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

Question

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

Quizplus · Accepted Answer

The equilibrium constant for the reverse reaction is the reciprocal of the given equilibrium constant, and for the reaction with coefficients doubled, it is the square of the reciprocal. Given K = 0.190 for COBr2 ⇄ CO + Br2, the K for the reverse reaction is 1/0.190. For 2CO + 2Br2 ⇄ 2COBr2, it's (1/0.190)^2 = 27.7.

The Equilibrium Constant, Kc, for the Decomposition of COBr2 COBr2(g)

The Equilibrium Constant, K_c, for the Decomposition of COBr₂ COBr₂(g)