The initial concentration of $N_2O_5$ in the following first order reaction $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + 1/2O_{2(g)}$ was $1.24 \times 10^{-2} \, mol \, L^{-1}$ at $318 \, K$. The concentration of $N_2O_5$ after $60 \, minutes$ was $0.20 \times 10^{-2} \, mol \, L^{-1}$. Calculate the rate constant of the reaction at $318 \, K$.

(N/A) For a first order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given:
$[A]_0 = 1.24 \times 10^{-2} \, mol \, L^{-1}$
$[A]_t = 0.20 \times 10^{-2} \, mol \, L^{-1}$
$t = 60 \, min$
Substituting the values:
$k = \frac{2.303}{60} \log \left( \frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} \right)$
$k = \frac{2.303}{60} \log(6.2)$
$k = \frac{2.303}{60} \times 0.7924$
$k \approx 0.0304 \, min^{-1}$