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Equilibrium Constants

Calculating Kc and the amounts present at equilibrium

  • 3 things to know
  • Component 1

Overview

Recall questions on the initial, change and equilibrium layout, why the changes follow the balancing numbers, converting moles to concentrations and when the volume cancels, worked calculations of Kc from initial amounts, and working back from Kc to an equilibrium amount.

What you need to know

Answer each one before you reveal it.

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Key term

How is Kc worked out from initial amounts, and when does the volume cancel?

Key term

Which three quantities are tabulated for each species when Kc is worked out from initial amounts?

Key term

Why must amounts in moles be converted into concentrations before being used in a Kc expression?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Which three quantities are tabulated for each species when Kc is worked out from initial amounts?

    Model answer

    The initial amount, the change that occurs as equilibrium is reached, and the amount present at equilibrium.

  2. 2

    Why must amounts in moles be converted into concentrations before being used in a Kc expression?

    Model answer

    Kc is defined in terms of concentrations, so every amount must first be divided by the volume of the container.

  3. 3

    When can the volume be left out of a Kc calculation?

    Model answer

    When the total number of moles is the same on both sides of the equation, because the volume terms then cancel.

  4. 4

    One mole of ethanoic acid is mixed with one mole of ethanol and 0.333 mol of the acid remains at equilibrium. Calculate Kc for the esterification.

    Model answer

    0.667 mol of ester and of water have formed and 0.333 mol of ethanol is left, so Kc = (0.667 × 0.667) ÷ (0.333 × 0.333) = 4.01, with no units.

  5. 5

    One mole of hydrogen and one mole of iodine are placed in a 1.00 dm³ vessel and 1.56 mol of hydrogen iodide is present at equilibrium. Calculate Kc.

    Model answer

    0.78 mol of each reactant has been used, leaving 0.22 mol dm⁻³ of each, so Kc = 1.56² ÷ (0.22 × 0.22) = 50.3, with no units.

  6. 6

    Two moles of phosphorus(V) chloride are placed in a 2.00 dm³ vessel and 0.50 mol remains at equilibrium as it dissociates into PCl₃ and Cl₂. Calculate Kc.

    Model answer

    1.50 mol of each product has formed, so the concentrations are 0.25, 0.75 and 0.75 mol dm⁻³, giving Kc = (0.75 × 0.75) ÷ 0.25 = 2.25 mol dm⁻³.

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