JEE MAIN - Chemistry (2024 - 31st January Evening Shift - No. 2)
$$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{g})$$ The correct relationship between $$\mathrm{K}_{\mathrm{P}}, \alpha$$ and equilibrium pressure $$\mathrm{P}$$ is
$$K_P=\frac{\alpha^{1 / 2} P^{3 / 2}}{(2+\alpha)^{3 / 2}}$$
$$K_P=\frac{\alpha^{3 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}$$
$$K_P=\frac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{3 / 2}}$$
$$K_P=\frac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}}$$
Explanation
$$\begin{aligned} & \mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g}) \\ & \mathrm{t}=\mathrm{t}_{\mathrm{eq}} \quad(1-\alpha) \quad \alpha \quad \frac{\alpha}{2} \end{aligned}$$
$$\mathrm{P}_{\mathrm{B}}=\frac{\alpha}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{A}}=\frac{(1-\alpha)}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}_{\mathrm{C}}=\frac{\frac{\alpha}{2}}{\left(1+\frac{\alpha}{2}\right)} . \mathrm{P}$$
$$\mathrm{K_P=\frac{P_B \cdot P_C^{\frac{1}{2}}}{P_A}}$$
$$\mathrm{=\frac{(\alpha)^{\frac{3}{2}}(P)^{\frac{1}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}}}$$
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