Part 1: Mole fractions, partial pressures and Kp

Turn an equilibrium gas composition into partial pressures and construct a Kp expression with consistent units.

Each gas contributes to the total pressure

For an ideal gas mixture, a gas’s partial pressure is the pressure it would exert alone in the same volume at the same temperature. Its mole fraction is its amount divided by the total amount of gas. Mole fractions have no units, add to one and lie between zero and one. Partial pressures add to the total pressure.

Use all gases when calculating the total amount, including any inert gas. Only reacting species appear in the equilibrium expression. AQA 3.1.10 focuses on homogeneous gas equilibria, so all reacting species in the examples here share the gas phase.

Worked mixture calculation

An illustrative mixture contains 0.600 mol H₂, 0.300 mol N₂ and 0.100 mol Ar at 240 kPa. Total gas amount = 1.000 mol. Their mole fractions are 0.600, 0.300 and 0.100, so the partial pressures are 144, 72.0 and 24.0 kPa. Check: 144 + 72 + 24 = 240 kPa.

Using only the two potentially reactive gases in the denominator would give the wrong partial pressures. A mass fraction is not a mole fraction: convert masses to moles first if that is how the mixture is specified.

The balanced equation sets the powers

Write products over reactants using equilibrium partial pressures, each raised to its stoichiometric coefficient. Use p(X) notation so pressure cannot be mistaken for the concentration notation [X]. Kp is constant at a fixed temperature for a specified reaction equation and pressure convention.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the denominator contains p(H₂) cubed. Do not multiply the pressure by three instead of cubing it.

Pressure units must stay consistent

In the A-level pressure-based convention, units come from pressure raised to the total gaseous product coefficients minus the total gaseous reactant coefficients. For ammonia formation, the power is 2 − 4 = −2, so pressures in kPa give Kp in kPa⁻².

A thermodynamic equilibrium constant defined using pressure ratios to a standard pressure is dimensionless; that is a different convention. For these AQA calculations, use the pressure-based expression and units requested by the question. Changing from kPa to Pa changes the numerical Kp when the powers do not cancel. Do not change pressure units without changing Kp consistently.

Kp belongs to the equation as written

Reversing a reaction takes the reciprocal of Kp. Multiplying every coefficient by two squares Kp. Its units change accordingly. Compare constants only when the chemical equation and unit convention are the same.

For H₂ + I₂ ⇌ 2HI with Kp = 36.0, the reverse reaction has Kp = 1/36.0 = 0.0278. For ½H₂ + ½I₂ ⇌ HI, Kp = √36.0 = 6.00. A large numerical Kp can suggest product-favoured equilibrium for a fixed convention, but it is not directly a percentage yield.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources