AQA A-Level Chemistry: Kinetics (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 3.1.5 Kinetics of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: 3.1.5.1 to 3.1.5.5 and Required practical 3. All of it is AS content, so nothing here is A-level only. The topic is examined on Paper 2 at AS and A-level, and A-level Paper 3 can test any content.

Learn the ideas first in the kinetics study guide and the kinetics revision notes. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. A calculator is allowed for every question.

Questions

1. Define the term activation energy. [1]

2. Explain why most collisions between reactant particles do not lead to a reaction. [2]

3. State three features of a Maxwell–Boltzmann distribution curve for a sample of gas at a fixed temperature. [3]

4. A sealed sample of a gas mixture is warmed from 300 K to 310 K.

(a) Describe how the Maxwell–Boltzmann distribution at 310 K differs from the one at 300 K. [3] (b) Explain why this small rise in temperature causes a large increase in the rate of reaction. [2]

5. In the decomposition of hydrogen peroxide, the concentration of H₂O₂ falls from 1.20 mol dm⁻³ to 0.84 mol dm⁻³ in 90 s. Calculate the mean rate of reaction over this time. Give the units. [2]

6. A student investigates how temperature affects the rate of reaction between sodium thiosulfate and hydrochloric acid. She times how long it takes for a cross under the flask to disappear.

Temperature / °C	25	35	45	55
Time, t / s	96	50	26	14

(a) Calculate 1/t for each temperature, to 3 significant figures. [2] (b) Calculate how many times faster the reaction is at 55 °C than at 25 °C. [1] (c) State two variables she must keep the same. [2] (d) Explain why 1/t can be used as a measure of the rate. [1] (e) Suggest why both solutions are warmed separately before they are mixed. [1] (f) Identify one hazard in this experiment and a precaution to reduce the risk. [2]

7. Explain how increasing the concentration of hydrochloric acid affects the rate of its reaction with magnesium. [3]

8. Excess calcium carbonate powder reacts with 25 cm³ of dilute hydrochloric acid. The carbon dioxide is collected in a gas syringe.

Time / s	0	20	40	60	80	100	120
Volume of CO₂ / cm³	0	36	60	76	86	90	90

(a) Calculate the mean rate of reaction between 0 and 40 s. [1] (b) Calculate the mean rate of reaction between 40 and 80 s. [1] (c) A tangent drawn to the curve at t = 0 passes through (0, 0) and (15 s, 40 cm³). Calculate the initial rate. [2] (d) Explain why the rate decreases as the reaction proceeds. [2] (e) Explain why the volume stops changing after 100 s. [1]

9. Use a Maxwell–Boltzmann distribution to explain how a catalyst increases the rate of a reaction between gases at constant temperature. [4]

10. A 2.00 g sample of calcium carbonate reacts with excess hydrochloric acid in a flask on a balance. After 50 s the mass has fallen by 0.55 g. Mr of CO₂ = 44.0; Mr of CaCO₃ = 100.1.

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

(a) Calculate the amount, in moles, of CO₂ released in the first 50 s. [1] (b) Calculate the mean rate of CO₂ release in mol s⁻¹ over this time. [1] (c) Calculate the maximum mass of CO₂ that the sample can release. [2] (d) Calculate the percentage of the calcium carbonate that has reacted after 50 s. [1]

11. A reaction between two gases is carried out in a closed reactor. Explain, in terms of collisions and the Maxwell–Boltzmann distribution, the effect on the rate of each of these changes made separately: increasing the pressure at constant temperature; increasing the temperature at constant pressure; adding a catalyst. [6]

12. A student writes: “Doubling the concentration of a reactant and adding a catalyst both increase the rate for the same reason: each one increases the proportion of particles with energy above the activation energy.” Evaluate this statement. [3]

Answers

1. The minimum energy that particles must have when they collide for a reaction to occur [1]. [1] Examiner insight: “Minimum” is the key word; “the energy needed to start a reaction” usually earns nothing.

2. Most particles have energy less than the activation energy [1]; so when they collide no bonds break and the particles move apart unchanged, an unsuccessful collision [1]. [2] Examiner insight: An answer about collision frequency does not score here; the mark depends on comparing particle energies with Ea.

3. Any three: the curve starts at the origin (no molecule has zero energy) [1]; it has a single peak at the most probable energy, with the mean energy to its right [1]; it approaches but never meets the energy axis, as there is no maximum energy [1]. Also allow: the area under the curve equals the total number of molecules. [3] Examiner insight: Each feature must be a separate point for its own mark; “it is a curve with a peak” alone is too vague for credit.

4. (a) The peak moves to higher energy (to the right) [1]; the peak is lower [1]; the area under the curve is unchanged because the number of molecules is the same [1]. (b) Many more molecules now have energy equal to or greater than Ea [1]; so the frequency of successful collisions rises sharply, by much more than the small rise in collision frequency [1]. [5] Examiner insight: In (b), “particles move faster and collide more often” alone gets at most one mark; the proportion with E ≥ Ea must be the main reason.

