What is the entropy of vaporization for water if the enthalpy of vaporizing #"1 mol"# of water is #"40.66 kJ/mol"# at its normal boiling point?

1 Answer
Truong-Son N.
Oct 13, 2017

A two-phase equilibrium indicates a constant-pressure, constant-temperature process. Since the change in Gibbs' free energy is a function of pressure and temperature,

#DeltaG = 0 = DeltaH - TDeltaS#

at constant temperature and pressure. Therefore, we can write:

#DeltaH_"vap" = T_bDeltaS_"vap"#

where #T_b# is the boiling point in #"K"#. As a result,

#color(blue)(DeltaS_"vap") = (DeltaH_"vap")/T_b = "40.66 kJ/mol"/"373.15 K"#

#= "0.1090 kJ/mol"cdot"K"#

#= color(blue)("109.0 J/mol"cdot"K")#

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