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) Leftrightarrow PCl_(3) + Cl_(2)`
`{:(,"Initial",1,0,0),(,"Final",1-alpha,alpha,alpha):}`
Partial `=(1-alpha)/(1+alpha) P=(alpha)/(1+alpha)P=(alpha)/(1+alpha)P`
Pressure
Total moles `=1-alpha+alpha+alpha=1+alpha`