Equilibrium and Le Chatelier's Principle

Equilibrium and Le Chatelier’s principle are Supplement topics that carry 3-6 marks on Paper 4. The key skill is predicting which direction the equilibrium shifts when conditions change, and applying this to industrial processes like the Haber process and Contact process.

Reversible reactions

A reversible reaction can go in both directions. The symbol ⇌ is used instead of a single arrow.

Example: N2 + 3H2 ⇌ 2NH3 (the Haber process)

The forward reaction produces ammonia. The backward reaction decomposes ammonia back into nitrogen and hydrogen. Both reactions can occur.

Another example used in practicals: hydrated copper(II) sulfate ⇌ anhydrous copper(II) sulfate + water

CuSO4.5H2O (blue) ⇌ CuSO4 (white) + 5H2O

Heating drives the reaction forward (white powder). Adding water reverses it (blue crystals reform). See water of crystallisation for details.

Dynamic equilibrium

When a reversible reaction occurs in a closed system (nothing can enter or leave), it eventually reaches dynamic equilibrium:

1. The rate of the forward reaction equals the rate of the backward reaction
2. The concentrations of reactants and products are constant (not necessarily equal)
3. Both reactions are still occurring (it is dynamic, not static)

The word “dynamic” means reactions are still happening. The word “equilibrium” means the rates are balanced, so there is no net change.

Le Chatelier’s principle

If a system at equilibrium is disturbed by a change in conditions, the position of equilibrium shifts to partially counteract the change.

Effect of concentration changes

Change	Equilibrium shifts to…	Reason
Increase concentration of a reactant	Right (toward products)	System opposes the increase by using up the added reactant
Decrease concentration of a reactant	Left (toward reactants)	System opposes the decrease by making more of the removed reactant
Increase concentration of a product	Left (toward reactants)	System opposes the increase by converting product back to reactants
Remove a product	Right (toward products)	System opposes the removal by making more product

Effect of temperature changes

For temperature, you need to know whether the forward reaction is exothermic or endothermic:

For N2 + 3H2 ⇌ 2NH3 (forward reaction is exothermic, Delta H = -92 kJ/mol):

Change	Equilibrium shifts to…	Reason
Increase temperature	Left (endothermic direction)	System absorbs excess heat by shifting to the endothermic direction
Decrease temperature	Right (exothermic direction)	System releases heat to oppose cooling

Key rule: Increasing temperature favours the endothermic direction. Decreasing temperature favours the exothermic direction.

Temperature changes actually change the position of equilibrium AND the yield at equilibrium.

Effect of pressure changes (gaseous reactions)

Pressure only affects reactions involving gases. Count the moles of gas on each side:

N2 + 3H2 ⇌ 2NH3 — 4 moles of gas on the left, 2 on the right

Change	Equilibrium shifts to…	Reason
Increase pressure	Side with fewer moles of gas (right)	System reduces pressure by decreasing total moles of gas
Decrease pressure	Side with more moles of gas (left)	System increases pressure by producing more moles of gas

If both sides have equal moles of gas, pressure changes have no effect on equilibrium position.

Effect of a catalyst

A catalyst has no effect on the position of equilibrium. It speeds up both forward and backward reactions equally, so equilibrium is reached faster but the proportions of reactants and products at equilibrium are unchanged.

Application: the Haber process

N2(g) + 3H2(g) ⇌ 2NH3(g), Delta H = -92 kJ/mol

Condition	Effect on yield of NH3	Effect on rate
High pressure (200 atm)	Increases yield (fewer moles on right)	Increases rate
Low temperature	Increases yield (exothermic direction)	Decreases rate (too slow)
Iron catalyst	No effect on yield	Increases rate

The compromise: The Haber process uses about 450 C and 200 atmospheres. A lower temperature would give a higher yield but the reaction would be too slow. The temperature used is a compromise between yield and rate. The iron catalyst helps achieve a reasonable rate at this temperature.

Application: the Contact process

2SO2(g) + O2(g) ⇌ 2SO3(g), Delta H = -197 kJ/mol

Condition	Effect on yield of SO3
High pressure	Increases yield (3 moles -> 2 moles)
Low temperature	Increases yield (exothermic forward)

Conditions used: 450 C, 1-2 atm, vanadium(V) oxide (V2O5) catalyst. Again, a compromise between yield and rate.

Common exam mistakes

1. “The catalyst shifts the equilibrium” — no. A catalyst speeds up both reactions equally and does not change the position of equilibrium.
2. “At equilibrium, the amounts of reactants and products are equal” — they are constant, but not necessarily equal. The position of equilibrium determines the ratio.
3. “Increasing pressure always increases yield” — only if the forward reaction produces fewer moles of gas. If both sides have the same number of moles, pressure has no effect.
4. Confusing rate and yield: High temperature increases rate but may decrease yield. The Haber process compromise exists because of this conflict.
5. Forgetting the closed system requirement: Equilibrium can only be established if nothing enters or leaves the system.

Worked exam questions

The Haber process: N2 + 3H2 ⇌ 2NH3. The forward reaction is exothermic. Predict and explain the effect of increasing the temperature on the yield of ammonia. [3 marks]

• The yield of ammonia decreases [1]
• Increasing temperature favours the endothermic direction / the backward reaction [1]
• The equilibrium shifts to the left, converting ammonia back to nitrogen and hydrogen [1]

Explain why a compromise temperature is used in the Haber process rather than a very low temperature. [2 marks]

• A lower temperature would give a higher yield of ammonia [1]
• But the rate of reaction would be too slow to be economically viable / practical [1]

The reaction 2SO2 + O2 ⇌ 2SO3 is at equilibrium. Predict the effect of increasing pressure on the position of equilibrium. Explain your answer. [3 marks]

• The equilibrium shifts to the right / toward products [1]
• There are 3 moles of gas on the left and 2 on the right [1]
• The system reduces pressure by shifting to the side with fewer moles of gas [1]