(a) For first order reaction the integral rate law is: $k_t=\ln (\;{a_0}/{a_1})$ Given, $\;\; a_0=1.6 × 10^{-2} {mol} {L}^{-1}$ For $t=300 {\ s}, \;\; a_t=0.8 × 10^{-2} {mol} {L}^{-1}$ For $t=600 {\ s}, \;\; a_t=0.4 × 10^{-2} {mol} {L}^{-1}$ Using first set of data in the rate law, $k × 300\;=\ln \;{1.6 × 10^{-2}}/{0.8 × 10^{-2}}$ $k\;=0.00231 {\ s}^{-1}$ Using second set of data in the rate law, $k × 600\;=\ln \;{1.6 × 10^{-2}}/{0.4 × 10^{-2}}$ $k\;=0.00231 {\ s}^{-1}$ The value of $k$ is consistent,therefore it follows first order reaction. (b) The half-life of first order reaction is given by the following equation : $t_{1 / 2}\;=\;{\ln 2}/{k}=2.303 × \;{\log 2}/{k}$ $∴ \;\; t_{1 / 2}\;=2.303 × \;{\log 2}/{0.00231}$ $=300.08 {\ s} .$