Dear student,

I will give u a hint regarding this:

The degree of dissociation of PCl₅ at a certain temperature and under atmospheric pressure is 0.2. Calculate the pressure at which it will be half dissociated at the same temperature.

PCl₅ dissociates as:
PCl₅ ? PCl₃ Cl₂
If α is the degree of dissociation at certain temperature under atmospheric pressure, then
Initial concentration:
PCl₅ = 1
PCl₃ = 0
Cl₂ = 0
At Equilibrium:
PCl₅ = 1 – α
PCl₃ = α
Cl₂ = α
Total number of moles at equilibrium = 1 – α + α + α = 1 + α
Partial pressures of PCl₅, PCl₃ and Cl₂ will be:
p (PCl₃) = α p / 1 + α
p (Cl₂) = α p / 1 + α
p (PCl₅) = (1 – α) p / 1 + α
K_p = p (PCl₃) X p (Cl₂) / p (PCl₅)
K_p = [(α p / 1 + α) X (α p / 1 + α)]/[ (1 – α) p / (1 + α)]
= α² p / (1 – α) ²
Substituting p = 1 atm and α = 0.2
K_p = (0.2) ² X 1 / (1 – (0.2)) ²
= 0.041
When α = 1/2 = 0.5, let pressure is p^’
K_p = α² p^’ / (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

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All the best.

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