Edexcel IAL Chemistry · Unit 4
Chemical Equilibria
Big idea: when a reversible reaction reaches equilibrium, there's a fixed
"scorecard" ratio of products to reactants — called the equilibrium constant — and that
number only changes if you change the temperature. Everything else
(concentration, pressure, catalysts) can shift the equilibrium's *position*, but never
the constant itself.
Summary — What This Chapter Covers
- Kc — the equilibrium constant written in terms of concentrations (mol dm⁻³), used for reactions in solution.
- Kp — the equilibrium constant written in terms of partial pressures (Pa or kPa), used for gaseous reactions.
- Solids (and liquids, for Kp) are left out of equilibrium expressions entirely.
- Calculating Kc or Kp often needs an ICE table (Initial, Change, Equilibrium) to find missing quantities.
- Kp calculations sometimes require converting moles → mole fraction → partial pressure first.
- Only a change in temperature changes the value of Kc or Kp. Concentration, pressure, and catalysts shift the position of equilibrium but leave K unchanged.
- Entropy links to equilibrium through
ΔStotal = R ln K — a big K means the reaction is strongly product-favoured entropically.
1. The Equilibrium Constant, Kc
1.1 Writing the Kc Expression
For any reversible reaction that has settled into equilibrium, we can write an expression that
links the equilibrium constant to how much of each substance is hanging around. For a general reaction:
the equilibrium constant expression is:
Think of it like a recipe ratio at the finish line: once the reaction stops changing overall (even though
forward and backward reactions are still happening — it's dynamic equilibrium), you measure how
much of everything is present and plug it into this fraction.
Golden Rule
Solids are never included in the K
c expression. Only aqueous, liquid, and gaseous
species that actually have a meaningful "concentration" that changes get a slot in the expression. A solid's
concentration doesn't really change (its "concentration" is just a fixed density), so it's treated as a
constant and folded into K
c itself.
Worked Example — Deducing Kc
Write the Kc expression for: Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq)
Answer:
Kc = [Fe³⁺(aq)] ÷ ([Fe²⁺(aq)][Ag⁺(aq)])
Notice Ag(s) is completely missing — it's a solid, so it never appears in the expression at all, not even as a "1".
Practice Question
Deduce the Kc expression for: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
1.2 Calculating Kc from Moles and Volume
Exam questions rarely hand you concentrations directly — instead they usually give you the
number of moles of each substance at equilibrium plus the total volume of the mixture.
You need to convert moles into concentration first:
Worked Example — Calculating Kc
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
500 cm³ of the equilibrium mixture contains 0.235 mol ethanoic acid, 0.035 mol ethanol,
0.182 mol ethyl ethanoate, and 0.182 mol water. Find Kc.
p