GCE A2ChemistrySummer 2008

Mark SchemesIssued: October 2008

iii

NORTHERN IRELAND GENERAL CERTIFICATE OF SECONDARY EDUCATION (GCSE) AND NORTHERN IRELAND GENERAL CERTIFICATE OF EDUCATION (GCE)

MARK SCHEMES (2008)

Foreword

Introduction

Mark Schemes are published to assist teachers and students in their preparation for examinations. Through the mark schemes teachers and students will be able to see what examiners are looking for in response to questions and exactly where the marks have been awarded. The publishing of the mark schemes may help to show that examiners are not concerned about finding out what a student does not know but rather with rewarding students for what they do know.

The Purpose of Mark Schemes

Examination papers are set and revised by teams of examiners and revisers appointed by the Council. The teams of examiners and revisers include experienced teachers who are familiar with the level and standards expected of 16- and 18-year-old students in schools and colleges. The job of the examiners is to set the questions and the mark schemes; and the job of the revisers is to review the questions and mark schemes commenting on a large range of issues about which they must be satisfied before the question papers and mark schemes are finalised.

The questions and the mark schemes are developed in association with each other so that the issues of differentiation and positive achievement can be addressed right from the start. Mark schemes therefore are regarded as a part of an integral process which begins with the setting of questions and ends with the marking of the examination.

The main purpose of the mark scheme is to provide a uniform basis for the marking process so that all the markers are following exactly the same instructions and making the same judgements in so far as this is possible. Before marking begins a standardising meeting is held where all the markers are briefed using the mark scheme and samples of the students’ work in the form of scripts. Consideration is also given at this stage to any comments on the operational papers received from teachers and their organisations. During this meeting, and up to and including the end of the marking, there is provision for amendments to be made to the mark scheme. What is published represents this final form of the mark scheme.

It is important to recognise that in some cases there may well be other correct responses which are equally acceptable to those published: the mark scheme can only cover those responses which emerged in the examination. There may also be instances where certain judgements may have to be left to the experience of the examiner, for example, where there is no absolute correct response – all teachers will be familiar with making such judgements.

The Council hopes that the mark schemes will be viewed and used in a constructive way as a further support to the teaching and learning processes.

v

CONTENTS

Page

A2 1: Module 4 1

A2 2: Module 5 7

A2 3A: Module 6A 15

ADVANCEDGeneral Certificate of Education

2008

MARKSCHEME

Chemistry

Assessment Unit A2 1assessing

Module 4: Further Organic, Physical and Inorganic Chemistry

[A2C11]

THURSDAY 22 MAY, MORNING

1

AVAILABLEMARKS

2

Section A

1 D

2 A

3 C

4 D

5 C

6 D

7 A

8 D

9 D

10 D

[2] for each correct answer [20] 20

Section A 20

3

AVAILABLEMARKS

Section B

11 (a) red precipitate [1]

(b) (i) rotates plane of polarised light [1] (ii) 4 [1]

(c) RCH = NNH

NO2

O

NO2 [2]

(d) (i) C6H12O6 → 2C2H5OH + 2CO2 [2]

(ii) it lowers the activation energy [1] 8

12 (a) 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O [2]

(b) (i) NO = 2 [1] O2 = 1 [1] Rate = k[NO]2[O2] [1] [3]

(ii)

[2]

(iii) Rate = k[NO]2[O2] 6 × 10–4 = k × (1 × 10–3)2(1 × 10–3) k = 6 × 10–4/1 × 10–9

k = 6 × 105 dm6 mol–2 s–1 [3]

(c) (i) pushes equilibrium to right [1] (dependent on explanation) to increase number of molecules (and increase pressure) [1] [2]

(ii) mole fraction N2O4 = 0.2 partial pressure N2O4 = 0.2 × 125 = 25 kPa mole fraction NO2 = 0.8 partial pressure NO2 = 0.8 × 125 = 100 kPa

