AQA A Level Chemistry

Revision Notes

1.8 4 Calculations Involving the Equilibrium Constant

Kc Calculations

Calculations involving Kc

  • In the equilibrium expression each figure within a square bracket represents the concentration in mol dm-3
  • The units of Kc therefore depend on the form of the equilibrium expression
  • Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture
  • The concentrations of the reactants and products can then be calculated from the number of moles and total volume

Equation to calculate concentration from number of moles and volume

Worked Example

Calculating Kc of ethanoic acid

In the reaction:

CH3COOH(I) + C2H5OH(I) ⇌ CH3COOC2H5(I) + H2O(I)

ethanoic acid     ethanol          ethyl ethanoate       water

500 cm3 of the reaction mixture at equilibrium contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water.

Use this data to calculate a value of Kc for this reaction.

Answer

Step 1: Calculate the concentrations of the reactants and products

Step 2: Write out the balanced chemical equation with the concentrations of beneath each substance

Step 3: Write the equilibrium constant for this reaction in terms of concentration

Equilibrium Constant Calculations WE Step 1 equation 3

Step 4: Substitute the equilibrium concentrations into the expression

Equilibrium Constant Calculations WE Step 1 equation 4

Step 5: Deduce the correct units for Kc

All units cancel out

Therefore, Kc = 4.03

Note that the smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures

  • Some questions give the initial and equilibrium concentrations of the reactants but not the products
  • An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation

Worked Example

Calculating Kc of ethyl ethanoate

Ethyl ethanoate is hydrolysed by water:

CH3COOC2H5(I) + H2O(I) ⇌ CH3COOH(I) + C2H5OH(I)

ethyl ethanoate       water        ethanoic acid     ethanol

0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1dm3. At equilibrium 0.0654 mol of water are present.

Use this data to calculate a value of Kc for this reaction.

Answer

Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table

Equilibria Calculating Kc of ethyl ethanoate table

Step 2: Calculate the concentrations of the reactants and products

Step 3: Write the equilibrium constant for this reaction in terms of concentration

Equilibrium Constant Calculations WE Step 3 equation

Step 4: Substitute the equilibrium concentrations into the expression

Equilibrium Constant Calculations WE Step 4 equation

= 0.28

Step 5: Deduce the correct units for Kc

Equilibrium Constant Calculations WE Step 5 equation

All units cancel out

Therefore, Kc = 0.28

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