Chemistry Q57 – JEE-MAIN 2026

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}

)

Clue 1 of 3: Ideal Gas Law and Equilibrium

Recognize that He is an inert gas and its partial pressure can be calculated directly using the ideal gas law, while the partial pressures of A and B depend on the equilibrium shift.

🥷

Ninja Strategy•Calculate Inert Gas First

First calculate the partial pressure of the inert gas (He) using the ideal gas law, which immediately eliminates options with incorrect He pressure. Then, proceed to calculate the partial pressure of B(g) to distinguish between remaining options.

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Step 1: Calculate Partial Pressure of He✦ Active

Helium (He) is an inert gas and does not participate in the reaction. Its partial pressure can be calculated directly using the ideal gas law, $P_{H e} V = n_{H e} R T$ .

P_{H e} = \frac{n_{H e} R T}{V} = \frac{(1 mol) (0.082 {L atm K}^{- 1} {mol}^{- 1}) (400 K)}{10 L} = 3.28 atm

Step 2: Determine Equilibrium Moles of A and B○ Expand

Set up an ICE (Initial, Change, Equilibrium) table for the reaction $A (g) ⇌ B (g)$ . Let $x$ be the moles of A that react.

Initial moles: $n_{A} = 1$ , $n_{B} = 0$ . Change: $- x$ for A, $+ x$ for B. Equilibrium moles: $n_{A} = (1 - x)$ , $n_{B} = x$ . The equilibrium constant $K_{c} = \frac{[B]}{[A]} = \frac{x / V}{(1 - x) / V} = \frac{x}{1 - x}$ . Given $K_{c} = 4.0$ , solve for $x$ :

\frac{x}{1 - x} = 4.0 ⟹ x = 4.0 - 4.0 x ⟹ 5.0 x = 4.0 ⟹ x = 0.8 mol

Thus, at equilibrium, $n_{B} = 0.8$ mol.

💡 Teacher's Secret Hint

Remember that for gas-phase reactions with $Δ n_{g} = 0$ , $K_{p} = K_{c}$ and the volume term cancels out in the concentration ratio.

Step 3: Calculate Partial Pressure of B(g)○ Expand

Using the ideal gas law for B(g) at equilibrium:

P_{B} = \frac{n_{B} R T}{V} = \frac{(0.8 mol) (0.082 {L atm K}^{- 1} {mol}^{- 1}) (400 K)}{10 L} = 2.624 atm

The partial pressures of He and B(g) are $3.28 atm$ and $2.624 atm$ respectively.

💡 Teacher's Secret Hint

Ensure the units are consistent throughout the calculation and that the final answer matches the requested order (He then B(g)).

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Chemistry Question 57 – JEE-MAIN 2026

Clue 1 of 3: Ideal Gas Law and Equilibrium

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