The degree of dissociation of PCl₅ at 1 atm is 0.2. Calculate the pressure at which PCl₅ is dissociated to 50%

PCl₅↔ PCl₃ + Cl₂
If α is the degree ofdissociation , then
Initial concentration:
PCl₅= 1
PCl₃= 0
Cl₂= 0
At Equilibrium:
PCl₅= 1 – α
PCl₃= α
Cl₂= α
Totalnumberofmolesat equilibrium = 1 – α + α + α = 1 + α
Partial pressuresof PCl₅, PCl₃and Cl₂will be:
p (PCl₃) = α p / 1 + α
p (Cl₂) = α p / 1 + α
p (PCl₅) = (1 – α) p / 1 + α
Kp= p (PCl₃) X p (Cl₂) / p (PCl₅)
Kp=[(α p / 1 + α) X (α p / 1 + α)]/[(1 – α) p / (1 + α)]
= α²p / (1 – α)²
Substituting p = 1 atm and α = 0.2
Kp= (0.2)²X 1 / (1 – (0.2))²
= 0.041
When α = 1/2 = 0.5, let pressure is p’
Kp= α2p’/ (1 – α)2
0.041 = (0.5) sup>2 p’/ (1 – (0.5))2
p’= (0.041) [1 – (0.5)2] / (0.5)2
p’= 0.125 atm