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 ……….

Options

A. `-441`
B. `-100`
C. `+441`
D. `+100`

Answer

B

Explanation

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)`