A system at equilibrium is balanced, but it is easy to disturb: add a substance, squeeze the container, warm it up. Le Châtelier's principle predicts what happens next: when a system at equilibrium is disturbed, it shifts in the direction that partly counteracts the change. This page applies the principle to each kind of change and shows which changes alter K and which do not. Topic 7.10 then shows how to justify every one of these shifts with Q and K, which is what the exam's free-response questions require.
Changing a concentration
Add a reactant, and the system shifts toward the products, using up some of what you added. Remove a reactant, and it shifts toward the reactants to replace some. The same goes for products, in the opposite direction.
At the particle level this is topic 7.2 again: adding PCl3 or Cl2 to PCl5 ⇌ PCl3 + Cl2 makes PCl3-Cl2 collisions more frequent, so the reverse rate jumps above the forward rate until the two are equal again.
The key word is partly. The shift never fully undoes the change: if you add Cl2, [Cl2] ends higher than it was before, though lower than right after you added it.
Worked example. PCl5(g) ⇌ PCl3(g) + Cl2(g) is at equilibrium with [PCl5] = 0.128 M and [PCl3] = [Cl2] = 0.072 M. Cl2 is added to raise [Cl2] to 0.200 M. At the new equilibrium, [PCl3] = 0.039 M. Find the new [PCl5] and [Cl2].
1. Direction: Cl2, a product, was added, so the shift is toward reactants. PCl3 fell by 0.072 − 0.039 = 0.033 M.
2. All coefficients are 1, so PCl5 rose by 0.033 M: 0.128 + 0.033 = 0.161 M.
3. Cl2 fell by the same amount from 0.200 M: 0.200 − 0.033 = 0.167 M, higher than the original 0.072 M.
Check: (0.039)(0.167) / 0.161 = 0.040, the same K as before.
Pure solids and liquids are not in K, so adding or removing them causes no shift (as long as some solid remains). Removing an ion from solution by precipitating it, as Ag+ does to Cl−, does count: it lowers that ion's concentration.
Changing the volume of a gas mixture
Squeeze a gas mixture into half the volume and every concentration (and partial pressure) doubles at once. The system responds by shifting toward the side with fewer moles of gas, which lowers the total pressure. Expand it and the shift goes toward more moles of gas.
- N2(g) + 3 H2(g) ⇌ 2 NH3(g): 4 mol gas → 2 mol gas. Compressing favors NH3.
- H2(g) + I2(g) ⇌ 2 HI(g): 2 mol → 2 mol. A volume change causes no shift.
- C(s) + H2O(g) ⇌ CO(g) + H2(g): count only gases, 1 mol → 2 mol. Expanding favors the products.
Adding an inert gas such as argon at constant volume raises the total pressure but changes no reacting gas's concentration or partial pressure, so there is no shift.
For a reaction in solution, adding water (dilution) lowers every dissolved concentration, and the equilibrium shifts toward the side with more dissolved particles.
Changing the temperature
Temperature is different: it is the only change that alters K. A handy way to predict the direction is to treat heat as if it were a reactant or product:
- Endothermic forward reaction (ΔH > 0): heat is a "reactant". Heating shifts the equilibrium toward products, and K increases. Cooling does the opposite.
- Exothermic forward reaction (ΔH < 0): heat is a "product". Heating shifts toward reactants, and K decreases.
The pink and blue cobalt equilibrium shows this in a test tube. Co(H2O)62+ (pink) + 4 Cl− ⇌ CoCl42− (blue) + 6 H2O is endothermic: a hot-water bath turns it blue, an ice bath pink.
Changes that do not shift an equilibrium
A catalyst speeds the forward and reverse reactions by the same factor (topic 7.2), so it causes no shift; it just helps the system reach equilibrium sooner. An inert gas at constant volume and extra pure solid or liquid cause no shift either. Readers regularly see answers that suggest a catalyst or an inert gas to "increase the yield"; neither works.
