For the decomposition of azoisopropane into hexane and nitrogen at 543 K, the following data is obtained.
${\left({\mathrm{CH}}_3\right)}_2\mathrm{CHN}=\mathrm{NCH}{\left({\mathrm{CH}}_3\right)}_2⟶{\mathrm{N}}_2+{\mathrm{C}}_6{\mathrm{H}}_{14}$

Experiment	Time (s)	Total pressure (mm Hg)
I	0	35.0
II	360	54.0

The rate constant at 360 is

Let the initial pressure of azoisopropane is p_o and,after (r ) time, its pressure reduces by x.
So,         ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CHN}=\mathrm{NCH}{\left({\mathrm{CH}}_3\right)}_2⟶{\mathrm{N}}_2+{\mathrm{C}}_6{\mathrm{H}}_{14}$
$\begin{array}{lccc}\text{ At }\mathrm{t}=0& {\mathrm{p}}_0& 0& 0\\ \text{ After }360\mathrm{s}& {\mathrm{p}}_0-\mathrm{x}& \mathrm{x}& \mathrm{x}\end{array}$
Total pressure after 360s $\left({\mathrm{p}}_{\mathrm{t}}\right)={\mathrm{p}}_0-\mathrm{x}+\mathrm{x}+\mathrm{x}=35+\mathrm{x}=54$
                                            $\begin{array}{l}\mathrm{x}=54-35=19\mathrm{mmHg}\\ \mathrm{x}=19\mathrm{mmHg}\end{array}$
as,                             $\begin{array}{l} {\mathrm{p}}_0=35\mathrm{mmHg}\\ {\mathrm{p}}_0-\mathrm{x}=35-19=16\mathrm{mmHg}\end{array}$
For first order reaction,
rate constant, $\left(\mathrm{k}\right)=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{p}}_0}{{\mathrm{p}}_0-\mathrm{x}}$
                            $=\frac{2.303}{360}\log \frac{35}{16}=\frac{2.303}{360}\log 2.187$
                            $=\frac{2.303}{360}\times 0.34=2.17\times {10}^{-3}{\mathrm{s}}^{-1}$