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?
- If p value is extremely high compared to $\mathrm{K}_{\mathrm{p}}, \alpha \approx 1$
- When $p$ increases $\alpha$ decreases
- If $k_p$ value is extremely high compared to $p, \alpha$ becomes much less than unity
- When $p$ increases $\alpha$ increases
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
$\begin{aligned}
& X_A=\frac{a-a \alpha}{a+a \alpha}=\frac{1-\alpha}{1+\alpha} \\
& X_B=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha} \\
& X_C=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha}
\end{aligned}$
Now if $P$ is total pressure then partial pressure o $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g})$ and $\mathrm{C}(\mathrm{g})$ are
$\begin{aligned}
& \mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P} \\
& \mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\
& \mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\
& \mathrm{~K}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}}{\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P}} \\
& \mathrm{~K}_{\mathrm{P}}=\frac{\alpha^2 \mathrm{P}}{1-\alpha^2}
\end{aligned}$
As $K_P$ is only function of temperature.
So as P $\uparrow \quad \alpha \downarrow$
Asked in: JEE Main 2025 (02 Apr Shift 2)