct answer.

19 B Steam distillation is used to remove a volatile oil from a complex mixture. Oils are insoluble in water. The higher the boiling temperature of the oil, the smaller is the proportion of the oil in the mixture of oil and water distilling over, so the less efficient the process. However, it works well with oils that boil up to about 200°C, so option C is not correct.

20 B The hydrogen bromide in step 1 must be gaseous. This rules out options C and D. Step 2 is nucleophilic substitution and takes place with an aqueous base. This is a further reason for ruling out option D. Step 3 requires oxidation of a secondary alcohol to a ketone and requires acidified Cr2O72− ions. This rules out options A and C. Step 4 needs both HCN and CN− ions, which again rules out options A and D. Perhaps the best way to work out the correct answer is to put a tick or a cross for each step beside each response. Thus: Step 1

Step 2

Step 3

Step 4

   

   

   

   

A B C D

Section B 21 a The first step is the reduction of S2O82– ions by Fe2+ ions: 2Fe2+(aq) + S2O82–(aq) → 2Fe3+(aq) + 2SO42–(aq) E° for step 1 = +2.01 + (–0.77) = +1.24 V  This value is positive, so the reaction is feasible  and Fe2+ ions will reduce persulfate ions. The second step is the reduction, by I – ions, of the Fe3+ ions formed in step 1 back to Fe2+ ions: 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(s) E° for step 2 = +0.77 + (–0.54) = +0.23 V This value is positive, so the second step is also feasible . The Fe2+ ions are regenerated and act as a catalyst for the reaction . e

In the first step, the Fe2+ ions reduce the persulfate, S2O82–, ions. The value of E° is obtained by reversing the third equation and adding it to the first: 2Fe2+(aq)  2Fe3+(aq) + 2e– 2–(aq)

S2O8

+

2e–

 2SO4

2–(aq)

E° = –(+0.77) = –0.77 V E° = +2.01 V

In the second step, the iodide ions reduce the Fe3+ ions back to Fe2+ ions. The standard potential for this process is obtained by adding the third half-equation to the reversed second equation: 2Fe3+(aq) + 2e–  2Fe2+(aq) 2I–(aq)

 I2(s) +

2e–

E° = +0.77 V E° = –(+0.54) = –0.54 V

b Fe2+: [Ar] 3d6  Zn2+: [Ar] 3d10  The d-orbitals in the Fe2+ ions are partially filled, whereas those in the Zn2+ are fully occupied .

© Philip Allan Updates

Edexcel A2 Chemistry 115

Practice Unit Test 5

e

A transition metal is defined as a metal whose ions have d-orbitals that are not fully filled. Do not say that it is the atoms that have incompletely filled d-orbitals.

c (i) [Fe(H2O)6]3+ + H2O [Fe(H2O)5OH]2+ + H3O+ . e

In this question, the simpler equation below would be allowed, as you are asked for an equation showing the acidity of the solution, not the type of reaction: [Fe(H2O)6]3+  [Fe(H2O)5OH]2+ + H+ Remember that it is the formation of H3O+ (or H+) ions that makes a solution acidic.

(ii) [Fe(H2O)6]3+ + 6CN– → [Fe(CN)6]3– + 6H2O  (iii) The d-orbitals are split by the water ligands into three of lower energy and two of higher energy . When white light is shone into the solution, an electron is promoted from one of the lower levels to a higher level , absorbing some of the frequencies  in the white light. This makes the solution appear

the complementary colour to the colour of the light absorbed . e

You must not state that the colour is caused by the electron falling to a lower energy level and emitting light. This answer would earn no marks.

(iv) Iron is used as the catalyst in the Haber process, in which nitrogen and hydrogen react to form ammonia

. e

The equation N2 + 3H2  2NH3 would score the mark. Just the words ‘Haber process’ would not be sufficient.

