If the rate constant of a reaction is $0.03\ \text{s}^{-1}$, how much time does it take for $7.2\ \text{mol L}^{-1}$ concentration of the reactant to get reduced to $0.9\ \text{mol L}^{-1}$? (Given: $\log 2 = 0.301$)

Solution & Explanation

\textbf{Answer:} (C) $69.3$ s \textbf{Solution:} The unit of the rate constant ($\text{s}^{-1}$) indicates a first-order reaction. Using the integrated first-order rate law: $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$ Here $k = 0.03\ \text{s}^{-1}$, $[A]_0 = 7.2\ \text{mol L}^{-1}$, $[A] = 0.9\ \text{mol L}^{-1}$, so $\frac{[A]_0}{[A]} = \frac{7.2}{0.9} = 8$. $t = \frac{2.303}{0.03}\log 8 = \frac{2.303}{0.03}\times 3\log 2 = \frac{2.303}{0.03}\times 3\times 0.301 \approx 69.3\ \text{s}$