Question

Pcl5 decomposes as pcl3 and cl2 if at equilibrium total pressure is p and density of a gaseous mixture is d at temperature t then degree of association

Answer 1

Explanation:

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Solution: PCl5(g) → PCl3(g) + Cl2(g)

We are given the vapour densities at equilibrium at 200oC and 250oC.

The initial vapour density will be the same at both the temperatures as

it would be MPCl5 / 2.

∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25

Vapour density at equilibrium at 200oC = 70.2

∴ Total moles at equilibrium / Total moles initial = 1 + α = Vapour density initial / Vapour density at equilibrium = 104.25 / 70.2 = 1.485

∴ a = 0.485

At 250oC, 1 + α = 104.25 / 57.9 = 1.8

∴ a = 0.8