OxfordAQA International GCSE Chemistry: Rate of reaction (9202) – 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.

Every question here is based on the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards. Together they test section 3.8.1 Rate of reaction, statements a to h, and the required practical on factors affecting rate. There are no tiers. Where needed, use relative atomic masses C = 12, O = 16 and Ca = 40, and take one mole of any gas to occupy 24 dm³ at room temperature and pressure.

Links: rate of reaction study guide · rate of reaction revision notes · OxfordAQA chemistry hub · topic checklist · 10-minute diagnostics

Questions

1. This question is about collision theory.

(a) Define the term activation energy. [1] (b) State the two conditions that must be met when particles collide for a reaction to take place. [2]

2. Magnesium carbonate reacts with dilute hydrochloric acid in an open conical flask on a balance: MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂

Time after mixing (s)	0	20	40	60	80	100	120	140
Balance reading (g)	98.60	98.30	98.08	97.94	97.86	97.82	97.81	97.81

(a) Explain why the balance reading decreases. [1] (b) Calculate the mean rate of reaction over the first 40 s, in g/s. [2] (c) Calculate the mean rate of reaction between 80 s and 120 s, in g/s. [2] (d) State the time at which the reaction stopped. [1] (e) Use collision theory to explain why the reaction slows down over time. [2]

3. Explain, in terms of particles, how a rise in temperature affects the rate of a reaction. [3]

4. Nitrogen monoxide gas reacts with oxygen gas to form nitrogen dioxide. It is compressed into a smaller volume at constant temperature. Explain why the rate increases. [2]

5. A piece of limestone is a cuboid measuring 6 cm × 2 cm × 1 cm.

(a) Calculate the total surface area of the cuboid. [1] (b) The cuboid is cut into twelve 1 cm cubes. Calculate the total surface area of the cubes and the increase in surface area. [2] (c) Explain why the cubes react faster with acid than the single cuboid. [2]

6. A student plans to investigate how the concentration of sulfuric acid affects the rate of its reaction with zinc granules. She will use acid concentrations of 0.25, 0.50, 0.75 and 1.00 mol/dm³ and collect the hydrogen.

(a) Name the independent and dependent variables. [2] (b) State two control variables. [2] (c) Describe how she could obtain a value for the rate of each reaction. [2] (d) Give one hazard and say how she could reduce the risk from it. [1]

7. A student adds 0.20 g of three different powders, P, Q and R, to separate 25 cm³ samples taken from one bottle of hydrogen peroxide solution. Hydrogen peroxide decomposes to water and oxygen. He measures the oxygen formed in 60 s.

Powder added	none	P	Q	R
Oxygen formed in 60 s (cm³)	3	4	52	18

(a) Which powder is the most effective catalyst for this reaction? [1] (b) Do the results show that P is a catalyst for this reaction? Give a reason. [1] (c) Describe how the student could show that powder Q is not used up in the reaction. [3] (d) Another student says Q will speed up every reaction. Comment on this claim. [1]

8. Many industrial processes use catalysts.

(a) Which metal is the catalyst in the Haber process? [1] (b) Explain how using a catalyst reduces the cost of an industrial process. [3] (c) Suggest why a factory can keep using the same catalyst for a long time. [1]

9. A student times how long a cross takes to disappear when sodium thiosulfate solution reacts with dilute hydrochloric acid at different temperatures. At 25 °C her three runs took 104 s, 98 s and 101 s.

Temperature (°C)	15	25	35	45	55
Mean time to hide cross (s)	190	see above	95	31	18

(a) Calculate the mean time at 25 °C. [1] (b) Calculate 1/time at 25 °C and at 55 °C. [2] (c) Identify the anomalous result and suggest one cause of it. [2] (d) Write a conclusion and explain it in terms of particles. [2] (e) Suggest one way to make the end point less dependent on the observer. [1]

10. A student adds 1.75 g of calcium carbonate lumps to an excess of dilute hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ The reaction finishes after 175 s.

(a) Calculate the maximum volume of carbon dioxide, in cm³, at room temperature and pressure. [3] (b) Calculate the mean rate of reaction over the 175 s, in cm³/s and in mol/s. [2] (c) The experiment is repeated with 1.75 g of calcium carbonate powder. State and explain the effect on the initial rate and on the final volume of gas. [3]

11. Two experiments, A and B, use the same reactants. The volume of gas is recorded.

Time (s)	0	10	20	30	40	60	80
Volume in A (cm³)	0	32	50	57	60	60	60
Volume in B (cm³)	0	14	26	36	44	54	60

(a) Which experiment had the faster rate at the start? [1] (b) Calculate the mean rate over the first 20 s for each experiment. [2] (c) Suggest one change to the conditions that could make A faster than B, and explain your choice. [2] (d) Explain why both experiments finish with the same volume of gas. [1]

Answers

1. (a) The minimum energy particles need for a reaction to happen [1]. (b) The particles must collide [1] with sufficient energy, at least the activation energy [1]. Examiner insight: Two marks need two separate ideas; “they must collide hard” is vague about energy.

2. (a) Carbon dioxide gas is formed and escapes from the open flask [1]. (b) Mass lost = 98.60 − 98.08 = 0.52 g [1]; 0.52 ÷ 40 = 0.013 g/s [1] (c) Mass lost = 97.86 − 97.81 = 0.05 g [1]; 0.05 ÷ 40 = 0.00125 g/s [1] (d) 120 s [1] (e) The reactants are used up, so their concentration falls [1]; collisions become less frequent [1]. Examiner insight: In (d), 140 s is a common slip; the reaction stopped at the first time the reading stopped changing.

