Chemical equilibria: reversible reactions, dynamic equilibrium

AS Level Chemistry · Equilibria. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

Test this topic freeAll AS Level Chemistry topics

The idea

A reversible reaction can go in both directions. In a closed system it reaches dynamic equilibrium: the forward and reverse reactions continue at equal rates, so the concentrations of reactants and products stay constant.

Le Chatelier's principle says that if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise the change. Use it to predict the effect of changing temperature, concentration or pressure.

The equilibrium constant, Kc or Kp, tells you how far the reaction goes. Its expression comes from the balanced equation: products on top, reactants underneath, each raised to the power of its number of moles. Only a change in temperature changes its value.

Rules to remember

Common mistake

Students use starting amounts in the Kc expression. Work out the equilibrium amounts first, then divide by the volume to get concentrations.

Worked example

1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed. At equilibrium for CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O, 0.67 mol of ester is present. What is Kc?

  1. 0.67 mol of ester forms, so 0.67 mol of water forms and 0.67 mol of each reactant is used up.
  2. Equilibrium amounts: acid = 1.00 − 0.67 = 0.33 mol; ethanol = 0.33 mol; ester = 0.67 mol; water = 0.67 mol.
  3. Kc = [ester][water] ÷ ([acid][ethanol]); there are two terms on top and two underneath, so the volume cancels.
  4. Kc = (0.67 × 0.67) ÷ (0.33 × 0.33) = 0.4489 ÷ 0.1089 = 4.1.

Answer: Kc = 4.1 (no units)

Practice questions

Try each one, then open the answer.

1. 1.0 mol of A and 1.0 mol of B are placed in a 2.0 dm3 vessel. At equilibrium for A + B ⇌ C, 0.40 mol of C is present. What is Kc?

  1. A
    0.44 dm3 mol−1
  2. B
    1.11 dm3 mol−1
  3. C
    2.22 dm3 mol−1
  4. D
    4.44 dm3 mol−1
Show answer

Answer: C. At equilibrium A = B = 1.0 − 0.40 = 0.60 mol. Concentrations (÷ 2.0 dm3): A = B = 0.30, C = 0.20 mol dm−3. Kc = 0.20/(0.30 × 0.30) = 2.22 dm3 mol−1. Using moles instead of concentrations gives 1.11.

2. Which change alters the value of the equilibrium constant Kc for a reaction?

  1. A
    adding a catalyst
  2. B
    changing the temperature
  3. C
    increasing the pressure
  4. D
    increasing the concentration of a reactant
Show answer

Answer: B. Kc depends only on temperature. Changing concentration or pressure may shift the position of equilibrium, but the system readjusts until the same Kc value is restored; a catalyst changes neither.

3. An equilibrium mixture contains 2.0 mol of N2, 6.0 mol of H2 and 2.0 mol of NH3 at a total pressure of 200 kPa. What is the partial pressure of hydrogen?

  1. A
    60 kPa
  2. B
    150 kPa
  3. C
    40 kPa
  4. D
    120 kPa
Show answer

Answer: D. Total amount = 10.0 mol, so the mole fraction of H2 = 6.0/10.0 = 0.60. Partial pressure = mole fraction × total pressure = 0.60 × 200 = 120 kPa.

More questions on this topic

Read the full notes

Keep going

← Redox processes: electron transfer and changes in oxidation numberBrønsted–Lowry theory of acids and bases →All rules on one pageStuck? Ask a question