A first order reaction takes 40 min for 30% decomposition. What is its half-life? (log(10/7) = 0.155)

1. After 30% decomposition, $[R] = 0.70[R]_0$, so $\dfrac{[R]_0}{[R]} = \dfrac{1}{0.7} = \dfrac{10}{7}$. 2. $k = \dfrac{2.303}{40}\log\dfrac{10}{7} = \dfrac{2.303}{40}\times0.155 = 8.92\times10^{-3}\ \text{min}^{-1}$. 3. $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{8.92\times10^{-3}}$. 4. $= 77.7\ \text{min}$. 5. Distractor A uses a wrong log (e.g. treats it as 50%); distractor C miscomputes k. _Source: syllabus Chemistry Ch 3 "Chemical Kinetics", p.25_