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^2P)`
B. `K_p=(alpha^2P^2)/(1-alpha^2)`
C. `K_p=(alphaP^2)/(1-alpha^2)`
D. `K_p=(alpha^2P)/(1-alpha^2)`

Answer

D

Explanation

`{:(,PCl_5 hArr, PCl_3 + , Cl_2),("Initial conc.", 1,0,0),("At equ.",1-alpha,alpha,alpha):}`
Total = `1-alpha +alpha + alpha =1+alpha`
`p(PCl_3)=alpha/(1+alpha).P`
`p(Cl_2)=alpha/(1+alpha).P`
`therefore K_p=(p(PCl_3)xxp(Cl_2))/(p(PCl_5))`
`=((alpha/(1+alpha).P)(alpha/(1+alpha).P))/((1-alpha)/(1-alpha).P)`
`=(alpha^2P)/(1-alpha^2)`