Part 2: Partial pressures, Kp and heterogeneous equilibria

Convert equilibrium amounts into mole fractions and partial pressures, then construct an expression with the correct powers and units.

Each gas occupies the whole container

The partial pressure of a gas is the pressure it would exert alone at the same temperature in the volume occupied by the mixture. For an ideal gas mixture, p(A) = x(A)Ptotal, where x(A) = n(A)/ntotal. The mole fraction has no units; a partial pressure has pressure units. Edexcel specifies atmospheres for Topic 11 Kp calculations.

The total amount includes every gas in the vessel, including inert gases, but excludes solids and liquids. The sum of all mole fractions is one and the sum of all partial pressures is the total pressure. These give quick checks before a high power magnifies a numerical error.

For 0.400 mol A, 0.800 mol B and 0.800 mol inert gas at 5.00 atm, x(A) = 0.200 and p(A) = 1.00 atm. Omitting the inert gas would give an incorrect partial pressure even though that inert gas does not appear in the equilibrium expression.

Write the expression before calculating

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kp = p(NH₃)²/[p(N₂)p(H₂)³]. Coefficients become powers. Units simplify to atm²/atm⁴ = atm⁻². Use partial pressures, not the same total pressure in every position.

In a heterogeneous equilibrium, include the gaseous species in the Kp expression. For NH₄HS(s) ⇌ NH₃(g) + H₂S(g), Kp = p(NH₃)p(H₂S), with units atm². For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂), with units atm. Pure solids contribute no varying pressure term.

A numerical constant depends on how its concentration or pressure units are specified. Convert all pressures to the required unit before substituting. The dimensionless thermodynamic constant used inside ln K in Topic 13 has standard-state normalisation; do not insert a quantity carrying arbitrary units into a logarithm without the convention supplied.

Worked example: gas amount changes during reaction

Initially 1.20 mol N₂ and 3.60 mol H₂ are placed in a vessel. At equilibrium, 0.400 mol N₂ has reacted and total pressure is 8.00 atm. The equation forms 0.800 mol NH₃ and consumes 1.20 mol H₂, leaving 0.800 mol N₂ and 2.40 mol H₂. Total equilibrium gas amount is 4.00 mol.

Mole fractions are N₂ 0.200, H₂ 0.600 and NH₃ 0.200. Partial pressures are therefore 1.60, 4.80 and 1.60 atm. They sum to 8.00 atm. Kp = 1.60²/(1.60 × 4.80³) = 0.0144676 atm⁻² = 0.0145 atm⁻² to three significant figures.

The stoichiometric ratios belong in the change row, while the mole-fraction denominator uses the equilibrium total. Using the initial 4.80 mol total would produce partial pressures that do not sum correctly to 8.00 atm.

Worked example: a solid cannot supply a partial pressure

A sealed vessel contains excess solid NH₄HS at a temperature where Kp = 0.0900 atm². Starting with no product gas and no other gas source, the decomposition makes equal amounts of NH₃ and H₂S, so their partial pressures are equal. Let each pressure be p: p² = 0.0900 and p = 0.300 atm. Total gas pressure is 0.600 atm.

Equal product pressures here follow from equal amounts formed and the specified initial state. They would not follow if NH₃ had already been added. Adding more solid while solid remains and temperature is fixed does not change these equilibrium pressures.

The June 2023 Pearson 9CH0/03 report, Q10(b)(iii), identified mistakes involving mole fractions, partial pressures and inclusion of solid nickel. The transferable method is to label every intermediate quantity and derive the expression from states and coefficients; the exact credit in that question is not a universal mark allocation.

Quick checks

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

Sources