JEE MAIN - Chemistry (2024 - 5th April Evening Shift - No. 25)
Consider the following single step reaction in gas phase at constant temperature.
$$2 \mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \rightarrow \mathrm{C}_{(\mathrm{g})}$$
The initial rate of the reaction is recorded as $$\mathrm{r}_1$$ when the reaction starts with $$1.5 \mathrm{~atm}$$ pressure of $$\mathrm{A}$$ and $$0.7 \mathrm{~atm}$$ pressure of B. After some time, the rate $$r_2$$ is recorded when the pressure of C becomes $$0.5 \mathrm{~atm}$$. The ratio $$\mathrm{r}_1: \mathrm{r}_2$$ is _________ $$\times 10^{-1}$$. (Nearest integer)
Explanation
$$2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})$$
As this is single step reaction,
$$\mathrm{r}=\mathrm{k}_{\mathrm{f}}[\mathrm{A}]^2[\mathrm{~B}]^2$$
$$\begin{array}{lllll} & 2 \mathrm{~A}(\mathrm{~g})+ & \mathrm{B}(\mathrm{g}) \rightarrow & \mathrm{C}(\mathrm{g}) \\ \mathrm{t}=0 & 1.5 & 0.7 & 0 \\ \mathrm{t}=\mathrm{t}^{\prime} & 1.5-2 \mathrm{x} \quad & 0.7-\mathrm{x} & \mathrm{x} \end{array}$$
$$\begin{aligned} & \Rightarrow x=0.5 \mathrm{~atm} \\ & \mathrm{r}_1=\mathrm{k}_{\mathrm{f}}(1.5)^2(0.7) \\ & \mathrm{r}_2=\mathrm{k}_{\mathrm{f}}(0.5)^2(0.2) \\ & \frac{r_1}{r_2}=\frac{9 \times 7}{2}=31.5 \\ & =315 \times 10^{-1} \end{aligned}$$
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