CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Chemistry Advanced Level Paper 3 (9CH0/03)
Question
Number
1(a)(i)
Acceptable Answer
Additional Guidance

suitable scale and axes labelled including units
(1)

all points plotted correctly
(1)

line of best fit
(1)
Mark
Plotted points use at least half the available space in
both directions
±½ a square
Smooth line, no dot to dot
NOTE: Points at t = 2000 and 6000 s should not be on
the line of best fit.
3
1
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
1(a)(ii)
Acceptable Answer
Additional Guidance

1st order with respect to SO2Cl2 as half-life is constant

values of two half-lives shown in graph in range 2000 –
2400 (s)
(1)

rate = k[SO2Cl2]
Mark
Example of graph
(1)
(1)
3
2
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
1(a)(iii)
Acceptable Answer
Additional Guidance
Mark
Example of calculation

tangent drawn at 0.200 mol dm-3 SO2Cl2
(1)

value of gradient
(1)
e.g. gradient = (-) 0.375/6400 = (-) 5.859 x 10−5
must be consistent with tangent shown on graph
Question
Number
1 (b)

k = rate / 0.200
(1)

value of k to 2 or 3 SF and units
(1)
Acceptable Answer
k = 5.859 x 10−5/ 0.200
= 2.9 x 10−4 s−1 / 2.93 x 10−4 s−1
Allow TE from incorrect rate equation / allow TE from
incorrect gradient
Units must be consistent with rate equation
4
Additional Guidance
Mark

RDS Step 1 SO2Cl2 → SO2Cl + Cl
(1)
The RDS must involve one mol of SO2Cl2

Step 2 SO2Cl + Cl → SO2 + Cl2
(1)
The equations taken together must add up to the
overall equation
SO2Cl2 → SO2 + Cl2
e.g. SO2Cl2 → SO2 + 2Cl
2Cl → Cl2
Ignore state symbols
2
(Total for Question 1 = 12 marks)
3
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
2(a)
Question
Number
2(b)(i)
Question
Number
2(b)(ii)
Question
Number
2(b)(iii)
Acceptable Answer
Additional Guidance
Mark
Example of calculation

mass of C
(1)
mass of C = 2.83 x 12/44 = 0.772 g

mass of H
(1)
mass of H = 0.693 x 2/18 = 0.077 g

mass of oxygen
and
mols of C, H and O
mass of oxygen = 0.952 – (0.772 + 0.077) = 0.103 g
(1)

ratio
(1)

empirical and use of molar mass to
determine molecular formula
(1)
Acceptable Answer
An answer that makes reference to the
following point:
 contains benzene ring / arene
Acceptable Answer

is an aldehyde or ketone / is a
carbonyl compound
Acceptable Answer

(Q is a) ketone
Moles
ratio
C
0.772/12=0.0643
0.0643/0.0063=10
H
0.077/1=0.077
0.077/0.00643=12
O
0.103/16=0.0064
0.0064/0.0064=1
Empirical formula = C10H12O, this has mass =148
so molecular formula = C10H12O
Additional Guidance
5
Mark
Ignore comments about alkenes / unsaturated
Additional Guidance
1
Mark
Both aldehyde and ketone required for first alternative
Additional Guidance
1
Mark
Allow ‘not an aldehyde’ if first alternative given in
(b)(ii)
1
4
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
2(c)
Acceptable Answer
Additional Guidance
An answer that makes reference to the following points:
Mark
Award 1 mark for any structure of C10H12O which
includes a benzene ring and a ketone
e.g.
(2)

Chemical shift
(1)
peak at 7.5 ppm indicates a benzene ring and peak at 2.55
ppm indicates H on carbon adjacent to C=O and peak at 1.25
ppm indicates alkyl hydrogens

Splitting pattern
(2)
Any 2 from 3:

complex multiplet indicates a benzene ring

septuplet indicates 6 equivalent protons on adjacent carbons

doublet indicates one hydrogen on adjacent carbon

Area under peak
(2)
Any 2 from 3:

five H’s indicate C6H5 / monosubstituted benzene ring

one H indicates one H on C next to carbonyl group

six H’s indicates six H’s on two methyl groups
7
(Total for Question 2 = 15 marks)
5
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
3(a) (i)
Acceptable Answer
Mark
An answer that makes reference to the following points:
EITHER
 the solution of manganate(VII) ions
OR
Question
Number
3(a) (ii)
Additional Guidance
(1)

as distilled water/other solutions remaining in burette will dilute
the solution of manganate(VII) ions
(1)

