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Question

The equilibrium constant kc of the reaction A2 (g) + B2 (g) 2AB (g) is 50. If 1 mol of A2 and 2 mol of B2 are mixed, the amount of AB at equilibrium would be:


A

.934 mol

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B

.467 mol

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C

1.4 mol

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D

1.867 mol

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Solution

The correct option is D

1.867 mol


A2(g)+B2(g)2AB(g)At start120At equ.1x2x2x

Let us take the volume of the vessel (at equilibrium) to be V litres
Kc=4x2(1x)(2x)=50
There won't be a volume term since Δn = 0
Solving the above quadratic, we get x = .943

But we need (2x) since in the question, the amount of AB formed is required. Hence d) 1.867 mol


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