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R2.2 Exam Practice

Test your knowledge on Rate of Change

Section B: Data Analysis (Paper 1B Style)

Calculator and Data Booklet permitted. Show all working clearly.

Question 1: Rate of Decomposition Calculate

5 marks

A student investigates the decomposition of hydrogen peroxide:
2H_2O_2(aq) → 2H_2O(l) + O_2(g)
The following data were recorded:

Time / s[H₂O₂] / mol dm⁻³
00.500
100.350
200.245
300.171

(a) Describe the shape of the graph if [H₂O₂] is plotted against time. [1]

(b) Calculate the average rate of reaction between t = 10 s and t = 20 s. Include units. [2]

(c) Explain, using collision theory, why the rate of reaction decreases over time. [2]

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(a) A curve with a negative gradient that becomes less steep over time / a decreasing exponential curve [1]

(b) Change in concentration = 0.350 − 0.245 = 0.105 mol dm⁻³ [1]

Average rate = 0.105 / 10 = 0.0105 mol dm⁻³ s⁻¹ [1]

(c) As the reaction proceeds, [H₂O₂] decreases [1]; therefore the frequency of successful collisions between reactant particles decreases [1]

Examiner tip: Always give the rate as a positive value. The negative sign in Δ[reactant]/Δt indicates the concentration is decreasing, but rate is defined as positive.

Question 2: Maxwell-Boltzmann Distribution Explain

4 marks

(a) Sketch how the Maxwell-Boltzmann distribution curve changes when the temperature increases from T₁ to T₂. [1]

(b) Explain why a higher temperature increases the rate of reaction, referring to collision theory and activation energy. [2]

(c) State how a catalyst affects the distribution differently from an increase in temperature. [1]

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(a) T₂ curve: flatter (lower peak) AND shifted to the right (higher average energy). Both features required. [1]

(b) At higher temperature, a greater proportion of particles have kinetic energy ≥ Eₐ [1]; this results in a higher frequency of successful collisions [1]

(c) A catalyst does not change the shape or position of the distribution curve; it provides an alternative pathway with a lower Eₐ [1]

Examiner tip: A common error is stating that a catalyst "gives particles more energy"; this is incorrect. A catalyst lowers the Eₐ so more particles already have sufficient energy.
Links to: R1.1.1 Energy Profile Diagrams

Section C: Structured Questions (Paper 2 Style)

Show all working. State answers with appropriate significant figures and units.

Question 3: Activation Energy and Catalysis Define

5 marks

(a) Define activation energy (Eₐ). [1]

(b) Explain the role of a catalyst in a chemical reaction. [2]

(c) Distinguish between homogeneous and heterogeneous catalysis. [2]

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(a) The minimum energy required for a reaction to occur / the energy needed to reach the transition state [1]

(b) A catalyst provides an alternative reaction pathway [1] that has a lower activation energy [1]

(c) Homogeneous: catalyst is in the same phase as the reactants [1]; Heterogeneous: catalyst is in a different phase from the reactants [1]

Examiner tip: Never say a catalyst "speeds up" the reaction without explaining the mechanism. Always state: alternative pathway → lower Eₐ → greater proportion of particles with sufficient energy.

Question 4: Rate Expression HL Deduce

4 marks

The reaction between compounds A and B was studied at constant temperature:

Experiment[A] / mol dm⁻³[B] / mol dm⁻³Initial rate / mol dm⁻³ s⁻¹
10.100.102.0 × 10⁻³
20.200.104.0 × 10⁻³
30.100.208.0 × 10⁻³

(a) Deduce the order of reaction with respect to A and with respect to B. [2]

(b) Write the rate expression for this reaction. [1]

(c) Calculate the rate constant k using data from Experiment 1. State its units. [1]

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(a) Order w.r.t. A: first order (1). [A] doubles (Exp 1→2), rate doubles. [1]

Order w.r.t. B: second order (2). [B] doubles (Exp 1→3), rate quadruples (×4). [1]

(b) Rate = k[A][B]² [1]

(c) k = Rate/[A][B]² = 2.0 × 10⁻³/(0.10)(0.10)² = 2.0 mol⁻² dm⁶ s⁻¹ [1]

Examiner tip: Always show the method for deducing orders: compare experiments where only one concentration changes. State what happens to the rate when the concentration doubles/triples.

Question 5: The Arrhenius Equation HL Calculate

5 marks

The Arrhenius equation is k = Ae⁻Eₐ/RT.

(a) Explain the significance of the pre-exponential factor A and the exponential term e⁻Eₐ/RT. [2]

(b) For a reaction, k = 1.25 × 10⁻³ s⁻¹ at 300 K and k = 3.41 × 10⁻³ s⁻¹ at 310 K. Calculate Eₐ in kJ mol⁻¹. (R = 8.31 J K⁻¹ mol⁻¹) [3]

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(a) A (pre-exponential factor) represents the frequency of collisions and the probability of favourable orientation [1]

e⁻Eₐ/RT represents the fraction of particles with kinetic energy ≥ Eₐ [1]

(b)

  • ln(k_2/k_1) = Eₐ/R(1/T_1 - 1/T_2) [1]
  • ln(3.41 × 10⁻³/1.25 × 10⁻³) = Eₐ/8.31(1/300 - 1/310) → 1.004 = Eₐ/8.31 × 1.075 × 10⁻⁴ [1]
  • Eₐ = 77 600 J mol⁻¹ = 77.6 kJ mol⁻¹ [1]
Examiner tip: The two-point Arrhenius equation avoids needing to know A. Always check your units: if R is in J K⁻¹ mol⁻¹, Eₐ comes out in J mol⁻¹; don't forget to convert to kJ.
Links to: R2.2.3 Arrhenius Equation
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