Chemistry - Chemical Kinetics Question with Solution | TestHub

Given, rate $=k{\left[{\mathrm{H}}^{\ast }\right]}^n$ Initial $\mathrm{pH}=3\mathrm{so}\left[{\mathrm{H}}^{+}\right]=1\times 10^{-3}$, initial rate $={\mathrm{r}}_1$ Final $\mathrm{pH}=1$ so $\left[{\mathrm{H}}^{+}\right]=1\times 10^{-1}$ Final rate $=r_2=100r_1$ On substituting values, we get $$\begin{array}{l}{r}_1=k{\left[1\times 10^{-3}\right]}^n\\ r_2=100r_1=k{\left[10^{-1}\right]}^n\end{array}$$ Dividing eqs (i) by (ii). $$\frac{r_1}{100r_1}={\left[\frac{1\times 10^{-3}}{1\times 10^{-1}}\right]}^n$$ $\therefore$ $$\begin{array}{c}\frac{1}{100}={\left[\frac{10^{-2}}{1}\right]}^n\Rightarrow {\left[\frac{1}{100}\right]}^1={\left[\frac{1}{100}\right]}^n\\ n=1\end{array}$$ Thus, the reaction is of first order.