Chemistry - Chemical Kinetics Question with Solution | TestHub

Order of reaction $=0$

Half lifer 

$$\begin{gathered} {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} = \frac{\left[{\mathrm{A}}_0\right]}{2\mathrm{K}} \\ {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\propto \left[{\mathrm{A}}_0\right] \end{gathered}$$

$\therefore$ Initial concentration doubled then half-life i.e  ${\text{t}}_{\frac{1}{2}}$ becomes double