2022 H2 Chem P1 Q13 — Chemical equilibrium Solution

The answer is B.

\(2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}\)

Amounts (mol):
• Initial: \(\text{SO}_2\) 2.00, \(\text{O}_2\) 2.00, \(\text{SO}_3\) 0
• Change: \(\text{SO}_2\) \(-1.80\), \(\text{O}_2\) \(-0.90\), \(\text{SO}_3\) +1.80
• Equilibrium: \(\text{SO}_2\) 0.20, \(\text{O}_2\) 1.10, \(\text{SO}_3\) 1.80

Forming 1.80 mol of \(\text{SO}_3\) uses 1.80 mol of \(\text{SO}_2\) and 0.90 mol of \(\text{O}_2\) (mole ratio 2 : 1 : 2).

Equilibrium concentrations (volume 0.500 dm\(^3\)): \([\text{SO}_2] = 0.40\), \([\text{O}_2] = 2.20\), \([\text{SO}_3] = 3.60\) mol dm\(^{-3}\).

\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \frac{(3.60)^2}{(0.40)^2 \times 2.20} = \frac{12.96}{0.352} = 36.8\) mol\(^{-1}\) dm\(^3\)

• A is wrong: 4.09 is \(\frac{3.60}{0.40 \times 2.20}\), obtained by leaving out the squares on \([\text{SO}_3]\) and \([\text{SO}_2]\).
• C is wrong: 73.6 is \(\frac{(1.80)^2}{(0.20)^2 \times 1.10}\), obtained by using amounts in mol instead of concentrations; because the powers do not cancel, the volume must be included.
• D is wrong: 203 comes from taking the \(\text{O}_2\) used as 1.80 mol instead of 0.90 mol, giving \([\text{O}_2] = 0.40\) mol dm\(^{-3}\) and \(\frac{12.96}{(0.40)^2 \times 0.40} = 203\).