Part 1: Dynamic equilibrium and changing conditions

Separate the speed of reaching equilibrium from the composition at equilibrium, and predict responses to changing conditions.

Equal rates, constant concentrations

A dynamic equilibrium is established in a closed system when forward and reverse reactions continue at equal rates. Concentrations remain constant, but the concentrations of reactants and products need not be equal. “Dynamic” means reactions have not stopped. A homogeneous equilibrium has all reacting species in the same phase.

On a rate–time graph starting with reactants, the forward rate often falls and reverse rate rises until they meet. On a concentration–time graph, concentrations level off at possibly different heights. Do not confuse equality of rates with equality of concentrations.

Explain how the equal rates are reached

Imagine a closed vessel initially containing reactants only. At first there are no products to undergo the reverse reaction. Forward reaction creates products while using reactants; typically the forward rate decreases and the reverse rate increases. Eventually the two rates match. Individual particles keep reacting in both directions, but there is no net change in the amounts.

A concentration graph levels off at equilibrium. It does not require the reactant and product curves to meet. A rate graph instead shows the forward and reverse curves meeting at the same non-zero value. A graph where both rates become zero describes a stopped reaction, not the usual dynamic-equilibrium model.

The system must retain the reacting substances for this equilibrium model. Opening a vessel so that a gaseous product escapes can prevent the original equilibrium from being maintained. “Closed” does not mean temperature cannot be controlled by exchanging energy with the surroundings.

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Apply Le Chatelier to the stated change

When conditions change, an equilibrium responds in the direction that opposes the imposed change. Increasing a reactant concentration favours its consumption; removing a product favours its replacement. This is a partial response, not a claim that the original concentrations are fully restored.

Increasing pressure by decreasing volume favours the side with fewer gas molecules; count gaseous coefficients only. If gaseous mole totals are equal, this pressure change does not shift position. A temperature increase favours the endothermic direction; a decrease favours the exothermic direction.

A catalyst accelerates both directions and allows equilibrium to be reached sooner without changing its position at fixed temperature. H032/01 June 2025 Q13 and Q23(a) assess this distinction and the meaning of dynamic equilibrium.

Count gas particles and state how pressure was changed

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), decreasing volume compresses the mixture. Both sides contain gases, but three stoichiometric moles on the left correspond to two on the right. The equilibrium shifts towards SO₃, reducing some of the imposed pressure increase. Say which side is favoured, not only “the side with fewer moles”.

For H₂(g) + I₂(g) ⇌ 2HI(g), each side contains two gas moles, so compression does not change equilibrium composition in the ideal model. Pressure may still change rates because gas concentrations change. Equal gas totals mean no position shift, not “no effect of any kind”.

An inert gas added at fixed volume raises total pressure but leaves the reacting gases’ concentrations unchanged in the ideal model, so no composition shift is predicted. This is a useful application of the concentration argument, not a new AS partial-pressure calculation. State “compression” when that is the change you mean.

Use the sign of the forward enthalpy change

For N₂O₄(g) ⇌ 2NO₂(g), the forward direction is endothermic. Heating therefore favours NO₂; cooling favours N₂O₄. If the equation is written the other way round, the forward sign reverses, but the physical prediction remains the same. Do not memorise “heating shifts left”.

Pressure and temperature can oppose each other. Heating this equilibrium favours two gas moles of NO₂, while compression favours one mole of N₂O₄. If both are changed together, a qualitative rule does not determine which effect dominates without further information.

Observe a reversible response

In a supervised equilibrium demonstration, use equal samples and change one condition. For the exothermic association 2NO₂(g) ⇌ N₂O₄(g), heating a sealed prepared tube makes the equilibrium mixture browner (more NO₂), while cooling makes it paler. These toxic gases require sealed teacher-prepared apparatus and appropriate controls.

For concentration, I₂(aq) + 2OH⁻(aq) ⇌ I⁻(aq) + IO⁻(aq) + H₂O(l) becomes paler as added hydroxide favours colourless products; removing OH⁻ by adding acid can restore brown iodine. Use small amounts under the centre’s risk assessment. A control sample helps distinguish chemical shifts from dilution alone.

Distinguish the imposed change from the later response

Adding more of a coloured reactant immediately changes its concentration and can change colour before the equilibrium has responded. The subsequent net reaction partially opposes that addition. It does not generally remove every extra molecule and restore all original concentrations.

For temperature demonstrations, use equal initial samples, one reference and controlled hot/cold baths. Allow time to reach the new temperature and compare under the same lighting. For concentration changes, consider dilution as well as the reagent’s chemical role; a paler solution alone does not establish the direction of an equilibrium shift.

A strong observation statement names the change, such as “the brown colour becomes more intense”. A supported inference then links it to a named species and direction. “Equilibrium moves” is an inference, not a directly visible observation.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources