JEE MAIN - Chemistry (2025 - 2nd April Evening Shift - No. 20)
Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
If $p$ being the total pressure, $K_p$ is the pressure equilibrium constant and $\alpha$ is the degree of dissociation, then which of the following is true at equilibrium?
Explanation
_2nd_April_Evening_Shift_en_20_1.png)
a moles of $A(g)$ taken initially and at time
Now moles fraction of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are
$$\begin{aligned} & \mathrm{X}_{\mathrm{A}}=\frac{\mathrm{a}-\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{1-\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{B}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \\ & \mathrm{X}_{\mathrm{C}}=\frac{\mathrm{a} \alpha}{\mathrm{a}+\mathrm{a} \alpha}=\frac{\alpha}{1+\alpha} \end{aligned}$$
Now if P is total pressure then partial pressure of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are
$$\begin{aligned} & \mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\ & \mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \end{aligned}$$
$$\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}}{\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P}} \\ & \mathrm{~K}_{\mathrm{P}}=\frac{\alpha^2 \mathrm{P}}{1-\alpha^2} \end{aligned}$$
As $K_P$ is only function of temperature.
So as $\mathrm{P} \uparrow \quad \alpha \downarrow$
Comments (0)
