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 $\frac{[A{]}_0}{[A]}=\frac{7.2}{0.9}=8$. $t=\frac{2.303}{0.03}\log 8=\frac{2.303}{0.03}\times 3\log 2=\frac{2.303}{0.03}\times 3\times 0.301\approx 69.3\text{s}$