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Q. 8 mol of $AB_3$(g) are introduced into a 1.0 $dm^3$ vessel. If it dissociates as
$2AB_{3}\left(g\right) {\rightleftharpoons} A_{2}\left(g\right) + 3B_{2}\left(g\right).$ At equilibrium, 2 mol of $A_2$ are found to be present. The equilibrium constant of this reaction is

AIEEEAIEEE 2012Equilibrium

Solution:

$2AB_{3}\left(g\right) {\rightleftharpoons} A_{2}\left(g\right) + 3B_{2}\left(g\right).$
$\begin{matrix}at\,t = 0&8&0&0\\ at\,eq.&\left(8-2\times2\right)&2&3\times2\\ &= 4&2&6\end{matrix}$
now $K_{C} = \frac{\left[A_{2}\right]\left[B_{2}\right]^{3}}{\left[AB_{3}\right]^{2}} = \frac{2/1\times\left[6/1\right]^{3}}{\left[4/1\right]^{2}} = 27$