5 exam-style question sets on Reaction rate & energy (Paper 2), each with a hint and a fully worked answer. The app holds 25 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.
Practise this section in the app →
Question 1
Pure zinc reacts with EXCESS dilute sulphuric acid according to the balanced equation:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
A learner uses this reaction to investigate factors that affect reaction rate. Three experiments are carried out, each using 2 g of zinc:
| Experiment | State of zinc (2 g) | Concentration of H₂SO₄ (in excess) | Temperature (°C) |
|---|---|---|---|
| 1 | Granules | 1,5 mol·dm⁻³ | 25 |
| 2 | Granules | 1,5 mol·dm⁻³ | 40 |
| 3 | Powder | 1,5 mol·dm⁻³ | 25 |
The volume of hydrogen gas is recorded as a function of time in each experiment.
The Maxwell–Boltzmann distribution curves shown apply to the reaction mixture: curve M applies to Experiment 1, and curve N was obtained after ONE reaction condition was changed.
- aDefine the term reaction rate. (2)
- bIn Experiment 1 the average rate of formation of H₂(g) over the first 4 minutes is 0,056 dm³·min⁻¹. Calculate the mass of zinc present in the flask at t = 4 minutes. Take the molar gas volume at 25 °C as 24,5 dm³·mol⁻¹. (6)
- cUse the collision theory to explain why the reaction rate in Experiment 2 is higher than that in Experiment 1. (4)
- dSketch, on the same set of axes, the volume of H₂(g) versus time curves for Experiment 1 and Experiment 3. Label the curves 1 and 3. NO numerical values are required. (2)
- eHow will the total volume of H₂(g) produced in Experiment 2 compare to that in Experiment 1? Write down GREATER THAN, LESS THAN or EQUAL TO. (1)
- fWhat change in reaction conditions was made to obtain curve N? (1)
- gGive a reason for the answer to QUESTION f by referring to the shape of curve N. (1)
This question has a diagram, shown in the app.
Hint
Convert the gas volume to moles with n = V/Vm, use the 1 : 1 mole ratio back to zinc, and remember the zinc (not the excess acid) fixes the final volume.
Worked answer
a) Reaction rate is the change in concentration (or amount/mass/volume) of a reactant or product per unit time. ✓✓
b) V(H₂) = 0,056 × 4 = 0,224 dm³ ✓
mol ✓✓
Mole ratio Zn : H₂ = 1 : 1, so n(Zn) reacted = 9,14 × 10⁻³ mol ✓
m(Zn) reacted = 9,14 × 10⁻³ × 65 = 0,59 g ✓
Mass remaining = 2 − 0,59 = 1,41 g ✓
c) At the higher temperature the particles have a higher average kinetic energy. ✓ More particles have kinetic energy equal to or greater than the activation energy. ✓ Therefore there are more effective collisions per unit time ✓ and the reaction rate increases. ✓
d) Full marks: curve 3 has a STEEPER initial gradient than curve 1 ✓; both curves level off (plateau) at the SAME final volume ✓ (same 2 g Zn is the limiting reagent; acid in excess).
e) EQUAL TO ✓ (temperature changes the rate, not the amount of limiting zinc).
f) The temperature was increased. ✓
g) Curve N has a lower peak that is shifted to higher kinetic energy — the total number of particles is unchanged but a larger fraction of particles has high kinetic energy, which is the effect of a temperature increase. ✓
Modelled on Nov 2024 Paper 2, Question 5 — same skills, new scenario.
Question 2
The potential energy profile for the hypothetical reaction P + Q ⇌ R is shown. The products are at a HIGHER potential energy than the reactants. Eₐ(f) and Eₐ(r) are the activation energies of the forward and reverse reactions. Which ONE of the following combinations is possible for this reaction?
This question has a diagram, shown in the app.
Hint
Use ΔH = Eₐ(forward) − Eₐ(reverse), and check the sign against the diagram.
Worked answer
A. The profile is endothermic, so ΔH must be positive and Eₐ(f) > Eₐ(r). Check: ΔH = Eₐ(f) − Eₐ(r) = 250 − 100 = +150 kJ·mol⁻¹ ✓. B is internally consistent but gives a negative ΔH, contradicting the diagram. C gives Eₐ(f) − Eₐ(r) = −150 kJ·mol⁻¹, which does not equal +150. D gives a difference of +150, not +300, so the three values are inconsistent. Modelled on Nov 2024 Paper 2, Question 1.4 — same skill, new scenario.
