JEE Main 2026 Apr 6 Shift 1, Chemistry Q7: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2026Apr 6, Shift 1Chemistry

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.

$A (g) ⇌ B (g)$

$K_{C}$ 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 $^{- 1}$ )

Solution

Let $x$ moles of $A$ convert to $B$ . Then at equilibrium: $[A] = \frac{1 - x}{10}$ , $[B] = \frac{x}{10}$ .

$K_{C} = \frac{[ B ]}{[ A ]} = \frac{x}{1 - x} = 4 \Rightarrow x = 0.8$ .

He is inert; its $1$ mole is unchanged. Apply $P V = n R T$ for each gas in the $10$ L flask at $400$ K:

$P_{He} = \frac{1 \cdot 0.082 \cdot 400}{10} = 3.28$ atm.

$P_{B} = \frac{0.8 \cdot 0.082 \cdot 400}{10} = 2.624$ atm.

Concept behind this question

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