AQA A-Level Chemistry Paper 2, 2024: Question 1

11 marks · Medium difficulty · State/Explain/Numerical

Kinetics of Propanone with Bromine: Rate Equations and Mechanism

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Question text

01 Propanone reacts with bromine in alkaline conditions.

CH COCH + Br + OH– ⟶ CH COCH Br + Br– + H O

33 2 3 2 2

The rate equation for this reaction is

Rate = k [CH COCH ] [OH–]

01.1 Sketch a graph on the axes provided to show how, at constant temperature, the

concentration of bromine changes during this reaction.

[1 mark]

01.2 Table 1 shows the initial rate of this reaction for experiments using different mixtures

containing propanone, bromine and hydroxide ions.

Table 1

[CH COCH ] [Br ] [OH–] Initial rate

Experiment 3 3 2

/ mol dm–3 / mol dm–3 / mol dm–3 / mol dm–3 s–1

11.50 × 10–2 2.50 × 10–2 2.50 × 10–2 2.75 × 10–11

21.50 × 10–2 2.50 × 10–2 8.25 × 10–11

33.75 × 10–3 5.00 × 10–2 1.00 × 10–1

Complete Table 1.

Use the data from experiment 1 to calculate the rate constant k for this reaction.

Give the units for the rate constant.

[5 marks]

k 4 Units

01.3 Figure 1 shows an incomplete mechanism for this reaction.

Figure 1

Complete the mechanism in Figure 1 by adding four curly arrows and any relevant

lone pair(s) of electrons.

[4 marks]

01.4 Use evidence from the rate equation to explain why Step 1 is the

rate determining step.

[1 mark]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidelines Mark

Curve eg Curve with decreasing gradient

01.1 A straight line with a negative gradient was also

(1 x AO1)

allowed.

(This is only appropriate if the concentrations of the

other reagent are in such large excess that they are

effectively constant, but, since the concentrations

of the other reagents are not specified, credit was

given for taking this approach.)

– A-LEVEL CHEMISTRY – – JUNE 2024

M1 For [OH–] = 7.50 × 10–2

M2 For rate = 2.75 × 10–11

rate M3 For rearranging rate equation

M3 k = [ ][ –]

CH3 COCH3 OH

01.2 –11

2.75 × 10 (5 x AO2)

OR k = –2 –2 For inserting correct numbers in rearranged

(1.5 × 10 ) × (2.5 × 10 ) equation 11

M4 k = 7.3(3) × 10–8

If rearrangement upside down lose M3

but can score M4 for 1.36 × 107 as ECF

M5 units = mol–1dm3s–1

M5 for mol dm-3 s as ECF

– A-LEVEL CHEMISTRY – –

01.3

(4 x AO2)

M1 Arrow from C–H bond to C-C

M2 Arrow from C=O bond to O

M3 Arrow from lone pair on O to C–O bond

Dipoles must be correct if shown for M4

M4 Arrow from Br–Br bond to Br

– A-LEVEL CHEMISTRY – – JUNE 2024

Step 1 includes CH COCH and OH– and these are also in the rate Br not in step 1 and not in rate equation so it has

33 2

equation to be step 1 1

01.4

Or (1 x AO2)

Step 1 contains all the species in the rate equation

How to answer it

Propane–Bromine in Alkaline Conditions: Rate Equation, k & Mechanism

What this question tests

Understanding kinetics of a second-order reaction, including sketching concentration–time graphs, using initial rate data to deduce a rate constant and its units, interpreting and completing a curly-arrow mechanism, and explaining why step 1 is the rate-determining step using evidence from the rate equation.

Part (a)

Sketching the concentration–time graph for Br₂

✅ Correct feature

A curve with decreasing gradient, starting at the initial [Br₂] and approaching zero as reaction proceeds.

1 mark

❌ Common error

Drawing a straight line (constant rate) instead of a curved line.

Part (b)

Using initial rate data to calculate k

📐 Calculation steps

Rate law: rate = k [CH₃COCH₃][OH⁻]
Substitute experiment 1 data:
2.75 × 10⁻¹¹ = k (1.50 × 10⁻²)(2.50 × 10⁻²)
Solve for k:
k = 7.3 × 10⁻⁸
Units: mol⁻¹ dm³ s⁻¹

✅ Correct answer

k = 7.3 × 10⁻⁸ mol⁻¹ dm³ s⁻¹

5 marks

🧠 Exam technique

Use initial rates and only include reactants present in the rate equation (ignore Br₂, zero order).

Part (c)

Curly-arrow mechanism

✅ Steps

Arrow from C–H bond to adjacent C–C bond.
Arrow from C=O π bond to O atom.
Arrow from lone pair on O⁻ to C–O bond reforming.
Arrow from Br–Br bond to Br atom (to form Br⁻).

4 marks

❌ Common errors

Incorrect dipoles on Br₂.
Forgetting one curly arrow (only 3 given).

Part (d)

Why step 1 is rate-determining

✅ Explanation

Step 1 involves CH₃COCH₃ and OH⁻, which are the only species in the rate equation. Since Br₂ is not included, it must react in a later, faster step.

1 mark

Final takeaways

💡 Key knowledge

Zero-order species do not affect the rate, even if present in the reaction.
Rate constants must always include correct units.

🧠 Exam technique

Always check which reactants appear in the rate law.
Curly arrows must show electron pair movement — wrong direction = no marks.

❌ Common errors

Straight-line graph instead of decreasing curve.
Using Br₂ concentration in the rate law.
Incomplete curly-arrow mechanism.

Topics

Physical Chemistry · Organic Chemistry · 3.1.9 Rate Equations · 3.3.1 Introduction to Organic Chemistry

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.