At some temperature and under a pressure of `4` atm, `PCl_(5)` is `10%` dissociated. Calculated the pressure at which `PCl_(5)` will be `20%` dissociated temperature remaining same.

Answer

D

Explanation

When `PCl_(5)` is `10%` dissociated
`alpha=0.1`
`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",1,,0,,0),("At equilibrium",-0.1,,0.1,,0.1),(,=0.9 "mol",,,,):}`
Total number of moles at equilibrium `=0.9+0.1+0.1`
`=1.1` "mol"
`p_(PCl_(5))=0.9/1.1xx4 "atm"=(0.9xx4)/1.1`
`p_(PCl_(3))=0.1/1.1xx4 "atm" =0.4/1.1`
`p_(Cl_(2))=0.1/1.1xx4 "atm"=0.4/1.1`
`K_(p)=(p_(PCl_(3))xxp_(Cl_(2)))/p_(PCl_(5))=(0.4/1.1xx0.4/1.1)/((0.9xx4)/1.1)=0.0404`
Second case: When `PCl_(5)` is `20%` dissociated:
Suppose total pressure =P atm, then
`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial",1 "mol",,0,,0),("At equilibrium",1-0.2,,0.2,,0.2),(,=0.8,,,,):}`
Total number of "moles" `=0.8+0.2+0.2=1.2 "mol"`
`p_(PCl_(3))=0.8/1.2xxP "atm"`
`p_(PCl_(3))=0.2/1.2xxP "atm"`
`p_(Cl_(2))=0.2/1.2xx P "atm"`
`K_(p)=((0.2P)/1.2xx(0.2P)/1.2)/((0.8 P)/1.2)=0.2/1.2xx0.2/0.8xxP`
`(=0.404 "calculate above")`
`P=0.97 "atm"`