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)=(alphaP^(2))/(1-alpha^(2))`
D. `K_(p)=(a^(2)P)/(1-alpha^(2))`

Answer

D

Explanation

`underset(1-alpha)underset(1)(PCl_(5(g)))hArrunderset(alpha)underset(0)(PCl_(3(g)))+underset(alpha)underset(0)(Cl_(2(g)))`
`K_(p)=(n_(PCl_(3))xxn_(Cl_(2)))/(n_(PCl_(5)))xx[(P)/(sum n)]^(1)`
`=(alpha^(2))/((1-alpha))*(P)/((1+alpha))=(alpha^(2)P)/((1-alpha^(2)))`