Kp = ΡNO2

2/PN2O4

= 1002/25 = 10 000/25 = 400 kΡa [4]

[NO]

time

rate

[NO]

AVAILABLEMARKS

4

(iii) Kp value increases (as equilibrium moves to right) [1] (dependent on explanation) to remove excess thermal energy/ drive towards endothermic change [1] [2]

(d) (i) nitric acid is stronger nitrogen is in +5 oxidation state in nitric acid [1] nitrogen is in +3 oxidation state in nitrous acid [1] nitrogen has a higher oxidation state in HNO3 than in HNO2 [1] Any two from three [2]

(ii) 3V3+ + H2O + NO3– → 3VO2+ + 2H+ + NO [2]

(iii) acidified iron(II) sulphate iron + 2 → +3 is oxidised [1] nitrogen +3 → +2 is reduced [1]

acidified potassium manganate(VII) manganese +7 → +2 reduced [1] nitrogen +3 → +5 oxidised [1] [4] 26

13 (a) (i) Al2Cl6 [1]

(ii) phosphorus [1]

(iii) silicon [1]

(b) (i) reacts with acids and alkalis [1]

(ii) Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O [2] 6

14 (a) (i)

[1]

(ii) CH2 — O — COC17H31

CH— O — COC17H31

CH2— O — COC17H31 [2]

(iii) sunflower oil [1] is more unsaturated/contains more C C double bonds [1] iodine value reflects degree of unsaturation [1] [2]

CH2

H H

C C

H H

C C

CH3(CH2)4 (CH2)7 COOH

5

AVAILABLEMARKS

(iv) known mass [1] add Wij’s solution and place in dark, add KI [1] prepare a blank [1] titrate with standard thiosulphate solution using starch indicator (blue black to colourless) [1] [4]

Quality of written communication [2]

(b) (i)

[2]

(ii) moles NaOH = 17.5 × 0.1/1000 = 1.75 × 10–3

moles citric acid = 1.75 × 10–3/3 = 5.83 × 10–4 = 0.000583 moles in 1 litre = 40 × 0.000583 = 0.0233 RMM citric acid = 192 concentration = 0.0233 × 192 = 4.48 (gdm–3) [4] 17

15 (a) C20H27O11N + 2H2O → C6H5CHO + 2C6H12O6 + HCN [2]

(b) (i)

[3]

(ii) nucleophilic addition [2]

(c) Ka = [H+][C6H5COO–]/[C6H5COOH] = [H+]2/[C6H5COOH] [H+]2 = 6.3 × 10–5 × 0.001 [H+] = 2.51 × 10–4 moldm–3

pH = –log[H+] pH = 3.6 [3]

(d) (i) propanone is more soluble [1] (dependent on explanation) propanone has smaller (hydrophobic) hydrocarbon chain [1] (or vice versa for heptan-2-one) [2]

(ii) heptan-2-ol [1] 13

Section B 70

Total 90

CH2CO2Na

HO C CO2Na + 3H2O

CH2CO2Na

+ 3NaOH →

CH2COOH

HO C COOH

CH2COOH

HC

O

O

+ CN– →

H C

O

CN

O–

H C

O

CN

OH

H+

→

ADVANCEDGeneral Certificate of Education

2008

MARKSCHEME

ChemistryAssessment Unit A2 2

assessing

Module 5: Analytical, Transition Metals andFurther Organic Chemistry

[A2C21]

WEDNESDAY 28 MAY, AFTERNOON

7

Quality of Written Communication:

2 marks The candidate expresses ideas clearly and fluently through well-linked sentences and paragraphs. Arguments are generally relevant and well-structured. There are few errors of grammar, punctuation and spelling.

1 mark The candidate expresses ideas clearly, if not always fluently. Arguments may sometimes stray from the point. There may be some errors of grammar, punctuation and spelling, but not such as to suggest a weakness in these areas.

