JEE Main 2026 — Chemical Equilibrium Question with 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:
$K_c=\frac{n_B}{n_A}=\frac{x}{1-x}=4.0$

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

At equilibrium, the moles of the gases are:
$n_{He}=1$
$n_B=0.8$

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

Partial pressure of He:
$P_{He}=\frac{1\times 0.082\times 400}{10}=3.28$ atm

Partial pressure of $B(g)$:
$P_B=\frac{0.8\times 0.082\times 400}{10}=2.624$ atm

Answer: $3.28,2.624$