Thermal decomposition for N₂O₅(g) follows 1st order kinetics as per N₂O₅(g)→ 2NO₂(g)+ $\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}$Initial presure of N₂O₅ is 100mm. After 10 minutes, the pressure developed is 130mm, what is rate constant?

$$\begin{gathered} {\mathrm{N}}_2{\mathrm{O}}_{5_{\left(\mathrm{g}\right)}}+2{\mathrm{NO}}_{2_{\left(\mathrm{g}\right)}}+\frac{1}{2}{\mathrm{O}}_{2_{\left(\mathrm{g}\right)}} \\ \mathrm{At} \mathrm{t}=0 \min 100 \mathrm{mm} 0 0 \\ \mathrm{At} \mathrm{t}=10 \min 100-\mathrm{x} 2\mathrm{x} \frac{1}{2}\mathrm{x} \\ \mathrm{so}, \mathrm{at} \mathrm{t}=10 \min , 100-\mathrm{x}+2\mathrm{x}+\frac{1}{2}\mathrm{x}=130 \mathrm{mm} \\ \mathrm{x}=20 \mathrm{mm} \\ \mathrm{Given}, {\mathrm{p}}^{°}=100 \mathrm{mm} \\ \mathrm{p}=100-20=80 \mathrm{mm} \\ \mathrm{According} \mathrm{to} 1\mathrm{st} \mathrm{order} \mathrm{kinetics}, \mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{p}}_{°}}{\mathrm{p}} \\ =\frac{2.303}{10}\log \frac{100}{80} \\ =0.693 {\min }^{-1} \end{gathered}$$