OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 17

6 marks · Hard difficulty · Extended Response

Explain how reaction orders can be determined from experimental initial rates data, and determine the rate equation and rate constant for the reaction.

Practise this question

Question

An exam question showing the chemical equation for the production of bromine from bromide, bromate, and hydrogen ions. A table provides initial concentrations of Br-, BrO3-, and H+ alongside initial rates for four experiments. The student is asked to explain how reaction orders are determined from the results, and to determine the rate equation and rate constant.
Question text

Bromine, Br2, can be produced by the reaction:

5Br–(aq) + BrO –(aq) + 6H+(aq) 3Br (aq) + 3H O(l)

32 2

A student investigates the rate of this reaction by carrying out four experiments at the same

temperature. The student’s results are shown below.

[Br –] [BrO –] [H+] Initial rate

Experiment 3

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

12.00 × 10–2 1.20 × 10–1 8.00 × 10–2 2.52 × 10–4

26.00 × 10–2 1.20 × 10–1 8.00 × 10–2 7.56 × 10–4

34.00 × 10–2 6.00 × 10–2 8.00 × 10–2 2.52 × 10–4

42.00 × 10–2 6.00 × 10–2 4.00 × 10–1 3.15 × 10–3

Explain how the reaction orders can be determined from the student’s results, and determine the

rate equation and rate constant for this reaction.

… [6]

Additional answer space if required.

Mark scheme

Show the mark scheme A mark scheme showing a 6-mark level-based response grid. Level 3 (5-6 marks) requires all 3 correct orders linked to explanations, rate equation, and rate constant. Guidance lists indicative points including 1st order with respect to Br-, 1st order with respect to BrO3-, 2nd order with respect to H+, the rate equation, and calculations for the rate constant k with units dm9 mol-3 s-1.

AO

Question Answer Marks Guidance

element

17 Please refer to the marking instructions on page 5 of this 6 Indicative scientific points may include

mark scheme for guidance on marking this question. Orders

AO3.1 • 1st order wrt Br–

Level 3 (5–6 marks) 3 • 1st order wrt BrO –

ALL 3 correct orders linked to explanations • 2nd order wrt H+

AND rate equation AND rate constant AO3.2

3 Rate equation

There is a well-developed line of reasoning which is • rate = k [Br–] [BrO –] [H+]2

clear and logically structured.

Calculation of k from any row of data, e.g.

Level 2 (3–4 marks) Rate

Three correct orders k =

[Br⁻][BrO₃⁻][H⁺]²

AND two out of: 2.52 x 10⁻⁴

some evidence of an explanation linked to an order k = 0.020 x 0.120 x (0.080)² = 16.4(0625)

rate equation

rate constant

OR 12

Three correct orders

with an attempt at:

Some evidence of an explanation link to an order ------------------------------------------------------------------

rate equation Explanations from results e.g.

rate constant Br– [Br–] 3 rate 3 Expts 1 and 2

OR

BrO – [Br–] 2 AND [BrO –] 2

Two correct orders linked to explanations 3 3

AND rate equation rate: no change Expts 1 and 3

AND rate constant consistent with the OR

[Br–] 2/3 AND [BrO –] 2

candidate’s orders 3

rate: 1/3 Expts 2 and 3

H+ [BrO –] 2 AND [H+] 5

There is a line of reasoning with some structure and rate 12.5 Expts 1 and 4

supported by some evidence. OR

[Br–] 3 and [BrO –] 2 and [H+] 5

rate 4.17 Expts 2 and 4

Level 1 (1–2 marks) OR

Two correct orders [Br–] 2 and [H+] 5

OR rate 12.5 Expts 3 and 4

One correct order

AND attempts to determine rate equation OR rate ALLOW a sequential approach where they

constant. apply known orders first

OR

One correct order ALLOW minor slips as we are looking for an

AND attempts an explanation. holistic approach to LoR marking

There is an attempt at a logical structure with a

reasoned conclusion from the evidence. NOTE: A clear and logically structured

response would link orders to the experiment

0 marks No response worthy of credit. and experimental results provided. They could

provide units

Units

dm9 mol–3 s–1

ALLOW any order, e.g. mol–3 dm9 s–1

How to answer it

Difficulty: Hard

Determining Reaction Orders, Rate Equation, and Rate Constant

What this question tests

This 6-mark extended-response question assesses your ability to interpret initial rate-concentration data. You must logically deduce the individual orders with respect to three separate reactants (Br⁻, BrO₃⁻, and H⁺), construct an overall rate equation, and calculate the rate constant ( k ) including its correct derived units. Level-of-response marking means your explanation must be structured, clear, and mathematically justified.

Complete Examination Solution & Examiner Guidance

Question 17: Kinetics Analysis (6 Marks)

✅ Final Correct Answers

  • Order wrt Br⁻: 1st order
  • Order wrt BrO₃⁻: 1st order
  • Order wrt H⁺: 2nd order
  • Rate Equation: rate = k [Br⁻][BrO₃⁻][H⁺]²
  • Rate Constant ( k ): 16.4
  • Units of k : dm⁹ mol⁻³ s⁻¹

💡 Key Knowledge

  • Initial Rates Method: Compare experiments where the concentration of only one reactant changes while others remain constant.
  • Proportionality: If concentration doubles and rate doubles, order = 1. If concentration doubles and rate quadruples (2²), order = 2.
  • Units Derivation: Always cancel units algebraically from the rearranged rate equation: k = rate / ([A][B][C]²) .