Distilled water, followed by the solution of manganate(VII) ions
(1)

So the distilled water will not dilute the solution of
manganate(VII) ions
(1)
Acceptable Answer
2
Additional Guidance
Mark
An answer that makes reference to the following points:

measure from the bottom of the meniscus (of the solution of
manganate(VII) ions)
(1)

with eyes / head at same level (as meniscus)
or
ensuring the burette is vertical
or
use of white background to emphasise meniscus
or
ensuring funnel is removed from top of burette
or
ensuring there is no air bubble in the tip of the burette
2
(1)
6
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
3(a) (iii)
Acceptable Answer
Mark
An explanation that makes reference to the following point:

Question
Number
3(b)
Additional Guidance
it will not change the amount / moles (of reactants) in the flask
Acceptable Answer

1
Additional Guidance
Mark
Ignore state symbols
H2C2O4 → 2CO2 + 2H+ + 2e− / C2O42- → 2CO2 + 2e−
and
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O

−
+
2+
5H2C2O4 + 2MnO4 + 6H → 2Mn + 10CO2 + 8H2O / 5C2O4
2MnO4− + 16H+ → 2Mn2+ + 10CO2 + 8H2O
(1)
2-
+
Allow for 1 mark:
5H2C2O4 + 2MnO4− + 16H+ → 2Mn2+ + 10CO2
+ 8H2O + 10H+
(2)
Question
Number
3(c)
Acceptable Answer

mean = 22.7 / 22.70 (cm3)
3
Additional Guidance
(1)
Mark
1
7
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
3(d)
Acceptable Answer
Additional Guidance
Mark
Example of calculation

calculates amount of MnO4−
(1)
Amount of MnO4− = 22.7/1000 x 0.0105 =
2.3835 x 10−4 mol
TE on mean in (c)

3
calculates amount of H2C2O4 (in 25 cm )
(1)
Amount of H2C2O4 (in 25 cm3)
= 2.3835 x 10−4 x 2.5
= 5.95875 x 10−4 mol
TE on mol ratio in equation in (b)

calculates amount of H2C2O4 (in 250 cm3)
(1)
Amount of H2C2O4 (in 250 cm3)
= 5.95875 x 10−4 x 10
= 5.95875 x 10−3 mol

calculates relative molecular mass of acid
(1)
Relative molecular mass of acid
= 0.747 / 5.95875 x 10−3 = 125.4

calculates x
(1)
So relative mass of water of
crystallisation = 125.4 – 90 = 35.4
35.4/18 = 1.97 so x = 2
Do not award x = 1.97
5
8
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
3 (e)
Acceptable Answer
Additional Guidance
Mark
An answer that makes reference to the following point:

more MnO4− would show greater amount of H2C2O4

so molecular mass of xH2O would seem to be less, so value for x
would be less
(1)

more MnO4− means lower molecular mass

therefore mass of water of crystallisation is less so x is less than
2
(1)
(1)
or
(1)
2
(Total for Question 3 = 16 marks)
9
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
4 (a)
Question
Number
4 (b)
Acceptable Answer
Mark
An explanation that makes reference to the following point:

production of CO2
(1)

NaHCO3 + HCl → NaCl + CO2 + H2O
(1)
Acceptable Answer
An answer that makes reference to the following points:
 drying agent

