1.8.3Physical Chemistry
The Equilibrium Constant, Kc — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 5 parts Lesson 47 of 102
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5The Expression
General form
- General form For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ ÷ [A]ᵃ[B]ᵇ, using equilibrium concentrations in mol dm⁻³.What the symbols mean: Kc = equilibrium constant in terms of concentration · [ ] = concentration at equilibrium (mol dm−3) · each power is the number in front of that substance in the balanced equation
Powers
- Powers Each concentration is raised to the power of its coefficient in the balanced equation.
Products over reactants
- Products over reactants Concentrations of products are placed on the top and those of reactants on the bottom.
Homogeneous
- Homogeneous AS level work uses homogeneous equilibria, where all species are in the same phase.
Size of Kc
- Size of Kc A large Kc means the position lies well to the right with mostly products, and a small Kc means mostly reactants.
Part 2 of 5Units and Behaviour
Units and Behaviour
Units
- Units Substitute the units of concentration into the expression and cancel, which may leave mol dm⁻³ to some power or no units.
Temperature
- Temperature Kc changes only with temperature and is fixed at a given temperature.
Concentration and pressure
- Concentration and pressure Changing concentration or pressure shifts the position of equilibrium but Kc stays the same, unless temperature changes.
Catalyst
- Catalyst A catalyst has no effect on Kc, since forward and reverse rates are changed by the same factor.
| Equilibrium | Expression for Kc | Units |
|---|---|---|
| N₂O₄ ⇌ 2NO₂ | [NO₂]² ÷ [N₂O₄] | mol dm⁻³ |
| H₂ + I₂ ⇌ 2HI | [HI]² ÷ [H₂][I₂] | None |
| N₂ + 3H₂ ⇌ 2NH₃ | [NH₃]² ÷ [N₂][H₂]³ | mol⁻² dm⁶ |
Part 3 of 5Calculation Method
1
Write the balanced equation and the expression for Kc.
2
Convert the equilibrium amounts in moles to concentrations by dividing by the volume in dm³.
3
If given initial amounts, use an initial, change and equilibrium table to find the unknowns.
4
Substitute the concentrations into the expression, remembering to apply the powers.
5
Calculate Kc, state its units and round sensibly.
Part 4 of 5Worked Example
Question
- For N₂O₄(g) ⇌ 2NO₂(g) in a 2.00 dm³ vessel at equilibrium there are 0.0400 mol N₂O₄ and 0.0300 mol NO₂.
- Calculate Kc.
- [N₂O₄] = 0.0400 ÷ 2.00 = 0.0200 mol dm⁻³.
- [NO₂] = 0.0300 ÷ 2.00 = 0.0150 mol dm⁻³.
- Kc = [NO₂]² ÷ [N₂O₄] = (0.0150)² ÷ 0.0200.
- Kc = 2.25 × 10⁻⁴ ÷ 0.0200 = 0.01125 mol dm⁻³.
Answer Kc = 1.13 × 10⁻² mol dm⁻³.
Part 5 of 5Exam Checklist & Tips
1.8.3 – Key Points for the Exam
- Equilibrium constant, Kc The ratio of product to reactant concentrations, each raised to the power of its coefficient, at equilibrium.
- Homogeneous equilibrium An equilibrium in which all species are in the same phase.
- Equilibrium concentration The concentration of a species once equilibrium has been reached.
- Dynamic equilibrium A state in which forward and reverse reactions occur at equal rates.
Common Mistakes
- Using initial concentrations instead of equilibrium concentrations in the expression.
- Omitting the powers from the coefficients in the balanced equation.
- Forgetting to divide by the volume to convert moles into concentrations.
- Writing that a change in pressure or a catalyst alters the value of Kc.
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3 & IGCSE —
Kc
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