(v) A ligand must have a lone pair of electrons that it can use to form a dative bond with a transition metal ion. If a ligand is bidentate, it has two sites, each with a lone pair of electrons. 1,2-diaminoethane has the formula H2NCH2CH2NH2. Each of the two nitrogen atoms has a lone pair of electrons  and so can form a bond at each site with an Fe3+ ion. The two nitrogen atoms are separated by two carbon atoms and so a five-membered ring is formed with the iron ion, allowing it to be a bidentate ligand . The geometry of this ring is stable. e

This question is marked with an asterisk, which shows that quality of written communication is being tested. You must ensure that you have expressed yourself clearly or else you will lose the ‘chemistry’ mark.

d amount (moles) of sodium thiosulfate = concentration × volume in dm3 = 0.200 × 27.3/1000 = 0.00546 mol  amount of I2 produced = –12 × 0.00546 = 0.00273 mol amount of Fe3+ = 2 × 0.00273 = 0.00546 mol  molar mass of FeCl3 = 55.8 + (3 × 35.5) = 162.3 g mol–1  mass of FeCl3 = moles × molar mass = 0.00546 × 162.5 = 0.886 g  purity = 0.886 × 100/1.00 = 88.6% 

22 a (i) The hydrogen atoms of the –NH2 group are δ+ and can form hydrogen bonds with the δ− oxygen atoms of water molecules ; the lone pair of the electrons on the δ− nitrogens of the –NH2 groups can also

form hydrogen bonds with the δ+ hydrogens of water . In addition, ethylamine is a stronger base than ammonia and will react to some extent to form ions , as in the equation. C2H5NH2 + H2O  C2H5NH3+ (aq) + OH−(aq) e

You must state not only that hydrogen bonds are formed between ethylamine and water, but also explain how the bonds arise.

© Philip Allan Updates

Edexcel A2 Chemistry 116

Practice Unit Test 5

(ii) At first, ethylamine (deprotonates the hydrated copper ions) and a pale blue precipitate  (of hydrated copper hydroxide) is formed. With excess ethylamine, this precipitate dissolves to form a deep blue solution  (of the copper(II) ethylamine complex). e

The question only asked for observations. In the answer above, the reasons are given in brackets

b Step 1: heat the phenylamine with a mixture of concentrated hydrochloric acid  and tin . This will reduce the nitrobenzene to phenylammonium ions, C6H5NH3+. Step 2: add aqueous sodium hydroxide . This deprotonates the phenylammonium ions and phenylamine is formed.

c (i) Sodium nitrite and dilute acid . (ii) At a higher temperature the diazonium compound will decompose . If the temperature falls below 0°C, the rate of the reaction is too slow and an insignificant amount of the diazonium salt will be produced .

(iii) N

OH

N

O–)

( allow

Correct N=N  rest of molecule 

23 a (i)

Step 1

CH2

H2C

✔ (both arrows)

Br

+ CH2

H2C

Br

✔

✔

Br Step 2

+ – CH2 + Br

H2C

•Br– •

BrH2C

CH2Br

Br

(ii) Production of electrophile: Br2 + AlBr3 → Br+ + AlBr4–  Step 1

H

✔ Br+

✔

Step 2 +

e

H

+

✔

Br

AlBr4–

Br

Br + HBr + AlBr 3

It is sensible to write each mechanism as two separate steps. Do not forget to add the equation for the generation of the electrophile in benzene reaction mechanisms.

b The first reaction step is similar for ethene and for benzene because it is the addition of a Br+ electrophile. The second step with ethene is further addition, resulting in an overall addition reaction. With benzene the second step is the loss of H+. The result is a substitution reaction. The reason for the difference is that by losing H+, the π-electrons in the ring become fully delocalised again

© Philip Allan Updates

, thus regaining the resonance Edexcel A2 Chemistry 117

Practice Unit Test 5

stabilisation energy . There cannot be any delocalisation with ethene , so the addition of Br– is energetically more favourable as the second step. e

Do not say that benzene does not react by addition because of the stability of the benzene ring. The first step is addition, which is followed by elimination to regain this stability.

c (i) C6H6 + HNO3 → C6H5NO2 + H2O  (ii)

O C

O +

H 3C

CH3

C

✔ + HCl

Cl

(iii)

SO3H + SO3

✔

or SO3H + H2S2O7

e

+ H2SO4

Fuming sulfuric acid can be regarded as a solution of sulfur trioxide, SO3, in sulfuric acid or as oleum, H2S2O7, which then acts as a source of sulfur trioxide. C6H6 can be used as the formula for benzene and C6H5 in the product (as in (i)).

