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Question

The Kc for, A2(g)+B2(g)2AB(g) at 100 is 50. If one litre flask containing one mole of A2 is connected with a two litre flask containing 2 mole of B2, how many mole of AB will be formed at 100, rounded to the nearest integer?

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Solution

The equilibrium expression, the initial number of moles and the equilibrium number of moles are as given below. A2(g)+B2(g)2AB(g)
Initial mole 120
Equilibrium mole (1x)(2x)2x
The volume becomes 3 litre on joining two containers.
At equilibrium
[A2]=1x3;[B2]=2x3;[AB]=2x3
The expression for the equilibrium constant becomes
Kc=[AB]2[A2][B2]
=(2x/3)2(1x)3(2x)3=4x2(23x+x2)
or 50=4x2(23x+x2) .(i)
Solving equation (i) x=0.935 and 2.326; the later value is not valid since x2
[AB]=2x=2×0.935=1.87
Hence, the number of moles of AB formed at equilibrium is 1.87.

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