Rate of reaction, collision theory & catalysts (RP5) – GCSE Chemistry exam questions with mark scheme

1Grade 1–3Foundation3 marks

A student reacts marble chips (calcium carbonate) with hydrochloric acid. The carbon dioxide produced is collected in a gas syringe.

(a) 48 cm³ of gas is collected in the first 60 s. Calculate the mean rate of reaction in cm³/s. [2]

(b) Give one other piece of apparatus that could be used to collect and measure the volume of gas. [1]

(a) mean rate = 48 ÷ 60(1)
(a) 0.8 (cm³/s) allow 0.80; correct answer with no working scores 2(1)
(b) measuring cylinder / burette allow upturned / inverted over water (in a trough); ignore beaker; do not accept gas syringe (already given) or balance(1)
Answer: (a) 0.8 cm³/s (b) upturned measuring cylinder (or burette) filled with water
Examiner tip: Mean rate = quantity ÷ time — writing 60 ÷ 48 loses both marks.

2Grade 4–54 marks

Magnesium ribbon reacts with dilute hydrochloric acid.

Explain, in terms of particles, why the reaction is faster at 40 °C than at 20 °C. [4]

particles move faster / have more (kinetic) energy ignore 'vibrate more'(1)
so they collide more frequently / more collisions per second do not accept 'more collisions' with no reference to time/frequency(1)
more particles have energy greater than or equal to the activation energy allow more particles have enough energy to react / collisions are more energetic(1)
so there are more frequent successful collisions / a greater proportion of collisions are successful allow more effective collisions per second(1)
Answer: Particles have more energy, collide more often, and more collisions have energy ≥ activation energy, so the frequency of successful collisions increases.
Examiner tip: 'More collisions' alone is not enough — you must say more frequent collisions (per second) and more with energy above the activation energy.

3Grade 6–7Higher3 marks

The graph shows the volume of gas produced when a metal carbonate reacts with acid. A tangent has been drawn to the curve at 20 s.

Use the tangent to determine the rate of reaction at 20 s. Give your answer to 2 significant figures. [3]

uses the tangent: change in volume ÷ change in time, e.g. (60 − 16) ÷ (40 − 0)(1)
answer in range 1.0–1.2(1)
unit cm³/s (allow cm³ s⁻¹)(1)
Answer: 1.1 cm³/s (accept 1.0–1.2)
Examiner tip: Read two points far apart on the tangent, not on the curve — using one point (37.9 ÷ 20) gives the mean rate, not the rate at 20 s.

4Grade 4–54 marks

The diagram shows the reaction profile for a reaction carried out with and without a catalyst.

(a) Which path, A or B, is the reaction with a catalyst? Give a reason for your answer. [2]

(b) Is the reaction exothermic or endothermic? Give a reason. [1]

(a) B(1)
(a) (B has a) lower activation energy / lower peak allow B needs less energy for the reaction to start(1)
(b) exothermic and products have less energy than reactants (both needed)(1)
(c) provides a different (alternative) reaction pathway with a lower activation energy allow 'lowers the activation energy'; ignore 'gives particles more energy'; do not accept 'catalyst is used up'(1)
Answer: (a) B — lower activation energy (b) exothermic; products lower in energy than reactants (c) provides an alternative pathway with lower activation energy
Examiner tip: Say the catalyst provides a different pathway — catalysts are not 'used up' and do not give particles more energy.

5Grade 6–7Higher4 marks

A student investigated the reaction between calcium carbonate and hydrochloric acid by measuring the mass of the flask and contents. The mass decreased by 0.66 g in 120 s.

(a) Explain why the mass decreased. [1]

(b) Relative formula mass (M_r) of CO₂ = 44

Assume all the mass lost is carbon dioxide.

Calculate the mean rate of reaction in mol/s, in terms of the carbon dioxide produced. Give your answer in standard form. [3]

(a) carbon dioxide (gas) is produced and escapes (from the flask) allow 'a gas is given off and lost to the air'(1)
(b) moles of CO₂ = 0.66 ÷ 44 = 0.015 (mol) allow alternative route: 0.66 ÷ 120 = 5.5 × 10⁻³ (g/s) for this mark, then ÷ 44 for the second mark(1)
(b) 0.015 ÷ 120 allow ecf from their moles(1)
(b) 1.25 × 10⁻⁴ (mol/s) allow 1.3 × 10⁻⁴; correct answer with no working scores 3; 0.000125 (not in standard form) scores 2(1)
Answer: (a) CO₂ gas escapes (b) 1.25 × 10⁻⁴ mol/s
Examiner tip: Convert mass to moles first — 0.66 ÷ 120 = 5.5 × 10⁻³ is a rate in g/s, not mol/s.

