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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.
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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
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Page 3 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9701 42
© Cambridge International Examinations 2016
Question Answer Mark
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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P
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3
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4
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on
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P
Q
4
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(b)
(c)
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5
tio
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Page 6 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9701 42
© Cambridge International Examinations 2016
Question Answer Mark
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
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Page 7 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9701 42
© Cambridge International Examinations 2016
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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P
Q
7
Pag
Qu
ge
es
(a)
(b)
(c)
8
tio
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(i
(ii
) (
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(ii
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on
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ii)
M1wegroM2rin
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be
M1M2
CH
CH
C
eit
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ng
ect
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P
Q
8
Pag
Qu
ge
es
(a)
(b)
(c)
(d)
(e)
9
tio
)
)
) (
(i
)
) (
(i
(ii
on
(i)
ii)
(i)
ii)
ii)
P Q R Q
se
LiA
(m
C
CH
C
aks
= c
ee l
a
2,4
Al H
mus
CH
H3
Cam
amketsec
car
ine
alk
u
4-d
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st b
H3
mb
midetonco
rbo
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kali
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To
be s
C
CH
brid
e ne nd
ony
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ne
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ersa
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e In
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and
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Page 10 Mark Scheme Syllabus Paper
Cambridge International AS/A Level – March 2016 9701 42
© Cambridge International Examinations 2016
Question Answer Mark
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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