Equilibrium Terms (OCR A Level Chemistry A): Revision Note

Calculating concentrations

Ethanoic acid and ethanol react according to the following equation:

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

At equilibrium, 500 cm³ of the reaction mixture 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.

Calculate the concentration of each chemical at equilibrium.

Answer

• [CH₃COOH] $=\frac{0.235}{0.500}=$0.470 mol dm^-3 

  • [C₂H₅OH] $=\frac{0.035}{0.500}=$0.070 mol dm^-3 

  • [CH₃COOC₂H₅] $=\frac{0.182}{0.500}=$0.364 mol dm^-3 

  • [H₂O] $=\frac{0.182}{0.500}=$0.364 mol dm^-3 

Ethyl ethanoate is hydrolysed by water:

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

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 the equilibrium concentrations of each chemical.

Answer

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

Step 2: Calculate the concentrations of the reactants and products

• [CH₃COOC₂H₅] $=\frac{0.0654}{1.00}=$0.0654 mol dm^-3 

  • [H₂O] $=\frac{0.0654}{1.00}=$0.0654 mol dm^-3 

  • [CH₃COOH] $=\frac{0.0346}{1.00}=$0.0346 mol dm^-3 

  • [C₂H₅OH] $=\frac{0.0346}{1.00}=$0.0346 mol dm^-3 

Working out mole fractions

A sample of  0.25 mole of nitrogen and 0.75 mole of hydrogen were reacted together to form ammonia. The equilibrium amount of nitrogen was 0.16 mol.

N₂ (g) + 3H₂ (g) ⇌  2NH₃ (g)

Calculate the mole fractions of nitrogen, hydrogen and ammonia.

Answer

Write out the equation and record the initial, the change and the equilibrium amounts:

You can check you have the mole fractions correct by adding them up and making sure they come to 1: 0.195 + 0.585 + 0.220 = 1

The total pressure for the reaction, described above, of nitrogen and hydrogen to form ammonia was 150 kPa.

Calculate the partial pressure of each gas.

Answer

• Partial pressure of N₂ (g) = 0.195 x 150 = 29.25 kPa

  • Partial pressure of H₂ (g) = 0.585 x 150 = 87.75 kPa

  • Partial pressure of NH₃ (g) = 0.220 x 150 = 33.0 kPa

• You can check that your partial pressures are correct by adding them up:

  • 29.25 + 87.75 + 33.0 = 150

• They should add up to the total pressure - if they don't, then there is at least one calculation wrong somewhere!