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What controls how fast a chemical reaction goes?

Collision theory, activation energy, the Maxwell-Boltzmann distribution, and the effects of temperature, concentration, surface area and catalysts on the rate of reaction.

An Edexcel 9CH0 Topic 9 answer covering collision theory, activation energy, the Maxwell-Boltzmann distribution, and how temperature, concentration, surface area and catalysts affect reaction rate.

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What this topic is asking

Edexcel Topic 9 wants you to use collision theory and the Maxwell-Boltzmann distribution to explain reaction rates, and to explain the effects of temperature, concentration, pressure, surface area and catalysts. The marks come from precise language ("proportion of molecules with EEaE \ge E_a", "frequency of successful collisions") rather than loose phrases.

The answer

Collision theory

For a reaction to occur, particles must collide, with energy at or above the activation energy (EaE_a) and with the correct orientation (geometry that lets the right bonds break and form). Most collisions are unsuccessful because they are too gentle or wrongly oriented. The rate of reaction depends on the frequency of successful collisions per unit time.

The Maxwell-Boltzmann distribution

Raising the temperature shifts the curve to the right and lowers and broadens its peak (the total area, the number of molecules, is unchanged). The crucial effect is on the shaded area beyond EaE_a: because EaE_a sits on the steep tail, even a modest temperature rise sharply increases the proportion of molecules with EEaE \ge E_a. Adding a catalyst does not change the curve at all; it moves EaE_a to a lower value, so a larger area of the same curve now lies beyond the (lower) barrier.

Factors affecting rate

  • Temperature. A higher temperature increases the proportion of molecules with EEaE \ge E_a and makes collisions slightly more frequent, so the rate of successful collisions rises sharply. This energy effect dominates the frequency effect.
  • Concentration (or pressure for gases). More particles per unit volume means collisions are more frequent, so successful collisions per second increase.
  • Surface area. Breaking a solid into smaller pieces exposes more particles at the surface, increasing the frequency of collisions with the other reactant.

Catalysts

On a reaction profile the catalysed route has a lower hump; on the Maxwell-Boltzmann diagram the lower EaE_a means a greater proportion of molecules can react. Catalysts can be heterogeneous (different phase, e.g. iron in the Haber process, working by adsorption) or homogeneous (same phase, e.g. acid catalysis of esterification).

Examples in context

Example 1. Catalytic converters in cars. A catalytic converter contains platinum, palladium and rhodium on a honeycomb support. It catalyses the conversion of toxic CO\text{CO} and NO\text{NO} into CO2\text{CO}_2 and N2\text{N}_2, for example 2CO+2NO2CO2+N22\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2. The metals work heterogeneously: gases adsorb onto the surface, bonds weaken, the reaction proceeds with a much lower EaE_a, and the products desorb. The catalyst is not consumed, which is why a converter lasts the life of the car (provided it is not poisoned by lead).

Example 2. Why flour mills risk dust explosions. Solid flour burns slowly, but finely dispersed flour dust has an enormous surface area, so collisions between fuel and oxygen are vastly more frequent. The initial rate becomes so high that combustion is effectively instantaneous, which is why grain silos and flour mills enforce strict dust control. This is the surface-area factor from collision theory taken to an industrial extreme.

Try this

Q1. State the two conditions needed for a successful collision. [2 marks]

  • Cue. Energy at or above EaE_a, and the correct orientation.

Q2. Explain why increasing temperature increases the rate of reaction. [3 marks]

  • Cue. The Maxwell-Boltzmann distribution shifts right, so a greater proportion of molecules have energy at or above EaE_a, and collisions are more frequent, giving more successful collisions per second.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksUsing the Maxwell-Boltzmann distribution, explain why a small rise in temperature causes a large increase in the rate of a reaction.
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Refer to the shape change of the curve and to the activation energy.

Raising the temperature shifts the Maxwell-Boltzmann curve to the right and lowers and broadens the peak (1). A greater proportion of molecules now have energy at or above the activation energy EaE_a (1). The fraction beyond EaE_a rises sharply because EaE_a lies on the steep tail of the curve, so even a small temperature rise more than doubles it (1). Collisions are also slightly more frequent, so the rate of successful collisions increases markedly (1).

Markers reward "proportion of molecules with EEaE \ge E_a", not simply "molecules have more energy", and credit the dominance of the energy effect over the frequency effect.

Edexcel 20215 marksA student investigated the reaction of 0.50 g0.50\ \text{g} of marble chips (calcium carbonate) with excess dilute hydrochloric acid. (a) Write the equation. (b) Calculate the volume of CO2\text{CO}_2 produced at room temperature and pressure (24.0 dm3mol124.0\ \text{dm}^3\,\text{mol}^{-1}). (c) State and explain one change that would increase the initial rate without changing the final volume of gas.
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Combine a mole calculation with a kinetics explanation.

(a) CaCO3+2HClCaCl2+H2O+CO2\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 (1).

(b) M(CaCO3)=100.1 g mol1M(\text{CaCO}_3) = 100.1\ \text{g mol}^{-1}, so n=0.50/100.1=5.0×103 moln = 0.50 / 100.1 = 5.0 \times 10^{-3}\ \text{mol} (1). The mole ratio CaCO3:CO2\text{CaCO}_3 : \text{CO}_2 is 1:11:1, so n(CO2)=5.0×103 moln(\text{CO}_2) = 5.0 \times 10^{-3}\ \text{mol}. Volume =5.0×103×24.0=0.12 dm3= 5.0 \times 10^{-3} \times 24.0 = 0.12\ \text{dm}^3 (120 cm3120\ \text{cm}^3) (1).

(c) Use smaller chips / powder it (1): this increases the surface area, so collisions between acid particles and the solid are more frequent and the initial rate rises, but the amount of CaCO3\text{CaCO}_3 (and so the CO2\text{CO}_2 produced) is unchanged (1).

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