Question
Number
4 (c)
Additional Guidance
mixture becomes transparent / clear (from cloudy)
Allow HCO3− + H+ → CO2 + H2O
Ignore state symbols
Additional Guidance
(1)
Allow to remove water
(1)
Allow drying agent stops clumping together
Acceptable Answer
Additional Guidance
2
Mark
2
Mark
An explanation that makes reference to the following points:

they will reduce the yield
(1)

as some of the 2-chloro-2-methylpropane will react with
the water (and reform the alcohol)
(1)
Allow ‘as some of the 2-chloro-2-methylpropane
might dissolve in the water’
2
10
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
4(d)
Acceptable Answer
Additional Guidance

breaking of CO bond in protonated alcohol
(1)

structure of intermediate
(1)

correct charge on carbon in intermediate
(1)

attack of chloride ion with lone pair
(1)
Mark
Arrow must originate from bond and finish on
oxygen
Arrow must originate from lone pair on Cl−
4
(Total for Question 4 = 10 marks)
11
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
*5 (a)
Acceptable Answer
Additional Guidance
This question assesses a student’s ability to show a coherent and logically
structured answer with linkages and fully-sustained reasoning.
Marks are awarded for indicative content and for how the answer is
structured and shows lines of reasoning.
The following table shows how the marks should be awarded for indicative
content.
Number of
Number of
indicative
marks awarded
marking
for indicative
points seen
marking points
in answer
6
4
5–4
3
3–2
2
1
1
0
0
The following table shows how the marks should be awarded for structure
and lines of reasoning.
Mark
Guidance on how the mark scheme should
be applied:
The mark for indicative content should be
added to the mark for lines of reasoning.
For example, an answer with five indicative
marking points which is partially structured
with some linkages and lines of reasoning
scores 4 marks (3 marks for indicative
content and 1 mark for partial structure
and some linkages and lines of reasoning).
If there are no linkages between points,
the same five indicative marking points
would yield an overall score of 3 marks (3
marks for indicative content and no marks
for linkages).
Number of marks awarded for
structure of answer and
sustained line of reasoning
Answer shows a coherent and logical
structure with linkages and fully
sustained lines of reasoning
demonstrated throughout
Answer is partially structured with
some linkages and lines of reasoning
Answer has no linkages between
points and is unstructured
2
1
0
12
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Indicative Content:

technique A allows heating for a longer time period / technique B is
suitable for a shorter time period

technique A ensures oxidation / technique B may result in loss of
reactant before oxidation

technique A minimises release of flammable or harmful vapours /
technique B allows release of flammable or harmful vapours, unless
carried out in fume cupboard

technique A involves more expensive equipment or is more complex
to set up / technique B is simpler to set up or uses less expensive
equipment

as complete oxidation is not necessary to qualitatively show it is an
alcohol that can be oxidised

technique B is most appropriate as simpler / technique A is most
appropriate as oxidation is guaranteed
6
13
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
5 (b)
Acceptable Answer
Additional Guidance
Mark
An explanation that makes reference to the following pairs:
either

no, as chloride ions could be oxidised by dichromate(VI) ions
(reducing amount of oxidant / producing toxic chlorine)

Eo values for dichromate(VI) and chloride ions being oxidised to
chlorine are sufficiently close (even though Eo is negative
= -0.03 V) for chlorine to be formed
(1)

yes, as reaction is dependent on H+ ions (not anions)
(1)

Eo value is negative (-0.03 V)
(1)
(1)
or
Question
Number
5 (c)
Acceptable Answer
2
Additional Guidance
Mark
1
(Total for Question 5 = 9 marks
14
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
6 (a)
Acceptable Answer
Additional Guidance

suitable extrapolation
(1)

∆T in the range of 29 – 33(oC)
(1)
Mark
Vertical at 3 minutes and smooth curve of best fit
back from 10 minutes to 3 minutes
2
Question
Number
6 (b) (i)
Acceptable Answer
Additional Guidance
Mark
An explanation that makes reference to the following point:


as reaction is not instantaneous / takes time to release
heat
(1)
so some heat is lost as mixture heats up
2
(1)
15
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
6 (b) (ii)
Question
Number
6 (c)
Acceptable Answer
Additional Guidance
An explanation that makes reference to the following points:
A greater surface area of zinc

by using more finely powdered zinc
(1)

because it will increase the rate of reaction
(1)
Less time for heat loss
Acceptable Answer

