A Level Chemistry: Rate Equations and Catalysis — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — 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

---

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]

---

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]

---

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].

---

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.