AQA A-Level Chemistry: Rate equations (7405) – 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 3.1.9 Rate equations of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for exams from June 2016 onwards: 3.1.9.1, 3.1.9.2 and Required practical 7. All of it is A-level only and is placed on A-level Paper 2; Paper 3 can test any content.

Learn the ideas first in the rate equations study guide and the rate equations revision notes. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. Use R = 8.31 J K⁻¹ mol⁻¹ wherever it is needed.

Questions

1. Define the terms order of reaction with respect to a reactant and rate constant. [2]

2. For the reaction between P and Q, Rate = k[P][Q]². At 300 K, k = 25 mol⁻² dm⁶ s⁻¹. Show that these are the correct units for k, then calculate the rate when [P] = 0.040 mol dm⁻³ and [Q] = 0.20 mol dm⁻³. [3]

3. For a reaction, Rate = k[X]²[Y]. The concentration of X is tripled and the concentration of Y is halved at the same temperature. Calculate the factor by which the rate changes. [2]

4. The initial rate of the reaction G + H + J → products was measured at 298 K.

Exp	[G] / mol dm⁻³	[H] / mol dm⁻³	[J] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.020	0.050	0.10	3.0 × 10⁻⁵
2	0.040	0.050	0.10	6.0 × 10⁻⁵
3	0.040	0.10	0.10	2.4 × 10⁻⁴
4	0.020	0.10	0.30	1.2 × 10⁻⁴

(a) Deduce the order with respect to G, H and J. Explain each. [3] (b) Write the rate equation. [1] (c) Calculate k and give its units. [2] (d) Calculate the initial rate when [G] = 0.050, [H] = 0.020 and [J] = 0.50 mol dm⁻³. [1]

5. The concentration of reactant M was measured during a reaction at constant temperature.

t / s	0	50	100	150	200
[M] / mol dm⁻³	0.500	0.465	0.430	0.395	0.360

(a) Explain how the data show that the reaction is zero order with respect to M. [1] (b) Calculate the rate constant, with units. [2]

6. A concentration–time curve is drawn for reactant N. The tangent at t = 0 passes through (0 s, 0.400 mol dm⁻³) and (40 s, 0.080 mol dm⁻³). Later, when [N] = 0.200 mol dm⁻³, the tangent passes through (50 s, 0.320 mol dm⁻³) and (150 s, 0.120 mol dm⁻³). Use these tangents to deduce the order with respect to N. [4]

7. A first-order reaction has an Arrhenius constant A = 6.0 × 10¹¹ s⁻¹ and Ea = 92 kJ mol⁻¹. The equation k = Ae^(−Ea/RT) applies.

(a) Calculate k at 350 K. [3] (b) Explain the qualitative effect of a rise in temperature on the rate constant. [2]

8. The rate constant of a second-order reaction was measured at four temperatures.

T / K	290	310	330	350
k / mol⁻¹ dm³ s⁻¹	0.112	0.404	1.25	3.40

The equation ln k = −Ea/(RT) + ln A applies.

(a) Calculate 1/T and ln k for the first and last temperatures. [1] (b) A plot of ln k against 1/T is a straight line. Use the first and last points to find its gradient, then calculate Ea in kJ mol⁻¹. [3] (c) Calculate the Arrhenius constant A, with units. [2]

9. The reaction 2R + S → R₂S has the rate equation Rate = k[R]². Two mechanisms are suggested.

• Mechanism 1: R + S → RS (slow); RS + R → R₂S (fast)
• Mechanism 2: R + R → R₂ (slow); R₂ + S → R₂S (fast)

(a) Explain which mechanism is consistent with the rate equation. [3] (b) State what the rate equation shows about the role of S. [1]

10. In Required practical 7, a student uses a clock reaction between J and K. She varies the volume of J solution and adds water, keeping the volume of K solution and the temperature the same.

Volume of J / cm³	4.0	8.0	12.0	16.0
Volume of water / cm³	16.0	12.0	8.0	4.0
Time for colour change, t / s	144	36	16	9

(a) Explain why water is added in different volumes. [1] (b) Explain why 1/t can be used as a measure of initial rate. [1] (c) Use the data to deduce the order with respect to J. [2] (d) State two other variables that must be controlled. [2]

11. A coloured reactant W decomposes. It is followed with a colorimeter (continuous monitoring). At 298 K, tangents to the concentration–time curve give these rates.

[W] / mol dm⁻³	0.060	0.030
Rate / mol dm⁻³ s⁻¹	4.2 × 10⁻⁴	2.1 × 10⁻⁴

(a) Deduce the order with respect to W. Explain your answer. [2] (b) Calculate k at 298 K, with units. [2] (c) At 318 K, k is four times its value at 298 K. Use ln k = −Ea/(RT) + ln A to calculate Ea. [4]

Answers

1. Order: the power to which the concentration of that reactant is raised in the rate equation [1]. Rate constant: the constant of proportionality in the rate equation (fixed at a given temperature) [1]. [2] Examiner insight: Saying an order is “how much the reactant affects the rate” is too vague; the mark needs “power” and “rate equation”.

2. k = Rate ÷ [P][Q]² = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹ [1]. Rate = 25 × 0.040 × (0.20)² [1] = 0.040 mol dm⁻³ s⁻¹ [1]. [3] Examiner insight: A rate with no units, or with the units of k, usually loses the final mark.

3. Factor = 3² × ½ [1] = 4.5 [1]. [2] Examiner insight: Squaring the factor for the second-order reactant is the method mark; 3 × ½ = 1.5 scores nothing.

