Le Chatelier's principle (SACE Stage 2 Chemistry)

Syllabus dot point

“Apply Le Chatelier's principle to predict the effect of changes in concentration, pressure and temperature on the position of equilibrium.”

9 min readUpdated 2 Jun 2026

Quick answer

Le Chatelier's principle states that if a system at equilibrium is disturbed, it shifts to partly oppose the change and restore equilibrium. Adding a reactant shifts the system to consume it (toward products); increasing pressure shifts it toward the side with fewer gas moles; raising temperature shifts it in the endothermic direction. Only temperature changes change $K_c$ ; concentration, pressure and catalysts shift the position but leave $K_c$ unchanged.

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What this dot point is asking
Lead worked calculation
The principle
Concentration changes
Pressure and volume changes (gases only)
Temperature changes
Catalysts and inert gases
Why it matters for managing processes
Exam-style questions

What this dot point is asking

SACE expects you to predict the direction of shift for each disturbance, justify it, and state correctly whether $K_c$ changes.

Lead worked calculation

Predicting shifts for the Haber equilibrium

For $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ , $\Delta H = -92\ \text{kJ mol}^{-1}$ , predict the effect of (i) removing ammonia, (ii) decreasing the volume, (iii) lowering the temperature, on both the position and $K_c$ .

Removing ammonia (a product)

The system opposes the loss by making more $\text{NH}_3$ , so it shifts right. $K_c$ unchanged (concentration change).

Decreasing the volume (raising pressure)

Left has $4$ gas moles, right has $2$ . The system shifts toward fewer gas moles, to the right. $K_c$ unchanged (pressure change).

Lowering the temperature

The forward reaction is exothermic, so cooling shifts the equilibrium in the exothermic (forward) direction, to the right, and increases $K_c$ .

Conclude

All three changes favour ammonia, but only the temperature change alters $K_c$ . This explains why industrial ammonia synthesis uses high pressure and a moderate, compromise temperature.

The principle

Le Chatelier's principle

If a system at equilibrium is subjected to a change in concentration, pressure, volume or temperature, the position of equilibrium shifts in the direction that partly counteracts the change. The system never fully cancels the disturbance; it moves to a new equilibrium that reduces its effect.

Concentration changes

Adding a reactant or product makes the system shift to use up the added substance. Removing a species makes the system shift to replace it.

Add reactant or remove product, the equilibrium shifts to the right (more product).
Add product or remove reactant, the equilibrium shifts to the left (more reactant).

$K_c$ is unchanged: the ratio is restored once the new equilibrium is reached.

Pressure and volume changes (gases only)

Pressure changes only affect equilibria with a different number of gas moles on each side.

Increasing pressure (decreasing volume): the system shifts toward the side with fewer gas moles to reduce the pressure.
Decreasing pressure (increasing volume): the system shifts toward the side with more gas moles.
If both sides have equal gas moles, pressure has no effect on the position.

$K_c$ is unchanged by pressure or volume changes.

Temperature changes

Temperature is the one factor that changes $K_c$ , because it changes the rate constants of the forward and reverse reactions by different amounts. Treat heat as a reactant or product.

Temperature and the direction of an enthalpy-driven shift

For an exothermic forward reaction (heat is a "product"): raising the temperature shifts the equilibrium to the left (endothermic direction) and lowers $K_c$ ; lowering the temperature shifts it right and raises $K_c$ . For an endothermic forward reaction the effects reverse. Temperature is the only change that moves the value of $K_c$ .

Catalysts and inert gases

A catalyst speeds the forward and reverse reactions equally, so it reaches equilibrium faster but does not shift the position or change $K_c$ . Adding an inert gas at constant volume changes total pressure but not the partial pressures or concentrations of the reacting species, so it has no effect on the position.

Saying concentration or pressure changes alter Kc

The single most penalised SACE error is claiming that adding a reactant, or increasing the pressure, changes $K_c$ . These shift the position of equilibrium but $K_c$ is restored at the new equilibrium, so its value is unchanged. Only temperature changes $K_c$ . A second error is applying a pressure argument to an equilibrium with equal gas moles on both sides (no shift) or to one involving only solids and liquids (pressure has no effect).

Why it matters for managing processes

Le Chatelier reasoning lets chemists and engineers steer industrial equilibria toward higher yield by choosing concentration, pressure and temperature deliberately. It is the conceptual basis for optimising processes such as the Haber and Contact processes, where every condition is selected for its effect on the equilibrium position.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style5 marks

Consider

2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)

\Delta H = -198\ \text{kJ mol}^{-1}

. Predict and justify, using Le Chatelier's principle, the effect of each change on the position of equilibrium and on

K_c

: (a) increasing the pressure; (b) increasing the temperature; (c) adding a catalyst.

Show worked answer →

(a) There are $3$ moles of gas on the left and $2$ on the right. Increasing the pressure shifts the equilibrium toward the side with fewer gas moles, i.e. to the right (more $\text{SO}_3$ ). $K_c$ is unchanged. (2 marks)

(b) The forward reaction is exothermic. Increasing the temperature shifts the equilibrium in the endothermic (reverse) direction, to the left (less $\text{SO}_3$ ), and decreases $K_c$ . (2 marks)

(c) A catalyst speeds the approach to equilibrium equally in both directions, so the position is unchanged and $K_c$ is unchanged. (1 mark)

SACE 2019-style4 marks

For the equilibrium

\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)

(deep red), predict and explain the colour change when (a) more

\text{Fe}^{3+}

is added, and (b) the solution is diluted with water.

Show worked answer →

(a) Adding $\text{Fe}^{3+}$ increases a reactant concentration. The system opposes this by shifting to the right, consuming $\text{Fe}^{3+}$ and $\text{SCN}^-$ to make more $\text{FeSCN}^{2+}$ , so the solution becomes a deeper red. (2 marks)

(b) Dilution lowers all concentrations. The system shifts toward the side with more dissolved particles to partly restore concentration; there are $2$ aqueous species on the left and $1$ on the right, so it shifts left, breaking down $\text{FeSCN}^{2+}$ and making the colour paler. (2 marks)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board