5. Change in concentration = 1.20 − 0.84 = 0.36 mol dm⁻³; rate = 0.36 ÷ 90 [1] = 0.0040 mol dm⁻³ s⁻¹ (4.0 × 10⁻³) with units [1]. [2] Examiner insight: The units mark is lost for mol dm⁻³ alone; rate units must include per unit time (s⁻¹).

6. (a) 1/96 = 0.0104 s⁻¹; 1/50 = 0.0200 s⁻¹ [1]; 1/26 = 0.0385 s⁻¹; 1/14 = 0.0714 s⁻¹ [1]. (b) 96 ÷ 14 = 6.86 times faster [1]. (c) Any two: volume and concentration of the thiosulfate solution [1]; volume and concentration of the acid [1]; also allow the same cross, flask or total volume. (d) The same amount of sulfur forms each time before the cross is hidden, so rate is proportional to 1/t [1]. (e) So the mixture is at the recorded temperature from the moment the reaction starts [1]. (f) Sulfur dioxide is produced, which is toxic and an irritant [1]; work in a well-ventilated room (or fume cupboard) and use small volumes [1]. [9] Examiner insight: For (c), “the amount of solution” is too vague; a variable needs a named quantity such as volume or concentration.

7. At higher concentration there are more acid particles in the same volume [1]; so collisions between acid particles and the magnesium surface are more frequent [1]; so there are more collisions with energy ≥ Ea per unit time and the rate increases [1]. [3] Examiner insight: Writing that the particles “have more energy” at higher concentration contradicts the answer and can cost the final mark.

8. (a) 60 ÷ 40 = 1.5 cm³ s⁻¹ [1]. (b) (86 − 60) ÷ 40 = 26 ÷ 40 = 0.65 cm³ s⁻¹ [1]. (c) Gradient = 40 ÷ 15 [1] = 2.7 cm³ s⁻¹ (2.67) [1]. (d) The acid is used up so its concentration falls [1]; so collisions become less frequent and there are fewer successful collisions per second [1]. (e) All the hydrochloric acid has reacted (it is the limiting reagent) [1]. [7] Examiner insight: Allow error carried forward in (c) from a tangent you have drawn yourself, provided the gradient is read from two well-separated points on the tangent.

9. A catalyst provides an alternative reaction route with a lower activation energy [1]; the distribution curve itself does not change, as the temperature is constant [1]; with Ea(cat) to the left of Ea, a greater proportion of molecules have energy ≥ the activation energy [1]; so there are more successful collisions per unit time and the rate increases [1]. [4] Examiner insight: A diagram that shifts or reshapes the curve for the catalysed reaction loses the “curve unchanged” mark even if the words are right.

10. (a) 0.55 ÷ 44.0 = 0.0125 mol [1]. (b) 0.0125 ÷ 50 = 2.5 × 10⁻⁴ mol s⁻¹ [1]. (c) Amount of CaCO₃ = 2.00 ÷ 100.1 = 0.01998 mol, which gives 0.01998 mol CO₂ (1 : 1) [1]; mass = 0.01998 × 44.0 = 0.879 g [1]. (d) 0.55 ÷ 0.879 × 100 = 62.6 % [1]. [5] Examiner insight: Error carried forward is normally allowed from (a) into (b) and from (c) into (d), so show each value you use.

11. Pressure: more molecules per unit volume [1]; so collisions are more frequent, more successful collisions per second, rate increases; the proportion with E ≥ Ea is unchanged [1]. Temperature: the Maxwell–Boltzmann curve shifts to higher energy, so a much greater proportion of molecules have E ≥ Ea [1]; so many more collisions are successful per second and the rate increases greatly [1]. Catalyst: provides an alternative route of lower activation energy [1]; the curve is unchanged but more molecules have E ≥ the lower Ea, so more collisions succeed per second and the rate increases [1]. [6] Examiner insight: This is marked as two points for each change; an answer that explains two changes well but skips the third cannot reach full marks.

12. The statement is partly correct: for a catalyst, a greater proportion of particles have energy above the (lower) activation energy of the alternative route [1]. It is wrong for concentration: doubling the concentration increases the frequency of collisions, not the proportion of particles with E ≥ Ea [1]. So the two changes increase the rate for different reasons, and the statement is not valid [1]. [3] Examiner insight: “Evaluate” needs a judgement; listing facts about each change without concluding whether the statement is right usually loses the last mark.

Where marks are usually lost

• Leaving “minimum” out of the activation energy definition.
• Drawing Maxwell–Boltzmann curves that touch the energy axis, start above the origin, or have different areas.
• Making the higher-temperature curve taller instead of lower.
• Explaining temperature only through collision frequency.
• Claiming that higher concentration or pressure gives particles more energy.
• Moving the distribution curve for a catalyst, or omitting “alternative route”.
• Using t instead of 1/t for rate in a disappearing-cross experiment.
• Leaving out rate units, or giving mol dm⁻³ with no s⁻¹.
• Reading a tangent gradient from two points too close together.
• Naming a vague control variable such as “amount” without saying volume or concentration.

Next steps

• Kinetics revision notes
• Kinetics 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 3.1.5 Kinetics (3.1.5.1 to 3.1.5.5), with Required practical 3.