In a reaction at equilibrium ‘ x ’ moles of the reactant A decomposes to give 1 mole each of C and D. If the fraction of A decomposed at equilibrium is independent of initial concentration of A then the value ‘ x ’ is

Q: In a reaction at equilibrium ‘ x ’ moles of the reactant A decomposes to give 1 mole each of C and D. If the fraction of A decomposed at equilibrium is independent of initial concentration of A then the value ‘ x ’ is

(A) 1

(B) 3

(C) 2

(D) 4

Solution: $\large x A \rightleftharpoons C + D $

     a    0    0 at initial

 a(1 – α)   aα/x    aα/x  at equilibrium

$\large K = \frac{[C][D]}{[A]^x}$

$\large K = \frac{[a \alpha/x][a \alpha/x]}{[a(1-\alpha)]^x}$

$\large K = \frac{a^2 \alpha^2}{x^2 a^x (1-\alpha)^x}$

$\large K = \frac{a^{2-x} \alpha^2}{x^2 (1-\alpha)^x}$

To become independent of a

2 -x = 0 ⇒ x = 2

Ans: (C)