Enthalpy of vapourization of benzene is `+35.3 kJ mol^(-1)` at its boiling point of `80^(@)C`. The entropy change in the transition of the vapour to liquid at its boiling points [in `K^(-1) mol^(-1)`] is

Options

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

Answer

D

Explanation

`C_(6)H_(6)(l)toC_(6)H_(6)(g)` , `DeltaH=+35.3 kJ mol^(-1)`
Reverse reaction
`C_(6)H_(6)(g)toC_(6)H_(6)(l)`, `DeltaH=-35.3 kJ mol^(-1)`
`T=80+273=353 K`
`DeltaS=(DeltaH)/(T)=(-35.5xx1000)/(350)=-100 JK^(-1) mol^(-1)`