Calculation of Q using Q = mc∆T
Mark
Additional Guidance
(1)
2
Mark
Using ∆T from (a)
Example of calculation
25 x 4.18 x 30.5 = 3 187 (J)
Allow 4.2 for 4.18

amount of CuSO4(aq)
(1)
= 25/1000 x 0.720 = 0.018

energy per mol = Q/0.018
(1)
3 187/0.018

∆H = -Q/1000 kJ mol−1
(1)
= −177 kJ mol−1 / −180 kJ mol−1 /
−177 000 J mol−1 / −180 000 J mol−1
Use of 4.2 gives −178 kJ mol−1
Must have correct sign and units for 4th mark
4
16
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
6 (d) (i)
Question
Number
6 (d) (ii)
Question
Number
6 (d) (iii)
Acceptable Answer

Additional Guidance
(217 – answer to (c) / 217) x 100
Example of calculation
(217 – 177 / 217) x 100 = 18.4%
Acceptable Answer
Additional Guidance
An answer that makes reference to two of the following points:

no lid is used
(1)

SHC of solution may not be 4.18
(1)

energy absorbed by zinc / energy absorbed by
thermometer / polystyrene cup
(1)

mass of solution is greater than 25 grams
(1)
Mark
1
Mark
If answer in (c) is less negative than -217, points
one and three are valid, and for point two
candidates must state less than 4.18 for SHC
If answer in (c) is more negative than -217, only
point two can be awarded for candidates stating
SHC is greater than 4.18
Do not award marks for just ‘heat loss to
surroundings’
2
Acceptable Answer
Additional Guidance
Mark
An explanation that makes reference to the following points:

∆T would be less, so ∆rH / enthalpy change calculated
would be less negative
(1)

because heat energy is transferred to greater volume of
water than expected volume
(1)
2
(Total for question 6 = 15 marks)
17
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
7 (a)
Question
Number
7 (b) (i)
Question
Number
7 (b) (ii)
Acceptable Answer


Mark
+5
1
Acceptable Answer
Additional Guidance
choice of suitable reducing agent
Mark
From Data Booklet of the standard electrode
potentials, the right hand substance in numbers 1624
Acceptable Answer
1
Additional Guidance
Mark
An answer that makes reference to the following point:

Question
Number
7 (b) (iii)
Additional Guidance
Allow reaction may be too slow
may be kinetically hindered
or
activation energy may be too high
1
Acceptable Answer
Additional Guidance
Mark
An explanation that makes reference to the following points:

(yellow to) green then blue
(1)

when the blue colour initially forms, it mixes with the yellow
unreacted VO2+ (to turn green)
(1)
Ignore colours caused by contamination with
reducing agent chosen
2
18
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
7 (b) (iv)
Acceptable Answer

justification of choice
correct equation for reaction
Acceptable Answer

SO2 + ½O2 → SO3
Mark
Eo value is between +0.34 V and −0.26 V
An explanation that makes reference to the following point:

Question
Number
7 (c) (i)
Additional Guidance
(1)
(1)
Allow by comparison of electrode potentials /
calculation of Ecell / use of anti-clockwise rule
S2O32− + VO2+ + 2H+ → V3+ + H2O + ½S4O62−
or multiples
2
Additional Guidance
Mark
Allow multiples
Ignore state symbols
1
19
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
*7 (c)
(ii)
Acceptable Answer
Additional Guidance
This question assesses a student’s ability to show a coherent and
logically structured answer with linkages and fully-sustained
reasoning.
Marks are awarded for indicative content and for how the answer is
structured and shows lines of reasoning.
The following table shows how the marks should be awarded for
indicative content.
Number of
Number of marks
indicative marking
awarded for indicative
points seen in
marking points
answer
6
4
5–4
3
3–2
2
1
1
0
0
The following table shows how the marks should be awarded for
structure and lines of reasoning.
Mark
Guidance on how the mark scheme should be
applied:
The mark for indicative content should be added to
the mark for lines of reasoning. For example, an
answer with five indicative marking points which is
partially structured with some linkages and lines of
reasoning scores 4 marks (3 marks for indicative
content and 1 mark for partial structure and some
linkages and lines of reasoning).
If there are no linkages between points, the same
five indicative marking points would yield an overall
score of 3 marks (3 marks for indicative content and
no marks for linkages).
Number of marks
awarded for structure of
answer and sustained
line of reasoning
Answer shows a coherent and
logical structure with linkages and
fully sustained lines of reasoning
demonstrated throughout
Answer is partially structured with
some linkages and lines of
reasoning
Answer has no linkages between
points and is unstructured
2
1
0
20
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Indicative Content:

