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