For the reaction A(g) B(g) at 495 K, ΔrGo = -9.478kJ/mol. 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
K; ln 10 = 2.303]
For the reaction A(g) B(g) at 495 K, ΔrGo = -9.478kJ/mol. 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 K; ln 10 = 2.303]
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1 Answer
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ΔG° = -9.478 kJ/mol
Using ΔG° = -2.303 RT log K_p
-9.478 × 10³ = -2.303 × 8.314 × 495 log K_p
1 = log K_p ⇒ K_p = 10
Here, for the given reaction A (g)? B (g), K_p = K_c
Initial A = 22 mmol
At equilibrium A = 22 - x mmol; B = x mmol
K_c = [B] / [A] = (x/V) / (22-x)/V) = x / (22-x) = 10
x = 10 (22-x) ⇒ x = 220 - 10x ⇒ 11x = 220 ⇒ x = 20
So, mmol of B at equilibrium are 20.
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