For the process `H_(2)O(l)hArrH_(2)O(g),DeltaH=40.8kJ*mol^(-1)` at the boiling poinit of water. Calculate molar entropy change for vaporisation from liquid phase.

Explanation

Molar entropy change for vaporisation `DeltaS=(DeltaH_("vap"))/(T_(b))`
Given `DeltaH_("vap")=40.8kJ*mol^(-1),` for water `T_(b)=373K`,
`therefore DeltaS=(40.8xx10^(3))/(373)=109.38J*K^(-1)*mol^(-1)`
`therefore `Molar entropy change for vaporisation of water
`=109.38J*K^(-1)*mol^(-1)`