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Equilibrium Constants

Calculating Kc and equilibrium amounts

  • 3 things to know
  • A2 Unit 3

Overview

Recall questions on Calculating Kc and equilibrium amounts.

What you need to know

Answer each one before you reveal it.

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Evaluation

Why must the acid catalyst be allowed for when the ethanoic acid remaining in an equilibrium mixture is found by titration?

Process

Explain why a catalyst has no effect on the value of Kc.

Key term

Why must equilibrium concentrations, rather than starting concentrations, be used in a Kc expression?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why must equilibrium concentrations, rather than starting concentrations, be used in a Kc expression?

    Model answer

    The constant describes the composition of the mixture once the forward and reverse reactions are proceeding at equal rates. Starting concentrations describe a state that is not at equilibrium, so using them would give a meaningless number.

  2. 2

    Describe how the equilibrium amount of each species is found when the initial amounts and the amount of one product formed are known.

    Model answer

    A table is set out with the initial moles, the change in moles and the equilibrium moles for each species. The change for each species is the amount of the known product multiplied by the ratio of its coefficient to that of the product, subtracted for reactants and added for products.

  3. 3

    An equilibrium mixture in a 1.00 dm³ vessel contains 0.0300 mol of N₂O₄ and 0.0200 mol of NO₂ for the equilibrium N₂O₄(g) ⇌ 2NO₂(g). Calculate Kc with its units.

    Model answer

    In a 1.00 dm³ vessel the concentrations are numerically equal to the moles, so [N₂O₄] = 0.0300 and [NO₂] = 0.0200 mol dm⁻³. Kc = (0.0200)² / 0.0300 = 4.00 × 10⁻⁴ / 0.0300 = 1.33 × 10⁻² mol dm⁻³.

  4. 4

    In a 5.00 dm³ vessel an equilibrium mixture contains 0.50 mol PCl₅, 1.50 mol PCl₃ and 1.50 mol Cl₂. Calculate Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

    Model answer

    Dividing by the volume gives [PCl₅] = 0.50/5.00 = 0.10, [PCl₃] = 1.50/5.00 = 0.30 and [Cl₂] = 1.50/5.00 = 0.30 mol dm⁻³. Kc = (0.30 × 0.30)/0.10 = 0.090/0.10 = 0.90 mol dm⁻³.

  5. 5

    In a 2.00 dm³ vessel an equilibrium mixture contains 0.20 mol SO₂, 0.10 mol O₂ and 0.80 mol SO₃. Calculate Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

    Model answer

    The concentrations are [SO₂] = 0.10, [O₂] = 0.050 and [SO₃] = 0.40 mol dm⁻³. Kc = (0.40)² / ((0.10)² x 0.050) = 0.16 / (0.010 × 0.050) = 0.16 / (5.0 × 10⁻⁴) = 320 mol⁻¹ dm³.

  6. 6

    Hydrogen and iodine, 1.00 mol of each, are sealed in a 2.00 dm³ vessel and reach equilibrium containing 1.56 mol of hydrogen iodide. Calculate Kc for H₂(g) + I₂(g) ⇌ 2HI(g).

    Model answer

    Forming 1.56 mol of HI uses 0.78 mol of each reactant, so 0.22 mol of each remains. The concentrations are [HI] = 1.56/2.00 = 0.78 and [H₂] = [I₂] = 0.22/2.00 = 0.11 mol dm⁻³. Kc = (0.78)² / (0.11 × 0.11) = 0.6084/0.0121 = 50.3, with no units.

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