Question

For the reaction a(g) 2 b(g), a reaction vessel initially contains only a at a pressure of pa = 1.32 atm. at equilibrium, pa = 0.25 atm. calculate the value of kp. (assume no changes in volume or temperature.)

Answer 1

When there is no change in volume and temperature:
So we can assume that:
P1/n1 = P2/n2
and p1/p2 = n1/n2
So we can assume that the coefficients in the balanced equation as pressure ratios = mole ratios
so from the reaction:
A(g) ↔ 2 B (g)
∴ 1 mole A → 2 mole B
when we have PA(initial) = 1.32 atm & PB (initial) = 0 atm & PA(equ) = 0.25 atm 
∴PB (intial) = (PA(intial))-PA(equ) * 2 mol
                  = (1.32 - 0.25) *2 = 2.14 atm

SO we can get Kp from this formula:
Kp = p(B)^2 / [P(A)
     = 2.14^2 / 0.25
     = 18.32

Answer 2

                         A(g)      ⇄    2 B(g)
Initial               1.32                0
Change          -1.07           + 2.14       
Equilibrium      0.25              2.14
Kp = P²$_{B}$ / P$_{A}$
     = (2.14)² / 0.25 = 18.3  
Note that:
Change of A is calculated by (1.32 - 0.25 = 1.07)
also change in B is calculated by multiplying change of A times 2