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

Options

A. `K_(p)=alpha^(2)/(1+alpha^(2)P)`
B. `K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`
C. `K_(p)=P^(2)/(1-alpha^(2))`
D. `K_(p)=(alpha^(2)P)/(1-alpha^(2))`

Answer

D

Explanation

`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",1,,0,,0),("Final",1-alpha,,alpha,,alpha),("Partial pressure"=,(1-alpha)/(1+alpha).P,,alpha/(1+alpha).P,,alpha/(1+alpha).P):}`
Total "mole" `=1-alpha+alpha+alpha=1+alpha`
`K_(p)=(alpha/(1+alpha)P.alpha/(1+alpha).P)/((1-alpha)/(1+alpha)P)=(alpha^(2)P)/(1-alpha^(2))`