CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER … · Chemistry Advanced Level Paper 3 (9CH0/03) Question Number Acceptable Answer Additional Guidance Mark . 1(a)(i) suitable scale

Chemistry Advanced Level Paper 3 (9CH0/03)

Question

Number Acceptable Answer Additional Guidance Mark

1(a)(i) suitable scale and axes labelled including units (1)

all points plotted correctly (1)

line of best fit (1)

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

CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER 32 MARK SCHEME

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Text Box

Question


1(a)(ii) 1st order with respect to SO2Cl2 as half-life is constant

(1)

values of two half-lives shown in graph in range 2000 –

2400 (s) (1)

rate = k[SO2Cl2] (1)

Example of graph

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2

(Total for Question 1 = 12 marks)

Question


1(a)(iii)

tangent drawn at 0.200 mol dm-3 SO2Cl2 (1)

value of gradient (1)

k = rate / 0.200 (1)

value of k to 2 or 3 SF and units (1)

Example of calculation

e.g. gradient = (-) 0.375/6400 = (-) 5.859 x 10−5

must be consistent with tangent shown on graph

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

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Question


1 (b) RDS Step 1 SO2Cl2 → SO2Cl + Cl (1)

Step 2 SO2Cl + Cl → SO2 + Cl2 (1)

The RDS must involve one mol of SO2Cl2

The equations taken together must add up to the overall equation SO2Cl2 → SO2 + Cl2

e.g. SO2Cl2 → SO2 + 2Cl

2Cl → Cl2

Ignore state symbols

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Question Number

Acceptable Answer Additional Guidance Mark

2(a)

mass of C (1)

mass of H (1)

mass of oxygen

and mols of C, H and O (1)

ratio (1)

empirical and use of molar mass to determine molecular formula

(1)


mass of C = 2.83 x 12/44 = 0.772 g

mass of H = 0.693 x 2/18 = 0.077 g

mass of oxygen = 0.952 – (0.772 + 0.077) = 0.103 g

C H O

Moles 0.772/12=0.0643 0.077/1=0.077 0.103/16=0.0064

ratio 0.0643/0.0063=10 0.077/0.00643=12 0.0064/0.0064=1

Empirical formula = C10H12O, this has mass =148

so molecular formula = C10H12O

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Question


2(b)(i) An answer that makes reference to the following point:

contains benzene ring / arene

Ignore comments about alkenes / unsaturated 1

Question Number


2(b)(ii) is an aldehyde or ketone / is a

carbonyl compound

Both aldehyde and ketone required for first alternative 1

Question Number


2(b)(iii) (Q is a) ketone Allow ‘not an aldehyde’ if first alternative given in

(b)(ii)

1


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Question


2(c) An answer that makes reference to the following points:

(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

Award 1 mark for any structure of C10H12O which

includes a benzene ring and a ketone

e.g.

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5

Question


3(a) (i) An answer that makes reference to the following points:

EITHER the solution of manganate(VII) ions (1)

as distilled water/other solutions remaining in burette will dilute

the solution of manganate(VII) ions (1)

OR 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)

2

Question Number


3(a) (ii) 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

(1)

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Question


3(a) (iii) An explanation that makes reference to the following point:

it will not change the amount / moles (of reactants) in the flask

1

Question


3(b) H2C2O4 → 2CO2 + 2H+ + 2e− / C2O4

2- → 2CO2 + 2e−

and

MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (1)

5H2C2O4 + 2MnO4− + 6H+ → 2Mn2+ + 10CO2 + 8H2O / 5C2O4

2- +

2MnO4− + 16H+ → 2Mn2+ + 10CO2 + 8H2O

(2)


Allow for 1 mark: 5H2C2O4 + 2MnO4

− + 16H+ → 2Mn2+ + 10CO2

+ 8H2O + 10H+

3

Question


3(c) mean = 22.7 / 22.70 (cm3) (1)

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Question


3(d)

calculates amount of MnO4− (1)

calculates amount of H2C2O4 (in 25 cm3) (1)

calculates amount of H2C2O4 (in 250 cm3) (1)

calculates relative molecular mass of acid (1)

calculates x (1)


Amount of MnO4− = 22.7/1000 x 0.0105 =

2.3835 x 10−4 mol

TE on mean in (c)

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)

Amount of H2C2O4 (in 250 cm3)

= 5.95875 x 10−4 x 10 = 5.95875 x 10−3 mol

Relative molecular mass of acid

= 0.747 / 5.95875 x 10−3 = 125.4

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

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Question Number


3 (e) An answer that makes reference to the following point:

more MnO4− would show greater amount of H2C2O4 (1)

so molecular mass of xH2O would seem to be less, so value for x

would be less (1)

or

more MnO4− means lower molecular mass (1)

therefore mass of water of crystallisation is less so x is less than

2 (1)

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Question


4 (a) An explanation that makes reference to the following point:

production of CO2 (1)

NaHCO3 + HCl → NaCl + CO2 + H2O (1)

Allow HCO3

− + H+ → CO2 + H2O


2

Question Number


4 (b) An answer that makes reference to the following points:

drying agent (1)

mixture becomes transparent / clear (from cloudy) (1)

Allow to remove water

Allow drying agent stops clumping together

2

Question


4 (c) 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


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Question Number


4(d)

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)

Arrow must originate from bond and finish on oxygen

Arrow must originate from lone pair on Cl−

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Question


*5 (a) 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

indicative marking

points seen in answer

Number of

marks awarded for indicative

marking points

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.

