Edexcel A-Level Chemistry Paper 3, June 2022: Question 10

12 marks · Hard difficulty · Practical Techniques and Data Analysis

Follow the rate of the iodine-propanone reaction using a titrimetric method, plot the data, determine the order with respect to iodine, and deduce the rate-determining step and catalyst role.

Practise this question

Question

Question text

10 The progress of the reaction between iodine and propanone with an acid catalyst can

be followed in an experiment using a titrimetric method.

Procedure

Step 1 Mix 25 cm3 of 1 mol dm–3 aqueous propanone with 25 cm3 of 1 mol dm–3

sulfuric acid in a beaker. Both these reactants are in excess.

Step 2 Start the stop clock as 50 cm3 of 0.02 mol dm–3 iodine solution is added to the

beaker. Mix the reactants thoroughly.

Step 3 Withdraw a 10.0 cm3 sample of the reaction mixture, using a pipette, and

transfer it to a conical flask.

Step 4 Add a spatula measure of sodium hydrogencarbonate, noting the exact time.

Step 5 Titrate the iodine present in the 10.0 cm3 sample with

0.01 mol dm–3 sodium thiosulfate solution, using starch indicator.

Step 6 Continue to withdraw 10.0 cm3 samples about every two minutes, repeating

Steps 4 and 5 with each sample.

(a) (i) Explain why sodium hydrogencarbonate is added in Step 4.

(2)

(ii) Write the ionic equation for the reaction that takes place during Step 4.

State symbols are not required.

(1)

(b) Some data from the experiment are shown.

Time sodium hydrogencarbonate is added / min 2.0 5.0 6.5 8.0 10.5 12.0

Volume of sodium thiosulfate / cm3 19.2 15.5 14.0 12.1 9.5 7.2

(i) Plot a graph of the volume of sodium thiosulfate against the time the

sodium hydrogencarbonate is added.

(2)

*P67095A02836*

(ii) Explain how the graph of volume of thiosulfate against time confirms the

reaction is zero order with respect to iodine, I2.

(3)

*P67095A02936*3 3

A student researching the mechanism for the reaction found this example.

O OH

Step 1 + H+ +

+ H

C C

H3C CH3 H2C CH3

OH +

Step 2 + I + I–

C 2

I C

H2C CH3 C

CH3

+O O

Step 3 + H+

I C I C

C CH3 C CH3

H2 H2

O O

Overall acidic + –

reaction + I2 + H + I

C conditions C

H3C CH3 IH2C CH3

(i) Predict which of the three steps is the rate-determining step.

Justify your answer.

(2)

… 30

… *P67095A03036*

(ii) The student stated that

‘The hydrogen ions cannot be acting as a catalyst.

One hydrogen ion is a reactant in Step 1 but two hydrogen ions are formed as

products in Steps 1 and 3.’

Explain whether or not this statement is valid.

*P67095A03136* (2)

(Total for Question 10 = 12 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

10(a)(i) An explanation that makes reference to the following points: (2)

• to stop / freeze / quench the reaction (1) Allow ‘to allow time for the titration to be carried

out’

Ignore just ‘slows down the reaction’

• by neutralising the (remaining sulfuric) acid / H+ (1) Allow by reacting with the acid / removing the

acid

Allow catalyst for acid

Do not award if incorrect acid specified

Question

Answer Additional Guidance Mark

Number

10(a)(ii) • ionic equation Examples of equations (1)

NaHCO + H+ → CO + H O + Na+

32 2

NaHCO + H O+ → CO + 2H O + Na+

33 2 2

Allow

HCO − + H+ → CO + H O

32 2

HCO − + H+ → H CO

32 3

Allow multiples

Allow balanced equations with H O+

Allow Na+ and SO 2− in equations, provided they

are crossed through

Ignore state symbols, even if incorrect

Do not award

CO 2− + 2H+ → CO + H O

32 2

Question

Answer Additional Guidance Mark

Number

10(b)(i) Example of graph (2)