0 marks The candidate expresses ideas satisfactorily, but without precision. Arguments may be of doubtful relevance or obscurely presented. Errors in grammar, punctuation and spelling are sufficiently intrusive to disrupt the understanding of the passage.

8

9

AVAILABLEMARKS

Section A

1 A

2 B

3 B

4 C

5 D

6 A

7 C

8 B

9 D

10 D [20] 20

Section A 20

AVAILABLEMARKS

10

Section B

11 (a) HNO3 + 2H2SO4 → NO2+ + H3O

+ + 2HSO4– [2]

(b) removes protons from H2SO4 [1]

(c) regenerated [1] 4

12 (a) binding or complexation of a multidentate ligand [1] donates more than one pair of electrons/more than one coordination bond [1] octahedral [1] (any two from three) [2]

(b) titrate known volume of Mg2+ solution [1] Buffer to pH 10 [1] indicator [1] eriochrome black solachrome black Colour change [1] (eriochrome) red to blue (solachrome) red to blue ratio (edta to Mg2+) is 1:1 [1] (any four from five) [4] 6

13 (a) (i) Zn + 2Fe3+ → + 2Fe2+ + Zn2+ [1]

(ii) DE

φ

= +0.32 (so reaction should proceed) [1]

(b) (i) colourless to pink [2]

(ii) 5Fe2+ + MnO4– + 8H+ → Mn2+ + 5Fe3+ + 4H2O [2]

(iii) mole of manganate(VII) solution 21.6 × 0.1 × 10–3

= 2.16 × 10–3

= 0.00216 in 25 cm3 moles of Fe2+ = 0.00216 × 5 = 0.0108 in 1 dm3 moles of Fe2+ = 40 × 0.0108 = 0.432 mol [1]

(iv) FeSO4 = 56 + 32 + 64 = 152 mass iron(II) sulphate = 0.432 × 152 g = 65.66 g [1]

(v) volume of manganate(VII) solution ≡ iron(III) ions = 22.8 – 21.6 = 1.2 cm3 moles of manganate(VII) = 1.2 × 0.1 × 10–3

= 1.2 × 10–4

moles of Fe2+ = 5 × 1.2 × 10–4

= 6 × 10–4 mol

moles of Fe2+ in 1 dm3

= 40 × 6 × 10–4 mol = 2.4 × 10–2 mol

11

AVAILABLEMARKS

Fe2(SO4)3 = 2 × 56 + 3 × 32 + 12 × 16 = 112 + 96 + 192 = 400

2Fe3+ ≡ 1 Fe2(SO4)3 ∴ 1.2 × 10–2 = 0.012 mol Fe2(SO4)3

Mass of iron(III) sulphate = 0.012 × 400 = 4.8 g [3] 11

14 (a) (i) Primary structure is the order of its amino acids [1] secondary structures: alpha-helix or beta-pleated sheet [1] tertiary structure is the (full 3-dimensional) folded structure of the polypeptide chain [1] [3]

(ii) COOH/COO– [1] NH2 [1] [2]

(iii) place mixture at origin of chromatographic paper or plate develop using solvent 1 [1] turn paper of plate through 90˚ and develop using solvent 2 [1] locate spots using (ninhydrin) [1] Comparison of Rf values for leucine [1] [4]

Quality of written communication [2]

(b) (i)

Four different groups attached to a carbon atom [1] [3]

(ii) Transfer of H+ from COOH to NH2 [1]

(iii) Strong attraction between + and – charges [2]

(iv)

[2]

(v) bubbles (of nitrogen) [1] 20

CH3

* CH — OH

+H3N — CH — C — O–

* O [2]

CH3

CHOH

NH2 — CH — COOH + HNO2 →

CH3

CHOH

HO — CH — COOH + N2 + H2O

AVAILABLEMARKS

12

15 (a) (i) nucleophilic substitution [1]

(ii) C4H9Br + NH3 → C4H9NH2 + HBr or C4H9NH3+ Br– [1]

