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

B

Explanation

According to Gibb.s formula, `Delta G = Delta H- T Delta S`
At equilibrium, `Delta G=0` And entropy decreases on going from vapour to liquid. i.e., `Delta H= T Delta S`
`Delta S= - (Delta H)/(T)= - (35300)/(80+273) = -100JK^(-1) mol^(-1)`