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Question

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 D α=0.6814
PCl5PCl3+Cl2
1 0 0 Moles before dissociation
(1α) α α Moles after dissociation
Given = 0.2 at 1 atm pressure
KP=nPCl3×nCl2nPCl5×[Pn]n
=α.α(1+α)[P1+α]=Pα21α2=1×(0.2)21(0.2)2
KP=0.0416atm
Again when α is desired at 0.5, KP remains constant and thus,
KSP=Pα21α2
0.0416=P×(0.5)21(0.5)2
P=0.1248atm

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