1 mole of `PCl_(5)` taken at 5 atm, dissociates into `PCl_(3)` and `Cl_(2)` to the extent of 50%
`PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)` Thus `K_(p)` is `:`

Options

A. 2.5
B. 1.67
C. 0.5
D. 2

Answer

A

Explanation

`{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Initial",1"mol",,,,),("At eqm.",1-0.5,,0.5,,0.5):}`
Total no. of moles at eqm. `=1.5`
Moles of ,
`PCl_(5)=0.5`
`PCl_(3)=0.5`
`Cl_(2)=0.5`
Initially `PV=nRT`
`5xxV=1xxRT`
`V=(RT)/(5)`
At eqm. `PV=1.5 RT`
`(PRT)/(5)=1.5RT`
`P=7.5 "atm"`
`:. p_(PCl_(5))=(0.5)/(1.5)xx7.5=2.5"atm"`
`p_(PCl_(3))=2.5 "atm"`
`p_(Cl_(2))=2.5"atm"`
`K_(p)=(p_(PCl_(3))xxp_(Cl_(2)))/(p_(PCl_(5)))=(2.5xx2.5)/(2.5)=2.5"atm"`