• y axis labelled with volume and cm3

and Do not award time in seconds

x axis labelled with time and min Suitable scale so that points cover at least half the

and available space along the x axis and at least 2 large squares

suitable scale (1) on y axis (as shown)

• all points plotted correctly and line of best fit (1) ± ½ a small square

Allow M2 as TE if axes wrong way around

Ignore extrapolations

Question

Answer Additional Guidance Mark

Number

10(b)(ii) An explanation that makes reference to the following points: Ignore references to half-life (3)

• [I2] is proportional to the volume (of sodium thiosulfate) (1) Allow description of proportional

• gradient does not change / is constant / the graph shows a Allow decreases at a constant rate

straight line / is linear (as [I2] decreases) (1) Ignore volume (of sodium thiosulfate) / [I2] is

proportional to time

• which means the rate doesn’t change / increase or decrease Allow [I2] does not affect the rate (of reaction) /

(as [I2] increases or decreases) (1) rate is independent of [I2]

Question

Answer Additional Guidance Mark

Number

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

• Step 1 is the rate determining step (1) Stand alone

Allow RDS / slow step

• as it involves (1 mol of) both propanone and hydrogen ions Conditional on M1

(which matches the rate equation) (1) Allow it does not involve I2 (which is zero order)

Allow it involves both species in the rate

equation

Allow I2 is not involved in the RDS so RDS must

be before Step 2

Question

Answer Additional Guidance Mark

Number

10(c)(ii) An explanation that makes reference to the following points: Ignore reference to specific steps. (2)

(The statement is not valid because)

• one hydrogen ion is regenerated / reformed (so is acting as a Do not award M1 if candidate states that it is

catalyst) (1) valid

Ignore it is an autocatalyst

• the other hydrogen ion is lost from the propanone (when

replaced by iodine) / is a (by-)product of the reaction / is

used to form HI (1)

(Total for Question 10 = 12 marks)

TOTAL FOR PAPER = 120 MARKS

How to answer it

Kinetics, Titrimetric Methods & Reaction Mechanisms

Edexcel A-Level Chemistry • Exam Question Study Guide

What this question tests

Experimental Design & Quenching: Understanding how to stop a reaction mid-course (quenching) using chemical reagents like sodium hydrogencarbonate.
Graphical Analysis of Kinetics: Plotting experimental data and interpreting linear graphs to deduce reactant orders (specifically zero order with respect to iodine).
Reaction Mechanisms & Rate Equations: Linking a multi-step reaction mechanism to an experimentally determined rate equation to identify the rate-determining step (RDS).
Catalyst Definition: Rigorous evaluation of whether a species acts as a catalyst by examining its consumption and regeneration across reaction steps.

Part (a): Quenching and Quenching Reagents

(i) Explain why sodium hydrogencarbonate is added in Step 4. (2 marks)

✅ Correct Answer

To stop / freeze / quench the reaction by neutralising the (remaining sulfuric) acid / H⁺ ions.

🧠 Exam Technique

Key terms are essential here. You must state both what it does to the reaction (quench/stop) and how it achieves this chemically (neutralising the acid catalyst).

❌ Common Errors

Writing that it "just slows down the reaction" or failing to specify neutralisation of the acid catalyst. Vague statements do not gain credit.

Mark Breakdown: 1 mark for stating it quenches/stops the reaction; 1 mark for neutralising the acid / H⁺ ions.

(ii) Write the ionic equation for the reaction taking place during Step 4. State symbols are not required. (1 mark)

✅ Correct Answer

NaHCO₃ + H⁺ → CO₂ + H₂O + Na⁺

Alternative valid forms:
NaHCO₃ + H₃O⁺ → CO₂ + 2H₂O + Na⁺
HCO₃⁻ + H⁺ → CO₂ + H₂O

💡 Key Knowledge

Step 4 involves reacting the acid catalyst with a mild base (sodium hydrogencarbonate). Carbonates and hydrogencarbonates react with acids to produce a salt, water, and carbon dioxide gas.

