Edexcel A-Level Chemistry Paper 2, June 2023: Question 6

12 marks · Hard difficulty · Practical Techniques and Data Analysis

Investigate the kinetics of the iodine-propanone reaction using a titrimetric method and determine rate equations and mechanisms.

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Question

Question text

6 Iodine reacts with propanone in acidic conditions.

H+(aq)

I (aq) + CH COCH (aq) CH COCH I(aq) + H+(aq) + I–(aq)

23 3 3 2

A student was asked to investigate the kinetics of this reaction.

The student predicted that the rate equation for the reaction would be

rate = k[I (aq)][CH COCH (aq)][H+(aq)]0

23 3

because the balanced equation shows that one molecule of iodine reacts with

one molecule of propanone and the acid is a catalyst.

(a) The student first determined the order of reaction with respect to iodine by

keeping the concentrations of propanone and acid constant.

The student used the outline procedure shown.

• mix 25 cm3 of aqueous propanone with 25 cm3 of dilute sulfuric acid in a

conical flask

• add 25 cm3 of aqueous iodine, immediately start a stopwatch and swirl the

mixture in the conical flask

• use a pipette to remove a 10.0 cm3 sample of the solution and place it in a

clean conical flask

• add a spatula measure of sodium hydrogencarbonate and note the exact time

it is added

• take four more 10.0 cm3 samples of the mixture and add

sodium hydrogencarbonate to each of them at regular time intervals

• titrate the unreacted iodine in the samples with sodium thiosulfate solution

using starch indicator.

(i) State how the student could ensure that the concentrations of propanone

and acid are effectively constant throughout the experiment.

(1)

(ii) Explain why sodium hydrogencarbonate is added.

(2)

*P71913A01328*

(b) The student obtained these results.

Time / min 5 10 15 20 25

Volume of thiosulfate / cm3 15.0 13.8 12.6 11.4 10.2

(i) Give a reason why it is not necessary to calculate the concentration of iodine

at each time to work out the order of reaction with respect to iodine.

(1)

(ii) Plot a graph to show that the order of reaction with respect to iodine is zero.

(2)

(iii) Give a reason why the graph shows that the order of reaction with respect to

iodine is zero.

(1)

rate = k[CH COCH (aq)][H+(aq)][I (aq)]0

33 2

(i) Deduce a possible rate determining step in the mechanism of this reaction.

Curly arrows are not required.

(2)

(ii) Data from two experiments carried out at the same temperature are shown.

Experiment [CH COCH (aq)] / mol dm–3 [H+(aq)] / mol dm–3 [I (aq)] / mol dm–3 Rate / mol dm–3 s–1

33 2

13.0 0.4 0.02 3.36 × 10–5

24.0 0.2 0.04

What is the rate, in mol dm–3 s–1, in Experiment 2?

(1)

A 2.24 × 10–5

B 3.36 × 10–5

C 4.48 × 10–5

D 8.96 × 10–5

(iii) The experiment in (a) is repeated but using aqueous bromine instead of

aqueous iodine. All other conditions are kept the same.

Explain how you would expect the rate of reaction of bromination of

propanone to compare with the rate of iodination of propanone.

Assume that the reaction between bromine and propanone in acidic

conditions has the same rate equation as that between iodine and propanone

in acidic conditions.

(2)

(Total for Question 6 = 12 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

6(a)(i) An answer that makes reference to the following point: (1)

• use (a much) higher concentration (of propanone and acid (1) Allow use a high concentration (of propanone and acid)

(than iodine)

Allow use a (large) excess (of propanone and acid)

Ignore use a higher volume of propanone and acid

Question

Answer Additional Guidance Mark

Number

6(a)(ii) An explanation that makes reference to the following point: (2)

• to neutralise the acid / H+ (1) Allow to remove / react with the acid / catalyst

• so that the reactions is quenched / stopped (1) Allow to freeze the reaction

Ignore ‘slow down’

Question

Answer Additional Guidance Mark

Number

6(b)(i) An answer that makes reference to the following point: (1)

• ([I2] / concentration (of iodine) is (directly) proportional to the Allow reverse argument

volume of (sodium) thiosulfate / Na S O / S O 2− Allow they react in a specific ratio / quoted ratio / 2:1

22 3 2 3

Do not award equal to or inversely proportional to

Do not award I− / iodide (ions)

Question

Answer Additional Guidance Mark

Number

6(b)(ii) Example of graph (2)

• axes with time on x axis

and

labelled, including units

and

suitable scale so that points cover at least half the

axes in both directions (1) Allow x axis to start at 5

• points plotted correctly to (±½ square) Allow M2 if axes wrong way around in M1

and

straight line through points (1)

Question

Answer Additional Guidance Mark

Number

6(b)(iii) (1)

• (zero order) because graph is a straight line / linear Allow the gradient is constant

rate is constant Do not award if their graph is a straight horizontal line

Ignore through the origin

This mark is dependent on the graph being a straight line with

a constant gradient

Question

Answer Additional Guidance Mark

Number

6(c)(i) Example of equation (2)

• an equation that has only H+ and CH COCH on the left (1) CH COCH + H+ → CH C(O+H)CH

33 3 3 3 3

CH COCH + H+ → CH C+(OH)CH

33 3 3

• correct product (1) Allow structural / skeletal formulae e.g. product shown as

Allow C H O+

Do not award CH COCH +

Question Answer Mark

number

6(c)(ii) −5 (1)

The only correct answer is A (2.24 × 10 )

B is incorrect because 3.36 × 10−5 would be the rate if the reaction was zero order with respect to CH COCH and first order

with respect to H+ and I

C is incorrect because 4.48 × 10−5 would be the rate if the reaction was first order with respect to CH COCH H+ and I

3 3, 2

D is incorrect because 8.96 × 10−5 would be the rate if the reaction was first order with respect to CH COCH and I and zero

33 2

order with respect to H+

Question

Answer Additional Guidance Mark

Number

6(c)(iii) An explanation that makes reference to the following points: (2)

• rate will be the same (1) Do not allow the rate would be constant.

