Missing Kc value

Question

The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant.

A(g) + 2B(g) --> AB2(g) Kc = 59

AB2(g) + B(g) --> AB3(g) Kc = ?

A(g) + 3B(g) --> AB3(g) Kc = 478

Answer 1

AB₂(g) + B(g) --> AB₃(g) .. Kc = ?

Given

(1) A(g) + 2B(g) --> AB₂(g) .. Kc = 59

(2) A(g) + 3B(g) --> AB₃(g) .. Kc = 478

reverse (1): AB₂ --> A + 2B .. Kc = 1/59 = 0.0169

copy (2) : A(g) + 3B(g) --> AB₃(g) .. Kc = 478

Add them: AB₂(g) + A(g) + 3B(g) --> A(g) + 2B(g) + AB₃(g)

Combine/cancel like terms to get AB_2(g) + B(g) --> AB₃(g)

When adding two reactions, the K for the resulting reaction is the product of the individual K values. Thus...

K = (0.0169)(478) = 8.10