2020 H2 Chem P1 Q13 — Kinetics Solution

The answer is D (2.9 \(\text{mol dm}^{-3}\)).

\(n(\text{O}_2) = 6.00 \div 24.0 = 0.250\) mol
\(n(\text{H}_2\text{O}_2)\) reacted \(= 2 \times 0.250 = 0.500\) mol (2 mol \(\text{H}_2\text{O}_2\) give 1 mol \(\text{O}_2\))

The half-life is 40 min, so 120 min = 3 half-lives. The fraction of \(\text{H}_2\text{O}_2\) left is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\), so the fraction that has reacted is \(\frac{7}{8}\).
\(\frac{7}{8} \times n_\text{initial} = 0.500\), so \(n_\text{initial} = 0.571\) mol
Initial \([\text{H}_2\text{O}_2] = \frac{0.571}{200/1000} = 2.86\ \text{mol dm}^{-3}\), which is 2.9 \(\text{mol dm}^{-3}\).

• A is wrong: 0.57 is the initial amount of \(\text{H}_2\text{O}_2\) in mol, not divided by the volume of 0.200 \(\text{dm}^3\).
• B is wrong: 1.4 comes from forgetting that 2 mol \(\text{H}_2\text{O}_2\) form 1 mol \(\text{O}_2\): \(\frac{7}{8} \times n_\text{initial} = 0.250\) gives 0.286 mol, and \(0.286 \div 0.200 = 1.43\ \text{mol dm}^{-3}\).
• C is wrong: 2.5 comes from assuming all the \(\text{H}_2\text{O}_2\) has decomposed (\(0.500 \div 0.200 = 2.50\ \text{mol dm}^{-3}\)), but after 3 half-lives \(\frac{1}{8}\) of it is still left.