At 540 K, 0.10 mole of PCl₅ is heated in an 8.0 L flask. The pressure of the equilibrium mixture is found to be 1.0 atm. Calculate K_p and _c for the reaction.

The reaction will be given as:
PCl₅ ⇌ PCl₃ Cl₂

Initial moles:
PCl₅ = 0.10
PCl₃ = 0
Cl₂ = 0

Equivalent moles:
PCl₅ = 0.10 – α
PCl₃ = α
Cl₂ = α

Total number of moles = 0.10 – α + α + α = 0.10 + α
Now, we know that pV = nRT
And it is given that V = 8.0 litre, p = 1 atm, T = 540 K and R = 0.082
Therefore,
p= nRT / V
1 = (0.10 + α) X 0.082 X 540 / 8.0
(0.10 + α) = 8 X 1 / 0.082 X 540 = 0.180

Therefore, α = 0.180 – 0.10 = 0.8

Now
K_p = p (PCl₃) X p (Cl₂) / p (PCl₅)
p (PCl₃) = α p / 0.1 + α
p (Cl₂) = α p / 0.1 + α
p (PCl₅) = (0.1 – α) p / 0.1 + α
Therefore,

K_p = (.008) ² X 1 / (0.1 + 0.08) X (0.1 – 0.08)
= 0.0064 / 0.18 X 0.02
K_p = 1.78 atm
Relation between K_p and K_c is given as:
K_p = K_c (RT) ^∆n
Therefore,
K_c = K_p / (RT) ^∆n
∆n = 1 + 1 – 1 = 1
Hence,
K_c = 1.78 / (0.082 X 540) ¹
= 0.042 mol litre^-1
Hence the value of K_p = 1.78 atm and K_c = 0.042 mol litre^-1