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5.3.2Additional lessons

The Equilibrium Law — IB Chemistry SL Revision Notes & Topical Questions

5 min read 5 parts Lesson 74 of 98

📚 IB Chemistry SL Hub / Topical Revision
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Part 1 of 5The Equilibrium Law Expression

5.3.2 The Equilibrium Law

The equilibrium law links the concentrations of species at equilibrium through one constant at a fixed temperature.

General form
  • For aA + bB ⇌ cC + dD at equilibrium, K = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), with products on top and reactants below.
Square brackets
  • [X] means the equilibrium concentration of X in mol dm⁻³; round brackets would not be accepted.
Powers
  • Each concentration is raised to the power of its coefficient in the balanced equation.
Equation specific
  • K applies only to the equation as written, so changing the coefficients or reversing the reaction changes K.

Part 2 of 5Which Species Are Included

Which Species Are Included
Gases
  • Gaseous species are included, using their concentrations in mol dm⁻³.
Aqueous species
  • Dissolved ions and molecules are included because their concentrations can change.
Pure solids
  • Solids are left out because their concentration is effectively constant, for example Ag(s) in Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq).
Pure liquids
  • Pure liquids are also left out of the expression, since their concentration does not vary.

Part 3 of 5Writing an Expression Step by Step

1
Write the balanced equation with state symbols and the ⇌ sign.
2
Identify the gaseous and aqueous species, because only these appear in K.
3
Put the products in the numerator, each in square brackets raised to its coefficient.
4
Put the reactants in the denominator in the same way.
5
Substitute equilibrium concentrations and work out K, then decide whether units are needed from the powers.
EquationExpression for K
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)[NH₃]² / ([N₂][H₂]³)
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)[SO₃]² / ([SO₂]²[O₂])
Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq)[Fe³⁺] / ([Ag⁺][Fe²⁺])

Part 4 of 5Worked Example

Question

  • At a certain temperature an equilibrium mixture of 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) contains [SO₂] = 0.20 mol dm⁻³, [O₂] = 0.10 mol dm⁻³ and [SO₃] = 0.40 mol dm⁻³.
  • Calculate K.

  1. Write the expressionK = [SO₃]² / ([SO₂]² [O₂]).
  2. Substitute the valuesK = (0.40)² / ((0.20)² × 0.10).
  3. Calculate numerator 0.16 and denominator 0.040 × 0.10 = 0.0040.

Answer K = 0.16 / 0.0040 = 40 (units mol⁻¹ dm³).

Part 5 of 5Exam Checklist & Tips

5.3.2 – Key Points for the Exam

  • Equilibrium law The statement that at constant temperature the ratio of product to reactant concentrations, each raised to its coefficient, is constant.
  • Equilibrium constant, K The value of the equilibrium expression at a given temperature.
  • Equilibrium expression The ratio of product concentrations to reactant concentrations, each to the power of its coefficient.
  • Heterogeneous equilibrium An equilibrium involving species in more than one phase, where pure solids and liquids are omitted.

Common Mistakes

  • Using round brackets instead of square brackets in the K expression.
  • Putting reactants on top; products always go in the numerator.
  • Including solids such as Ag(s) or CaCO₃(s) in the expression.
  • Forgetting to raise each concentration to the power of its coefficient.

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3 & IGCSE —

Kc

  • : Kc! (ΔH < 0) Kc. (ΔH > 0) Kc.
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