AS Chemistry: Reaction Kinetics — 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 available free from your board.

Related: Reaction Kinetics revision notes

---

Questions

1. State the two conditions required for a collision between particles to result in a reaction. [2]

2. Define activation energy. [2]

3. Explain, in terms of collision theory, the effect on rate of:

(a) increasing the concentration of a solution [2] (b) increasing the pressure of a gas [2] (c) increasing the surface area of a solid [2] (d) increasing the temperature [3]

4. A Boltzmann distribution curve is drawn for a gas at temperature T₁.

(a) Describe how the curve changes when the temperature is raised to T₂. [3]

(b) Explain, using the curve, why a small temperature rise produces a large rate increase. [3]

(c) Sketch or describe how a catalyst is shown on the same diagram, and explain its effect. [3]

5. Explain why a catalyst does not change the position of equilibrium in a reversible reaction. [2]

6. Iron is used as a catalyst in the Haber process (a gas-phase reaction), while an enzyme in solution catalyses a reaction between two dissolved reactants.

(a) State which of these is a homogeneous catalyst and which is heterogeneous, giving a reason for each. [2] (b) Describe the three-step mechanism by which a heterogeneous catalyst such as iron increases the rate of a gas-phase reaction. [3]

7. A student claims that adding a catalyst to a reaction increases the overall yield of product obtained.

(a) State whether this claim is correct, and explain why. [2] (b) State one property of the catalyst itself that remains unchanged after the reaction has finished. [1]

---

8. State and explain the effect of adding a catalyst on the Boltzmann distribution curve and on the activation energy of a reaction at constant temperature. [2]

Answers

1. The particles must collide with energy greater than or equal to the activation energy [1] and in the correct orientation [1].

2. The minimum energy that colliding particles must possess [1] for a reaction to occur [1].

3. (a) More particles per unit volume [1], so more frequent collisions, and therefore a greater number of successful collisions per second [1] (concentration does not change the proportion of collisions with sufficient energy and correct orientation – that fraction depends on temperature and activation energy, not concentration). (b) The particles are closer together [1], so collisions are more frequent [1]. (c) More particles are exposed at the surface [1], so there are more collisions per second with the other reactant [1]. (d) Particles have more kinetic energy and move faster, so collisions are more frequent [1]. More importantly, a greater proportion of particles have energy ≥ E_a [1], so a much larger fraction of collisions are successful [1].

4. (a) The peak moves to the right (higher energy) and becomes lower [1]; the curve broadens [1]; the area under the curve stays the same, since the total number of particles is unchanged [1].

(b) The area under the curve to the right of E_a represents the particles able to react [1]. Because the curve’s tail rises steeply, a small shift produces a large increase in that area [1], so the proportion of successful collisions rises sharply [1].

(c) A vertical line drawn at a lower energy than the original E_a [1]. The catalyst provides an alternative route of lower activation energy [1], so a greater proportion of particles have sufficient energy to react [1].

5. It increases the rate of the forward and reverse reactions equally [1], so equilibrium is reached sooner but at the same position [1].

6. (a) The enzyme is homogeneous — it is in the same phase (aqueous solution) as the dissolved reactants [1]. Iron in the Haber process is heterogeneous — it is a solid, a different phase from the gaseous reactants [1]. (b) Reactant gas molecules adsorb onto the catalyst’s surface [1]; while adsorbed, the reaction takes place, with the surface providing an alternative pathway of lower activation energy [1]; the product molecules then desorb, freeing the surface to adsorb further reactant molecules [1].

7. (a) No [1] — a catalyst only increases the rate at which equilibrium (or completion) is reached; it does not alter the thermodynamics of the reaction, so the same equilibrium position and the same maximum yield are eventually obtained with or without it [1]. (b) It is chemically unchanged at the end of the reaction (not consumed) [1].

8. The catalyst provides an alternative route with a lower activation energy; the Boltzmann distribution curve itself does not change, because the temperature is the same [1]. A greater proportion of molecules now have energy equal to or greater than the (lower) activation energy, so there are more successful collisions per unit time [1].

Examiner insight (June 2025): weaker answers drew a new distribution curve for the catalyst instead of keeping the curve and moving the activation energy to a lower value. Try the real question next: Cambridge International AS & A Level Chemistry 9701, June 2025, Paper 22, Question 3.

---

Where marks are usually lost

• Giving only “more collisions” for a temperature increase — the energy effect is the dominant one.
• Saying a catalyst “lowers the activation energy” of the original reaction; it provides an alternative route.
• Changing the area under a Boltzmann curve when temperature changes.
• Drawing the Boltzmann curve touching the y-axis (it starts at the origin).
• Confusing homogeneous with heterogeneous catalysis — the test is simply whether the catalyst is in the same phase as the reactants or not.
• Omitting the desorption step when describing heterogeneous catalysis, leaving the surface permanently blocked in the answer’s own logic.
• Claiming a catalyst increases the yield of a reaction, rather than only the rate at which the (unchanged) yield is reached.

Homogeneous vs heterogeneous — the phase test

The distinction between the two types of catalyst comes down to a single question: is the catalyst in the same phase as the reactants, or a different one? A homogeneous catalyst (same phase, commonly a catalyst dissolved alongside aqueous reactants) works by forming a temporary intermediate species with a reactant, which then reacts further to release the product and regenerate the catalyst. A heterogeneous catalyst (different phase, most often a solid catalysing a gas-phase or solution reaction) works through adsorption, reaction at the surface, and desorption, as described in question 6. Both types share every general property of a catalyst — unchanged at the end, no effect on ΔH, yield, or equilibrium position — the phase relationship only determines the specific mechanism by which the rate increase is achieved. For the full comparison and Boltzmann distribution diagrams, see the Reaction Kinetics revision notes.