Chemistry - Chemical Kinetics Question with Solution | TestHub

$\underset{{\mathrm{P}}_{\mathrm{i}}=363\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{H}\mathrm{O}}\to \mathrm{C}\mathrm{O}+\mathrm{C}{\mathrm{H}}_4$

Now we know that

$\mathrm{r}\propto {\mathrm{P}}_{\mathrm{R}}^{\mathrm{n}}$...(i)

where${\mathrm{P}}_{\mathrm{R}}=$reactant pressure

$\mathrm{n}=$order of reaction

Now rate of reaction is 1.00 torr${\mathrm{s}}^{-1}$, when reactant pressure is$\left(363-363\times \frac{5}{100}\right)\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}=344.85\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r},$similarly rate of reaction is 0.5 torr${\mathrm{s}}^{-1}$, when reactant pressure is$\left(363-363\times \frac{33}{100}\right)\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}=243.21\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}$

Therefore, applying equation (i)

$\frac{1}{0.5}={\left(\frac{344.85}{243.21}\right)}^{\mathrm{n}}$

$2=(1.418{)}^{\mathrm{n}}$

$2^{\frac{1}{\mathrm{n}}}=1.418$

So$\mathrm{n}\approx 2$