Plot of $[\mathrm{N}_2\mathrm{O}_5]$ vs time(ii) Initial conc. of$\mathrm{N}_2\mathrm{O}_5 = 1.63 \times 10^{-2} \mathrm{M}$ Half of this concentration $= 0.815 \times 10^{-2} \mathrm{M}$ Time corresponding to this concentration = 1440 s. Hence, $t_{1/2} = 1440 \mathrm{s}$ (iii) Plot of log $[\mathrm{N}_2\mathrm{O}_5]$ vs time (iv) As plot of log $[\mathrm{N}_2\mathrm{O}_5]$ vs time is a straight line. Hence it is a reaction offirst order, i.e. rate law is: Rate $=k \ [\ \mathrm{N}_2 \mathrm{O}_5\ ]$ (v) For first order reaction, $\log R = - \frac{k}{2.303} t + \log R_0$ Therefore slope of the graph drawn between log R and t will be$-\frac{k}{2.303}$. $\therefore\;$ Slope of the line $= - \frac{k}{2.303} = \frac{y_{2} - y_{1}}{t_{2} - t_{1}}$ or, Slope $= \frac{-2.46 - (-1.79)}{3200 - 0} = - \frac{0.67}{3200}$ or, $\frac{k}{2.303} = \frac{0.67}{3200}$ or, $k = \frac{0.67}{3200} \times 2.303 = 4.82 \times 10^{-4} \mathrm{s}^{-1}$ (vi) $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4.82 \times 10^{-4}} = 1438 \mathrm{s}$ The value of $t_{1/2}$ calculated from the value of k is very close to that obtainedfrom graph.