JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Equilibrium

For the reaction

A (g) ⇌ B (g)

495 K, Δ_{r} G^{\circ} = - 9.478 kJ mol^{- 1}

. If we start the reaction in a closed container at 495 K with 22 millimoles of A , the amount of B in the equilibrium mixture is
____ millimoles
(Round off to the nearest Integer).

[R = 8.314 J mol^{- 1} K^{- 1}

; In

10 = 2.303]

Answer: 20

Solution

The standard free energy change fixes the equilibrium constant through

Δ_{r} G^{\circ} = - R T ln K

ln K = \frac{9478}{8.314 \times 495} = 2.303 \Rightarrow K = 10

Let

x

millimoles of A have been converted when equilibrium is reached. The vessel then holds

(22 - x)

millimoles of A and

x

millimoles of B, and since one mole of gas gives one mole of gas the ratio of amounts equals

K

\frac{x}{22 - x} = 10 \Rightarrow 11 x = 220 \Rightarrow x = 20

The equilibrium mixture contains 20 millimoles of B.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Equilibrium

Solution

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