For a first-order reaction, the time required for 99.9% completion () is approximately how many time — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

For a first-order reaction, $t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{100 - 99.9} = \frac{2.303}{k} \log 10^3 = \frac{3 \times 2.303}{k} = \frac{6.909}{k}$. Since $t_{1/2} = \frac{0.693}{k}$, the ratio is $6.909 / 0.693 \approx 10$. Thus $t_{99.9\%} \approx 10 \times t_{1/2}$.