For the reactionA(g)⇌2B(g), a reaction vessel initially contains only A at a pressure of PA=1.11 atm...

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For the reactionA(g)⇌2B(g), a reaction vessel initially contains only A at a pressure of PA=1.11 atm...

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

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Answer 1

Answer #1

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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Answer 2

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