Equilibrium Constant Kc for Homogeneous Systems (Oxford AQA International A Level (IAL) Chemistry): Revision Note

Deduce the equilibrium expression for the following reactions:

1. Ag⁺ (aq) + Fe²⁺ (aq) $\rightleftharpoons$ Ag (s) + Fe³⁺ (aq)

2. N₂ (g) + 3H₂ (g) $\rightleftharpoons$ 2NH₃ (g)

3. 2SO₂ (g) + O₂ (g) $\rightleftharpoons$ 2SO₃ (g)

Answers:

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

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

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

At equilibrium, 500 cm³ of the following reaction mixture contains 0.235 mol of ethanoic acid, 0.0350 mol of ethanol, 0.182 mol of ethyl ethanoate and 0.182 mol of water.

CH₃COOH (l) + C₂H₅OH (l) $\rightleftharpoons$ CH₃COOC₂H₅ (l) + H₂O (l)

Use this information 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 mol dm^-3 

  • [C₂H₅OH (l)] = $\frac{0.0350}{0.500}$ = 0.070 mol dm^-3 

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

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

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

• Step 3: 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 4: Substitute the equilibrium concentrations into the expression:
  

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

  • K_c = 4.03

• Step 5: Deduce the correct units for K_c:

  • K_c = $\frac{\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)}{\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)}$

  • 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

Ethyl ethanoate is hydrolysed in water.

CH₃COOC₂H₅ (l) + H₂O (l) $\rightleftharpoons$ CH₃COOH (l) + C₂H₅OH (l) 

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

At equilibrium, 0.0654 mol of water are present.

Use this information to calculate a value of K_c for this reaction.

Answer:

• Step 1: Write the balanced chemical equation, with the concentrations beneath each substance, into an initial, change and equilibrium (ICE) table:

• Step 2: Calculate the concentrations of the reactants and products:

  • [H₂O (l)] = $\frac{0.0654}{1.000}$ = 0.0654 mol dm^-3 

  • [CH₃COOC₂H₅ (l)] = $\frac{0.0654}{1.000}$ = 0.0654 mol dm^-3 

  • [C₂H₅OH (l)] = $\frac{0.0346}{1.000}$ = 0.0346 mol dm^-3 

  • [CH₃COOH (l)] = $\frac{0.0346}{1.000}$ = 0.0346 mol 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 = $\frac{0.0346\times 0.0346}{0.0654\times 0.0654}$

  • K_c = 0.28

• Step 5: Deduce the correct units for K_c:

  • K_c = 

  • All units cancel out

  • Therefore, K_c = 0.28

You need to be able to predict the qualitative effects of changes of temperature on the value of K_c.

To do this:

• Write the K_c expression

• Use the balanced chemical equation and enthalpy change information to talk about the concentrations of each species in the expression

• Then talk about the overall effect of a temperature change on the value of K_c.

Factors affecting K_c 

An equilibrium is established in the following reaction:

AB (aq) + CD (aq) $\rightleftharpoons$ AC (aq) + BD (aq)   ΔH = +180 kJ mol^-1

Which factors would affect the value of K_c in this equilibrium?

Answer:

• Only a change in temperature will affect the value of K_c

• Any other changes in conditions would result in the position of the equilibrium moving to oppose this change

• Adding a catalyst increases the rate of reaction meaning the state of equilibrium will be reached faster but has no effect on the position of the equilibrium and, therefore, K_c is unchanged