For a reaction A(g) + B(g) → 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

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)