Le Châtelier's Principle

You’ll be able to

State Le Châtelier’s principle and predict shifts from concentration changes
Predict the effect of pressure and volume changes on gaseous equilibria
Predict how temperature changes shift equilibrium and change the value of K

A system pushes back against a stress

Le Châtelier’s principle: if a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance. Add a reactant and the system consumes some of it (shifts right); remove a product and the system makes more of it (shifts right). Adding or removing a species changes the position of equilibrium but, crucially, does not change the value of K — only temperature does that.

Pressure and volume: follow the gas moles

For reactions involving gases, increasing pressure by decreasing the volume shifts the equilibrium toward the side with fewer moles of gas — that relieves the pressure. Decreasing pressure (larger volume) shifts toward more moles of gas. If both sides have equal gas moles, a volume change causes no shift. Adding an inert gas at constant volume changes nothing, because no partial pressures of the reacting gases change.

Temperature: treat heat as a reagent

Temperature is the one stress that actually changes K. Write heat into the equation: for an exothermic reaction (ΔH < 0), heat is a product, so raising the temperature shifts the reaction left and lowers K. For an endothermic reaction (ΔH > 0), heat is a reactant, so raising the temperature shifts it right and raises K. A catalyst speeds both directions equally and does not shift equilibrium at all.

Temperature and K (heat-as-reagent rule)

Exothermic (ΔH < 0): heat is a product → raising T shifts left, K decreases. Endothermic (ΔH > 0): heat is a reactant → raising T shifts right, K increases.

Only a temperature change alters the numerical value of K. Concentration, volume, pressure, and catalyst changes never do.

Worked example

For the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ (exothermic). Predict the effect of (a) adding more N₂, (b) compressing the mixture to a smaller volume, and (c) raising the temperature.

1.(a) Adding N₂ raises a reactant concentration. The system consumes some N₂ by shifting right → more NH₃. K unchanged.
2.(b) Count gas moles: 4 mol gas on the left (1 + 3), 2 mol on the right. Compression favors the side with fewer moles → shift right, more NH₃. K unchanged.
3.(c) Exothermic means heat is a product. Raising T adds "product," so the system shifts left → less NH₃, and K decreases.

Answer: (a) shifts right, (b) shifts right, (c) shifts left. Industrially this is the classic tension: high pressure boosts yield, but the needed high temperature (for speed) lowers it, so a compromise temperature plus a catalyst is used.

Checkpoint

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the volume of the container is suddenly increased (pressure decreased). Which way does the equilibrium shift?

Checkpoint

The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic. If the temperature is increased, what happens?

Checkpoint

In the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) (deep red), extra SCN⁻ is added. What is observed and why?

Watch out

Do not confuse rate with position. A catalyst and (for gases) an added inert gas at constant volume both leave the equilibrium position untouched. And only temperature changes K — if a question claims adding a reactant "increased K," it is wrong.

Answer the 3 checkpoints as you read.