Equilibrium Constant, Kc (Edexcel International A Level (IAL) Chemistry): Revision Note

Deducing equilibrium expressions

Deduce the equilibrium constant expression for the following reactions

1. Ag⁺ (aq) + Fe²⁺ (aq) ⇌ Ag (s) + Fe³⁺ (aq)

2. N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

3. 2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

Answer 1:

   $K=\frac{\left[{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)\right]}{\left[{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\right]\left[{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\right]}$

• [Ag (s)] is not included in the equilibrium constant expression as it is a solid

Answer 2:

   $K=\frac{{\left[{\mathrm{NH}}_3\left(\mathrm{g}\right)\right]}^2}{\left[{\mathrm{N}}_2\left(\mathrm{g}\right)\right]{\left[{\mathrm{H}}_2\left(\mathrm{g}\right)\right]}^3}$

Answer 3: 

   $K=\frac{{\left[{\mathrm{SO}}_3\left(\mathrm{g}\right)\right]}^2}{{\left[{\mathrm{SO}}_2\left(\mathrm{g}\right)\right]}^2\left[{\mathrm{O}}_2\left(\mathrm{g}\right)\right]}$

Calculating K_c 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 K_c for this reaction.

Answer

Step 1: Calculate the concentrations of the reactants and products

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

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

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

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

Step 2: Write the equilibrium constant for this reaction in terms of concentration

• K_c = $\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]}{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

Step 3: Substitute the equilibrium concentrations into the expression

• K_c = $\frac{[0.364]\times [0.364]}{[0.070]\times [0.470]}=4.03$

Step 4: Deduce the correct units for K_c 

• K_c = $\frac{\left[mold{m}^{-3}\right]\times \left[mold{m}^{-3}\right]}{\left[mold{m}^{-3}\right]\times \left[mold{m}^{-3}\right]}$

  • All units cancel out

Therefore, K_c = 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

Calculating K_c 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 1.00 dm³. At equilibrium, 0.0654 mol of water are present.

Use this data to calculate a value of K_c 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

Step 2: Calculate the concentrations of the reactants and products

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

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

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

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

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

• K_c = $\frac{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]}$

Step 4: Substitute the equilibrium concentrations into the expression

• K_c = $\frac{[0.346]\times [0.346]}{[0.654]\times [0.654]}=0.28$

Step 4: Deduce the correct units for K_c 

• K_c = $\frac{\left[mold{m}^{-3}\right]\times \left[mold{m}^{-3}\right]}{\left[mold{m}^{-3}\right]\times \left[mold{m}^{-3}\right]}$

  • All units cancel out

Therefore, K_c = 0.28