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In the dissociation of PCl(5) as PCl(5...

In the dissociation of `PCl_(5)` as
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
If the degree of dissociation is `alpha` at equilibrium pressure `P`, then the equilibrium constant for the reaction is

A

`K_(p)=(alpha^(2))/(1+alpha^(2)P)`

B

`K_(p)=(alpha P^(2))/(1-alpha^(2))`

C

`K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`

D

`K_(p)=(alpha^(2)P)/(1-alpha^(2))`

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial moles",1,,-,,-),("Equilibrium moles :",1-alpha,,alpha,,alpha):}`
Total moles `=1- alpha + alpha + alpha = (1+alpha)`
Total pressure `=P`
`:. P_(PCl_(5))=((1-alpha)P)/((1+alpha))`
`P_(PCl_(3))=(alphaP)/((1+alpha)),P_(Cl_(2))=(alphaP)/((1+alpha))`
Now `K_(p)=(P_(Cl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))=(alpha^(2)P^(2))/((1+alpha)(1-alpha)P)`
`=(alpha^(2)P)/(1-alpha^(2))`
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