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At 250 degree c and 1 atm pressure , the vapour density of PCl5 is 57.9 . What will be the dissociation of PCl5 ?
PCl5(g) → PCl3(g) + Cl2(g)We are given the vapour density at equilibrium at 250oC.The initial vapour density will be it would be MPCl5 / 2. ∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25 ∴ Total moles at equilibrium / Total moles initial = 1 + α = Vapour density initial / Vapour density at equilibriumAt 250oC, 1 + α = 104.25 / 57.9 = 1.8 ∴ a = 0.8
PCl5(g) → PCl3(g) + Cl2(g)
We are given the vapour density at equilibrium at 250oC.
The initial vapour density will be
it would be MPCl5 / 2.
∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25
∴ Total moles at equilibrium / Total moles initial = 1 + α = Vapour density initial / Vapour density at equilibrium
At 250oC, 1 + α = 104.25 / 57.9 = 1.8
∴ a = 0.8
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