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 Kp 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
Kp = 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,
reaction

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

Category: Free Energy and Chemical Equilibria

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