🧠 Exam Technique (Level of Response)

  • To achieve Level 3 (5–6 marks), state your pairings explicitly (e.g., "Comparing Experiments 1 and 2...").
  • Do not just write final numbers; show how changing one concentration impacts the rate while keeping others constant.
  • Ensure your rate constant matches your determined rate equation, even if your intermediate orders contained an error (error carried forward).

❌ Common Student Errors

  • Confusing Experiments: Failing to hold other reactant concentrations constant when deducing an order (e.g., trying to compare Experiments 1 and 3 without accounting for changes in both Br⁻ and BrO₃⁻).
  • Unit Blindness: Forgetting units for k or writing incorrect compound powers like mol⁻² dm⁶ .
  • Calculation Slips: Forgetting to square the hydrogen ion concentration [H⁺]² when calculating k .

📐 Step-by-Step Calculation Guide

Step 1: Determine the order with respect to Br⁻ (Compare Experiments 1 and 2)

  • From Exp 1 to Exp 2, [Br⁻] triples (from 2.00 × 10⁻² to 6.00 × 10⁻² ), while [BrO₃⁻] and [H⁺] stay constant.
  • The initial rate triples (from 2.52 × 10⁻⁴ to 7.56 × 10⁻⁴ ).
  • Since rate is proportional to concentration changes, the reaction is 1st order with respect to Br⁻.

Step 2: Determine the order with respect to BrO₃⁻ (Compare Experiments 1 and 3)

  • From Exp 1 to Exp 3, [Br⁻] doubles (from 2.00 × 10⁻² to 4.00 × 10⁻² ) and [BrO₃⁻] halves (from 1.20 × 10⁻¹ to 6.00 × 10⁻² ), while [H⁺] stays constant.
  • We know doubling [Br⁻] multiplies the rate by 2. However, the overall rate stays completely unchanged ( 2.52 × 10⁻⁴ ).
  • Therefore, the halving of [BrO₃⁻] must have cancelled out the doubling effect of [Br⁻] . Halving concentration changes rate by a factor of ( 1/2 )ⁿ = change factor. Since rate factor is 1 / 2 = 2 × (rate factor from Br⁻) , solving gives n = 1 . Thus, the reaction is 1st order with respect to BrO₃⁻. (Alternatively, compare Experiments 2 and 3 where [Br⁻] changes by 1.5 and [BrO₃⁻] halves).

Step 3: Determine the order with respect to H⁺ (Compare Experiments 1 and 4)

  • From Exp 1 to Exp 4, [Br⁻] is constant, [BrO₃⁻] halves (from 1.20 × 10⁻¹ to 6.00 × 10⁻² ), and [H⁺] is multiplied by 5 (from 8.00 × 10⁻² to 4.00 × 10⁻¹ ).
  • Because [BrO₃⁻] halves, it reduces the rate by a factor of 2 (since it is 1st order).
  • The expected rate from the BrO₃⁻ change alone would be 2.52 × 10⁻⁴ ÷ 2 = 1.26 × 10⁻⁴ .
  • However, the actual rate is 3.15 × 10⁻³ . The overall rate factor is 3.15 × 10⁻³ / 1.26 × 10⁻⁴ = 25 .
  • Since [H⁺] increased by a factor of 5, and 5² = 25 , the reaction is 2nd order with respect to H⁺.

Step 4: Calculate the Rate Constant ( k ) and Units

  • Rearrange the rate equation for k : k = rate / ([Br⁻][BrO₃⁻][H⁺]²)
  • Substitute values from Experiment 1:
    k = (2.52 × 10⁻⁴) / ((2.00 × 10⁻²) × (1.20 × 10⁻¹) × (8.00 × 10⁻²)²)
  • k = 2.52 × 10⁻⁴ / (2.00 × 10⁻² × 1.20 × 10⁻¹ × 6.40 × 10⁻³)
  • k = 2.52 × 10⁻⁴ / 1.536 × 10⁻⁵ = 16.4 (to 3 sig figs)
  • Units substitution:
    k = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)(mol dm⁻³)(mol dm⁻³)²) = mol dm⁻³ s⁻¹ / mol⁴ dm⁻¹² = dm⁹ mol⁻³ s⁻¹
Mark Scheme Allocation (6 Marks Total):
• Level 3 (5–6 marks): All 3 correct orders clearly justified with logical explanations, correct rate equation, and correct rate constant with units.
• Level 2 (3–4 marks): 3 correct orders with partial reasoning, or 2 correct orders with complete rate equation and consistent calculations.
• Level 1 (1–2 marks): 2 correct orders, or 1 correct order with valid attempts at the rate equation/constant.

Topics

Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.