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9476 / 2027
Reaction Kinetics

Lesson 4 of 5

Collisions, activation energy and temperature

Separate collision frequency from the fraction energetic enough to react.

A-Level 9476 (2026-2027)

  • I can explain the central kinetic terms.Syllabus

    Syllabus 9476, 8(a). Explain the central kinetic terms.

  • I can explain concentration effects through collision frequency.Syllabus

    Syllabus 9476, 8(f). Explain concentration effects through collision frequency.

  • I can explain activation energy using a Boltzmann distribution.Syllabus

    Syllabus 9476, 8(g). Explain activation energy using a Boltzmann distribution.

  • I can explain temperature effects on k and rate.Syllabus

    Syllabus 9476, 8(h). Explain temperature effects on k and rate.

  • I can explain catalytic lowering of a reaction barrier.Syllabus

    Syllabus 9476, 8(i). Explain catalytic lowering of a reaction barrier.

Reaction needs both encounters and sufficient energy

Heating changes the energy distribution; concentration mainly changes collision frequency.

Make a guess. You are not marked.

Warming a reaction from 300 K to 310 K raises the collision frequency by only about 2%. What happens to the rate?

  1. It rises by about 2%, the same as the collisions
  2. It does not change, because 10 K is small
  3. It rises far more, often roughly doubling
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It rises far more, often roughly doubling

Most of the increase comes from a larger fraction of collisions with energy at or above Ea, which the Boltzmann distribution shows grows quickly.

The activation energy, Ea, is the minimum energy that colliding particles need in order to react. Particles must collide with enough energy and in the right orientation. At a higher concentration, there are more particles in the same volume, so collisions happen more often. The rate equation tells you how strongly the rate responds.

Heating changes the Boltzmann energy distribution

At higher temperature the energy distribution has a lower, broader peak and a larger fraction of particles beyond the same activation-energy threshold. Both distributions are normalised; temperature does not move the threshold.

Schematic energy scale for the same number of particles. The area to the right of Ea is the energetic fraction. At higher temperature this fraction increases substantially; collision frequency also increases because particles move faster.

The Boltzmann distribution starts at zero and has a long tail. Particles have a spread of energies, and there is no sharp maximum energy. When the temperature rises, the curve shifts towards higher energies. A much larger fraction of particles now has enough energy to cross the barrier. Collisions also become a little more frequent. Both effects increase k, so the rate rises at the same concentrations. Heating does not lower the activation energy.

Adding more particles at the same temperature does not change the shape of the energy distribution. The fraction of particles with energy above Ea stays the same. What rises is the number of collisions that can succeed. So a temperature change and a concentration change work in different ways, even though both make a reaction faster.

Check your understanding

A sample of gas is heated at constant volume. How does its Boltzmann distribution change?

  1. The Ea line moves to lower energy, so more particles have enough energy to react
  2. The peak moves to higher energy and is higher, as more particles have high energy
  3. The peak moves to higher energy and is lower; the total area stays the same
  4. The total area increases, because the particles now have more energy than before
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The peak moves to higher energy and is lower; the total area stays the same

Heating spreads the same number of particles over a wider range of energies. The peak moves right, and it is lower because the area, the number of particles, is unchanged.

Check your understanding

A reaction is first order in A. Doubling [A] at constant temperature doubles the rate. Which statement explains this?

  1. Doubling [A] lowers the activation energy of the reaction pathway
  2. Doubling [A] doubles the rate constant k in rate = k[A]
  3. Doubling [A] raises the proportion of particles with energy at least Ea
  4. Twice as many A particles per unit volume, so twice as many effective collisions
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Twice as many A particles per unit volume, so twice as many effective collisions

At the same temperature, the same fraction of collisions succeeds. With twice as many A particles in each unit volume, effective collisions per second double.

Exam-style question

With the help of a sketch of the Boltzmann distribution, explain why the rate constant of a reaction increases when the temperature is raised. [3]

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A catalyst supplies another mechanism

More particles can react at the same temperature because the barrier is lower.

Make a guess. You are not marked.

A catalyst lowers the activation energy of a forward reaction by 40 kJ mol-1. What happens to the activation energy of the reverse reaction?

  1. It also falls by 40 kJ mol-1
  2. It rises by 40 kJ mol-1 to balance
  3. It stays the same, as the catalyst acts forwards
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It also falls by 40 kJ mol-1

The catalysed route has a lower peak but the same reactant and product levels, so both barriers drop by the same amount.

A catalyst changes the threshold, not the distribution

One energy distribution at a fixed temperature is crossed by an uncatalysed activation threshold and a lower catalysed threshold. The catalysed threshold admits a larger fraction of particles without changing their energies.

At fixed temperature the distribution stays the same. Lowering the relevant activation barrier increases the rate constant. Shading marks the fraction above the original higher barrier; catalysis also admits particles between the two thresholds.

A catalyst takes part in the reaction steps and is made again by the end of its cycle. It speeds up the forward and the reverse reactions, so equilibrium is reached sooner. It does not change ΔH, the equilibrium constant or the equilibrium composition at the same temperature. In practice a catalyst can still be poisoned or wear out, even though the overall equation does not use it up.

Use the same Boltzmann curve to explain the speed-up. The temperature has not changed, so the curve itself is the same. The catalyst gives an alternative pathway with a lower activation energy. The Ea line moves to the left on the energy axis. A greater proportion of particles now has energy equal to or greater than the new Ea, so the frequency of effective collisions increases. The rate constant k increases, and so does the rate. Each link in this chain usually earns its own mark.

Check your understanding

A catalyst is added to a reaction at constant temperature. Which quantity stays the same?

  1. The enthalpy change of reaction, delta H
  2. The fraction of particles with energy at least Ea
  3. The rate constant k at this temperature
  4. The activation energy, Ea, of the reaction
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The enthalpy change of reaction, delta H

delta H depends only on the energies of the reactants and products, and the catalyst changes neither.