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® IGCSE is the registered trademark of Cambridge International Examinations.

CAMBRIDGE INTERNATIONAL EXAMINATIONS

Cambridge International Advanced Subsidiary and Advanced Level

MARK SCHEME for the March 2016 series

9701 CHEMISTRY

9701/42 Paper 4 (A Level Structured Questions), maximum raw mark 100

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the March 2016 series for most Cambridge IGCSE

®

and Cambridge International A and AS Level components.

www.dynamicpapers.com

Page 2 Mark Scheme Syllabus Paper

Cambridge International AS/A Level – March 2016 9701 42

© Cambridge International Examinations 2016

Question Answer Mark

1 (a)

2

(b) (i) sp2 1

(ii) x = 60 / C60H60 1

(c) (i) reaction 1: Cl 2 and UV light; reaction 2: Al Cl 3, Cl 2 (NOT aqueous);

11

(ii) (free) radical substitution 1

(iii) CCl3

Cl

Cl

Cl

Cl

Cl

CCl3

or

1






2 (a) (i) Ca2+(g) + 2Cl –(g) → CaCl 2(s) (state symbols required) 1

(ii)

2

(iii) ∆Hlatt

o = –796 – 242 – 178 – 590 – 1150 + (2 × 349) = – 2258 kJ mol–1 3

(b) (higher temperature means that) particles have more energy; entropy (of the gas / system) increases because of an increase in the amount of disorder / randomness;

2

(c) (i)

reaction sign of ∆So

CO(g) + O2(g) → CO2(g) negative

Mg(s) + ½O2(g) → MgO(s) negative

CuSO4(s) + 5H2O(l) → CuSO4.5HsO(s) negative

NaHCO3(s) + H+(aq) → Na+(aq) + CO2(g) + H2O(l) positive

2

(ii) there is a reduction in the overall number of gaseous molecules 1

(d) ∆Sf

o = 386 – (192 + (3 × 131)) = –199 (J K–1

mol–1)

2

(e) (i) ∆Go = ∆H

o – T∆So

= 117 – ((298 × 175) / 1000) = (+) 64.85 (kJ mol–1)

2

(ii) ∆Go is positive and so the reaction is not spontaneous (at 298 K) 1

Ca2+(g) + 2Cl (g) (+ 2e–)

2nd I.E of Ca

1st I.E of Ca EA of Cl × 2

Atomisation/∆Hat of Ca

E(Cl-Cl)/2∆Hat of Cl ∆Hlatt

o

∆Hfo CaCl2(s)


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5 (a) (i) any metal with an Eo value more negative than –0.41 V, e.g. Fe, Mn, Zn, Mg, Cr, Al R: Li / Na / K / Ca / Ba

1

(ii) M1: value of Ecell correctly calculated (with correct sign) for metal named in (i) M2: Eo

cell is positive and so reaction is feasible 11

(b) M1: (Cr2O7

2– + 14H+ + 6e– ⇌ 2Cr3+ + 7H2O) Eo = +1.33 V

(H2O2 + 2H+ + 2e– ⇌ 2H2O) Eo = +1.77 V

Eo

cell = 0.44 (V) M2: E

o

cell (0.44 V) is positive (so the reaction is feasible) / Eo(Cr2O7

2– / Cr3+) is less

positive than Eo(H2O2 / H2O)

1

1

(c) M1:

Cr2O72–: ox.no Cr = +6 because –2 = 2 × ox.no(Cr) + (7 × –2)

CrO42–: ox.no Cr = +6 because –2 = ox.no(Cr) + (4 × –2)

M2: no change in oxidation number, so reaction is not redox

1

1

(d) M1: no. moles Cr deposited = 0.0312 / 52 = 6.0 × 10–4 moles M2: deduction that 6 moles of e– needed per mole of Cr /

reaction is Cr2O72- + 14H+ + 12e– → 2Cr + 7H2O

M3: no. moles of e– = 6 × 6.0 × 10–4 = (0.125 × t) / 96 500

so t = (6 × 6.0 × 10–4 × 96 500) / (0.125 × 60) = 46.3 min / 0.772 h / 2780 s

11

1





Question

Answer Mark

6 (a)

identity or value

V nitrogen or chlorine

X NO / NO2 Cl O2 / Cl O3

m 2, 3 1,2,3, or 4

W sulfur

Y SO2 or SO3

n 4, 3

3

(b) M1: (white precipitate is BaSO4)

descending the group ∆Hsol becomes more endothermic / positive; M2, M3 any two from:

∆Hlatt decreases / becomes more endothermic / becomes less exothermic

∆Hhyd decreases / becomes more endothermic / becomes less exothermic

∆Hhyd decreases more than ∆Hlatt

1

2


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mus

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be s

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1






9 (a) (i) polyester : Terylene / polylactic acid (PLA) / polyamide : nylon / Kevlar / Nomex

1

(ii) water or hydrochloric acid / hydrogen chloride 1

(b) (i)

polymer biodegradable

A yes

B yes

C no

D yes

2

(ii)

HOCH2CH2OH and or equivalent 1,4-diacyl chloride or equivalent 1,4-diester

2

(c) (i) V: it has two amine / NH2 groups (which can be protonated) or it has an amine / NH2 group on its side chain / R group

1

(ii) four (TT, TU, UT, UU) 1

(iii) hydrogen bonds; between the O / N atoms or named group (in the polypeptide) and water; or ion-dipole attractions; between NH3

+ / CO2

– and water;

2


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