OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2019: Question 21

6 marks · Hard difficulty · Extended Response

Explain how reaction orders are determined from students' results for the reaction between bromine and propanone, and determine the rate equation and rate constant.

Practise this question

Question

The question presents an investigation into the reaction between bromine and propanone in the presence of hydrochloric acid. Three sets of results are provided: Student 1 shows a graph of [Br2(aq)] against time with a straight decreasing line, Student 2 shows a graph of initial rate against [CH3COCH3(aq)] with a straight line through the origin, and Student 3 shows a table of initial concentrations and initial rates for two experiments. Students must explain how reaction orders are determined from these results and find the rate equation and rate constant.
Question text

Three students carry out a rates investigation on the reaction between bromine and propanone in

the presence of hydrochloric acid.

CH3COCH3(aq) + Br2(aq) CH3COCH2Br(aq) + HBr(aq)

Each student investigates the effect of changing the concentration of one of the reactants whilst

keeping the other concentrations constant.

Their results are shown below.

Results of student 1

[Br2(aq)]

/ mol dm–3

0 time/s

Results of student 2

initial rate

/ mol dm–3s–1

0 [CH COCH (aq)] / mol dm–3

Results of student 3

[Br (aq)] [CH COCH (aq)] [H+(aq)] Initial rate

Experiment 2 3 3

/ mol dm−3 / mol dm−3 / mol dm−3 / 10−5 mol dm−3 s−1

1 0.004 1.60 0.20 1.25

2 0.004 1.60 0.40 2.50

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

rate equation and rate constant. [6]

Additional answer space if required

Mark scheme

Show the mark scheme The mark scheme uses a levels-based response structure from Level 1 (1-2 marks) to Level 3 (5-6 marks) for determining all orders, the rate equation, and the rate constant. Indicative points include identifying zero order with respect to Br2 from the constant negative gradient, first order with respect to CH3COCH3 from the straight line through the origin, and first order with respect to H+ from Student 3's data. It also details calculations for the rate equation rate = k[CH3COCH3][H+] and the rate constant with units.

AO

Question Answer Marks Guidance

element

21 Please refer to the marking instructions on page 4 of this 6 3.1 ×4 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. 3.2 ×2 Orders

Student 1

Level 3 (5–6 marks) • zero order wrt Br

Most evidence used to determine the correct orders

Student 2

AND rate equation AND rate constant.

• 1st order wrt CH3COCH3

There is a well-developed line of reasoning which is clear Student 3

and logically structured. The information presented is • 1st order wrt H+

relevant and substantiated.

Explanations

Level 2 (3–4 marks) Student 1

Some evidence used to determine two orders correctly AND • constant gradient OR linear negative gradient

rate equation AND rate constant consistent with orders. OR constant rate OR rate independent of

OR

concentration OR decreasing half-life

Little evidence used to determine all three orders correctly

AND rate equation AND rate constant. Student 2

• straight line through 0,0

There is a line of reasoning presented with some structure. • OR rate directly proportional to [CH3COCH3]

The information presented is in the most part relevant and OR [CH COCH ] × 2, rate × 2

supported by some evidence.

Student 3

• [H+] × 2, rate × 2

Level 1 (1–2 marks)

Little evidence used to determine two orders correctly

OR Rate equation, rate constant and units

• rate = k [CH COCH ] [H+]

One order correct, with attempt to determine the rate 3 3

equation AND rate constant. ALLOW rate = k [Br ]0 [CH COCH ]1 [H+]1

23 3

rate 1.25 × 10–5

There is an attempt at a logical structure with a line of • k = + OR

[ CH3COCH3] [H ] 1.6 × 0.2

reasoning. The information is in the most part relevant.

• k = 3.9… × 10–5

0 marks • units: dm3 mol–1 s–1 (Any order, e.g.

No response or no response worthy of credit. –1 3 –1

mol dm s )

Total 6

How to answer it

Determining Orders, Rate Equation, and Rate Constant

OCR A-Level Chemistry • Rates of Reaction

What this question tests

This 6-mark extended response question tests your ability to interpret different graphical and tabular representations of kinetic data. You must deduce individual reactant orders from concentration-time graphs, initial rate-concentration graphs, and experimental data tables, combine them into an overall rate equation, and calculate the rate constant with correct units and significant figures.

Complete Examination Breakdown

✅ Full Mark Answers (Level 3)

  • Student 1 Order: Zero order with respect to Br₂ (constant gradient / linear negative correlation).
  • Student 2 Order: First order with respect to CH₃COCH₃ (straight line through origin / rate directly proportional).
  • Student 3 Order: First order with respect to H⁺ ([H⁺] doubles, initial rate doubles).
  • Rate Equation: rate = k[CH₃COCH₃][H⁺] (or including [Br₂]⁰ ).
  • Rate Constant & Units: k = 3.9 × 10⁻⁵ dm³ mol⁻¹ s⁻¹ .

💡 Key Knowledge

  • Graph 1 (Concentration-Time): A straight line with a constant negative gradient indicates zero order because the rate of reaction does not change as reactant is consumed.
  • Graph 2 (Initial Rate-Concentration): A straight line passing through the origin shows direct proportionality (first order).
  • Data Table: Comparing Experiment 1 and 2 shows that doubling [H⁺] from 0.20 to 0.40 doubles the initial rate from 1.25 to 2.50 × 10⁻⁵.

🧠 Exam Technique & Marking Strategy

  • This question is marked using a levels-of-response rubric (Level 1: 1–2 marks, Level 2: 3–4 marks, Level 3: 5–6 marks).
  • To secure Level 3, you must correctly identify all three orders, state the correct rate equation, and successfully calculate the numerical value and units of k with clear supporting workings.
  • Structure your answer clearly by addressing each student's findings individually before pulling it all together for the rate constant calculation.

❌ Common Student Errors

  • Misinterpreting the concentration-time graph for Student 1 as first-order due to it being a downward slope.
  • Forgetting to factor in the scale multiplier on the y-axis for Student 3's table ( × 10⁻⁵ ).
  • Dropping or scrambling the units for the second-order overall rate constant k .

📐 Step-by-Step Calculation of the Rate Constant (k)

  1. Rearrange the rate equation to make k the subject:
    k = rate / ([CH₃COCH₃][H⁺])
  2. Substitute values from Experiment 1 into the rearranged expression:
    k = (1.25 × 10⁻⁵) / (1.60 × 0.20)
  3. Calculate the numerical value:
    k = 3.90625 × 10⁻⁵ → rounds to 3.9 × 10⁻⁵ (matching the 2 significant figures used in the data table).
  4. Derive the units:
    k = (mol dm⁻³ s⁻¹) / (mol dm⁻³ × mol dm⁻³) = mol⁻¹ dm³ s⁻¹ (or dm³ mol⁻¹ s⁻¹ ).
Examiner Note: Top-level candidates communicated their logic fluently without leaving gaps in their working, ensuring every substitution for calculating k was explicitly matched to a defined experiment row.

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 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.