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4.4.2Unit 4. Rates, Equilibria and Further Organic Chemistry

Equilibrium Constant, Kp — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 30 of 56 Unit quiz

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Part 1 of 5Partial Pressure

4.4.2 Equilibrium Constant, Kp

For gaseous equilibria it is often easier to measure pressure than concentration.

Mole fraction
  • The mole fraction of a gas is its moles divided by the total moles of gas in the mixture.
Partial pressure
  • The partial pressure of a gas equals its mole fraction multiplied by the total pressure of the mixture.
Sum of parts
  • The partial pressures of all the gases in a mixture add up to the total pressure.
Symbol
  • The partial pressure of gas A is written p(A) or pA, and it carries units of pressure such as kPa.

Part 2 of 5Kp Expression

Kp Expression
Form
  • For aA(g) + bB(g) ⇌ cC(g), Kp = p(C)ᶜ / (p(A)ᵃ p(B)ᵇ), using equilibrium partial pressures.
    What the symbols mean: Kp = equilibrium constant in terms of partial pressures · p = partial pressure of each gas at equilibrium (Pa or kPa) · powers come from the balanced equation
Gases only
  • Only gases appear in Kp, so a solid in a heterogeneous equilibrium such as CaCO₃(s) ⇌ CaO(s) + CO₂(g) is omitted.
Units
  • Units follow from the powers;
    • 2SO₂ + O₂ ⇌ 2SO₃ has units of kPa⁻¹ because three pressures appear below and two above.
Link to Kc
  • Kp and Kc are different numbers for most reactions, but both change only with temperature.

Part 3 of 5Worked Method

1
Find the moles of each gas at equilibrium and add them to give the total moles.
2
Divide each amount by the total to give mole fractions.
3
Multiply each mole fraction by the total pressure to obtain the partial pressures.
4
Substitute the partial pressures into the Kp expression and calculate with units.

Part 4 of 5Worked Example

Question

  • An equilibrium mixture at a total pressure of 90 kPa contains 0.20 mol SO₂, 0.10 mol O₂ and 0.60 mol SO₃.
  • Calculate Kp for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

  1. Total moles = 0.20 + 0.10 + 0.60 = 0.90 mol, so mole fractions are 0.222, 0.111 and 0.667.
  2. Partial pressures = fraction × 90 kPa, giving p(SO₂) = 20 kPa, p(O₂) = 10 kPa and p(SO₃) = 60 kPa.
  3. Kp = p(SO₃)² / (p(SO₂)² × p(O₂)) = 60² / (20² × 10) = 3600 / 4000.

Answer Kp = 0.90 kPa⁻¹.

Part 5 of 5Exam Checklist & Tips

4.4.2 – Key Points for the Exam

  • Partial pressure The pressure a gas would exert if it alone occupied the container.
  • Mole fraction Moles of one gas divided by total moles of gas.
  • Kp Equilibrium constant expressed with partial pressures.
  • Total pressure The sum of all partial pressures in the mixture.

Common Mistakes

  • Putting concentrations into a Kp expression.
  • Forgetting to include the coefficients as powers.
  • Leaving out units or using the wrong units for pressure.
  • Including solid or liquid species in Kp.

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3 & IGCSE —

Kc

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