Le Châtelier’s principle
When a system at equilibrium is disturbed, the equilibrium shifts in the direction that partly counteracts the change. This is Le Châtelier’s principle勒夏特列原理 (2.11.5). It “partly counteracts,” not “cancels”: at the new equilibrium the quantity that was changed is still different from before.
| Change | Direction of shift |
|---|---|
| Add a reactant (or remove a product) | Forward |
| Add a product (or remove a reactant) | Reverse |
| Decrease the volume (raise the pressure), gases only | Toward fewer gas moles |
| Increase the volume (lower the pressure) | Toward more gas moles |
| Raise the temperature | In the endothermic direction |
| Lower the temperature | In the exothermic direction |
| Add a pure solid or liquid | No shift |
| Add a catalyst | No shift (equilibrium is just reached sooner) |
| Add an inert gas at constant volume | No shift (partial pressures don’t change) |
Temperature is the only change that alters K: if the forward reaction is endothermic, heating increases K; if it is exothermic, heating decreases K. Every other change leaves K alone and changes only Q (U7.10).
Choose a reaction and a change, and compare the concentrations at the original equilibrium, right after the change, and at the new equilibrium:
Shifts you can see
Color is a common way to watch equilibria shift in the lab. In a cobalt chloride solution, pink [Co(H₂O)₆]²⁺ and blue [CoCl₄]²⁻ are at equilibrium, and the forward reaction is endothermic. Do different things to the tube and see which way the color changes:
What do you do to the tube?
The size of each color change is schematic. Cobalt chloride solutions change color with temperature and are a common demonstration of Le Châtelier’s principle.
Example 1: The water-gas reaction
At high temperature, C(s) + H₂O(g) ⇌ CO(g) + H₂(g), ΔH = +131.3 kJ/mol (from NBS enthalpies of formation). Predict which way each change shifts the equilibrium: (1) adding more steam; (2) adding more coke (C); (3) compressing the volume; (4) raising the temperature; (5) adding a catalyst.
- (1) Add steam. A reactant concentration increases, so the equilibrium shifts forward.
- (2) Add coke. C is a pure solid that doesn’t appear in the K expression, so there is no shift.
- (3) Compress. The left side has 1 mol of gas and the right side 2 mol, so the equilibrium shifts toward fewer gas moles, to the left (reverse).
- (4)(5) Heat and catalyst. The forward reaction is endothermic, so heating shifts it forward and increases K; a catalyst causes no shift.
Example 2: The color of a copper–ammonia solution
Adding aqueous ammonia to a pale blue [Cu(H₂O)₄]²⁺ solution turns it deep blue: [Cu(H₂O)₄]²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq) + 4H₂O(l). Adding hydrochloric acid dropwise then turns it pale blue again. Explain both observations.
- Adding ammonia. [NH₃] increases, so the equilibrium shifts forward, forming more deep-blue [Cu(NH₃)₄]²⁺.
- Adding HCl. H⁺ reacts with NH₃ to form NH₄⁺, lowering [NH₃].
- The shift. With less reactant, the equilibrium shifts in reverse, forming more [Cu(H₂O)₄]²⁺, and the solution turns pale blue again.
- The pattern. Adding a reagent that “uses up” one substance in the equilibrium has the same effect as removing that substance.
- Add more H₂O(g)
A reactant concentration increases.
- Remove some H₂
A product concentration decreases.
- Add more C(s)
A pure solid doesn’t affect the equilibrium.
- Compress to half the volume
Toward fewer gas moles (1 mol on the left, 2 mol on the right).
- Raise the temperature
The forward reaction is endothermic.
- Add a catalyst
It speeds up the forward and reverse reactions equally.
- Add argon at constant volume
No partial pressure changes.
- Add more CO
A product concentration increases.
Common mistakes
Mistake 1: “A catalyst shifts the equilibrium forward and raises the yield.” A catalyst speeds up the forward and reverse reactions equally; it only shortens the time to reach equilibrium and doesn’t change the equilibrium composition.
Mistake 2: “Adding more of a solid reactant shifts the equilibrium forward.” Pure solids and liquids don’t appear in the K expression, so adding them doesn’t change Q and the equilibrium doesn’t shift.
Mistake 3: “The shift completely cancels the change.” It only partly counteracts it. After a reactant is added, for example, its concentration at the new equilibrium is still higher than before.
Quick check
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH > 0. Which change shifts the equilibrium forward?
The forward reaction is endothermic, so heating shifts it forward. Compressing shifts toward fewer gas moles, on the left; adding Cl₂ shifts it in reverse.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is at equilibrium. At constant temperature and volume, argon is added, raising the total pressure. The equilibrium:
Adding an inert gas at constant volume leaves the partial pressures (concentrations) of N₂, H₂, and NH₃ unchanged, so Q is unchanged.
After an exothermic reaction reaches equilibrium, the temperature is raised. Which statement is correct?
Heating an exothermic reaction shifts it in the endothermic, reverse direction, and K decreases.
The gas-phase reaction 2A(g) ⇌ B(g) + C(g) is at equilibrium, and the volume is decreased. The equilibrium:
There are 2 mol of gas on the left and 1 + 1 = 2 mol on the right, so changing the volume has no effect.
A little solid NaCl is added to a saturated solution in equilibrium with AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). What is most likely observed?
[Cl⁻] increases, so the equilibrium shifts in reverse and AgCl precipitates, lowering [Ag⁺] (the common-ion effect, U7.12).
After a reactant is added, the equilibrium shifts forward and a new equilibrium is established. Compared with the original equilibrium, the concentration of that reactant at the new equilibrium is:
The shift only partly counteracts the change; it can’t cancel it completely.
AP-style free response
Ethanol is made industrially from ethylene and steam over a catalyst: C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g), ΔH < 0.
(a) How does increasing the pressure (compressing the volume) affect the equilibrium yield of ethanol? Justify your answer.
(b) How does raising the temperature affect the equilibrium yield and K?
(c) The plant doesn’t run this reaction at a very low temperature but at about 300 °C. Explain why in terms of rate.
(d) What is the benefit of continually condensing the ethanol out of the mixture?
(e) Can a catalyst raise the equilibrium yield? Justify your answer.
Model answer and scoring
(a) The yield increases. There are 2 mol of gas on the left and 1 mol on the right, so compressing shifts the equilibrium forward, toward fewer gas moles.
(b) The forward reaction is exothermic, so heating shifts the equilibrium in reverse: the yield falls and K decreases.
(c) At too low a temperature the reaction is too slow, and reaching equilibrium takes too long. About 300 °C is a compromise between rate and yield.
(d) Removing product lowers Q, so the equilibrium keeps shifting forward and more ethylene is converted to ethanol.
(e) No. A catalyst speeds up the forward and reverse reactions equally; it only shortens the time to reach equilibrium and doesn’t change the equilibrium composition.
Scoring: (a) gas moles; (b) heating an exothermic reaction lowers K; (c) the rate–yield compromise; (d) removing product; (e) a catalyst doesn’t change the equilibrium.
Summary
- When conditions change, an equilibrium shifts in the direction that partly counteracts the change (it can’t cancel it).
- Concentration and volume changes alter only Q; only temperature alters K, and heating shifts the equilibrium in the endothermic direction.
- Pure solids, pure liquids, catalysts, and inert gases at constant volume don’t shift an equilibrium.
AP tip: On the AP exam, “by Le Châtelier’s principle” alone often isn’t enough. Say what quantity changed, whether Q or K changed and how, and therefore which way the equilibrium shifts (U7.10).