JEE Main 2025 — Chemical Equilibrium Question with Solution

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

$$\begin{array}{rl}& {\mathrm{X}}_{\mathrm{A}}=\frac{\mathrm{a}-\mathrm{a}\alpha }{\mathrm{a}+\mathrm{a}\alpha }=\frac{1-\alpha }{1+\alpha }\\ & {\mathrm{X}}_{\mathrm{B}}=\frac{\mathrm{a}\alpha }{\mathrm{a}+\mathrm{a}\alpha }=\frac{\alpha }{1+\alpha }\\ & {\mathrm{X}}_{\mathrm{C}}=\frac{\mathrm{a}\alpha }{\mathrm{a}+\mathrm{a}\alpha }=\frac{\alpha }{1+\alpha }\end{array}$$

Now if P is total pressure then partial pressure of $\mathrm{A}(\mathrm{g}),\mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are

$$\begin{array}{rl}& {\mathrm{P}}_{\mathrm{A}}=\left(\frac{1-\alpha }{1+\alpha }\right)\mathrm{P}\\ & {\mathrm{P}}_{\mathrm{B}}=\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}\\ & {\mathrm{P}}_{\mathrm{C}}=\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}\end{array}$$

$$\begin{array}{rl}& {\mathrm{K}}_{\mathrm{P}}=\frac{\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}\left(\frac{\alpha }{1+\alpha }\right)\mathrm{P}}{\left(\frac{1-\alpha }{1+\alpha }\right)\mathrm{P}}\\ & {\mathrm{K}}_{\mathrm{P}}=\frac{{\alpha }^2\mathrm{P}}{1-{\alpha }^2}\end{array}$$

As $K_P$ is only function of temperature.

So as $\mathrm{P}\uparrow \alpha \downarrow$