Solve this
Question:

$\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$

In the above first order reaction the initial concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ is $2.40 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$ at 318 K. The concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ after 1 hour was $1.60$ $\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. The rate constant of the reaction at $318 \mathrm{~K}$ is _________ $\times 10^{-3} \mathrm{~min}^{-1}$. (Nearest integer)

[Given : $\log 3=0.477, \log 5=0.699$ ]

Solution:

$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}}{\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]_{\mathrm{t}}}$

$=\frac{2.303}{60} \log \frac{2.4}{1.6}=6.76 \times 10^{-3} \mathrm{~min}^{-1} \approx 7 \times 10^{-3} \mathrm{~min}^{-1}$