Equilibrium constants

At equilibrium the forward and backward reactions go at the same rate, so the concentrations stay the same. The equilibrium constant, Kc, compares the concentrations of products and reactants at equilibrium; for gases, Kp does the same with partial pressures. Its value depends only on the temperature.

Key facts

  • For aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ), using equilibrium concentrations in mol dm⁻³.
  • Units: put the units into the expression and cancel. If the powers on the top and the bottom add up to the same number, Kc has no units.
  • To find Kc from starting amounts, work out the moles of everything at equilibrium (initial, change, equilibrium), then divide by the volume.
  • The partial pressure of a gas = mole fraction × total pressure, where mole fraction = moles of that gas ÷ total moles of gas. Kp uses partial pressures in place of concentrations.
  • Only temperature changes Kc or Kp. For an exothermic forward reaction, raising the temperature makes K smaller; for an endothermic one, it makes K bigger.
  • Changing a concentration or the pressure moves the position of equilibrium but leaves K the same; a catalyst changes neither.

Worked examples

Example 1

Write the expression for Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), and give its units.

  1. Products on top, reactants underneath, each to the power of its number in the equation. Kc = [NH₃]² ÷ ([N₂][H₂]³)
  2. Put in the units. (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = (mol dm⁻³)⁻²
  3. So the units are mol⁻² dm⁶.

Answer: Kc = [NH₃]² ÷ ([N₂][H₂]³); units mol⁻² dm⁶

Example 2

0.50 mol of PCl₅ is heated in a 2.0 dm³ container: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium there are 0.20 mol of Cl₂. Calculate Kc.

  1. Each mole of Cl₂ formed uses one mole of PCl₅ and makes one mole of PCl₃. PCl₅ 0.50 − 0.20 = 0.30 mol; PCl₃ 0.20 mol; Cl₂ 0.20 mol
  2. Divide by the volume, 2.0 dm³. [PCl₅] = 0.15, [PCl₃] = 0.10, [Cl₂] = 0.10 mol dm⁻³
  3. Put them into the expression. Kc = (0.10 × 0.10) ÷ 0.15 = 0.0667 mol dm⁻³

Answer: Kc = 0.0667 mol dm⁻³

Example 3

An equilibrium mixture for N₂O₄(g) ⇌ 2NO₂(g) contains 0.60 mol of N₂O₄ and 0.80 mol of NO₂ at a total pressure of 100 kPa. Calculate Kp.

  1. Total moles of gas. 0.60 + 0.80 = 1.40 mol
  2. Partial pressures = mole fraction × total pressure. p(N₂O₄) = 0.60 ÷ 1.40 × 100 = 42.9 kPa; p(NO₂) = 0.80 ÷ 1.40 × 100 = 57.1 kPa
  3. Put them into Kp. Kp = 57.1² ÷ 42.9 = 76.2 kPa

Answer: Kp = 76.2 kPa