Part 1: Equilibrium amounts, Kc and heterogeneous systems

Use stoichiometry to find what is present at equilibrium before calculating Kc. Separate a ratio of changes from the final composition.

Equilibrium is dynamic, not equal amounts

In a closed system at equilibrium, forward and reverse reactions continue at equal rates. Macroscopic concentrations remain constant because each species is formed as quickly as it is consumed. Equal rates do not require equal concentrations, and the equilibrium composition generally differs from the starting mixture.

Year 12 introduced Kc using supplied concentrations. Year 13 adds the preceding mole balance, heterogeneous systems and units. The expression belongs to a particular balanced equation at a specified temperature. Reversing that equation makes the new constant the reciprocal; doubling all coefficients squares the expression and hence squares its numerical value.

Worked example: some product was present initially

For N₂O₄(g) ⇌ 2NO₂(g), an original illustrative 2.00 dm³ vessel initially contains 0.600 mol N₂O₄ and 0.100 mol NO₂. At equilibrium there is 0.500 mol NO₂. The increase is 0.400 mol, not 0.500 mol, because some product was already present.

The 1:2 stoichiometric ratio means 0.200 mol N₂O₄ has dissociated. Complete the equilibrium row, then divide each amount by the whole vessel volume. Each gas occupies the entire container; do not divide the volume between gases.

Kc = 0.250²/0.200 = 0.3125 mol dm⁻³, reported as 0.313 mol dm⁻³. All final amounts are positive and the total nitrogen-atom inventory is conserved. Total molecular amount may change because one molecule forms two.

Which species belong in the expression?

A homogeneous equilibrium has one phase; a heterogeneous equilibrium has more than one. Include changing solution concentrations in Kc and gaseous partial pressures in Kp. Omit a pure solid or pure liquid phase because its activity is effectively constant while that phase remains present. Do not omit an aqueous solute just because it is dissolved in a liquid.

For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂] and Kp = p(CO₂). Adding more solid carbonate does not by itself change the equilibrium CO₂ pressure at fixed temperature if both solid phases remain. Removing all of one required phase invalidates that simple equilibrium situation.

Water is normally omitted as the solvent in dilute aqueous equilibria such as Ka. However, in a homogeneous liquid esterification mixture its concentration may change significantly and the supplied model can include it. “Leave out everything labelled liquid” is therefore too broad; distinguish a separate pure liquid phase from a component in a changing mixture.

Do the unit algebra alongside the numbers

For 2SO₂ + O₂ ⇌ 2SO₃, Kc = [SO₃]²/([SO₂]²[O₂]). Units simplify to (mol dm⁻³)⁻¹ = dm³ mol⁻¹. In a common-volume vessel, substituting moles directly only works numerically when the volume powers cancel. Here they do not.

If equilibrium concentrations of SO₂, O₂ and SO₃ are 0.400, 0.300 and 0.800 mol dm⁻³, Kc = 0.640/0.0480 = 13.3 dm³ mol⁻¹. If instead Kc and two concentrations are given, rearrange symbolically first. Squared quantities need a square root when isolated, not division by two. OCR does not require solving quadratic equations for these problems.

Quick checks

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

Sources