4. (a) G: Exps 1 and 2, [G] doubles and rate doubles, so first order [1]. H: Exps 2 and 3, [H] doubles and rate ×4, so second order [1]. J: Exps 1 and 4, [H] doubles (×4) and [J] triples, rate ×4, so the change is all due to H and J is zero order [1]. (b) Rate = k[G][H]² [1] (c) k = 3.0 × 10⁻⁵ ÷ (0.020 × 0.050²) = 0.60 [1] mol⁻² dm⁶ s⁻¹ [1] (d) Rate = 0.60 × 0.050 × 0.020² = 1.2 × 10⁻⁵ mol dm⁻³ s⁻¹ [1] Examiner insight: For J, two concentrations change, so show that the effect of H has been allowed for.

5. (a) [M] falls by the same amount (0.035 mol dm⁻³) in each 50 s, so the concentration–time graph is a straight line and the rate is constant [1]. (b) k = −gradient = (0.500 − 0.360) ÷ 200 [1] = 7.0 × 10⁻⁴ mol dm⁻³ s⁻¹ [1] Examiner insight: For a zero-order reaction k has the same units as rate; s⁻¹ alone is wrong here.

6. Initial rate = (0.400 − 0.080) ÷ 40 = 8.0 × 10⁻³ mol dm⁻³ s⁻¹ [1]. Rate at [N] = 0.200 = (0.320 − 0.120) ÷ 100 = 2.0 × 10⁻³ mol dm⁻³ s⁻¹ [1]. Concentration halves (0.400 → 0.200) and rate falls to a quarter [1], so second order [1]. [4] Examiner insight: The order mark depends on comparing the concentration factor with the rate factor.

7. (a) Ea = 92 000 J mol⁻¹ [1]; Ea/RT = 92 000 ÷ (8.31 × 350) = 31.63 [1]; k = 6.0 × 10¹¹ × e^(−31.63) = 1.10 × 10⁻² s⁻¹ [1] (b) k increases as the temperature rises [1]; because a much larger proportion of collisions have energy greater than or equal to Ea [1]. Examiner insight: Leaving Ea in kJ gives a k close to A, which is clearly wrong; expect to lose the conversion mark and the answer mark.

8. (a) 290 K: 1/T = 3.448 × 10⁻³ K⁻¹, ln k = −2.19; 350 K: 1/T = 2.857 × 10⁻³ K⁻¹, ln k = 1.22 [1] (b) Gradient = (1.22 − (−2.19)) ÷ (2.857 × 10⁻³ − 3.448 × 10⁻³) = 3.41 ÷ (−5.91 × 10⁻⁴) = −5.77 × 10³ K [1]; Ea = −gradient × R = 5.77 × 10³ × 8.31 [1] = 4.79 × 10⁴ J mol⁻¹ = 47.9 kJ mol⁻¹ [1] (c) ln A = ln k + Ea/RT = −2.19 + 47 900 ÷ (8.31 × 290) = 17.69 [1]; A = e^17.69 = 4.8 × 10⁷ mol⁻¹ dm³ s⁻¹ [1] Examiner insight: Keep at least four significant figures in 1/T; rounding to 3.4 × 10⁻³ and 2.9 × 10⁻³ throws Ea well outside an acceptable range.

9. (a) Mechanism 2 is consistent [1]. The rate equation shows two R particles and no S involved in the rate-determining step, which matches the slow step R + R → R₂ [1]. The slow step of Mechanism 1 would give Rate = k[R][S] [1]. (b) S is zero order, so it reacts only after the rate-determining step (in the fast step) [1]. Examiner insight: Naming the right mechanism without linking the slow step to the orders earns little.

10. (a) To keep the total volume constant, so [J] is proportional to its volume and the other concentrations stay the same [1]. (b) The same small, fixed amount of product forms before the colour change each time, so rate ∝ 1/t [1]. (c) Volume of J ×2 (4 → 8 cm³) and t falls from 144 s to 36 s, so 1/t ×4 [1]; rate ∝ [J]², so second order [1]. (d) Any two: temperature [1]; volume and concentration of K solution [1]. Examiner insight: “Amount of K” is often too vague; state volume or concentration for the mark.

11. (a) [W] halves and the rate halves [1], so first order [1]. (b) k = 4.2 × 10⁻⁴ ÷ 0.060 = 7.0 × 10⁻³ [1] s⁻¹ [1] (c) Subtracting the equation at the two temperatures: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) [1]; ln 4 = 1.386 [1]; 1/298 − 1/318 = 2.11 × 10⁻⁴ K⁻¹ [1]; Ea = 8.31 × 1.386 ÷ 2.11 × 10⁻⁴ = 5.46 × 10⁴ J mol⁻¹ = 54.6 kJ mol⁻¹ [1] Examiner insight: A correct k with wrong units still loses the units mark.

Where marks are usually lost

• Quoting an order without naming the two experiments compared.
• Crediting a rate change to one reactant when two concentrations changed.
• Cancelling units of k incorrectly for overall order 3.
• Using Ea in kJ mol⁻¹ or T in °C in the Arrhenius equation.
• Rounding 1/T too early, which wrecks a gradient taken from small differences.
• Giving Ea as a negative value after forgetting that the gradient is −Ea/R.
• Reading a tangent gradient from two points close together.
• Giving k of a zero-order reaction in s⁻¹.
• Vague control variables in a clock-reaction plan (“amount” instead of volume or concentration).

Next steps

• Rate equations revision notes
• Rate equations study guide
• AQA A-Level Chemistry exam preparation
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.1.9 Rate equations (3.1.9.1 and 3.1.9.2, A-level only), with Required practical 7.