For the vaporisation of water : `H_(2)O (I) hArr H_(2)O(g)` [1 atm. Pressure]
Given : `Delta S = 120 JK^(-1)` and `Delta H =+45.0 kJ`.
Evaluate the temperature at which liquid water and water vapour are in equilibrium at 1 atm. Pressure-

Explanation

We know that : at equilibrium `Delta G = 0`
`because " " Delta G = Delta H - T Delta S " " therefore " " T Delta S = Delta H rArr T=(Delta H)/(Delta S)=(45.0xx10^(3)J)/(120 JK^(-1))=375 K`