Le Chatelier's principle: the rule
If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, it shifts in the direction that reduces the effect of that change, and a new equilibrium is set up.
A simple way to predict it: after the change, find Q. If Q < K, the reaction moves forward. If Q > K, it moves backward. If K changes (only with temperature), compare Q with the new K.
Effect of concentration
Add a reactant or remove a product → Q becomes smaller than K → shifts forward.
Add a product or remove a reactant → Q becomes bigger than K → shifts backward.
Example: in Fe³⁺ + SCN⁻ ⇌ [Fe(SCN)]²⁺ (blood red), adding more SCN⁻ makes the colour deeper. Adding oxalic acid, which removes Fe³⁺, makes the colour fade.
Try it: at step 2 press “add N₂” and watch the arrow point forward.
Effect of pressure (change in volume)
Squeezing a gas mixture raises pressure. The system reduces pressure by moving to the side with fewer gas moles.
- N₂ + 3H₂ ⇌ 2NH₃: 4 → 2 moles, so high pressure favours NH₃.
- H₂ + I₂ ⇌ 2HI: 2 → 2 moles, so pressure has no effect.
- PCl₅ ⇌ PCl₃ + Cl₂: 1 → 2 moles, so high pressure favours PCl₅.
Solids and liquids are hardly affected by pressure; count only gas moles.
Effect of temperature
Temperature is the only factor that changes K.
- Exothermic forward reaction (ΔH < 0): heating shifts it backward and K decreases. Cooling favours products.
- Endothermic forward reaction (ΔH > 0): heating shifts it forward and K increases.
Colour test: 2NO₂ (brown) ⇌ N₂O₄ (colourless) is exothermic. In ice the gas turns pale; in hot water it turns dark brown.
Effect of catalyst and inert gas
A catalyst lowers the activation energy for both forward and backward reactions by the same amount. So equilibrium is reached sooner, but its position and K do not change.
Inert gas (like argon) that does not react:
- At constant volume: total pressure rises, but the concentrations (and partial pressures) of the reacting gases stay the same → no shift.
- At constant pressure: volume must increase, so reacting gases are diluted → shifts to the side with more gas moles.
Haber process: Le Chatelier in industry
N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol.
- High pressure (about 200 atm): fewer moles on the right → more NH₃.
- Low temperature favours NH₃, but the rate becomes too slow. So a middle temperature of about 700 K is used: a compromise.
- Iron catalyst: equilibrium comes faster.
- Liquefy and remove NH₃: product removed → keeps moving forward.
Try it at home: stir sugar into a small glass of water until some stays at the bottom. Warm the glass: more dissolves. Cool it: crystals come back. Dissolving takes in heat, so heat pushes it forward.
Key formulas and definitions
- Add reactant / remove product → forward (Q < K)
- Add product / remove reactant → backward (Q > K)
- Higher pressure → side with fewer gas moles; Δn(gas) = 0 → no effect
- Heat: exothermic → backward, K falls; endothermic → forward, K rises
- Catalyst → no shift, K same; inert gas at constant V → no shift; at constant P → side with more gas moles
Worked examples
1. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH < 0. What happens if we add more O₂?
Adding a reactant makes Q < K, so the equilibrium shifts forward. More SO₃ forms.
2. For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), what is the effect of increasing pressure?
Left side has 1 mole of gas, right side has 2. Higher pressure favours fewer moles, so it shifts backward and more PCl₅ forms.
3. For C(s) + H₂O(g) ⇌ CO(g) + H₂(g), ΔH = +131 kJ. Give two ways to get more H₂.
It is endothermic, so raise the temperature. Gas moles go from 1 to 2, so lower the pressure. Removing H₂ or adding steam also helps.
4. For H₂(g) + I₂(g) ⇌ 2HI(g), at equilibrium [H₂] = [I₂] = 0.1 M and [HI] = 0.8 M (Kc = 64). The volume is halved. Find Q just after and say which way it shifts.
All concentrations double: 0.2, 0.2, 1.6. Q = 1.6² / (0.2 × 0.2) = 2.56/0.04 = 64 = K. No shift: Δn = 0.
5. For N₂O₄ ⇌ 2NO₂, Kc = 0.36, at equilibrium [N₂O₄] = 0.74 M, [NO₂] = 0.52 M. The volume is doubled. Find Q and the direction.
New values: [N₂O₄] = 0.37, [NO₂] = 0.26. Q = 0.26² / 0.37 = 0.0676/0.37 = 0.183. Q < K (0.36), so it shifts forward (towards more gas moles), as Le Chatelier predicts for lower pressure.
6. For A ⇌ B (Kc = 4), at equilibrium [A] = 0.2 M, [B] = 0.8 M. We add 0.5 M of A. Find the new equilibrium concentrations.
Start: A = 0.7, B = 0.8. Shift forward by x: (0.8 + x)/(0.7 − x) = 4 → 0.8 + x = 2.8 − 4x → 5x = 2.0 → x = 0.4. New [A] = 0.3 M, [B] = 1.2 M. Check: 1.2/0.3 = 4 ✓. A added is partly used up, just as the principle says.
Common mistakes
- Saying a catalyst increases the yield. It only brings equilibrium faster; the amounts at equilibrium stay the same.
- Thinking pressure changes K. Pressure only shifts the position; only temperature changes K.
- Forgetting that pressure has no effect when gas moles are equal on both sides.
- Saying inert gas always shifts equilibrium. At constant volume it has no effect at all.