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AS · Practice questions · The equilibrium constant Kc

Put a number on it.

Six original Cambridge-style questions on Kc: expressions, units, calculations with equilibrium tables, and what changes the constant.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
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Products over reactants, powers from the equation.

01
Recall
[2 marks]

Write the Kc expression and deduce its units for N2O4(g) ⇌ 2NO2(g).

  • Kc = [NO2]2 ÷ [N2O4]. ✓
  • Units: (mol dm−3)2 ÷ mol dm−3 = mol dm−3. ✓
02
Calculation
[3 marks]

1.00 mol of N2O4 is placed in a 1.00 dm3 flask; at equilibrium 20.0% has dissociated to NO2. Calculate Kc.

  • 0.200 mol N2O4 reacts, forming 0.400 mol NO2; 0.800 mol N2O4 remains. ✓
  • In 1.00 dm3: [N2O4] = 0.800, [NO2] = 0.400 mol dm−3. ✓
  • Kc = 0.4002 ÷ 0.800 = 0.200 mol dm−3. ✓
03
Calculation
[3 marks]

1.00 mol of ethanoic acid and 1.00 mol of ethanol reach equilibrium with 0.667 mol of ethyl ethanoate: CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O. Calculate Kc and explain why the volume is not needed.

  • Equilibrium amounts: acid 0.333, alcohol 0.333, ester 0.667, water 0.667 mol. ✓
  • Kc = (0.667 × 0.667) ÷ (0.333 × 0.333) = 4.0. ✓
  • There are two concentration terms on top and bottom, so V cancels; Kc has no units. ✓
04
Calculation
[3 marks]

0.500 mol each of H2 and I2 are mixed in 1.00 dm3. Kc = 49.0 for H2 + I2 ⇌ 2HI. Calculate the equilibrium amount of HI.

  • Let x mol of H2 react: [HI] = 2x, [H2] = [I2] = 0.500 − x. ✓
  • (2x)2 ÷ (0.500 − x)2 = 49.0, so 2x ÷ (0.500 − x) = 7.00. ✓
  • 2x = 3.50 − 7.00x, so x = 0.389 and HI = 0.778 mol. ✓
05
Explain
[2 marks]

For 2SO2 + O2 ⇌ 2SO3, ΔH = −197 kJ mol−1. State and explain the effect on Kc of increasing the temperature.

  • The forward reaction is exothermic, so increasing the temperature shifts the position to the left. ✓
  • [SO3] falls while [SO2] and [O2] rise, so Kc decreases. ✓
06
Explain
[2 marks]

Explain why adding more O2 to the SO2/O2/SO3 equilibrium mixture at constant temperature does not change Kc.

  • Adding O2 momentarily makes the concentration ratio smaller than Kc. ✓
  • The position shifts right, forming SO3, until the ratio again equals the same Kc; only temperature changes Kc. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.