Verified Solution • 2025–26 Session moderate

QJEEM21Ch0fb3: JEE Main 2021 Chemistry - Class 11 JEEM21Ch0fb3 Solution: For the reaction at 495 K, r G = 9.478 kJ mol 1 . If we start…

For the reaction A(g)⇌B(g)A(g) \rightleftharpoons B(g) at 495 K, Δ\Delta rG ∘^\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; ln 10 = 2.303]

Step-by-Step Verified Solution

Δ\Delta Go = −- RT ln Keq

−- 9.478 ×\times 103 = −- 495 ×\times 8.314 ln Keq

ln Keq = 2.303 = ln 10

So, Keq = 10

Now, A(g) ⇌\rightleftharpoons B(g)

t=0220t=t22−xx

Keq=[B][A]=x(22−x)=10Keq = {{[B]} \over {[A]}} = {x \over {(22 - x)}} = 10

x = 20

So, millimoles of B = 20

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