6Grade 6–76 marks

Sodium thiosulfate solution reacts with hydrochloric acid to form a cloudy precipitate of sulfur.

Describe a method to investigate how the concentration of sodium thiosulfate solution affects the rate of this reaction. [6]

measure a fixed volume (e.g. 10 cm³) of sodium thiosulfate solution with a measuring cylinder into a conical flask(1)
place the flask on a paper marked with a black cross(1)
add a fixed volume (e.g. 10 cm³) of hydrochloric acid of the same concentration and start a stopwatch(1)
time how long until the cross can no longer be seen (from above)(1)
repeat using different concentrations of sodium thiosulfate (dilute with water, keeping total volume constant)(1)
control variables: same temperature, same volume/concentration of acid, same observer/cross(1)
repeat each concentration and calculate a mean(1)
rate is proportional to 1 ÷ time; plot concentration against 1/time (or time)(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that could be followed by someone else to get valid results; includes how concentration is varied, how the end point is judged and timed, and control of variables.
Level 2	3–4	Most of the key steps are present but some detail or order is missing, e.g. no control variables or unclear how concentration is changed.
Level 1	1–2	Some relevant steps identified but the method could not be followed to produce valid results.
No relevant content	0	No relevant content.

Answer: See indicative content: fixed volumes, black cross, time until cross disappears, vary concentration with constant total volume, control temperature, repeat.
Examiner tip: Keep the total volume the same when diluting, otherwise the depth of liquid over the cross changes and the test is not fair.

7Grade 1–3Foundation6 marks

Hydrogen peroxide solution slowly decomposes to form water and oxygen.

A student tested three powders, A, B and C, to see if they are catalysts for this reaction. For each test the student added 0.50 g of powder to 25 cm³ of hydrogen peroxide solution and timed how long it took to collect 40 cm³ of oxygen.

Powder added	Time to collect 40 cm³ of oxygen in s
none	410
A	85
B	22
C	410

(a) Which powder was the best catalyst for this reaction? Give a reason for your answer. [2]

(b) Which powder is not a catalyst for this reaction? Give a reason for your answer. [1]

(c) After the reaction with powder B, the student filtered off the powder, dried it and weighed it. The mass was still 0.50 g.

What does this show about a catalyst? [1]

(d) Calculate the mean rate of reaction with powder B. Give your answer in cm³/s to 2 significant figures. [2]

(a) B(1)
(a) shortest time (to collect 40 cm³ of oxygen) / oxygen produced fastest; allow highest rate; ignore 'made the most oxygen'. Reason mark depends on B being chosen(1)
(b) C and time is the same as with no powder (410 s) / it did not change the rate(1)
(c) a catalyst is not used up (in the reaction); allow it is not changed (chemically) by the reaction; ignore 'it does not react'(1)
(d) 40 ÷ 22(1)
(d) 1.8 (cm³/s); correct answer with no working scores 2 marks; do not accept 1.82 / 1.818 for this mark (not 2 significant figures)(1)
Answer: (a) B – shortest time; (b) C – same time as no powder; (c) catalysts are not used up; (d) 1.8 cm³/s
Examiner tip: In (b) you must link C to the 'no powder' result — just saying 'C was slowest' does not show it had no effect on the rate.

8Grade 4–55 marks

A student reacts a cube of calcium carbonate with dilute hydrochloric acid.

The cube has sides of length 3 cm.

(a) Calculate the surface area to volume ratio of the cube. [2]

(b) The student cuts the cube into 27 smaller cubes, each with sides of length 1 cm.

Calculate the total surface area of the 27 small cubes. [1]

Explain why. [2]

(a) surface area = 54 (cm²) and volume = 27 (cm³)(1)
(a) ratio = 2 : 1 (allow 54 : 27; allow 2); ecf from incorrect area or volume(1)
(b) 162 (cm²); allow 27 × 6(1)
(c) larger surface area (to volume ratio) / more particles of calcium carbonate exposed to the acid(1)
(c) (so) more frequent collisions; allow more collisions per second / in a given time; ignore 'more collisions' without reference to time; do not accept 'more energetic collisions'(1)
Answer: (a) 54 : 27 = 2 : 1; (b) 162 cm²; (c) greater surface area to volume ratio, so more frequent collisions
Examiner tip: Students lose the second mark in (c) by writing 'more collisions' — the examiner wants collisions per second (frequency), not just more of them.

