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. `-100`
B. `+100`
C. `+342`
D. `-342`

Answer

A

Explanation

Enthalpy of vaporisation of benzene
`=DeltaH_("vaporisation")=+35.3xx10^(3)J//mol`
Boiling point=`T_(b).pt`.=273+80=353 K
`therefore DeltaS_(vap).=(DeltaH_(vap))/(T_(b.pt))=(35.3xx10^(3))/(353)=100JK^(-1)mol^(-1)`
`therefore` Entropy change in the transition of vapour to liquid= `DeltaS_(cond.)=-DeltaS_(vap.)=-100JK^(-1)mol^(-1)`