For a first-order reaction, the integrated rate equation is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$

Given: Initial concentration, $[A]_0 = 0.1 \text{ M}$ Final concentration, $[A] = 0.001 \text{ M}$ Time, $t = 5 \text{ mins}$

Substituting the values in the equation: $k = \frac{2.303}{5} \log \left(\frac{0.1}{0.001}\right)$ $k = \frac{2.303}{5} \log (100)$ We know that $\log(100) = \log(10^2) = 2 \log(10) = 2 \times 1 = 2$ $k = \frac{2.303}{5} \times 2$ $k = \frac{4.606}{5} = 0.9212 \text{ min}^{-1}$