A first order reaction has a specific reaction rate of $10^{-2}{\text{s}}^{-1}$. How much time will it take for $20\text{g}$ of the reactant to reduce to $5\text{g}$?

Solution & Explanation

\textbf{Answer:} (B) $138.6\text{s}$ \textbf{Solution:} For a first order reaction $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{10^{-2}}=69.3\text{s}$. Reducing $20\text{g}\to 5\text{g}$ is a fall to one-fourth, i.e. two successive half-lives ($20\to 10\to 5$). Therefore $t=2\times t_{1/2}=2\times 69.3=138.6\text{s}$. Equivalently $t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R]}=230.3\times \log 4\approx 138.6\text{s}$.