3.

• Particles gain energy and move faster [1].
• They collide more frequently [1].
• Collisions are more energetic, so a greater proportion have at least the activation energy [1]. Examiner insight: “More frequent collisions” alone could describe concentration; the energy point makes it a temperature answer.

4. The same number of gas particles occupy a smaller volume, so the pressure is higher [1]; the particles collide more frequently [1]. Examiner insight: Do not say the particles move faster; the temperature is constant, so collision energy is unchanged.

5. (a) 2 × (6 × 2 + 6 × 1 + 2 × 1) = 40 cm² [1] (b) 12 × 6 × 1 × 1 = 72 cm² [1]; increase = 72 − 40 = 32 cm² [1] (c) The acid can reach more calcium carbonate particles at once [1], so collisions happen more often [1]. Examiner insight: “Bigger surface area” alone repeats the question; link exposed particles to collision frequency.

6. (a) Independent: concentration of the sulfuric acid [1]. Dependent: volume of hydrogen collected in a fixed time, or a value of rate from it [1]. (b) Any two [1] [1]: volume of acid; mass of zinc; size of zinc granules; temperature. (c) Collect the hydrogen in a gas syringe and record the volume after a fixed time [1]; divide the volume by the time to give a rate in cm³/s [1]. (d) Hydrogen is flammable, so keep flames away; or acid is an irritant, so wear eye protection [1]. Examiner insight: “Mass of zinc” is stronger than “amount of zinc”; name a quantity you can measure.

7. (a) Q [1] (b) No; 4 cm³ is almost the same as 3 cm³ with no powder, so P has no clear effect on the rate [1]. (c)

• Weigh the powder before adding it [1].
• After the reaction, filter off the powder, wash and dry it [1].
• Weigh it again: the mass is still 0.20 g [1].

(d) The claim is wrong: different reactions need different catalysts, so Q may not affect other reactions [1]. Examiner insight: In (c), skipping the drying step gives a mass that is too high, so name every step.

8. (a) Iron [1] (b)

• The catalyst increases the rate, so more product is made per hour [1].
• A useful rate is reached at a lower temperature or pressure [1].
• So less energy is used, lowering cost [1].

(c) It is unchanged at the end, so it keeps working [1]. Examiner insight: In (b), each mark needs a separate point; restating “it speeds the reaction up” counts once.

9. (a) (104 + 98 + 101) ÷ 3 = 101 s [1] (b) 1 ÷ 101 = 0.0099 s⁻¹ [1]; 1 ÷ 18 = 0.056 s⁻¹ [1] (c) 35 °C (95 s is far too long; it hardly differs from 25 °C) [1]. Cause: for example, the mixture had cooled below 35 °C, or the timer was started too early [1]. (d) As temperature increases, the rate increases [1]; particles move faster and collide more frequently and more energetically [1]. (e) Use a light sensor and data logger to detect when the cross is hidden [1]. Examiner insight: In (c), give a cause that would make the time longer; “human error” alone rarely gains credit.

10. (a) Mr of CaCO₃ = 40 + 12 + 3 × 16 = 100; moles = 1.75 ÷ 100 = 0.0175 mol [1] 1 mol CaCO₃ gives 1 mol CO₂, so 0.0175 mol CO₂ [1] Volume = 0.0175 × 24 000 = 420 cm³ [1] (b) 420 ÷ 175 = 2.4 cm³/s [1]; 0.0175 ÷ 175 = 1.0 × 10⁻⁴ mol/s [1] (c)

• The initial rate is higher [1].
• The final volume is the same, 420 cm³ [1].
• Powder has a larger surface area, so collisions are more frequent; the mass of calcium carbonate is unchanged, so the same amount of gas forms [1]. Examiner insight: Show the mole ratio step in (a) even when it is 1 : 1, so the route to 420 cm³ is clear.

11. (a) A [1] (b) A: 50 ÷ 20 = 2.5 cm³/s [1]; B: 26 ÷ 20 = 1.3 cm³/s [1] (c) A higher temperature in A [1]; the particles move faster and collide more frequently and more energetically [1]. A catalyst or a larger surface area, with a matching reason, also fits. (d) The same amounts of reactants were used, so the same amount of product forms [1]. Examiner insight: In (c), choose a change that fits equal final volumes; extra reactant could change the final volume.

Where marks are usually lost

• Missing “more energetic collisions” in a temperature explanation.
• Saying “more collisions” with no idea of how often they happen.
• Giving a rate without a unit, or the wrong unit for the quantity measured.
• Choosing the last time in a table as the end of the reaction.
• Saying a catalyst increases the yield or is used up.
• Control variables written as “amount” or “same equipment” without a measurable quantity.

Next steps

• Recap with the rate of reaction revision notes.
• Re-learn topics in the rate of reaction study guide.
• Practise moles with the quantitative practice set.
• Book a free trial class.
• Take a free 10-minute diagnostic.
• Browse every unit from the OxfordAQA chemistry hub and mark off rate of reaction on the topic checklist.

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards (OxfordAQA). Covered here: section 3.8.1 Rate of reaction, with its required practical.