2024 H2 Chem P1 Q9 — Kinetics Solution

The answer is C (32 hours).
A first-order reaction has a constant half-life.
0.016 to 0.008 to 0.004 to 0.002 mol \(\text{dm}^{-3}\) is 3 half-lives, which took 24 hours, so \(t_{1/2} = \frac{24}{3} = 8\) hours.
0.016 to 0.001 mol \(\text{dm}^{-3}\) is 4 half-lives (0.016, 0.008, 0.004, 0.002, 0.001), so the time needed is \(4 \times 8 = 32\) hours.
• A is wrong: 8 hours is only one half-life, the time to fall from 0.016 to 0.008 mol \(\text{dm}^{-3}\).
• B is wrong: 25.7 hours treats the reaction as zero order (constant rate): rate \(= \frac{0.014}{24}\) mol \(\text{dm}^{-3}\) per hour, time \(= \frac{0.015}{0.014} \times 24 = 25.7\) hours. For a first-order reaction the rate falls as the concentration falls.
• D is wrong: 48 hours doubles the 24 hours because the final concentration is halved; halving 0.002 to 0.001 mol \(\text{dm}^{-3}\) needs only one more half-life of 8 hours.