For a first order reaction, the time required for $99.9\%$ completion is how many times the time required for $50\%$ completion?

Solution

1. The half-life of a first order reaction is $t_{1/2}=\frac{0.693}{k}$.
2. The time for $99.9\%$ completion is $$\begin{array}{rl}t& =\frac{2.303}{k}\log (\frac{100}{100-99.9})\\ & =\frac{2.303}{k}\log (1000)\end{array}$$.
3. Since $\log (1000)=3$, we have $$\begin{array}{rl}t& =\frac{2.303\times 3}{k}\\ & =\frac{6.909}{k}\end{array}$$.
4. Dividing $t$ by $t_{1/2}$ gives $\frac{6.909/k}{0.693/k}\approx 10$.
5. Hence the answer is (A).