d (i)

H

H

O

O Br

Br + 3HBr

+ 3Br2

✔

Br e

Do not use C6H2(OH)Br3 for the formula of the product. Do not forget that this is a substitution reaction and that three hydrogen atoms are substituted by three bromine atoms. The result is that three moles of hydrogen bromide are also produced. The equation below is incorrect: H

H O

O Br

Br

+ 112 Br2

Br

© Philip Allan Updates

Edexcel A2 Chemistry 118

Practice Unit Test 5

(ii) In phenol, the lone pair of pz-electrons on the oxygen atom overlaps with the six delocalised pz-electrons (one from each carbon atom) to form a delocalised π-system of eight electrons over seven atoms .

The electron density in the π- cloud around each carbon atom is greater than it is in benzene. This makes phenol more susceptible to electrophilic attack , and so phenol reacts with bromine water whereas benzene reacts with liquid bromine only in the presence of a catalyst. e

You not only have to explain how the lone pair of electrons on the oxygen becomes part of the delocalised system, but also that this increases the electron density, making electrophilic attack easier.

Section C 24 a A condensation polymer is the product that is formed when monomers join together with loss of a small molecule , such as water or hydrogen chloride, at each link.

b (i) O

OH C

NH2

✔

✔ and O

C

NH2 OH

e

As a structural formula has been asked for, you should draw the carboxylic acid group showing the >C=O and –OH groups separately, not as COOH.

(ii)

O C

N

✔

H

e

C

N

O

H

The repeat unit of a polyamide has two oxygen atoms and two nitrogen atoms. The bonding in the peptide link should be clearly shown.

c (i) A chiral molecule is a molecule that is not superimposable on its mirror image . CH2OH H2N

C*

COOH

✔

H

(ii)

CH2OH +

H3N

C H

e

O

✔

C O–

Make sure that you put the ‘+’ sign on the nitrogen atom of the zwitterion.

© Philip Allan Updates

Edexcel A2 Chemistry 119

Practice Unit Test 5

The zwitterion has a positive charge at one end and a negative charge at the other. This means that, as well as quite strong London forces, stronger ionic forces of attraction exist between zwitterions .

As well as London forces , molecules of pentanoic acid have hydrogen bonds between them but these are weaker  than the ionic forces between zwitterions. More energy is needed to overcome  the

stronger attractive forces in serine than in pentanoic acid and so serine has the higher melting temperature. e

You must make it clear that the ionic forces are between zwitterions. Vague statements such as ‘there is strong ionic bonding in zwitterions’ will not score the first mark. Another common error is failure to state the relationship between stronger forces and the amount of energy needed to overcome these forces — the fourth mark.

(iii) I + H+(aq)

CH2OH +

H 3N

CH

O

✔

C OH

e

The –NH2 group is basic and so it accepts a proton (compare with the reaction of ammonia with H+ ions). Make sure that you put the ‘+’ sign on the nitrogen atom. I

+ OH−(aq)

CH2OH H 2N

CH

O

✔

C –

O e

The –COOH group is acidic and so is deprotonated by the base. I

+ PCl5

CH2Cl H2N

CH

✔

O

C Cl

e

✔

Phosphorus pentachloride reacts with compounds containing an –OH group. Serine has two, so the product has a –CH2Cl group and a –COCl group. A clue is that there are 2 marks for this answer.

d Step 1: oxidation of the primary alcohol to a carboxylic acid Reagents: potassium dichromate(VI) and dilute sulfuric acid  Product: CH2=CHCH2COOH  Step 2: addition of HCl to the double bond (Markovnikoff) Reagent: hydrogen chloride  Product: CH3CHClCH2COOH  Step 3: substitution of Cl by OH− Reagent: (aqueous) sodium hydroxide  Product: CH3CH(OH)CH2COOH e

If you cannot see the route straightaway think back from the product to the starting material. How do you make an acid? Answer — by oxidising a primary alcohol. This is present in the starting compound. How do you make an alcohol? Answer — by hydrolysing a halogenalkane. This can be made from an alkene by addition of H–halogen. You must oxidise the primary alcohol before putting an –OH group on carbon number 3.

© Philip Allan Updates

Edexcel A2 Chemistry 120