$\underset{\text{Azoisopropane}}{\mathrm{(CH_3)_2 CHN=NCH (CH_3)_{2(g)}}} \rightarrow \mathrm{N_{2(g)}} + \underset{\text{Hexane}}{\mathrm{C_6 H_{14(g)}}}$ $\begin{array}{cccc} \text{Initial pressure} & P_0 & 0 & 0 \\ \text{Pressure} & P_0-p & p & p \\ \text{after time t} & & & \end{array}$ Total pressure after time $t (P_t) = (P_0 - p) + p + p$$= P_0 + p$ or $p = P_t - P_0$$[R]_0 \propto P_0$ and $[R] \propto P_0 - p$On substituting the value of $p, [R] \propto P_0 - (P_t - P_0)$$\text{i.e. } [R] \propto 2P_0 - P_t$As decomposition of azoisopropane is a first order reaction$\therefore k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$$= \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}$When t = 360 sec,$k = \frac{2.303}{360} \log \frac{35.0}{2 \times 35.0 - 54.0}$$= \frac{2.303}{360} \log \frac{35.0}{16}$$= 2.175 \times 10^{-3} \mathrm{s}^{-1}$When $t = 720$ sec, $k = \frac{2.303}{720} \log \frac{35.0}{2 \times 35.0 - 63}$$= \frac{2.303}{720} \log 5$$= 2.235 \times 10^{-3} \mathrm{s}^{-1}$$\therefore$ Average value of $k = \frac{2.175 + 2.235}{2} \times 10^{-3} \mathrm{s}^{-1}$$= 2.20 \times 10^{-3} \mathrm{s}^{-1}$