A-Level Chemistry equilibria: Le Chatelier, Kc and Kp

Where equilibria sit in each specification

The key ideas

Le Chatelier's principle: if a change is made to a system at equilibrium, the position of equilibrium moves to oppose (minimise) the change. Adding reactant shifts it to the right. Increasing pressure shifts it towards fewer gas molecules. Increasing temperature shifts it in the endothermic direction. A catalyst speeds up both directions equally, so equilibrium is reached faster but its position does not change.

For aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d / ([A]^a[B]^b), using equilibrium concentrations in mol dm^-3. Kp uses partial pressures: the partial pressure of a gas is its mole fraction times the total pressure. Solids are left out of K expressions for heterogeneous systems. Work out the units each time from the expression.

Changes in concentration or pressure move the position of equilibrium but leave K unchanged; the concentrations adjust until the ratio equals K again. Only temperature changes K: for an exothermic forward reaction, raising the temperature lowers K; for an endothermic forward reaction, it raises K.

Worked example 1: calculating Kc from initial amounts

1. 1.00 mol of H2 and 1.00 mol of I2 are heated in a 1.00 dm^3 sealed flask. At equilibrium there are 1.56 mol of HI. H2(g) + I2(g) ⇌ 2HI(g).
2. Use an ICE table. Forming 1.56 mol of HI uses 0.78 mol of H2 and 0.78 mol of I2.
3. Equilibrium amounts: H2 = 0.22 mol, I2 = 0.22 mol, HI = 1.56 mol. The volume is 1.00 dm^3, so these are also the concentrations.
4. Kc = [HI]^2 / ([H2][I2]) = 1.56^2 / (0.22 × 0.22) = 2.4336 / 0.0484 = 50.3.
5. Units cancel (mol^2 dm^-6 over mol^2 dm^-6), so Kc has no units.

Worked example 2: calculating Kp from mole fractions

1. N2O4(g) ⇌ 2NO2(g). At equilibrium there are 0.70 mol of N2O4 and 0.60 mol of NO2, and the total pressure is 2.00 atm.
2. Total moles = 1.30. Mole fractions: N2O4 = 0.70 ÷ 1.30 = 0.538; NO2 = 0.60 ÷ 1.30 = 0.462.
3. Partial pressures: p(N2O4) = 0.538 × 2.00 = 1.077 atm; p(NO2) = 0.462 × 2.00 = 0.923 atm.
4. Kp = p(NO2)^2 / p(N2O4) = 0.923^2 ÷ 1.077 = 0.791 atm.
5. Cambridge questions may use Pa or kPa instead of atm; the method is the same, and the units follow from the expression.

Common mistakes that cost marks

• Using moles instead of concentrations in Kc when the volume is not 1 dm^3.
• Forgetting to raise concentrations to the power of their coefficients.
• Including solids in K for heterogeneous equilibria.
• Saying a catalyst shifts the equilibrium or increases the yield.
• Saying K changes when pressure or concentration changes.
• Leaving out units, or writing units when they cancel.

Exam technique and how a tutor helps

For every calculation, draw an ICE table: initial amounts, change, equilibrium amounts, then divide by volume. For Le Chatelier questions, name the change, state the direction of the shift, and give the reason in one sentence: 'Increasing the pressure shifts the equilibrium to the right, because there are fewer moles of gas on the right.'

Industrial questions such as the Haber process usually want the compromise: a lower temperature gives a higher yield for an exothermic reaction but a slower rate, and very high pressure gives a higher yield but costs more and needs stronger equipment.

In one-to-one lessons a tutor drills ICE tables with awkward stoichiometry and non-unit volumes until the student sets them up without hesitation, then turns to explanation questions marked against the official mark scheme so the wording earns full credit.

Self-check: can you do these?

• Write Kc for N2 + 3H2 ⇌ 2NH3 and give its units. (Answer: [NH3]^2/([N2][H2]^3); mol^-2 dm^6)
• What happens to K for an exothermic reaction when temperature rises? (Answer: it decreases)
• Does a catalyst change the equilibrium yield? (Answer: no, only the time taken to reach equilibrium)
• Write Kp for CaCO3(s) ⇌ CaO(s) + CO2(g). (Answer: Kp = p(CO2))
• Which side does increased pressure favour in 2SO2 + O2 ⇌ 2SO3? (Answer: the right, fewer gas moles)