In a system A(s) → 2B(g) + 3C(g). If the concentration of C at equilibrium is increased by a factor 2 , it will cause the equilibrium concentrations of B to change to

Q: In a system $ A(s) \rightleftharpoons 2B(g) + 3C(g) $ . If the concentration of C at equilibrium is increased by a factor 2 , it will cause the equilibrium concentrations of B to change to

(A) Two times of its original value

(B) One half of its original value

(C) 2 √2 time of its original value

(D) $\frac{1}{2\sqrt{2}}$ times of its original value

Solution: $ A(s) \rightleftharpoons 2B(g) + 3C(g) $

KC = [C]3 [B]2

If (C) becomes twice, then let the concentration of B is B’ then

KC = [2C]3 [B’]2 = [C]3 [B]2

$\large [B’]^2 = \frac{[B]^2}{8} $

$\large [B’] = \frac{[B]}{2\sqrt{2}} $

Ans: (D)