A Level Chemistry: Rate Equations and Catalysis — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Reaction Kinetics: Rate Equations and Catalysis study guide

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Section A

1. Define order of reaction with respect to a reactant. [2]

2. State how the half-life of a first-order reaction changes as the reaction proceeds. [1]

3. State the mode of action of a heterogeneous catalyst, in three stages. [3]

4. Explain why a homogeneous catalyst never appears in the overall balanced equation for the reaction it catalyses. [2]

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Section B

5. The following initial-rate data were obtained for the reaction P + Q → products:

Experiment	[P] / mol dm⁻³	[Q] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.050	0.050	1.2 × 10⁻³
2	0.100	0.050	2.4 × 10⁻³
3	0.100	0.100	2.4 × 10⁻³

(a) Deduce the order of reaction with respect to P and with respect to Q. [3]

(b) Write the rate equation and state the overall order. [2]

(c) Calculate the rate constant, including units. [2]

6. A first-order reaction has a rate constant k = 4.62 × 10⁻³ s⁻¹.

(a) Calculate the half-life of the reaction. [2]

(b) State how you could confirm from a concentration-time graph that the reaction is first order, without calculating k. [2]

7. A reaction 2X + Y → Z is thought to proceed by the mechanism:

Step 1 (slow): X + Y → XY

Step 2 (fast): XY + X → Z

(a) Deduce the rate equation implied by this mechanism. [2]

(b) Identify the intermediate. [1]

(c) The experimentally determined rate equation is rate = k[X][Y]. State whether this is consistent with the proposed mechanism, explaining your answer. [2]

8. Explain, using the Haber process (iron catalyst) as your example, how a heterogeneous catalyst increases the rate of a reaction. [4]

9. A catalytic converter contains platinum, palladium and rhodium, and catalyses 2CO + 2NO → 2CO₂ + N₂.

(a) State the type of catalysis involved, and outline the mechanism in general terms. [3]

10. The reaction between I⁻ and S₂O₈²⁻ is catalysed by Fe²⁺/Fe³⁺.

(a) Explain why the uncatalysed reaction between I⁻ and S₂O₈²⁻ is slow. [1]

(b) Write two equations showing how Fe²⁺/Fe³⁺ catalyses the reaction, and explain how this avoids the problem in (a). [3]

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11. The rate equation for a reaction is rate = k[A]²[B].

(a) State the factor by which the rate changes when [A] is doubled and [B] is halved at the same time. [1]

(b) Give the units of k, with concentrations in mol dm⁻³ and time in s. [1]

Answers

1. The power to which that reactant’s concentration is raised [1] in the experimentally determined rate equation [1].

2. It stays constant (the same throughout the reaction) [1].

3. Adsorption of reactant molecules onto active sites on the catalyst surface [1]; weakening of bonds within the adsorbed molecules, lowering the activation energy [1]; desorption of product molecules, freeing the active site [1].

4. It is consumed in one step of the mechanism [1] and regenerated in a later step, so it cancels out overall and does not appear in the final balanced equation, even though it genuinely takes part in the reaction pathway [1].

5. (a) Comparing 1 and 2: [P] doubles, [Q] constant, rate doubles — first order with respect to P [1]. Comparing 2 and 3: [Q] doubles, [P] constant, rate unchanged — zero order with respect to Q [1]. [1 for correct method/comparison shown].

(b) rate = k[P] [1]; overall order = 1 [1].

(c) 1.2 × 10⁻³ = k(0.050) [1] → k = 1.2 × 10⁻³/0.050 = 0.024 s⁻¹ [1].

6. (a) t½ = 0.693/k = 0.693/(4.62 × 10⁻³) = 150 s [2].

(b) Successive halvings of concentration take equal time intervals [1] — a constant half-life read directly from the graph confirms first order [1].

7. (a) rate = k[X][Y] (based on the rate-determining step, step 1) [2].

(b) XY [1].

(c) Yes, consistent [1] — the experimental rate equation matches the reactants of the proposed rate-determining step (step 1), and X’s second appearance (in the fast step 2) does not affect the rate equation since that step occurs after the rate-determining step [1].

8. N₂ and H₂ molecules adsorb onto active sites on the iron surface [1]. This adsorption weakens the N≡N and H–H bonds within the adsorbed molecules [1], lowering the activation energy needed for them to react compared with the uncatalysed gas-phase reaction [1]. The NH₃ formed then desorbs from the surface, freeing the active site for further N₂ and H₂ molecules to adsorb and react [1].

9. (a) Heterogeneous catalysis, since the precious metals are a different physical state (solid) from the gaseous reactants [1]. NO and CO adsorb onto the metal surface, where bond weakening lowers the activation energy for the reaction between them, before the CO₂ and N₂ products desorb, freeing the active site [1] [1].

10. (a) Both I⁻ and S₂O₈²⁻ are negatively charged, so they repel each other, making a direct collision between them slow and unlikely [1].

(b) 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻, then 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ [2]. Because Fe³⁺ is a cation, its reaction with I⁻ does not face the anion–anion repulsion that makes the direct route slow, and Fe²⁺ is regenerated at the end so it can catalyse further reaction [1].

11. (a) 2² × ½ = 2, so the rate doubles [1]. (b) k = rate ÷ ([A]²[B]) so the units are mol⁻² dm⁶ s⁻¹ [1].

Examiner insight (June 2025): common errors were multiplying by the wrong factor and writing overall equations that were not consistent with the orders in the rate equation. Apply each order to its own reactant. Try the real question next: Cambridge International AS & A Level Chemistry 9701, June 2025, Paper 42, Question 4.

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Where marks are usually lost

• Reading orders of reaction directly off the balanced equation instead of from the experimental data given.
• Forgetting that a constant half-life is unique to first-order reactions — stating it “gets shorter” or “gets longer” without checking which order applies.
• Including a homogeneous catalyst’s formula in the final balanced equation.
• In mechanism questions, forgetting that species appearing only after the rate-determining step do not affect the rate equation.