Edexcel A-Level Chemistry Paper 3, June 2023: Question 8

13 marks · Hard difficulty · Calculations

Deduce experimental techniques, write rate equations, sketch Maxwell-Boltzmann distributions, and use the Arrhenius equation to determine activation energy and calculate the effect of temperature and catalysts on reaction rates.

Practise this question

Question

A multipart question about reaction kinetics, Maxwell-Boltzmann distributions, and the Arrhenius equation. Part (a) asks for two experimental techniques for an iodine clock reaction, a reason for a slowest step, and a rate equation deduction. Part (b) asks to sketch a Maxwell-Boltzmann curve for a higher temperature and explain the rate increase. Part (c) provides Arrhenius equations, a graph of ln k against 1/T for determining activation energy, and a calculation involving the fraction of molecules with energy equal to or greater than the activation energy.
Question text

8 This question is about reaction kinetics and the Arrhenius equation.

(a) Different iodine clock reactions are often used to investigate reaction kinetics.

(i) The iodine clock reaction with hydrogen peroxide involves the

reaction shown.

H O + 2I– + 2H+ → I + 2H O

22 2 2

Deduce two possible experimental techniques which could be used to

monitor the progress of this reaction.

(2)

(ii) The iodate(V) reaction has the rate determining step

IO– + 3HSO– → I– + 3HSO–

33 4

Give a possible reason why this is the slowest step.

(1)

(iii) The chlorate(V) reaction has the rate determining step

ClO3– + 2H+ + I– → HIO + HClO2

Deduce the rate equation for this iodine clock reaction.

(1)

(b) The diagram shows a sketch of the Maxwell-Boltzmann curve for the distribution

of molecular energies of a reaction mixture at temperature 298K.

Number of

molecules with

energy E

*P71914A02132*

Energy E

(i) Add a curve to show the distribution at a temperature of 308K.

(1)

(ii) Explain why a temperature rise from 298K to 308K results in a large increase

in the rate of reaction.

Refer to the Maxwell-Boltzmann distribution in your answer.

(2)

(c) The Arrhenius equation may be written in a logarithmic or an exponential form.

E Ea

−

ln k = – a + ln A k = A e RT

RT

A is a constant.

(i) The rate constant, k, for the isomerisation of cyclopropane to propene was

measured at various temperatures.

The data obtained were used to draw the graph shown.

1 −1

/ K

22 T

1.1 × 10–3*P71914A02232*1.2×10–31.3×10–3 1.4 × 10–3

–2

–4

lnk

–6

–8

–10

Determine the activation energy, Ea, from the gradient of the graph.

Include units in your answer.

(3)

(ii) At a temperature T, the fraction of molecules with energy equal to or greater

than the activation energy is given by the expression

Ea

−

fraction of molecules = e RT

When a catalyst is added, the activation energy for a reaction is lowered.

Explain, using calculations, why lowering the activation energy from

50 000 J mol–1 to 25 000 J mol–1 at 298 K results in a large increase in the

rate of reaction.

(3) 23

*P71914A02332*

(Total for Question 8 = 13 marks)

Mark scheme

Show the mark scheme The mark scheme provides detailed answers and guidance for all parts of question 8, including accepted experimental methods for the clock reaction, Maxwell-Boltzmann sketching rules, gradient calculation steps for the Arrhenius graph, and exponential fraction calculations for uncatalysed and catalysed activation energies.

Question

Answer Additional Guidance Mark

Number

An answer that makes reference to any two of the following points: (2)

8(a)(i)

• colorimetry (1) Ignore just colour change

Do not award calorimetry

• (electrical) conductivity (1)

• quenching

and Allow cooling for quenching

titration with thiosulfate (1)

• quenching with excess carbonate COMMENT

and Allow cooling and titration with alkali

titration with acid (1)

• add fixed amount of sodium thiosulfate and a few drops of starch

solution and find the time until a blue-black colour is seen (1)

Allow dilatometry

Ignore pH

Question

Answer Additional Guidance Mark

Number

An answer that makes reference to following point: (1)

8(a)(ii)

• negative species / ions will repel (each other)

or

unlikely that four species / ions will simultaneously combine

Question

Answer Additional Guidance Mark

Number

(1)

8(a)(iii) • rate = k[ClO −][H+]2[I−] Accept species in any order

Allow rate = k[ClO −]1[H+]2[I−]1

Allow K for k

Allow r/R for rate

Ignore state symbols even if incorrect

Do not award missing charges

Do not award just

k[ClO −][H+]2[I−]

Question

Answer Additional Guidance Mark

Number

(1)

8(b)(i) • (increase in temperature) means peak shifts to the right Do not award the line crossing the other line

and twice

is lower Do not award the curve crossing the x-axis

Do not award a line which goes up on the

right or that plateaus high above the x axis,

e.g.

Question

Answer Additional Guidance Mark

Number

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

8(b)(ii)

• the area under the curve to the right of the Ea line has increased Allow answer/shading on the M-B sketch

(substantially) (1)

• so that a greater proportion of particles exceed the activation energy (1) Allow more molecules/particles have

energy greater than the activation energy

Do not award M2 if there is any reference

to the activation energy decreasing

Question Answer Additional Guidance Mark

Number

An example of calculation (3)

8(c)(i)

• gradient of slope expression (1) Allow gradient = ∆y ÷ ∆x or equivalent expression with values

• calculation of gradient (1) Gradient = ((−8 – −2) ÷ (1.31 × 10−3 – 1.13 × 10−3))

= (−) 33 333 / (−) 33 300 / (−) 33 000 (K)

