JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

One mole each of He and $A(g)$ are taken in a $10$ L closed flask and heated to $400$ K to establish the following equilibrium. $A(g) \rightleftharpoons B(g)$ . $K_c$ for this reaction at $400$ K is $4.0$ . The partial pressures (in atm) of He and $B(g)$ are respectively (at equilibrium) (Assume He, $A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082$ L atm K $^{-1}$ mol $^{-1}$ )

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$3.28, 2.624$
B
$2.624, 3.28$
C
$3.28, 0.656$
D
$0.656, 6.56$

Did you get this right?

Solution

The given reaction is $A(g) \rightleftharpoons B(g)$ .

Let the initial moles of $A$ be $1$ and at equilibrium, let $x$ moles of $A$ dissociate.

Moles at equilibrium:

$n_A = 1 - x$

$n_B = x$

Since $\Delta n_g = 0$ , the equilibrium constant $K_c$ can be written in terms of moles:

$K_c = \dfrac{n_B}{n_A} = \dfrac{x}{1 - x} = 4.0$

$x = 4 - 4x \Rightarrow 5x = 4 \Rightarrow x = 0.8$

At equilibrium, the moles of the gases are:

$n_{He} = 1$

$n_B = 0.8$

Using the ideal gas equation $P = \dfrac{nRT}{V}$ , the partial pressures are calculated as follows:

Partial pressure of He:

$P_{He} = \dfrac{1 \times 0.082 \times 400}{10} = 3.28$ atm

Partial pressure of $B(g)$ :

$P_B = \dfrac{0.8 \times 0.082 \times 400}{10} = 2.624$ atm

Answer: $3.28, 2.624$

Your note

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$3.28, 2.624$
B
$2.624, 3.28$
C
$3.28, 0.656$
D
$0.656, 6.56$

Did you get this right?

Solution

The given reaction is $A(g) \rightleftharpoons B(g)$ .

Let the initial moles of $A$ be $1$ and at equilibrium, let $x$ moles of $A$ dissociate.

Moles at equilibrium:

$n_A = 1 - x$

$n_B = x$

Since $\Delta n_g = 0$ , the equilibrium constant $K_c$ can be written in terms of moles:

$K_c = \dfrac{n_B}{n_A} = \dfrac{x}{1 - x} = 4.0$

$x = 4 - 4x \Rightarrow 5x = 4 \Rightarrow x = 0.8$

At equilibrium, the moles of the gases are:

$n_{He} = 1$

$n_B = 0.8$

Using the ideal gas equation $P = \dfrac{nRT}{V}$ , the partial pressures are calculated as follows:

Partial pressure of He:

$P_{He} = \dfrac{1 \times 0.082 \times 400}{10} = 3.28$ atm

Partial pressure of $B(g)$ :

$P_B = \dfrac{0.8 \times 0.082 \times 400}{10} = 2.624$ atm

Answer: $3.28, 2.624$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2026 — 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