1.

Enthalpy of vaporization of benzene is + 35.3 kJ mol^(-1) at its boiling point, 80^(@)C. The entropy change in the transition of the vapour to liquid at its boilling point [in JK^(-1) mol^(-1)] is ……….

Answer»

`-441`
`-100`
`+441`
`+100`

Solution :ACCORDING to Gibb.s formula
`DeltaG=DeltaH-TDeltaS`
At EUILIBRIUM, `DeltaG=0`
And entropy decreases on going from vapour to liquid. i.e., `DeltaH=TDeltaS`
`DeltaS=-(DeltaH)/(T)=-(35300)/(80+273)=-100"JK"^(-1)"MOL"^(-1)`


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