equilibrium

Initially, [A]o = [B]o = 0 atm
Both A(g) and B(g) are formed from the decomposition of X(s).

At equilibrium :
Assume that no of moles of A = m mol
Mole ratio A : B = 1 : 3
Hence, no. of moles of B = 3m mol
Mole fraction of A = m / (m + 3m) = 1/4
Mole fraction of B = 3m / (m + 3m) = 3/4
Partial pressure of A, PA = PT•XA = 0.8 x (1/4) = 0.2 atm
Partial pressure of B, PB = PT•XB+ = 0.8 x (3/4) = 0.6 atm

This is a heterogeneous equilibrium, and thus it is not necessary to put the concentrations all components of solids and/or liquids into the equilibrium expression for the modified equilibrium constant.
Modified equilibrium constant, KP
= (PA) x (PB)3
= (0.2) x (0.6)3
= 0.0432 atm4

2008-02-11 13:42:00 補充：
I have overlooked that the products are A2 and Bs, but not A and B.Therefore, in my answer, A should read as A2, and B should read as Bs.

2008-02-11 13:43:18 補充：
I have made the second mistake :A2 and Bs are not products, They are components on the right hand side instead.