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Question:

a moles of PCl5 are heated in a closed container to equilibriate PCl5(g)⇌PCl3(g)+Cl2(g) at a pressure of P atm. If x moles of PCl5 dissociate at equilibrium, then:

xa=KpKp+P

xa=(KpP)12

xa=(KpKp+P)12

xa=(Kp+PKp)12

Solution:

PCl5(g)⇌PCl3(g)+Cl2(g)
Moles of PCl5 dissociated=x
Moles of PCl5 left=(a−x)
Moles of PCl3 formed=x
Moles of Cl2 formed=x
Total moles=(a+x)
Degree of dissociastion α=xa
Pressure due to PCl5=(1−α)(1+α)P
Pressure due to PCl3=α(1+α)P
Pressure due to Cl2=α(1+α)P
Now, Kp=XPCl3.XCl2XPcl5P=α21−α2P
Kp(1−α2)=α2P
Kp−Kpα2=α2P
Kp=(Kp+P)α2→α=√KpKp+P=xa→xa=√KPKP+P