(iii) δ+ δ– C —Br polarised

lone pair on • •N/

• •N H3 attracted to

δ+C

bond formed C—N, bond broken C—Br [3]

(b) (i) lithal [1]

(ii) C3H7CN + 4 [H] → C3H7CH2NH2 [2]

(c) (i) reaction with water to split the molecule [1]

(ii)

(reaction coordinate) [3]

(iii) C3H7CN + NaOH + H2O → C3H7CO2Na + NH3 [2] 14

16 (a) (i) CuCO3.Cu(OH)2 = 222 % Cu = 128/222 × 102 = 57.7% [3]

(ii) CuCO3.Cu(OH)2 → 2CuO + H2O + CO2 [2]

(iii) CuO + C → Cu + CO [1] or 2CuO + C → 2 Cu + CO2

(b) (i) Cu(OH)2, [Cu(NH3)4(H2O)2] 2+ [2]

blue → deep blue [2] [4] ppt (solution)

(ii) [CuCl4]2– [1]

blue → green/yellow [1] [2]

enthalpy

products

reactants

uncatalysed

catalysed

13

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(iii) The overall stability constant is simply the equilibrium constant for the total reaction:

[Cu(H2O)6]2+ + 4NH3 [Cu(H2O)2(NH3)4]2+ + 2H2O [1]

is given by this expression:

overall Kstab = [Cu(NH3)4(H2O)2

2+][Cu(H2O)6

2+][NH3]4 [1]

which is large, so the formation of the product is favoured.

It is generally true that the higher the stability constant of a complex, the more readily the product (in this case [Cu(H2O)2(NH3)2]2+), forms [1]

The second reaction,

[Cu(H2O)2(NH3)4]2+ + 4Cl– [CuCl4]2- + 2H2O + 4NH3 [1]

…has an overall Kstab which again is large therefore yellow [CuCl4]2– forms readily [1]

To a maximum of [3] [3] 15

Section B 70

Total 90

ADVANCEDGeneral Certificate of Education

2008

ChemistryAssessment Unit A2 3A

assessing

Module 6A: Synoptic Paper

[A2C31]

FRIDAY 30 MAY, MORNING

MARKSCHEME

15

Quality of written communication:

2 marks The candidate expresses ideas clearly and fluently through well-linked sentences and paragraphs. Arguments are generally relevant and well-structured. There are few errors of grammar, punctuation and spelling.

1 mark The candidate expresses ideas clearly, if not always fluently. Arguments may sometimes stray from the point. There may be some errors of grammar, punctuation and spelling, but not such as to suggest a weakness in these areas.

0 marks The candidate expresses ideas satisfactorily, but without precision. Arguments may be of doubtful relevance or obscurely presented. Errors in grammar, punctuation and spelling are sufficiently intrusive to disrupt the understanding of the passage.

16

17

AVAILABLEMARKS

Section A

1 (a) bromine is corrosive [1] wear gloves [1] or bromine is poisonous etc. [1] fume cupboard [1] [2]

(b) (i) propanone very soluble in water/miscible [1]

(ii) reaction is exothermic/vigorous [1]

(c) (i) moles of propanone

grams = 0.5 × 0.79 = 0.395 g CH3COCH3 = C3H6O = 36 + 16 + 6 = 58 Mol = 0.395/58 = 0.0068 [1]

moles of bromine

grams = 3 × 3.1 = 9.3 g Br2 = 2 × 80 = 160 Mol = 9.3/160 = 0.058 [1]

moles of sodium hydroxide

Mol = 60 × 10–3 × 1.0 = 0.06 [1]

(ii) 1 CH3COCH3 = 0.0068 5 NaOH = 0.30 4 Br2 = 0.23 excess are NaOH and Br2 [1]

