JEE MAIN - Chemistry (2022 - 25th June Morning Shift - No. 21)
For a given chemical reaction
$$\gamma$$1A + $$\gamma$$2B $$\to$$ $$\gamma$$3C + $$\gamma$$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)
Answer
1
Explanation
Rate $=\frac{1}{r_{1}}\left(\frac{-d[A]}{d t}\right)=\frac{1}{r_{2}}\left(\frac{-d[B]}{d t}\right)=\frac{1}{r_{3}}\left(\frac{-d[C]}{d t}\right)$
$$ =\frac{1}{r_{4}}\left(\frac{d[D]}{d t}\right) $$
$\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{\mathrm{r}_{2}}\left(\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)$
$$ \frac{r_{4}}{r_{2}}=\frac{3}{2} $$
$\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\left(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right) \Rightarrow \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=2$
$r_{2}=2 r_{1}$
$$ \begin{aligned} &\mathrm{r}_{4}=1.5 \mathrm{r}_{2}=3 \mathrm{r}_{1} \\\\ &\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=1 \mathrm{~m} \cdot \mathrm{mol} ~\mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{r_{3}} \cdot \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \Rightarrow \frac{\mathrm{r}_{4}}{\mathrm{r}_{3}}=9 \\\\ &\mathrm{r}_{4}=9 \mathrm{r}_{3}=3 \mathrm{r}_{1} \\\\ &\Rightarrow \mathrm{r}_{1}=3 \mathrm{r}_{3} \\\\ &\begin{array}{rr} 3 \mathrm{r}_{3} \mathrm{~A}+6 \mathrm{r}_{3} \mathrm{~B} \rightarrow \mathrm{r}_{3} \mathrm{C}+9 \mathrm{r}_{3} \mathrm{D} \end{array} \\\\ &\begin{aligned} \therefore \text { rate of reaction } &=\frac{1}{9} \times 9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &=1 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \end{aligned} \end{aligned} $$
$$ =\frac{1}{r_{4}}\left(\frac{d[D]}{d t}\right) $$
$\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{\mathrm{r}_{2}}\left(\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)$
$$ \frac{r_{4}}{r_{2}}=\frac{3}{2} $$
$\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\left(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right) \Rightarrow \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=2$
$r_{2}=2 r_{1}$
$$ \begin{aligned} &\mathrm{r}_{4}=1.5 \mathrm{r}_{2}=3 \mathrm{r}_{1} \\\\ &\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=1 \mathrm{~m} \cdot \mathrm{mol} ~\mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{r_{3}} \cdot \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \Rightarrow \frac{\mathrm{r}_{4}}{\mathrm{r}_{3}}=9 \\\\ &\mathrm{r}_{4}=9 \mathrm{r}_{3}=3 \mathrm{r}_{1} \\\\ &\Rightarrow \mathrm{r}_{1}=3 \mathrm{r}_{3} \\\\ &\begin{array}{rr} 3 \mathrm{r}_{3} \mathrm{~A}+6 \mathrm{r}_{3} \mathrm{~B} \rightarrow \mathrm{r}_{3} \mathrm{C}+9 \mathrm{r}_{3} \mathrm{D} \end{array} \\\\ &\begin{aligned} \therefore \text { rate of reaction } &=\frac{1}{9} \times 9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &=1 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \end{aligned} \end{aligned} $$
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