JEE MAIN - Chemistry (2023 - 1st February Morning Shift - No. 20)
(i) $$\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 1}=3$$
(ii) $$\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 2}=1$$
If the degree of dissociation and initial concentration of both the reactants $$\mathrm{X}(\mathrm{g})$$ and $$\mathrm{A}(\mathrm{g})$$ are equal, then the ratio of the total pressure at equilibrium $$\left(\frac{p_{1}}{p_{2}}\right)$$ is equal to $$\mathrm{x}: 1$$. The value of $$\mathrm{x}$$ is _____________ (Nearest integer)
Answer
12
Explanation
_1st_February_Morning_Shift_en_20_1.png)
$$ \begin{aligned} & \mathrm{k}_{\mathrm{p}_1}=\frac{\left(\frac{\alpha}{1+\alpha} \times \mathrm{p}_1\right)^2}{\frac{1-\alpha}{1+\alpha} \mathrm{p}_1} \\\\ & 3=\frac{\alpha^2 \times \mathrm{p}_1}{1-\alpha^2} \end{aligned} $$
_1st_February_Morning_Shift_en_20_2.png)
$$ \begin{aligned} & \mathrm{k}_{\mathrm{p}_2}=\frac{\left(\frac{2 \alpha}{1+\alpha} \times \mathrm{p}_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times \mathrm{p}_2} \\\\ & 1=\frac{4 \alpha^2 \times \mathrm{p}_2}{1-\alpha^2} \\\\ & \frac{\mathrm{k}_{\mathrm{p}_1}}{\mathrm{k}_{\mathrm{p}_2}}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\ & \frac{3}{1}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\ & {\mathrm{p}_1} : {\mathrm{p}_2} = 12 : 1 \\\\ & \therefore \mathrm{x}=12 \end{aligned} $$
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