Chemistry - Chemical Kinetics Question with Solution | TestHub

$\text{Rate}\text{R}=\text{k}{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}$
From Exp. (i)${\mathrm{R}}_1=2\times 10^{–3}=\mathrm{k}{\left[0.3\right]}^{\mathrm{m}}{\left[0.4\right]}^{\mathrm{n}}$
From Exp. (ii)${\mathrm{R}}_2=8\times 10^{–3}=\mathrm{k}{\left[0.6\right]}^{\mathrm{m}}{\left[0.8\right]}^{\mathrm{n}}$
From Exp. (iii)${\mathrm{R}}_3=4\times 10^{–3}=\mathrm{k}\left[0.6{]}^{\mathrm{m}}\right[0.4{]}^{\mathrm{n}}$
$\frac{{\mathrm{R}}_3}{{\mathrm{R}}_1}\Rightarrow \frac{4\times 10^{–3}}{2\times 10^{–3}}=\frac{\mathrm{k}\left[0.6{]}^{\mathrm{m}}\right[0.4{]}^{\mathrm{n}}}{\mathrm{k}\left[0.3{]}^{\mathrm{m}}\right[0.4{]}^{\mathrm{n}}}$
$2\mathrm{\prime }=2^{\mathrm{m}}\therefore \mathrm{m}=1$
$\frac{{\mathrm{R}}_2}{{\mathrm{R}}_3}\Rightarrow \frac{8\times 10^{–3}}{4\times 10^{–3}}=\frac{\mathrm{k}\left[0.6{]}^{\mathrm{m}}\right[0.8{]}^{\mathrm{n}}}{\mathrm{k}\left[0.6{]}^{\mathrm{m}}\right[0.4{]}^{\mathrm{n}}}$
$2=2^{\mathrm{n}}\therefore \mathrm{n}=1$
$\therefore \text{Rate}\text{law}\left(\text{R}\right)=\text{k}\left[\text{A}\right]\left[\text{B}\right]$