For the reactionA(g)⇌2B(g), a reaction vessel initially contains only A at a pressure of PA=1.11 atm . At equilibrium, PA =0.34 atm . Calculate the Kp. assume no changes in volume or temperature
A(g)⇌2B(g)
Kp = PB^2 / PA
If PA0 = 1.11
then in equilibrium
Pa =0.34
find Kp
Pb0 = 0
Pbeq = 0 + 2x
Pa = 1.11-x = 0.34
x = 0.34-1.11 = 0.77
substitute in PBeq
Pbeq = 0 + 2x = 2*0.77= 1.54
Kp = PB^2 / PA = (1.54)^2/(0.34) = 6.97529
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