Verified Solution • 2025–26 Session moderate

QJEEM22Chac38: JEE Main 2022 Chemistry - Class 12 JEEM22Chac38 Solution: For a given chemical reaction 1 A + 2 B 3 C + 4 D Concentration …

For a given chemical reaction

γ\gamma 1A + γ\gamma 2B →\to γ\gamma 3C + γ\gamma 4D

Concentration of C changes from 10 mmol dm −- 3 to 20 mmol dm −- 3 in 10 seconds. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm −- 3 s −- 1. Therefore, the rate of reaction is _____________ mmol dm −- 3 s −- 1. (Nearest Integer)

Step-by-Step Verified Solution

Rate =1r1(−d[A]dt)=1r2(−d[B]dt)=1r3(−d[C]dt)=\frac{1}{r_{1}}\left(\frac{-d[A]}{d t}\right)=\frac{1}{r_{2}}\left(\frac{-d[B]}{d t}\right)=\frac{1}{r_{3}}\left(\frac{-d[C]}{d t}\right)

=1r4(d[D]dt)=\frac{1}{r_{4}}\left(\frac{d[D]}{d t}\right)

d[D]dt=r4r2(−d[B]dt)\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{\mathrm{r}_{2}}\left(\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)

r4r2=32\frac{r_{4}}{r_{2}}=\frac{3}{2}

−d[B]dt=r2r1(−d[A]dt)⇒r2r1=2\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\left(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right) \Rightarrow \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=2

r2=2r1r_{2}=2 r_{1}

amp;r4=1.5r2=3r1amp;d[C]dt=1 m⋅mol dm−3sec−1amp;d[D]dt=9 m⋅mol dm−3sec−1amp;d[D]dt=r4r3⋅d[C]dt⇒r4r3=9amp;r4=9r3=3r1amp;⇒r1=3r3amp;3r3 A+6r3 B→r3C+9r3Damp;∴ rate of reaction amp;=19×9 m⋅mol dm−3sec−1amp;=1 m⋅mol dm−3sec−1\begin{aligned}&\mathrm{r}_{4}=1.5 \mathrm{r}_{2}=3 \mathrm{r}_{1} \\\\&\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=1 \mathrm{~m} \cdot \mathrm{mol} ~\mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\&\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\&\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{r_{3}} \cdot \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \Rightarrow \frac{\mathrm{r}_{4}}{\mathrm{r}_{3}}=9 \\\\&\mathrm{r}_{4}=9 \mathrm{r}_{3}=3 \mathrm{r}_{1} \\\\&\Rightarrow \mathrm{r}_{1}=3 \mathrm{r}_{3} \\\\&\begin{array}{rr}3 \mathrm{r}_{3} \mathrm{~A}+6 \mathrm{r}_{3} \mathrm{~B} \rightarrow \mathrm{r}_{3} \mathrm{C}+9 \mathrm{r}_{3} \mathrm{D}\end{array} \\\\&\begin{aligned}\therefore \text { rate of reaction } &=\frac{1}{9} \times 9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\&=1 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1}\end{aligned}\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring