Q: For a reaction A_{(g)} + B_{(g)} $\rightleftharpoons $ C_{(g)} + D_{(g)}, the initial concentration of A and B are equals but the equilibrium concentration of C is twice that of equilibrium concentration of A. Then K is

(A) 4

(B) 9

(C) 1/4

(D) 1/9

Solution: A_{(g)} + B_{(g)} $\rightleftharpoons $ C_{(g)} + D_{(g)}

Initial conc. a a 0 0

Eq. conc. a – x a – x x x

Given , x = 2(a-x)

or, x = 2a/3

$\large K_c = \frac{x^2}{(a-x)^2} $

$\large K_c = \frac{(2a/3)^2}{(a-\frac{2a}{3})^2} = 4 $

Ans: (A)