Eight mole of a gas AB3 attain equilibrium in a closed container of volume 1 dm3 2AB3⇌A2(g)+3B2(g). If at equilibrium 2 mole of A2 are present then, equilibrium constant is:

t=0Ar equilibrium: 2AB38(8-a)⇌A20a/2(g)+3B203a2(g)Thus, KC=A2B23AB32; Also, a2=2 ∴a=4∴AB3=41;A2=21;B2=61Thus, Kc=2×6342=27mol2L−2

Eight mole of a gas AB3 attain equilibrium in a closed container of volume 1 dm3 2AB3⇌A2(g)+3B2(g). If at equilibrium 2 mole of A2 are present then, equilibrium constant is:

t=0Ar equilibrium: 2AB38(8-a)⇌A20a/2(g)+3B203a2(g)Thus, KC=A2B23AB32; Also, a2=2 ∴a=4∴AB3=41;A2=21;B2=61Thus, Kc=2×6342=27mol2L−2

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Eight mole of a gas AB₃ attain equilibrium in a closed container of volume 1 dm³ $2 {AB}_{3} ⇌ A_{2} (g) + 3 B_{2} (g)$ . If at equilibrium 2 mole of A₂ are present then, equilibrium constant is:

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36 mol²L^-2

72mol²L^-2

3 mol²L^-2

27 mol²L^-2

answer is D.

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Detailed Solution

$\underset{Ar equilibrium :}{\underset{t = 0}{}} \underset{(8 - a)}{\underset{8}{2 {AB}_{3}}} ⇌ \underset{a / 2}{\underset{0}{A_{2}}} (g) + \underset{\frac{3 a}{2}}{\underset{0}{3 B_{2}}} (g)$
Thus, $K_{C} = \frac{[A_{2}] {[B_{2}]}^{3}}{{[{AB}_{3}]}^{2}}; Also, \frac{a}{2} = 2 ∴ a = 4$
$∴ [{AB}_{3}] = \frac{4}{1}; [A_{2}] = \frac{2}{1}; [B_{2}] = \frac{6}{1}$
Thus, $K_{c} = \frac{2 \times 6^{3}}{4^{2}} = 27 {mol}^{2} L^{- 2}$