Chemistry - Chemical Kinetics Question with Solution | TestHub

It is known, $\frac{{\left({\mathrm{t}}_{1/2}\right)}_1}{{\left({\mathrm{t}}_{1/2}\right)}_2}={\left[\frac{{\mathrm{a}}_2}{{\mathrm{a}}_1}\right]}^{(\mathrm{n}-1)}$ Here, $\mathrm{n}=$ order of the reaction Given, ${\left({\mathrm{t}}_{1/2}\right)}_1=0.1\mathrm{s},{\mathrm{a}}_1=400$ ${\left({\mathrm{t}}_{1/2}\right)}_2=0.8\mathrm{s},{\mathrm{a}}_2=50$ On putting the values, $\frac{0.1}{0.8}={\left[\frac{50}{400}\right]}^{(\mathrm{n}-1)}$ Taking log on both sides $\begin{array}{l}\log \frac{0.1}{0.8}=(\mathrm{n}-1)\log \frac{50}{400}\\ \log \frac{1}{8}=(\mathrm{n}-1)\log \frac{1}{8}\\ \mathrm{n}-1=1\Rightarrow \mathrm{n}=2\end{array}$