(a) For first order reaction the integral rate law is:
${\mathit{k}}_{\mathit{t}}=\ln \left(\frac{{\mathit{a}}_0}{{\mathit{a}}_1}\right)$
Given, ${\mathit{a}}_0=1.6\times {10}^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$
For $\mathit{t}=300\mathrm{}\mathit{s},{\mathit{a}}_{\mathit{t}}=0.8\times {10}^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$
For $\mathit{t}=600\mathrm{}\mathit{s},{\mathit{a}}_{\mathit{t}}=0.4\times {10}^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$
Using first set of data in the rate law,
$\mathit{k}\times 300=\ln \frac{1.6\times {10}^{-2}}{0.8\times {10}^{-2}}$
$\mathit{k}=0.00231{\mathrm{}\mathit{s}}^{-1}$
Using second set of data in the rate law,
$\mathit{k}\times 600=\ln \frac{1.6\times {10}^{-2}}{0.4\times {10}^{-2}}$
$\mathit{k}=0.00231{\mathrm{}\mathit{s}}^{-1}$
The value of $\mathit{k}$ is consistent,therefore it follows first order reaction.
(b) The half-life of first order reaction is given by the following equation :
${\mathit{t}}_{1∕2}=\frac{\ln 2}{\mathit{k}}=2.303\times \frac{\log 2}{\mathit{k}}$
$\therefore {\mathit{t}}_{1∕2}=2.303\times \frac{\log 2}{0.00231}$ $=300.08\mathrm{}\mathit{s}.$