9Grade 6–7Higher8 marks

A student investigated how the concentration of hydrochloric acid affects the rate of its reaction with marble chips (calcium carbonate).

$$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$$

In each experiment the student used 20 cm³ of acid and an excess of marble chips of the same size. The carbon dioxide was collected in a gas syringe.

Curve A: 0.40 mol/dm³ hydrochloric acid
Curve B: 0.20 mol/dm³ hydrochloric acid

The graph shows the results.

(a) The final volume of gas for curve B is half the final volume for curve A.

Explain why. [2]

(b) Use the graph to determine the mean rate of reaction for curve A and for curve B over the first 20 s. [3]

Do your answers to part (b) support this hypothesis? Give a reason. [1]

(d) Explain, in terms of particles, why increasing the concentration of the acid increases the rate of reaction. [2]

(a) marble chips are in excess so the acid is the limiting reactant / the acid is all used up(1)
(a) acid B contains half the amount (moles) of HCl / half the number of acid particles, so half the amount of carbon dioxide is made; do not accept 'the reaction is slower' as the reason(1)
(b) reads 60 (cm³) for A and 30 (cm³) for B at 20 s(1)
(b) A: 60 ÷ 20 = 3.0 (cm³/s)(1)
(b) B: 30 ÷ 20 = 1.5 cm³/s; unit needed once for this mark; allow ecf from incorrect readings(1)
(c) yes, because the rate for A (3.0 cm³/s) is double the rate for B (1.5 cm³/s); allow ecf from (b) if consistent(1)
(d) (higher concentration means) more acid particles in the same volume(1)
(d) (so) more frequent collisions; allow more collisions per second; ignore 'more collisions' alone; do not accept particles move faster / have more energy(1)
Answer: (a) acid is the limiting reactant and B has half the moles of acid; (b) A = 3.0 cm³/s, B = 1.5 cm³/s; (c) yes, rate doubled; (d) more particles per volume, more frequent collisions
Examiner tip: In (a) 'B is slower' is not an explanation — the final volume depends on the amount of limiting reactant (the acid), not on the rate.

10Grade 6–76 marks

In an industrial process, two gases react together on the surface of a solid catalyst to make a gaseous product.

An engineer suggests three changes to make the reaction faster:

increase the pressure of the gases
increase the temperature
use the catalyst as small pellets instead of large lumps.

Explain how each change increases the rate of the reaction.

Use collision theory and ideas about activation energy in your answer. [6]

reactions happen when particles collide with enough energy / energy greater than or equal to the activation energy(1)
pressure: higher pressure means more gas particles in the same volume(1)
pressure: so collisions are more frequent (more collisions per second)(1)
temperature: particles move faster / have more kinetic energy(1)
temperature: so collisions are more frequent(1)
temperature: and a greater proportion of collisions have energy greater than or equal to the activation energy / more successful collisions per second(1)
small pellets: larger surface area to volume ratio of the catalyst(1)
small pellets: more catalyst surface available for the gas particles to collide with / react on, so more frequent (successful) collisions(1)
catalyst provides an alternative reaction pathway with a lower activation energy, so more collisions are successful(1)

Level	Marks	What it looks like
Level 3	5–6	Explains the effect of all three changes using collision theory. Correctly links pressure and surface area to collision frequency, and temperature to both collision frequency and the proportion of collisions with at least the activation energy. Reasoning is clear and logically linked.
Level 2	3–4	Explains the effect of at least two changes in terms of collisions, or all three with some confusion (e.g. treats temperature only as frequency, or says pressure gives more energetic collisions).
Level 1	1–2	Simple statements that the changes increase rate or cause more collisions, with little or no explanation of why.
No relevant content	0	No relevant content.

Answer: Pressure: more particles per volume, more frequent collisions. Temperature: faster particles, more frequent collisions and more collisions with energy ≥ activation energy. Small pellets: larger surface area of catalyst, more frequent collisions on its surface; catalyst lowers activation energy.
Examiner tip: Temperature is the only change here that increases the energy of collisions — saying pressure or surface area make collisions 'more energetic' caps you at Level 2.

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