At 473K, partially dissociated vapours of PCl5 are 62 times as heavy as H2. Calculate the degree of dissociation of PCl5.
A
α=19.367
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B
α=3.4684
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C
α=1.37
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D
α=0.6814
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Solution
The correct option is Dα=0.6814 PCl5⇌PCl3+Cl2 100 Moles before dissociation (1−α)ααMoles after dissociation
Given = 0.2 at 1 atm pressure KP=nPCl3×nCl2nPCl5×[P∑n]△n =α.α(1+α)[P1+α]=Pα21−α2=1×(0.2)21−(0.2)2 KP=0.0416atm Again when α is desired at 0.5, KP remains constant and thus, KSP=Pα21−α2 0.0416=P×(0.5)21−(0.5)2 ∴P=0.1248atm