If the half-life $t_{1/2}$ for a first order reaction is $1$ minute, then the time required for $99.9\%$ completion of the reaction is closest to

Solution & Explanation

\textbf{Answer:} (B) $10$ minutes \textbf{Solution:} For a first order reaction, $99.9\%$ completion means the reactant falls to $0.1\%$ of its initial value, a reduction by a factor of $1000\approx 2^{10}$, i.e. about $10$ half-lives. $t_{99.9}=10\times t_{1/2}=10\times 1=10$ minutes. Exactly: $k=\frac{0.693}{t_{1/2}}=0.693{\text{min}}^{-1}$, and $t=\frac{2.303}{k}\log \frac{100}{0.1}=\frac{2.303}{0.693}\times 3\approx 9.97\approx 10$ minutes.