JEE Main 2025 — Chemistry Physical Chemistry

2025mcqhard

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?

Official previous-year question

Held on 2 Apr 2025 · Verified 6 Jul 2026.

Options

A
If p value is extremely high compared to $\mathrm{K}_{\mathrm{p}}, \alpha \approx 1$
B
When $p$ increases $\alpha$ decreases
C
If $k_p$ value is extremely high compared to $p, \alpha$ becomes much less than unity
D
When $p$ increases $\alpha$ increases

Did you get this right?

Solution

JEE Main 2025 Chemistry, Chemical Equilibrium — solution figure

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 $\uparrow \quad \alpha \downarrow$

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Solution

JEE Main 2025 Chemistry, Chemical Equilibrium — solution figure

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 $\uparrow \quad \alpha \downarrow$

JEE Main 2025 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main 2025 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry