For the reaction $A (g) ⇌ B (g) a t 495 K, Δ_{r} G^{o} = - 9.478 k J m o l^{- 1}$ . If we start this 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 $8.314 J m o l^{- 1} K^{- 1};$ ] [ln10=2.303]

 JEE MAIN 2021

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Solution

$Δ G^{o} = - R T l n K_{e q}$

$Δ G^{o} = - 9.478 k J / m o l e$

$T = 495 K R = 8.314 J m o l^{- 1}$

$- 9.478 \times 10^{3} = - 495 \times 8.314 \times l n K_{e q}$

$l n K_{e q} = 2.303$

$l n K_{e q} = l n 10$

$S o K_{e q} = 10$

$N o w A (g) ⇌ B (g)$

$t = 0220$

$t = t 22 - x x$

$K_{e q} = \frac{[B]}{[C]} = \frac{x}{22 - x} = 10$
or x =20
So millmoles of B = 20

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 JEE MAIN 2021