Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : A(~g) leftharpoons B(~g)+C(~g) If p being the total pressure, K_p is the pressure equilibrium constant and α is the

Correct answer: B. When p increases α decreases a moles of A(g) taken initially and at time Now moles fraction of A(g), B(g) and C(g) are aligned & X_A=(a-a α)/(a+a α)=(1-α)/(1+α) & X_B=(a α)/(a+a α)=(α)/(1+α) & X_C=(a α)/(a+a α)=(α)/(1+α) aligned Now if P is total pressure then partial pressure o A(g), B(g) and C(g) are aligned & P_ A =((1-α)/(1+α)) P & P_ B =((α)/(1+α)) P & P_ C =((α)/(1+α)) P & ~K_ P = ((α)/(1+α)) P((α)/(1+α)) P ((1-α)/(1+α)) P & ~K_ P = α^2 P 1-α^2 aligned As K_P is only function of temperature. So as P α

JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2025 · 2 April · Evening Shift · Chemistry · Chemical Equilibrium

Consider the following chemical equilibrium of the gas phase...

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})

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? Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : A(~g) leftharpoons

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : A(~g) leftharpoons

Options

AIf p value is extremely high compared to $\mathrm{K}_{\mathrm{p}}, \alpha \approx 1$
BWhen $p$ increases $\alpha$ decreases
CIf $k_p$ value is extremely high compared to $p, \alpha$ becomes much less than unity
DWhen $p$ increases $\alpha$ increases

Correct answer

B. When $p$ increases $\alpha$ decreases

Step-by-step 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}<br />& X_A=\frac{a-a \alpha}{a+a \alpha}=\frac{1-\alpha}{1+\alpha} \\<br>& X_B=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha} \\<br>& X_C=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha}<br />\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}<br />& \mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P} \\<br>& \mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\<br>& \mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\<br>& \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}} \\<br>& \mathrm{~K}_{\mathrm{P}}=\frac{\alpha^2 \mathrm{P}}{1-\alpha^2}<br />\end{aligned}

K_P

is only function of temperature.
So as P

\uparrow \quad \alpha \downarrow

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