OxfordAQA A-Level Chemistry: Rate equations (9620) – 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 from your board or school.

These questions cover section 3.1.11 Rate equations of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: 3.1.11.1, 3.1.11.2 and Required practical 8. Being International A2 material, it is International A-level only. The Unit 4 paper (Organic 2 and Physical 2) tests it, and the Unit 5 paper (Practical and synoptic) may test any topic.

Learn the ideas first in the study guide and revision notes. See also the course hub, the printable checklist and the free 10-minute diagnostics. Use R = 8.31 J K⁻¹ mol⁻¹. Temperature is constant unless stated.

Questions

1. State what is meant by the order of a reaction with respect to one reactant, and by the rate constant. Explain why orders cannot be worked out from the balanced equation. [3]

2. For the reaction 2V + Z → products, Rate = k[V]²[Z].

(a) The rate is 3.0 × 10⁻⁶ mol dm⁻³ s⁻¹ when [V] = 5.0 × 10⁻³ mol dm⁻³ and [Z] = 2.0 × 10⁻³ mol dm⁻³. Calculate k, stating its units. [2] (b) [V] is reduced to one-third of its value and [Z] is tripled. Calculate the factor by which the rate changes. [2]

3. Initial rates were measured for the reaction B + C + 2D → products.

Run	[B] / mol dm⁻³	[C] / mol dm⁻³	[D] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.040	0.010	0.15	2.4 × 10⁻⁵
2	0.080	0.010	0.15	4.8 × 10⁻⁵
3	0.040	0.030	0.15	2.16 × 10⁻⁴
4	0.020	0.020	0.45	4.8 × 10⁻⁵
5	0.060	0.025	0.30	to be calculated

(a) Deduce the order with respect to B, C and D. Explain each deduction. [3] (b) Give the full rate equation. [1] (c) Find the value of k, including units. [2] (d) Work out the initial rate in run 5. [1]

4. The concentration of reactant Y was followed during a reaction.

t / s	0	5	10
[Y] / mol dm⁻³	0.1500	0.1488	0.1476

(a) Use these early readings to calculate the initial rate. [1] (b) Later, when [Y] = 0.075 mol dm⁻³, the tangent to the concentration–time curve passes through (230 s, 0.099 mol dm⁻³) and (630 s, 0.051 mol dm⁻³). Calculate the rate at that point. [1] (c) Deduce the order with respect to Y. Explain your answer. [2]

5. Compound D decomposes on a hot metal surface. A plot of rate against [D] is a horizontal straight line at a rate of 3.6 × 10⁻⁵ mol dm⁻³ s⁻¹.

(a) State the order with respect to D. [1] (b) Give the value of k with its units. [2] (c) Calculate the time taken for [D] to fall from 0.090 mol dm⁻³ to 0.018 mol dm⁻³. [1]

6. A second-order reaction has A = 8.0 × 10¹⁰ mol⁻¹ dm³ s⁻¹ and Ea = 82 kJ mol⁻¹. Use k = Ae^(–Ea/RT).

(a) Calculate k at 305 K. [3] (b) Calculate k at 315 K, and the factor by which k increases between the two temperatures. [2]

7. The rate constant of a first-order reaction was measured at four temperatures.

T / K	300	315	330	345
k / s⁻¹	0.00771	0.0317	0.115	0.370

(a) Use ln k = –Ea/RT + ln A to explain why ln k plotted against 1/T gives a straight line, and state what its gradient represents. [2] (b) The four points lie on a straight line. Use the points at 300 K and 345 K to calculate the gradient, then the activation energy in kJ mol⁻¹. [3] (c) Find the value of A, including its units. [2]

8. The reaction G + 3H → GH₃ has the rate equation Rate = k[G][H]².

(a) Suggest a two-step mechanism in which the first step is the rate-determining step. [2] (b) Explain why a single step, G + 3H → GH₃, is not consistent with the rate equation. [1] (c) Name the intermediate in your mechanism. [1]

9. For Required practical 8, a student runs a clock reaction, mixing a 0.200 mol dm⁻³ solution of reactant V with water and fixed volumes of the other solutions (total 50.0 cm³ each run).

Volume of V solution / cm³	5.0	10.0	15.0	20.0
Time to colour change, t / s	96	48	32	24

(a) Calculate the concentration of V in the mixture in the 5.0 cm³ run. [1] (b) Why does 1/t give a measure of the initial rate? Explain. [1] (c) Deduce the order of reaction with respect to V from these results. [2] (d) The student can judge the colour change only to within 1 s. Calculate the percentage uncertainty in the shortest time, and state which run is the least reliable. [2] (e) Suggest one change that would reduce this timing uncertainty. [1]

10. A continuous monitoring method was used for the reaction RCl + H₂O → ROH + H⁺ + Cl⁻, with [RCl] = 0.100 mol dm⁻³ at the start. At intervals, 5.00 cm³ samples were removed, run into ice-cold solvent and titrated with 0.0250 mol dm⁻³ sodium hydroxide.

(a) Explain why each sample is run into ice-cold solvent. [1] (b) The sample taken at 20 min needed 8.00 cm³ of sodium hydroxide. Calculate [RCl] at 20 min. [3] (c) Tangents to the concentration–time curve give a rate of 2.55 × 10⁻³ mol dm⁻³ min⁻¹ at the start and 1.53 × 10⁻³ mol dm⁻³ min⁻¹ at 20 min. Deduce the order with respect to RCl. [2] (d) Calculate k in s⁻¹. [2]

Answers

1. Order: the exponent on that reactant’s concentration in the rate equation [1]. Rate constant: the constant linking the rate to the concentration terms, which is fixed at a given temperature [1]. The rate depends on the mechanism (the rate-determining step), not the overall equation, so orders must be found by experiment [1]. [3] Examiner insight: An order definition without “power” and “rate equation” is usually too vague.

2. (a) k = 3.0 × 10⁻⁶ ÷ ((5.0 × 10⁻³)² × 2.0 × 10⁻³) = 60 [1] mol⁻² dm⁶ s⁻¹ [1] (b) Factor = (⅓)² × 3 [1] = ⅓ (the rate falls to one-third) [1] Examiner insight: Show each reactant’s factor separately so that any slip can still be followed.

3. (a) B: runs 1 and 2, [B] ×2 and rate ×2, so first order [1]. C: runs 1 and 3, [C] ×3 and rate ×9, so second order [1]. D: runs 1 and 4, [B] ×0.5 and [C] ×2 give ×0.5 × 4 = ×2, which is the whole rate change, so D is zero order even though [D] ×3 [1]. (b) Rate = k[B][C]² [1] (c) k = 2.4 × 10⁻⁵ ÷ (0.040 × 0.010²) = 6.0 [1] mol⁻² dm⁶ s⁻¹ [1] (d) Rate = 6.0 × 0.060 × 0.025² = 2.25 × 10⁻⁴ mol dm⁻³ s⁻¹ [1] Examiner insight: For D, show that the effects of B and C account for the whole rate change.

4. (a) Initial rate = (0.1500 – 0.1476) ÷ 10 = 2.4 × 10⁻⁴ mol dm⁻³ s⁻¹ [1] (b) Rate = (0.099 – 0.051) ÷ (630 – 230) = 1.2 × 10⁻⁴ mol dm⁻³ s⁻¹ [1] (c) [Y] halves (0.150 → 0.075) and the rate halves [1], so first order [1]. Examiner insight: Quote rates as positive values.

5. (a) Zero order [1] (b) Rate = k, so k = 3.6 × 10⁻⁵ [1] mol dm⁻³ s⁻¹ [1] (c) Time = (0.090 – 0.018) ÷ 3.6 × 10⁻⁵ = 2.0 × 10³ s [1] Examiner insight: Zero-order k has the units of rate, not s⁻¹.

6. (a) Ea = 82 000 J mol⁻¹ [1]; Ea/RT = 82 000 ÷ (8.31 × 305) = 32.35 [1]; k = 8.0 × 10¹⁰ × e^(–32.35) = 7.1 × 10⁻⁴ mol⁻¹ dm³ s⁻¹ [1] (b) At 315 K, k = 2.0 × 10⁻³ mol⁻¹ dm³ s⁻¹ [1]; factor = 1.99 × 10⁻³ ÷ 7.12 × 10⁻⁴ = 2.8 [1] Examiner insight: Carry unrounded k values into the factor; early rounding shifts the answer.

7. (a) Comparing with y = mx + c: ln k is y, 1/T is x, and –Ea/R and ln A do not change [1]. The gradient is –Ea/R [1]. (b) 300 K: 1/T = 3.3333 × 10⁻³ K⁻¹, ln k = –4.865; 345 K: 1/T = 2.8986 × 10⁻³ K⁻¹, ln k = –0.994 [1]; gradient = 3.871 ÷ (–4.347 × 10⁻⁴) = –8.90 × 10³ K [1]; Ea = 8.90 × 10³ × 8.31 = 7.40 × 10⁴ J mol⁻¹ = 74.0 kJ mol⁻¹ [1] (c) At 330 K: ln A = –2.163 + 74 000 ÷ (8.31 × 330) = 24.82 [1]; A = e^24.82 = 6.0 × 10¹⁰ s⁻¹ [1] Examiner insight: A negative Ea signals a sign error and is unlikely to be credited.

8. (a) Step 1 (slow): G + 2H → GH₂ [1]; Step 2 (fast): GH₂ + H → GH₃ [1] (b) A single step would put three H particles in the rate-determining step, giving Rate = k[G][H]³, but the order with respect to H is 2 [1]. (c) GH₂ [1] Examiner insight: The steps must also add up to the overall equation.

9. (a) [V] = 0.200 × 5.0 ÷ 50.0 = 0.020 mol dm⁻³ [1] (b) The same fixed amount of reaction has happened at each colour change, so rate ∝ 1/t [1]. (c) 1/t = 0.0104, 0.0208, 0.0313, 0.0417 s⁻¹, so 1/t is directly proportional to [V] [1]; the reaction is first order in V [1]. (d) 1 ÷ 24 × 100 = 4.2 % [1]; the 20.0 cm³ run (shortest time) is least reliable [1]. (e) Use more dilute solutions so that every run takes longer, or detect the colour change with a light sensor [1]. Examiner insight: “Repeat the experiment” does not shrink the timing uncertainty, so it is unlikely to gain the mark.

10. (a) Cooling and diluting slows the reaction almost to a stop, so [RCl] does not change before the titration [1]. (b) n(NaOH) = 0.0250 × 8.00 × 10⁻³ = 2.00 × 10⁻⁴ mol = n(H⁺) [1]; [H⁺] formed = 2.00 × 10⁻⁴ ÷ 5.00 × 10⁻³ = 0.0400 mol dm⁻³ [1]; [RCl] = 0.100 – 0.0400 = 0.0600 mol dm⁻³ [1] (c) [RCl] and the rate both fall to 0.60 of their start values (1.53 ÷ 2.55) [1], so first order [1]. (d) k = 2.55 × 10⁻³ ÷ 0.100 = 0.0255 min⁻¹ [1]; ÷ 60 = 4.25 × 10⁻⁴ s⁻¹ [1] Examiner insight: Each H⁺ formed means one RCl used; remember to subtract from 0.100.

Where marks are usually lost

• Naming an order without saying which two runs were compared.
• Writing units of k that do not match the overall order.
• Using kJ mol⁻¹ or °C inside k = Ae^(–Ea/RT).
• Rounding 1/T too early, which spoils a gradient taken from small differences.
• Reporting a negative rate or a negative activation energy.
• Suggesting “repeat” when the question asks how to reduce a specific uncertainty.

Next steps

• Rate equations revision notes
• Rate equations study guide
• OxfordAQA A-level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.11 Rate equations (3.1.11.1 Rate equations and 3.1.11.2 Determination of rate equation, International A2), with Required practical 8.