• because rate equation only involves propanone and acid / Allow the order with respect to bromine would be 0 / bromine

rate equation does not involve iodine / bromine (1) would not be in the rate determining step

M2 depends on correct M1

(Total for Question 6 = 12 marks)

How to answer it

Kinetics of the Iodine-Propanone Reaction

What this question tests

This comprehensive question assesses your practical and theoretical understanding of chemical kinetics. Key topics include: designing methods to maintain constant reactant concentrations (pseudo-order techniques), quenching reactions for titrimetric analysis, processing experimental data graphically to determine reaction orders, predicting rate-determining steps from rate equations, calculating rate constants and new rates, and applying reaction kinetics principles to alternative halogen systems.

Part (a): Experimental Design & Quenching

Parts (a)(i) and (a)(ii)

✅ Correct Answers

(a)(i): Use a much higher (or large excess) concentration of propanone and acid than iodine.
(a)(ii): To neutralise the acid/catalyst ( H⁺ ) and quench/stop the reaction.

💡 Key Knowledge

Constant concentration: Ingesting a large excess means any change in the reactant's concentration during the reaction is negligible, effectively keeping it constant.
Quenching: Sodium hydrogencarbonate ( NaHCO₃ ) reacts with the acid catalyst. Removing the catalyst halts the reaction instantly so the titration accurately reflects the iodine concentration at that exact timestamp.

❌ Common Errors

Saying use a higher volume instead of higher concentration in (a)(i). Volume changes total volume and dilutions; concentration is what matters.
In (a)(ii), stating that sodium hydrogencarbonate "slows down" the reaction. Examiners penalise this—it must completely stop or quench the reaction.

🧠 Exam Technique

Be precise with terminology: use terms like "large excess" and "quench".

Part (b): Processing Titration Data Graphically

Parts (b)(i), (b)(ii), and (b)(iii)

✅ Correct Answers

(b)(i): Concentration of iodine is directly proportional to the volume of sodium thiosulfate used.
(b)(ii): A well-scaled graph with axes correctly labelled with units ( Volume of thiosulfate / cm³ on y-axis, Time / min on x-axis), points plotted accurately within ±½ square, and a straight line of best fit drawn.
(b)(iii): Zero order, because the graph is a straight line (linear) or the rate is constant.

📐 Calculation & Data Insights

Proportionality: Because the titration volume of thiosulfate reacts directly with unreacted iodine, volume serves as a direct proxy for concentration without needing complex conversion calculations.
Linear trend: A constant gradient ( Δy/Δx ) on a concentration-time or volume-time graph signifies zero-order kinetics with respect to that reactant.

❌ Common Errors

Plotting axes the wrong way around or choosing awkward scales that occupy less than half the grid.
Attempting to force curves through points or claiming first order from a straight line.

Part (c): Rate Equations, Mechanisms, and Calculations

Parts (c)(i), (c)(ii), and (c)(iii)

✅ Correct Answers

(c)(i): CH₃COCH₃ + H⁺ → CH₃C(O⁺H)CH₃ (or carbocation structural variants showing addition of H⁺ to propanone carbonyl oxygen).
(c)(ii): Option A ( 2.24 × 10⁻⁵ ).
(c)(iii): The rate will be the same, because the rate equation only involves propanone and acid (it does not involve iodine or bromine).

💡 Key Knowledge: Mechanism Steps

The overall rate equation ( rate = k[CH₃COCH₃][H⁺] ) dictates that the rate-determining step must involve only one molecule of propanone and one hydrogen ion H⁺ . Iodine ( I₂ ) is zero order, so it cannot feature in the rate-determining step.

📐 Step-by-Step Calculation for (c)(ii)

Identify rate equation: rate = k[CH₃COCH₃][H⁺]
Analyze changes from Exp 1 to Exp 2:
- [CH₃COCH₃] changes from 3.0 to 4.0 (multiplied by 4/3 or 1.33 )
- [H⁺] changes from 0.4 to 0.2 (multiplied by 0.5 )
- [I₂] changes from 0.02 to 0.04 (zero order, so no effect on rate)
Calculate new rate:
Rate₂ = Rate₁ × (4/3) × 0.5
Rate₂ = 3.36 × 10⁻⁵ × 1.333 × 0.5 = 2.24 × 10⁻⁵ mol dm⁻³ s⁻¹

❌ Common Calculation Traps in (c)(ii)

Distractor B ( 3.36 × 10⁻⁵ ): Assumes zero order for propanone and first order for iodine.
Distractor C ( 4.48 × 10⁻⁵ ): Assumes all three reactants are first order.
Distractor D ( 8.96 × 10⁻⁵ ): Forgetting to account for the halved acid concentration correctly.

Topics

Physical Chemistry · Core Practicals · Core Practical 13a: Follow the rate of the iodine-propanone reaction by a titrimetric method · Topic 16: Kinetics II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.