Question

The enthalpy change for transition of liquid water to steam is 40.8 kJ `mol^(-1)` at 373K. Calculate `Delta S` for the process.

Answer 1

The transition under consideration is:
`H_(2)O(l)toH_(2)O(g)`
we know that, `DeltaS_("vapour")=(DeltaH_("vapour"))/(T)`
given, `DeltaH_("vapour")=40.8kJ" "mol^(-1)`
`=40.8xx1000J" "mol^(-1)`
`T=373K`
Thus, `DeltaS_("vapour")=(40.8xx1000)/(373)=109.38J" "K^(-1)mol^(-1)`