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Collision theory and equilibrium
Explaining Le Chatelier's principle with collision theory: forward and reverse rates after changes in concentration and pressure, the Maxwell–Boltzmann distribution, activation energies of the forward and reverse reactions, why temperature shifts equilibria, and why catalysts don't.
Le Chatelier's principle tells you which way an equilibrium will shift, but it doesn't say why. Molecules don't know about disturbances and can't decide to counteract anything. Everything they do comes down to collisions: how often particles meet and how hard they hit. Get that picture right and every Le Chatelier prediction explains itself.
Collision theory recap
For particles to react they must:
- collide,
- with at least the activation energy $E_a$, and
- in the right orientation, so the bonds that need to break and form are lined up.
The rate of a reaction depends on the number of successful collisions per second. That rises with concentration (or partial pressure) and surface area, which increase the collision frequency, and with temperature, which increases both the frequency and, far more importantly, the fraction of collisions with enough energy. A catalyst provides an alternative pathway with a lower $E_a$.
Concentration changes, explained by rates
At equilibrium the forward and reverse rates are equal. Suppose you add Fe³⁺ to the Fe³⁺/SCN⁻/FeSCN²⁺ equilibrium:
- More Fe³⁺ ions means more frequent collisions between Fe³⁺ and SCN⁻, so the forward rate jumps. The reverse rate hasn't changed yet.
- Because forward is now faster than reverse, there is a net forward reaction: [FeSCN²⁺] rises, and [Fe³⁺] and [SCN⁻] fall.
- As [FeSCN²⁺] rises the reverse rate increases; as reactants are used up the forward rate falls.
- When the two rates are equal again, a new equilibrium is established with more FeSCN²⁺ than before: a shift to the right.
On a rate–time graph the forward line jumps up at the moment of addition, then curves down, while the reverse line curves up to meet it. The new equilibrium rate is higher than the old one, because there are more particles of everything colliding.
Pressure changes, explained by rates
Halve the volume of N₂O₄ ⇌ 2NO₂ and every concentration doubles. Both rates increase, but not equally. The forward reaction needs only one N₂O₄ molecule, so its rate doubles. The reverse needs two NO₂ molecules to collide, so its rate depends on [NO₂]² and goes up four times. The reverse is now faster, so there is a net shift to the left, towards fewer gas molecules, until the rates match again. The side with more gas particles is always hit harder by a change in pressure.
Temperature and the Maxwell–Boltzmann distribution
The particles in a gas or solution don't all have the same energy. The Maxwell–Boltzmann distribution shows how many particles have each kinetic energy. Key features: the curve starts at zero (no particles are completely still), has a peak at the most likely energy, and has a long tail of very energetic particles. The total area under the curve is the total number of particles, so it stays the same when temperature changes.
Heating moves the peak to the right and lowers it, and the tail becomes fatter. The number of particles with energy above $E_a$ increases dramatically: often a 10 °C rise roughly doubles a reaction rate.
Now the key link. The forward and reverse reactions have different activation energies, and they differ by exactly ΔH:
$$E_a(\text{forward}) - E_a(\text{reverse}) = \Delta H$$
The endothermic direction always has the larger activation energy. When the temperature rises, the fraction of particles able to get over a high barrier grows proportionally more than the fraction able to get over a low barrier. So both reactions speed up, but the endothermic direction speeds up more. The rates are no longer equal, and the system shifts towards the endothermic side until they are equal again. Cooling slows the endothermic direction more, so the equilibrium shifts the exothermic way.
Because the two rates are affected differently, the ratio of rate constants changes, which is why K depends on temperature but not on concentration or pressure.
- With Forward exothermic selected, raise the temperature from 300 K to 400 K. By what factor does each fraction grow? Which direction gains more?
- Switch to Forward endothermic and repeat. Which way does the equilibrium shift now?
- Turn on the catalyst. Do both fractions increase? Does the comparison between forward and reverse change?
- Lower the temperature to 250 K. Explain, in terms of collisions, why cooling favours the exothermic direction.
This model uses small activation energies so the shaded areas are visible. Real activation energies are many times the average kinetic energy, so the real fractions are tiny, but the argument is the same.
Catalysts
A catalyst provides an alternative pathway with a lower activation energy. Because the forward and reverse reactions pass through the same transition state, the catalyst lowers both barriers by the same amount (look at the dashed path in the energy profile). Both rates increase by the same factor, so a system not yet at equilibrium reaches it faster, but the equilibrium concentrations and K don't change.
Test yourself
For a reaction at equilibrium, Eₐ(forward) = 80 kJ mol⁻¹ and Eₐ(reverse) = 130 kJ mol⁻¹. What is ΔH for the forward reaction, and how does heating affect the equilibrium?
ΔH = 80 − 130 = −50. Heating favours the endothermic reverse, which has the larger Eₐ.
Which feature of a Maxwell–Boltzmann distribution stays the same when the temperature increases?
The area represents the total number of particles.
The volume of an equilibrium mixture of N₂O₄ and NO₂ is halved. Immediately afterwards
Forward rate doubles; the reverse, which needs two NO₂ to collide, increases about fourfold.
Adding a catalyst to an equilibrium mixture
Both activation energies are lowered by the same amount.
Label the energy profile.
Put the events in order after extra SCN⁻ is added to an Fe³⁺/SCN⁻/FeSCN²⁺ equilibrium.
For a reaction to occur, particles must collide with at least the energy and the correct . The direction with the larger activation energy is the direction. When temperature rises, the of particles with enough energy increases. The difference between the forward and reverse activation energies equals .
Use collision theory and a Maxwell–Boltzmann distribution to explain why increasing the temperature of the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹, decreases the yield of SO₃.
The forward reaction is exothermic, so the reverse reaction is endothermic and has the higher activation energy: Eₐ(reverse) = Eₐ(forward) + 197 kJ mol⁻¹. When the temperature rises, the Maxwell–Boltzmann distribution flattens and spreads to higher energies, so a larger proportion of particles have energy at or above each activation energy, and both forward and reverse rates increase. However, the fraction of particles able to overcome the higher barrier increases by a much larger factor than the fraction able to overcome the lower one. The reverse rate therefore becomes greater than the forward rate, and there is a net reverse reaction: SO₃ is converted back into SO₂ and O₂. As [SO₃] falls and [SO₂] and [O₂] rise, the rates become equal again at a new equilibrium containing less SO₃, so the yield decreases.