Question 3
Nitrogen monoxide, NO(g), and oxygen, O₂(g), are mixed in a sealed 2 dm³ container at constant temperature. They react according to the balanced equation:
The concentration of O₂(g) in the container is recorded at regular time intervals. The results are shown in the table below.
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|---|
| [O₂] (mol·dm⁻³) | 0,50 | 0,35 | 0,26 | 0,20 | 0,17 | 0,16 | 0,16 |
The sketch shows Maxwell–Boltzmann distribution curves, P and Q, for the same reaction mixture at two different temperatures. Eₐ is the activation energy of the reaction.
- aDefine the term reaction rate. (2)
- bIs the reaction rate HIGHER at 10 s or at 40 s? Use the data in the table to give a reason for the answer. (2)
- cCalculate the average rate (in mol·s⁻¹) at which NO₂(g) is formed during the first 10 s. (5)
- dWhich reactant, NO or O₂, is in excess? (1)
- eThe experiment is repeated in a SMALLER sealed container at the same temperature, using the same initial amounts of both gases. How will the magnitude of the initial gradient of the concentration–time graph of O₂(g) change? Choose from INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer. (2)
- fThe original experiment is now repeated with a suitable solid catalyst present. Explain, in terms of the collision theory, why the reaction rate increases. (3)
- gIdentify the curve (P or Q) that represents the HIGHER temperature AND use the curves to explain why the reaction is faster at this temperature. (3)
This question has a diagram, shown in the app.
Hint
For the rate calculation, convert the change in [O₂] to moles using the container volume, then use the 2 : 1 mole ratio of NO₂ to O₂.
Worked answer
a) Reaction rate is the change in concentration (or amount/mass/volume) of a reactant or product per unit time. ✓✓
b) At 10 s. ✓ Around 10 s the concentration of O₂ changes much more per 10 s interval (0,15 mol·dm⁻³ in 0–10 s) than around 40 s (0,03 then 0,01 mol·dm⁻³ per interval), i.e. the gradient of the concentration–time graph is steeper at 10 s. ✓
c) mol·dm⁻³ ✓
mol ✓
mol (ratio 2 : 1) ✓
Rate mol·s⁻¹ ✓✓ (formula/ratio ✓ substitution ✓ answer+unit ✓)
d) O₂ ✓ (its concentration becomes constant at 0,16 mol·dm⁻³, which is not zero — the NO is used up first).
e) INCREASES ✓ — the same amounts of gas in a smaller volume give higher concentrations, so there are more particles per unit volume and more effective collisions per unit time. ✓
f) A catalyst provides an alternative reaction pathway of lower activation energy. ✓ More particles now have kinetic energy equal to or greater than the (lower) activation energy. ✓ Therefore there are more effective collisions per unit time and the rate increases. ✓
g) Curve Q ✓ (its peak is lower and shifted to higher kinetic energy). At the higher temperature the area under the curve beyond Eₐ is larger, so more particles have kinetic energy equal to or greater than Eₐ. ✓ More effective collisions occur per unit time, so the rate is higher. ✓
Modelled on Nov 2025 Paper 2, Question 5 — same skills, new scenario.
Question 4
Zinc granules react with EXCESS dilute sulfuric acid of concentration 0,2 mol·dm⁻³ at 25 °C:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
Which ONE of the following changes will NOT increase the initial rate of this reaction?
Hint
The acid is already in excess — does adding more of the same acid change its concentration?
Worked answer
C. The acid is in excess, and extra acid of the SAME concentration leaves the concentration unchanged, so the initial rate stays the same. A: powder exposes a larger surface area — faster. B: a higher temperature gives more effective collisions — faster. D: a higher concentration gives more collisions per second — faster.
Modelled on Nov 2025 Paper 2, Question 1.4 — same skills, new scenario.
Question 5
The reaction P(g) + Q(g) → PQ(g) is EXOTHERMIC. The activation energy of the forward reaction is 60 kJ·mol⁻¹. Which ONE of the following is possible for this reaction?
Hint
Use Eₐ(reverse) = Eₐ(forward) − ΔH and remember the sign of ΔH for an exothermic reaction.
Worked answer
B. For an exothermic reaction ΔH < 0, so Eₐ(reverse) = Eₐ(forward) − ΔH must be LARGER than 60 kJ·mol⁻¹. A value of 90 kJ·mol⁻¹ gives ΔH = 60 − 90 = −30 kJ·mol⁻¹ — possible. A and D: ΔH must be negative. C: Eₐ(reverse) = 45 gives ΔH = +15 kJ·mol⁻¹, i.e. endothermic.
Modelled on Nov 2025 Paper 2, Question 1.6 — same skills, new scenario.