(iii) 1 CH3COCH3 → 1 CBr4 0.0068 mol → 0.0068 mol CBr4 = 12 + 4 × 80 = 332 1.7 g = 1.7/332 mol = 0.0051 % yield = 0.0051/0.0068 × 100 = 75% [2]

(d) Buchner flask [1] Buchner funnel [1] suction applied to flask [1]

(e) wash further with water [1] add silver nitrate solution [1] no (cream) precipitate formed [1]

(f) melting point – lowered [1] boiling point – widened [1]

Quality of written communication [2] 20

Section A 20

18

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Section B

2 (a) (i) step 1 tin + concentrated hydrochloric acid [1] step 2 ethanoyl chloride [1] step 3 concentrated nitric and sulphuric acids [1] step 5 sodium nitrite and hydrochloric acid [1]

(ii)

[1] (b) below 5 ºC/ice [1] to prevent the decomposition of the diazo compound [1]

(c) extended conjugation/delocalised electrons/energy levels close together [1] electrons easily promoted/rise to a higher energy level/excited [1] part of (white) light removed [1]

(d) Van der Waals forces: large molecule/large mass/large number of electrons [1] polar/dipolar forces: nitro groups [1] hydrogen bonding: hydroxyl groups [1] 13

3 (a) butanedioic acid/butane 1, 4 dioic acid [2]

(b) (i) tetramethylsilane [1] Si(CH3)4 [1]

(ii) ratio of integration curve 2:1 signal at 2.8 is (CH2)2 = 4H signal at 4.9 is 2COOH = 2H [2]

(iii) in –COOH the H is next to oxygen O is electronegative/electron withdrawing H is deshielded [2]

(iv) the CH2 groups are chemically equivalent [1] in the same environment [1] the H on the –COOH has no adjacent Hs [1] (any two from three) [2] (c) (i) base peak [1]

(ii) 73 C3H5O2+ [1]

100: C4H4O3+ [1]

(d) (i) CH2COOH CH2COOCH3 | +2CH3OH → | +2H2O CH2COOH CH2COOCH3 [2]

(ii) catalyst/protonation of an oxygen atom [1]

(iii) a liquid is more volatile than a solid [1]

OH

19

AVAILABLEMARKS

(e) (i) produces two protons (in solution) [1]

(ii) pKa = 4.2 Ka = 6.3 × 10–5

K = [H+][X–]

[HX] = 6.3 × 10–5

[H+]2 = K[HX] = 6.3 × 10–5 × 0.1 = 6.3 × 10–6

[H+] = 2.5 × 10–3

pH = 2.6 [3] 21

4 (a) C19H16O4 [1]

(b) (i) four different groups attached to the same carbon atom or molecule not superimposable on its mirror image [1]

(ii) circle the carbon atom with one hydrogen attached to it [1]

(iii) lock and key mechanism [1] only one isomer fits [1] [2]

(c) (i) carbonyl/C O [1]

(ii)

[3]

(d) (i) hydroxyl group/OH [1]

(ii) hydrogen chloride [1] concentrated ammonia solution white fumes/smoke [1] [2] 12

5 (a) Na3[Co(NO2)6] = 3 × 23 + 59 + 6 × 46 = 69 + 59 + 276 = 404

% Co = 59/404 × 100 = 14.6% [3]

(b) (i) provides H+ ions (in the first equation) [1]

(ii) Co +2 goes to Co +3 N +3 goes to N +2 Cobalt is oxidised and nitrogen is reduced [3]

(iii) pink to yellow [1]

NH2NH

C O +

NO2

NO2

C NNH

NO2 + H2O

NO2

→

20

AVAILABLEMARKS

(c) (i) octahedral [1]

(ii)

[1]

(d) (i) K3[Co(NO2)6] [1]

(ii) flame test nichrome/platinum wire [1] concentrated hydrochloric acid/colourless or blue Bunsen flame [1] lilac/purple flame [1] [3] 14

Section B 60 Total 80

Co

NO2

NO2

NO2O2N

NO2

O2N