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

2

Answer is partially structured with some linkages and lines of reasoning

1

Answer has no linkages between

points and is unstructured 0

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).


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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

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Question Number


5 (b) 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) (1)

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)

or

yes, as reaction is dependent on H+ ions (not anions) (1)

Eo value is negative (-0.03 V) (1)

2

Question


5 (c)

1

(Total for Question 5 = 9 marks


14

Question


6 (a) suitable extrapolation (1)

∆T in the range of 29 – 33(oC) (1)

Vertical at 3 minutes and smooth curve of best fit

back from 10 minutes to 3 minutes

2

Question Number


6 (b) (i) 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 (1)

2


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Question


6 (b) (ii) An explanation that makes reference to the following points:

by using more finely powdered zinc (1)

because it will increase the rate of reaction (1)

A greater surface area of zinc

Less time for heat loss

2

Question Number


6 (c) Calculation of Q using Q = mc∆T (1)

amount of CuSO4(aq) (1)

energy per mol = Q/0.018 (1)

∆H = -Q/1000 kJ mol−1 (1)

Using ∆T from (a)

Example of calculation 25 x 4.18 x 30.5 = 3 187 (J)

Allow 4.2 for 4.18

= 25/1000 x 0.720 = 0.018

3 187/0.018

= −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

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Question


6 (d) (i) (217 – answer to (c) / 217) x 100


(217 – 177 / 217) x 100 = 18.4%

1

Question


6 (d) (ii) 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)

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

Question


6 (d) (iii) 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)


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Question


7 (a) +5 1

Question


7 (b) (i) choice of suitable reducing agent

From Data Booklet of the standard electrode potentials, the right hand substance in numbers 16-

24

1

Question Number


7 (b) (ii) An answer that makes reference to the following point:

may be kinetically hindered

or

activation energy may be too high

Allow reaction may be too slow

1

Question Number


7 (b) (iii) 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


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Question


7 (b) (iv) An explanation that makes reference to the following point:

justification of choice (1)

correct equation for reaction (1)

Eo value is between +0.34 V and −0.26 V

Allow by comparison of electrode potentials / calculation of Ecell / use of anti-clockwise rule

S2O3

2− + VO2+ + 2H+ → V3+ + H2O + ½S4O62−

or multiples

2

Question Number


7 (c) (i) SO2 + ½O2 → SO3 Allow multiples


1


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Question


*7 (c)

(ii)

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 indicative marking

points seen in answer

Number of marks awarded for indicative

marking points

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.

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

2

Answer is partially structured with

some linkages and lines of reasoning

1

Answer has no linkages between

points and is unstructured 0

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).


20

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

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Question


8(a) reagent for step 1 (1)

product of step 1 (1)

reagent for step 2 (1)

KCN and HCN or KCN and H+ or HCN

Allow structural or displayed formulae

Allow name or formula of any mineral acid

3

Question


8(b)

(1)

esterification (1)

2


22

Question


8(c)

NO2+ ion (1)

curly arrow from ring to electrophile (1)

formula of intermediate (1)

curly arrow from CH bond to ring and formation of H+ (1)

Ignore equation to produce electrophile

To any part of the electrophile including the plus sign

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

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23

Question


9(a)

For Mg allow 0 or 8 electrons in outermost shell

Ignore inner shell electrons

2

Question Number


9(b)(i)

amount of NaOH (1)

[NaOH] (1)

[NaF]eq (1)

Kc expression (1)

Value of Kc (1)


11.7/1000 x 0.100 = 0.00117 / 1.17 x 10−3

0.00117/(25/1000) = 0.0468 mol dm−3

0.250 – 0.0468 = 0.2032 mol dm−3

Kc = [NaOH]2 / [NaF]2

0.0530 (no units)

5

Question Number


9 (b) (ii) 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



24

Question Number


10(a)(i) orange to colourless

Allow brown / yellow to colourless

1

Question Number


10(a)(ii) 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

e.g.

2

Question


10(a)(iii) addition (1)

structure (1)

Example of structure:

2


25

Question Number


10(a)(iv)

Ignore addition of square brackets and n

1


26

Question


10(b) Raw materials (1)

wood is renewable

and polystyrene uses slightly more petroleum fractions

Energy use (1)

paper cup uses nearly 4x as much energy

Water use (1) paper cup uses nearly 50x as much water

Air emissions (1) paper cup produces chlorine and more

sulfur dioxide both of which are toxic / acidic and

polystyrene cup produces a lot of hydrocarbons

Justification (2) Selection of either cup with two valid reasons

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

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CHERRY HILL TUITION EDEXCEL CHEMISTRY A2 PAPER … · Chemistry Advanced Level Paper 3 (9CH0/03) Question Number Acceptable Answer Additional Guidance Mark . 1(a)(i) suitable scale