V2O5 acts as a heterogeneous catalyst

SO2 is adsorbed onto surface and reacts with the V2O5

SO3 forms and is only weakly bonded to surface

the SO3 desorbs from the surface

oxygen reacts with the VO2, regenerating the V2O5 catalyst

reaction follows different mechanism with lower activation
energy
Ignore comments related to collision theory
6
(Total for Question 7 = 14 marks)
21
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
8(a)
Acceptable Answer
Additional Guidance

reagent for step 1
(1)
KCN and HCN or KCN and H+ or HCN

product of step 1
(1)
Allow structural or displayed formulae

reagent for step 2
(1)
Allow name or formula of any mineral acid
Mark
3
Question
Number
8(b)
Acceptable Answer
Additional Guidance
Mark
(1)

esterification
(1)
2
22
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
8(c)
Acceptable Answer
Additional Guidance

NO2+ ion
(1)
Ignore equation to produce electrophile

curly arrow from ring to electrophile
(1)
To any part of the electrophile including the plus sign

formula of intermediate
(1)

curly arrow from CH bond to ring and formation of
H+
(1)
Mark
The horseshoe must cover at least three carbon atoms,
opening towards the carbon atom approached by the
electrophile and at least part of the plus charge within the
horseshoe
4
(Total for Question 8 = 9 marks)
23
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
9(a)
Acceptable Answer
Additional Guidance
Mark
For Mg allow 0 or 8 electrons in outermost shell
Ignore inner shell electrons
2
Question
Number
9(b)(i)
Question
Number
9 (b) (ii)
Acceptable Answer
Additional Guidance

amount of NaOH
(1)
Example of calculation
11.7/1000 x 0.100 = 0.00117 / 1.17 x 10−3

[NaOH]
(1)
0.00117/(25/1000) = 0.0468 mol dm−3

[NaF]eq
(1)
0.250 – 0.0468 = 0.2032 mol dm−3

Kc expression
(1)
Kc = [NaOH]2 / [NaF]2

Value of Kc
(1)
0.0530 (no units)
Acceptable Answer
Mark
5
Additional Guidance
Mark
An answer that makes reference to the following point:

No, as temperature is likely to be different on different
days and this affects value of Kc
Allow ‘yes, but only if the temperature is the same’
1
(Total for Question 9 = 8 marks)
24
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
10(a)(i)
Question
Number
10(a)(ii)
Question
Number
10(a)(iii)
Acceptable Answer

Additional Guidance
orange to colourless
Mark
Allow brown / yellow to colourless
1
Acceptable Answer
Additional Guidance
Mark
A explanation that makes reference to the following points:

becomes polar / forms temporary dipole
(1)

as electron density in C=C bond repels bond pair (of
electrons in bromine)
(1)
Polar bromine can be shown by using a diagram
Acceptable Answer
e.g.
2
Additional Guidance

addition
(1)

structure
(1)
Mark
Example of structure:
2
25
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
10(a)(iv)
Acceptable Answer
Additional Guidance
Mark
Ignore addition of square brackets and n
1
26
CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME
Question
Number
10(b)
Acceptable Answer


Raw materials
(1)
wood is renewable
and
polystyrene uses slightly more petroleum fractions


Energy use
paper cup uses nearly 4x as much energy
(1)


Water use
paper cup uses nearly 50x as much water
(1)


Air emissions
paper cup produces chlorine and more
sulfur dioxide both of which are toxic / acidic
and
polystyrene cup produces a lot of hydrocarbons
(1)


Justification
Selection of either cup with two valid reasons
(2)
Additional Guidance
Mark
For example:
polystyrene is more sustainable, as although it
uses up non-renewable raw materials, its
production requires much less energy which is
likely to save on fossil fuel use
or
paper is more sustainable, as no hydrocarbons are
emitted into the atmosphere and the raw materials
are renewable and use less petroleum fractions
6
(Total for Question 10 = 12 marks)
27