`PCl_(5)` dissociation a closed container as :
`PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g))`
If total pressure at equilibrium of the reaction mixture is `P` and degree of dissociation of `PCl_(5)` is `alpha`, the partial pressure of `PCl_(3)` will be:

Options

A. `((x)/(x-1))P`
B. `((x)/(1-x))P`
C. `((x)/(x+1))P`
D. `((2x)/(1-x))P`

Answer

C

Explanation

`{:(,PCl_(5)(g),hArr,PCL_(3)(g),+,Cl_(2)(l)),("Initial moles",1,,0,,0),("Moles at eqm.",1-x" "x,,x,,):}`
Total number of moles `=1+x`
`PCl_(3)=("No. of moles of" PCl_(3))/("Total no. of moles")xx"Total pressure"`
`=((x)/(1+x))P`