5.3.2Additional lessons
The Equilibrium Law — IB Chemistry SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 74 of 98
Section 1 of 5The Equilibrium Law Expression
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.
Section 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.
Section 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.
| Equation | Expression 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⁺]) |
Section 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.
- Write the expressionK = [SO₃]² / ([SO₂]² [O₂]).
- Substitute the valuesK = (0.40)² / ((0.20)² × 0.10).
- Calculate numerator 0.16 and denominator 0.040 × 0.10 = 0.0040.
Answer K = 0.16 / 0.0040 = 40 (units mol⁻¹ dm³).
Section 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 —
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- : ( ).
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