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Dynamic equilibrium and Le Chatelier's principleAQA A-Level Chemistry: Revision notes

Section 1

Reversible reactions and dynamic equilibrium

Many reactions are reversible, shown by the sign ⇌. In a closed system a reversible reaction reaches dynamic equilibrium when:

  • the forward and reverse reactions proceed at equal rates
  • the concentrations of reactants and products remain constant

It is dynamic because both reactions continue; nothing stops. The concentrations are constant but are not necessarily equal. Equilibrium needs a closed system (nothing can enter or leave).

Key termsreversible reactiondynamic equilibriumclosed system
Common mistake

Do not write that the reaction 'has stopped' or that concentrations are 'equal' at equilibrium. The rates are equal and the concentrations are constant.

Section 2

Le Chatelier's principle

Le Chatelier's principle: when a change is made to a system at equilibrium, the position of equilibrium moves to oppose the change.

It predicts the direction only. A shift to the right means more products are formed; a shift to the left means more reactants. It applies to homogeneous reactions (all species in the same phase) such as gases or solutions.

Key termsLe Chatelier's principleposition of equilibriumhomogeneous

Section 3

Concentration, pressure and temperature

  • Concentration: increasing a reactant concentration shifts the equilibrium to the right (to use it up). Removing a product also shifts it right.
  • Pressure (gases only): increasing the pressure shifts the equilibrium to the side with fewer moles of gas. If the moles are equal, there is no shift.
  • Temperature: increasing the temperature shifts the equilibrium in the endothermic direction (to absorb heat); decreasing it shifts it in the exothermic direction.

Example: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹. Higher pressure moves it right (4 mol → 2 mol); higher temperature moves it left (reverse is endothermic).

Key termsendothermic direction
Exam tip

Always link the direction of shift to the reason: 'moves to the side with fewer moles of gas to oppose the increase in pressure'.

Section 4

Catalysts and equilibrium

A catalyst does not affect the position of equilibrium. It lowers the activation energy of the forward and reverse reactions equally, so equilibrium is reached faster but the equilibrium yield is unchanged.

Key termscatalyst
Common mistake

A catalyst does not increase the yield at equilibrium. It only increases the rate at which equilibrium is reached.

Section 5

Industrial compromise conditions

For an exothermic reversible reaction such as the Haber process, low temperature gives a high equilibrium yield, but the rate is too slow. A compromise temperature (about 450 °C) gives an acceptable yield at an acceptable rate.

For pressure, a high pressure increases the yield where there are fewer moles on the product side. However, high pressure needs expensive equipment, more energy for compression and poses safety risks. A compromise pressure is used.

Products may be removed and unreacted gases recycled to raise overall yield and reduce waste.

Key termscompromise conditions
Exam tip

In a 'why a compromise' answer, give one effect on yield, one effect on rate or cost, and the compromise.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Dynamic equilibrium and Le Chatelier's principle

  1. Cobalt(II) ions in aqueous solution take part in the equilibrium [Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l). The [Co(H₂O)₆]²⁺ ion is pink and the [CoCl₄]²⁻ ion is blue, and the forward reaction is endothermic. A student prepares a purple mixture of the two ions in a sealed tube and leaves it at constant temperature.
    The purple mixture turns blue when it is heated. Explain this observation.2 marks
  2. Ammonia is manufactured by the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. The process uses an iron catalyst at a temperature of about 450 °C and a pressure of about 20 MPa.
    Explain why a compromise temperature of about 450 °C is used rather than a much lower temperature.2 marks
  3. In the Contact process, sulfur dioxide is oxidised to sulfur trioxide: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ mol⁻¹. A vanadium(V) oxide catalyst is used at about 450 °C and a pressure of about 200 kPa. Under these conditions about 99% of the sulfur dioxide is converted into sulfur trioxide.
    Predict and explain the effect of increasing the temperature on the equilibrium yield of sulfur trioxide.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).