JEE Main 2022 — Chemical Kinetics Question with Solution

$\frac{\mathrm{d}\left[\mathrm{c}\right]}{\mathrm{dt}}=\left(\frac{20-10}{10}\right)=1\mathrm{milimole}/{\mathrm{dm}}^3\sec$

${\mathrm{\gamma }}_1\mathrm{A}+{\mathrm{\gamma }}_2\mathrm{B}\to {\mathrm{\gamma }}_3\mathrm{C}+{\mathrm{\gamma }}_4\mathrm{D}$

$\frac{\mathrm{d}\left[\mathrm{D}\right]}{\mathrm{dt}}=1.5\left(-\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{B}\right]\right)=9 \mathrm{milimole}/{\mathrm{dm}}^3 \sec$

$\left(-\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{B}\right]\right)=2\left(-\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{A}\right]\right)$

Rate $=-\frac{1}{{\mathrm{\gamma }}_1}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=-\frac{1}{{\mathrm{\gamma }}_2}\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=\frac{1}{{\mathrm{\gamma }}_3}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}=\frac{1}{{\mathrm{\gamma }}_4}\frac{\mathrm{d}\left[\mathrm{D}\right]}{\mathrm{dt}}$

(i) $\frac{\mathrm{d}\left[\mathrm{D}\right]}{\mathrm{dt}}=\frac{{\mathrm{\gamma }}_4}{{\mathrm{\gamma }}_2}\left[-\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}\right]=1.5\left(-\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{B}\right]\right)$

$\frac{{\mathrm{\gamma }}_4}{{\mathrm{\gamma }}_2}=1.5$

(ii) $\frac{{\mathrm{\gamma }}_2}{{\mathrm{\gamma }}_1}\left[-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}\right]=\left(-\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{B}\right]\right)=2\left(-\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{A}\right]\right)$

$\frac{{\mathrm{\gamma }}_2}{{\mathrm{\gamma }}_1}=2$

(iii) $\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}=\frac{{\mathrm{\gamma }}_3}{{\mathrm{\gamma }}_4}\left[\frac{\mathrm{d}\left[\mathrm{D}\right]}{\mathrm{dt}}\right]$

$1=\frac{{\mathrm{\gamma }}_3}{{\mathrm{\gamma }}_4}\times 9$

$\frac{{\mathrm{\gamma }}_4}{{\mathrm{\gamma }}_3}=9$

So, if

 $$\begin{gathered} {\mathrm{\gamma }}_3 = 1 \mathrm{than} {\mathrm{\gamma }}_4 = 9 \\ {\mathrm{\gamma }}_2 = 6 \mathrm{and} {\mathrm{\gamma }}_1 = 3 \\ \mathrm{Rate} \mathrm{of} \mathrm{reaction} = \frac{1}{{\mathrm{\gamma }}_3}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}} = 1 \mathrm{m} \mathrm{mol} {\mathrm{dm}}^{-3} {\mathrm{s}}^{-1} \end{gathered}$$