A-level chemistry · Chapter 6

Equilibria

Le Chatelier, Kc and Kp

Dynamic equilibrium — forward and reverse rates are equal and concentrations are constant, in a closed system.

Le Chatelier's principle

When an equilibrium is disturbed, the position shifts to oppose the change. Raise the temperature and it shifts in the endothermic direction; raise the pressure and it shifts to the side with fewer gas moles. A catalyst causes no shift — equilibrium is simply reached faster.

The Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹) is the classic compromise: 400–450 °C, ~200 atm and an iron catalyst. A lower temperature would favour the yield, but the reaction becomes too slow to be worth it.

Equilibrium constant

Kc = [C]c[D]d / [A]a[B]b

K changes only with temperature — for an exothermic reaction, raising T decreases K. Changes in concentration or pressure move the position of equilibrium, but K itself stays put. Kp is the gas-phase version built from partial pressures, where p(X) = mole fraction × total pressure.

3.1.6.2Kc from equilibrium data — the ICE routine

Worked example. 1.00 mol H₂ and 1.00 mol I₂ reach equilibrium in a 1.00 dm³ flask; 0.80 mol of each has reacted. H₂ + I₂ ⇌ 2HI.
  1. Set out the ICE table: initial 1.00 / 1.00 / 0  ·  change −0.80 / −0.80 / +1.60  ·  equilibrium 0.20 / 0.20 / 1.60 mol.
  2. Substitute: Kc = [HI]² ÷ ([H₂][I₂]) = 1.60² ÷ (0.20 × 0.20) = 64 (no units — the concentrations cancel).

3.1.10Kp — the same idea with pressures

Mole fraction x = moles of gas ÷ total moles, and partial pressure p = x × P(total). Kp is built exactly like Kc but with the partial pressures of gases only — solids and pure liquids never appear in Kp expressions. Like Kc, Kp changes with temperature only: pressure shifts the position, not the constant.

Exam tip. If asked "explain why the position moves but Kc is unchanged" — the answer is always: Kc is constant at constant temperature; the concentration change alters the reaction quotient, so the system shifts until the ratio equals Kc again.

3.1.6.3The three industrial equilibria

ProcessReactionCatalystConditionsWhy the compromise
HaberN₂ + 3H₂ ⇌ 2NH₃ (ΔH −92)Fe400–450 °C, ~200 atmlow T = better yield but too slow; ~15% per pass, gases recycled
Contact2SO₂ + O₂ ⇌ 2SO₃V₂O₅~450 °C, ~2 atmyield already ~99.5% near atmospheric pressure — high p not worth it
EthanolC₂H₄ + H₂O ⇌ C₂H₅OHH₃PO₄300 °C, 60–70 atm~5% per pass; unreacted ethene recycled
You're reading the free preview

The full Equilibria chapter — every remaining section, the Mastery Vault, chapter quizzes and AI-marked exams — is in ChemLab.

Continue in the app