Allow range (−) 32475 to (−) 34159

• calculation of activation energy with units (1) E = − (−33 333 × 8.31 =) (+)276 997 J mol−1 / (+)277 000 J mol−1 /

a

(+)276.997 kJ mol−1 / (+)277 kJ mol−1

Allow any answer in the range 270 to 284

Ignore SF except 1 SF

Do not award M3 if Ea negative

TE from M2 to M3

Question

Answer Additional Guidance Mark

Number

(3)

8(c)(ii) • calculation of expression with uncatalysed Ea (1) 𝐸a 50000

𝑒− 𝑒−8.31 × 298 −9

𝑅𝑇 = = 1.70 × 10

𝐸a 25000

• calculation of expression with catalysed Ea (1) 𝑒− 𝑒− −5

𝑅𝑇 = 8.31 × 298 = 4.13 × 10

• about 24000 (times) increase −5 −9

Increase = (4.13 × 10 ÷ 1.70 × 10 ) = 24276

and

in the fraction of molecules now able to react (1) −5 −9

Allow reference to 4.13 × 10 >> 1.70 × 10

resulting in many more molecules able to react

Ignore just more molecules or bigger fraction

Allow a calculation involving the

determination of the difference/ratio between

the two values

(Total Question 8 = 13 marks)

How to answer it

Reaction Kinetics and the Arrhenius Equation

What this question tests

This comprehensive A-Level question evaluates your mastery of reaction kinetics, experimental monitoring techniques, rate-determining steps, Maxwell-Boltzmann distribution curves, and the Arrhenius equation. You will be tested on data extraction from graphs, recognizing collision theory factors, and executing multi-step mathematical calculations.

Question 8 (a)

Experimental Techniques, Rate-Determining Steps & Rate Equations

✅ Correct Answers

  • 8(a)(i): Any two from: colorimetry, electrical conductivity, quenching and titration with thiosulfate, quenching with excess carbonate and titration with acid, or adding a fixed amount of sodium thiosulfate and starch solution (time to blue-black colour).
  • 8(a)(ii): Negative species/ions will repel each other, OR it is statistically unlikely for four species/ions to simultaneously collide and combine.
  • 8(a)(iii): rate = k[ClO₃⁻][H⁺]²[I⁻]

💡 Key Knowledge

  • Monitoring progress: Choose methods that track changing concentrations (colour intensity via light absorption, conductivity via ion counts, or titrimetric quenching).
  • RDS Probability: Elementary steps involving 3 or more particles colliding simultaneously are extremely rare, making reactions proceed via multi-step mechanisms where the slowest step dictates overall rate.
  • Stoichiometry & Rate Equations: Reactants in the rate-determining step directly inform the orders in the rate equation.

❌ Common Errors

  • Writing vague answers like "measure colour change" instead of naming colorimetry.
  • Omitting charge signs or writing incorrect stoichiometric powers in part (iii).
  • Forgetting square brackets around concentration terms in the rate equation.
Total for 8(a): 4 marks
Question 8 (b)

Maxwell-Boltzmann Distribution & Temperature Effects

✅ Correct Answers

  • 8(b)(i): Sketch a curve at 308 K where the peak is shifted to the right (higher energy) and is lower in height than the 298 K curve.
  • 8(b)(ii): The area under the curve to the right of the activation energy (Eₐ) line increases substantially, meaning a greater proportion of particles exceed the activation energy.

🧠 Exam Technique

  • When sketching M-B curves for higher temperatures, ensure the curve never crosses the other line twice, never touches the x-axis, and does not arbitrarily plateau high up. The total area under the curve must remain constant (representing total molecules).
  • Always link temperature increases back to collision frequency and the proportion of successful collisions with energy ≥ Eₐ.

❌ Common Errors

  • Drawing a higher temperature curve that is taller and narrower (the exact opposite of correct behavior).
  • Failing to mention the "area under the curve" in explanations, losing accessible marks.
Total for 8(b): 3 marks
Question 8 (c)

The Arrhenius Equation & Activation Energy Calculations

📐 Step-by-Step Calculations

Part (i): Determining Activation Energy from a Graph

  1. Find Gradient: Use large coordinate triangles.
    Gradient = Δ(&ln; k) / Δ(1 / T) = (-8 - (-2)) / (1.31×10⁻³ - 1.13×10⁻³) = -6 / (1.8×10⁻⁴) = -33 333 K.
  2. Relate to Arrhenius Equation: Since &ln; k = (-Eₐ / R)(1 / T) + &ln; A, the gradient equals -Eₐ / R.
  3. Calculate Eₐ: Eₐ = -1 × gradient × R = -(-33 333) × 8.31 = +277 000 J mol⁻¹ (or 277 kJ mol⁻¹). Acceptable range: 270 to 284 kJ mol⁻¹.

Part (ii): Impact of Catalysts on Fraction of Molecules

  1. Uncatalysed fraction: e^(-50000 / (8.31 × 298)) = 1.70 × 10⁻⁹
  2. Catalysed fraction: e^(-25000 / (8.31 × 298)) = 4.13 × 10⁻⁵
  3. Determine increase: (4.13 × 10⁻⁵) / (1.70 × 10⁻⁹) ≈ 24 276 times increase (~24 000 times).

🧠 Exam Technique & Traps

  • Unit Watch: Activation energy is typically calculated in J mol⁻¹ but examiners often expect or accept kJ mol⁻¹. Always verify whether your final answer requires standard scientific notation or standard units.
  • Sign Convention: The gradient is negative, but activation energy Eₐ must always be expressed as a positive value. A negative Eₐ will lose marks immediately!
  • Significant Figures: Give your final answers to 2 or 3 significant figures unless specified otherwise.
Total for 8(c): 6 marks | Total for Question 8: 13 marks

Topics

Physical Chemistry · Core Practicals · Core Practical 13b: Use a clock reaction to determine a rate equation · Topic 9: Kinetics I · Topic 16: Kinetics II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.