Mark Breakdown: 1 mark for a balanced ionic or full equation representing the neutralisation of hydrogen ions by hydrogencarbonate. State symbols are ignored.

Part (b): Graphical Analysis and Reaction Order

(i) Plot a graph of the volume of sodium thiosulfate against the time the sodium hydrogencarbonate is added. (2 marks)

✅ Correct Answer

Axes correctly labelled with quantity and correct units ( Volume / cm³ on y-axis, Time / min on x-axis) with a sensible scale covering at least half the grid. All points accurately plotted with a clean straight line of best fit.

🧠 Exam Technique

Always check that your independent variable (time) is on the x-axis and your dependent variable (volume of thiosulfate) is on the y-axis. Ensure your scale uses linear increments and occupies most of the grid.

❌ Common Errors

Plotting time in seconds instead of minutes, or reversing the axes. Also, forcing a line through the origin when the data points do not support it will lose the line of best fit mark.

Mark Breakdown: 1 mark for correct axis labels, units, and suitable scale; 1 mark for accurate plotting and a correct line of best fit.

(ii) Explain how the graph of volume of thiosulfate against time confirms the reaction is zero order with respect to iodine, I₂. (3 marks)

✅ Correct Answer

The concentration of iodine is directly proportional to the volume of sodium thiosulfate used.
The graph is a straight line (shows a constant gradient).
This means the rate of reaction does not change as iodine concentration decreases, confirming zero order with respect to I₂.

💡 Key Knowledge

In kinetics, a straight-line graph of concentration (or a directly proportional proxy like titration volume) against time indicates a zero-order reaction because the rate is independent of that reactant's concentration ( rate = k[A]⁰ ).

Mark Breakdown: 1 mark linking [I₂] to thiosulfate volume; 1 mark noting the constant gradient / straight-line graph; 1 mark concluding that rate is independent of [I₂].

Part (c): Mechanisms and Catalysis

(i) Predict which of the three steps is the rate-determining step. Justify your answer. (2 marks)

✅ Correct Answer

Step 1 is the rate-determining step (RDS).
Justification: The overall rate equation is rate = k[H⁺][CH₃COCH₃] . Step 1 involves 1 molecule of propanone and 1 hydrogen ion, matching the stoichiometry of the rate equation.

🧠 Exam Technique

Always connect proposed steps back to the given rate equation. The molecularity of the rate-determining step (or steps prior to and including the RDS) must match the species present in the rate equation.

Mark Breakdown: 1 mark for identifying Step 1 as the RDS; 1 mark for justifying it based on involvement of 1 mol of propanone and 1 hydrogen ion matching the rate equation.

(ii) The student stated: 'The hydrogen ions cannot be acting as a catalyst. One hydrogen ion is a reactant in Step 1 but two hydrogen ions are formed as products in Steps 1 and 3.' Explain whether or not this statement is valid. (2 marks)

✅ Correct Answer

The statement is not valid.
One hydrogen ion is consumed in Step 1 but one is regenerated in Step 3 (net zero change in H⁺), meaning it is indeed acting as a catalyst.
The second hydrogen ion formed originates from the propanone molecule itself when replaced by iodine, not from the catalyst pool.

❌ Common Errors

Failing to track where each individual H⁺ goes across the multi-step mechanism. Students often get confused by counting total H⁺ on page-level summaries instead of tracing individual ionic species through the stepwise pathway.

Mark Breakdown: 1 mark for stating the statement is invalid and noting that one H⁺ is regenerated/reformed; 1 mark for explaining that the other hydrogen ion comes from the organic reactant (propanone) during substitution.

Topics

Core Practicals · Physical Chemistry · 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 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.