The half-life of a first-order reaction $\mathrm{N_2O_5(g)} \to 2\mathrm{NO_2(g)} + (1/2)\mathrm{O_2(g)}$ at $318\,\text{K}$ is determined experimentally. If the initial concentration of $\mathrm{N_2O_5}$ is $1.24\times 10^{-2}\,\text{mol L}^{-1}$ and falls to $0.20\times 10^{-2}\,\text{mol L}^{-1}$ in $60$ minutes, the rate constant $k$ is closest to

1. First-order integrated rate equation: $k = \dfrac{2.303}{t}\,\log_{10}\dfrac{[A]_0}{[A]}$. 2. Plug in $[A]_0 = 1.24\times 10^{-2}$, $[A] = 0.20\times 10^{-2}$, $t = 60\,\text{min}$. 3. Compute the ratio: $[A]_0/[A] = 1.24/0.20 = 6.2$. 4. $\log_{10}(6.2) = 0.7924$. 5. $k = \dfrac{2.303}{60} \times 0.7924 = 0.0384 \times 0.7924 = 0.0304\,\text{min}^{-1}$, i.e. about $3.0\times 10^{-2}\,\text{min}^{-1}$. 6. Option A uses log-base-e instead of log-base-10 without the conversion factor. Option C is roughly $k/6$ (likely confused with a half-life calc). Option D is two orders of magnitude too high. _Source: NCERT Class 12 Chemistry Part 1, Ch 3, Example 3.5 (the EXACT problem from the textbook), p. 14._