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1mol of Cl(2) and 3 mol of PCl(5) are pl...

`1`mol of `Cl_(2)` and `3` mol of `PCl_(5)` are placed in a `100 L` vessel heated to `227^(@)C`. The equilibrium pressure is `2.05` atm. Assuming ideal behaviour, calculate the degree of dissociation for `PCl_(5)` and `K_(p)` for the reaction.
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`

Text Solution

Verified by Experts

`{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("moles at t=0",3,,0,,1),("moles at equilibrium",(3-x),,x,,x):}`
`:.` Total moles present at equilibrium `=4+x`
Given, total pressure at equilibrium `=2.05`
Now `PV=nRT` at equilibrium
`2.05xx100=(4+x)xx0.0821xx500`
`:. x=0.9939`
Now the degree of dissociation for
`PCl_(5)=("moles dissociated")/("Total moles")`
`=0.9939/3=0.313` or `33.13%`
`:. K_(p)=[(n_(PCl_(3))xxn_(Cl_(2)))/n_(PCl_(5))]xx[P/(Sigman)]^(+1)`
`=x^(2)/((3-x))xx[2.05/(4+x)]`
`=((0.9939)^(2))/((3-0.9939))xx[2.05/(4+0.9939)]=0.20`
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