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Chemical Equilibria

Topic 3 of 4

Equilibrium expressions and calculations

Use equilibrium concentrations or partial pressures, with stoichiometry to find missing amounts.

A-Level 9476 (2026-2027)

Build K from the written equation and equilibrium quantities

Products go above reactants; stoichiometric coefficients become powers.

For aA + bB ⇌ cC + dD, the concentration expression is Kc = [C]c[D]d / ([A]a[B]b), using equilibrium concentrations. Kp uses equilibrium partial pressures of gaseous species in the same pattern. Pure solids and pure liquids are omitted because their activities are constant; a dissolved species is not omitted merely because it is in a liquid solution.

Expressions for common equilibrium forms
ReactionExpression
H2(g) + I2(g) ⇌ 2HI(g)Kc = [HI]2 / ([H2][I2]); Kp = p(HI)2 / (p(H2)p(I2)).
N2(g) + 3H2(g) ⇌ 2NH3(g)Kc = [NH3]2 / ([N2][H2]3); Kp = p(NH3)2 / (p(N2)p(H2)3).
CaCO3(s) ⇌ CaO(s) + CO2(g)Kp = p(CO2), with both pure solid phases present.
CH3COOH(aq) + C2H5OH(aq) ⇌ CH3COOC2H5(aq) + H2O(l)In a dilute aqueous treatment, water as the solvent is incorporated into the constant; retain the dissolved reacting species. Read the stated medium.

In conventional examination expressions, derive any units from the concentration or pressure powers and use one consistent pressure unit. For N2O4(g) ⇌ 2NO2(g), the concentration expression has units mol dm-3 and a pressure expression in kPa has units kPa. The numerical pressure-based value changes if the pressure unit changes, so report the unit used.

Worked example

Find Kp from a gas mixture

At equilibrium, a mixture contains 0.0200 mol N2O4 and 0.0400 mol NO2 at total pressure 150 kPa. Calculate Kp for N2O4 ⇌ 2NO2.

  1. Total gas amount = 0.0600 mol.
  2. p(N2O4) = (0.0200/0.0600)(150) = 50.0 kPa; p(NO2) = 100 kPa.
  3. Kp = p(NO2)2 / p(N2O4) = 1002/50.0.
Answer

Kp = 200 kPa for the stated pressure convention and temperature.

Reversing a reaction takes the reciprocal of K; multiplying all coefficients by a factor raises K to that power. A large K favours products relative to reactants in its expression, but gives no direct information about the speed of reaching equilibrium. The relationship converting Kp into Kc is not required in this syllabus.

Use stoichiometry to complete the equilibrium row

Amounts change in the coefficient ratio before they are converted to concentration or partial pressure.

Worked example

Calculate Kc from an observed equilibrium amount

A 1.00 dm3 vessel initially contains 0.100 mol H2 and 0.100 mol I2, with no HI. At equilibrium, it contains 0.160 mol HI. Find Kc for H2 + I2 ⇌ 2HI.

  1. Forming 0.160 mol HI consumes 0.0800 mol H2 and 0.0800 mol I2.
  2. Equilibrium amounts are 0.0200, 0.0200 and 0.160 mol respectively.
  3. Divide each amount by 1.00 dm3 to obtain the equilibrium concentrations.
  4. Kc = 0.1602/(0.0200 × 0.0200).
Answer

Kc = 64.0, with no net concentration unit for this expression.

An amount table for the same reaction, initially 0.100 mol of each reactant
StageH2 / molI2 / molHI / mol
Initial0.1000.1000
Change-x-x+2x
Equilibrium0.100 - x0.100 - x2x

Worked example

Find the equilibrium composition from K

For the same initial amounts in 1.00 dm3, suppose Kc = 64.0 is supplied instead. Calculate the equilibrium amounts without solving a quadratic.

  1. Substitute the equilibrium expressions: 64 = (2x)2/(0.100 - x)2.
  2. Concentrations are positive, so take the positive square root: 8 = 2x/(0.100 - x).
  3. 8(0.100 - x) = 2x; hence x = 0.0800 mol.
  4. Substitute back and check that no amount is negative.
Answer

H2 = I2 = 0.0200 mol and HI = 0.160 mol. Substitution into K verifies the result.

A useful table is not a substitute for checking the chemistry. Include any initial product, account for unequal coefficients, and use the actual equilibrium volume. When gaseous mole number changes, total pressure cannot be split using initial mole fractions. The required syllabus calculations do not need quadratic-equation solutions.