Calculate Kc for 2NOBr(g) <==> 2NO(g) Br2(g)

Question

The equilibrium constant for the reaction

2NO(g) Br2(g) <==> 2NOBr(g)

is Kc = 1.3 times 10^{ - 2} at 1000 K.

Can somebody explain how to do this?

Answer 1

since K = products / reactants

the reverse is the inverse

when they ask for the K of the reverse reaction, it has all the same concentrations, but inversed:



2NOBr(g) &lt;==&gt; 2NO(g) Br2(g)

K1 = [NO]2 [Br2] / [NOBr]2

&amp; the reverse reaction:

2NO(g) Br2(g) &lt;==&gt; 2NOBr(g)

K2 = [NOBr]2 /  [NO]2 [Br2]  

K2 is the inverse of K1

K2 = 1/K1

K2 = 1 / 1.3 times 10^{ - 2

K2 = 76.92

your answer is Kc = 77