The equilibrium constant, Kc , for the decomposition of COBr2?

Question

COBr2(g) <==> CO(g) + Br2(g)

is 0.190. What is Kc for the following reaction?

2CO(g) + 2Br2(g) <==> 2COBr2(g)

A. 0.0361

B. 2.63

C. 5.62

D. 10.5

E. 27.7

Answer 1

For this reaction :

CO + Br2 --&gt; COBr2

Kc = 1/0.190

And then, for the next reactions :

2CO + 2Br2 --&gt; 2COBr2

Kc = (1/0.190)^2

= 27.7

The answer is E.