Question

The equilibrium constant, Kc , for the decomposition of COBr2 COBr2(g) ↔ CO(g) + Br2(g) is 0.190. What is Kc for the following reaction? 2CO(g) + 2Br2(g) ↔ 2COBr2(g)

Answer 1

Answer:

$2CO(g)+2Br_2(g)\rightleftharpoons 2COBr_2(g)$

27.70 is $K_c'$ for the reaction.

Explanation:

$COBr_2(g)\rightleftharpoons CO(g) + Br_2(g)$

The equilibrium constant of the reaction = $K_c=0.190$

The expression of equilibrium constant is given by :

$K_c=\frac{[CO][Br_2]}{[COBr_2]}$..[1]

$2CO(g)+2Br_2(g)\rightleftharpoons 2COBr_2(g)$

The equilibrium constant expression for above reaction can be written as:

$K_c'=\frac{[COBr_2]^2}{[CO]^2[Br]^2}$

$K_c'=\frac{1}{(K_c)^2}$ ( from [1])

$K_c'=\frac{1}{(0.190)^2}=27.70$

27.70 is $K_c'$ for the reaction.