`n` mole of `PCl_(3)` and `n` mole of `Cl_(2)` are allowed to react at constant temperature `T` to have a total equilibrium pressure `P` , as `:`
`PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g)`
If `y` mole of `PCl_(5)` are formed at equilibrium , find `K_(p)` for the given reaction .

Options

A. `((2n-y)y)/((n-y)^(2)P)`
B. `(y)/((n-y)^(2)(2n-y)P)`
C. `((n-y)^(2).P)/((2n-y)y)`
D. `((n-y)^(2)(2n-y)P)/(y)`

Answer

1

Explanation

`{:(,PCl_(3),+,Cl_(2),hArr,PCl_(5)),(t=0,n,,n,,0),(t=teq.,n-y,,n-y,,y):}`
`K_(p)=(y)/((n-y)(n-y))[(P)/(2n-y)]^(-1)" "rArr" "K_(p)=((2n-y)y)/((n-y)^(2)P)`