In a pure first-order reaction, the time required for 99.9% completion of the reaction () is approxi — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

For 99.9% completion, $[A]_t = 0.1\% \text{ of } [A]_0 = [A]_0/1000$. Using the formula $t = \frac{2.303}{k}\log(1000) = \frac{3 \times 2.303}{k}$. Since $t_{1/2} = \frac{0.693}{k} \approx \frac{0.3 \times 2.303}{k}$, the ratio is $3 / 0.3 = 10$. Thus, $t_{99.9\%} \approx 10 \times t_{1/2}$.