Calculate the change in Gibbs' free energy and entropy for the freezing of liquid water at -5º C, if the vapor pressure above the ice is 3.012 mmHg and that above the water is 3.163 mmHg? #DeltabarH_"fus" = "5.85 kJ/mol"# at this temperature.

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1 Answer
Apr 23, 2018

For the supercooled water,

#DeltabarG_"fus" = -"0.1091 kJ/mol"#

#DeltabarS_"fus" = "22.22 J/mol"cdot"K"#


Well, normally phase changes occur at constant temperature and pressure, but apparently we're given a nonzero change in pressure... which makes sense because we're freezing at supercooled temperatures, for which the freezing is spontaneous and not a phase equilibrium.

The change in molar Gibbs' free energy should be nearly zero (if not zero). At constant temperature, the molar Gibbs' free energy Maxwell Relation shows:

#dbarG = cancel(-SdT)^(0) + barVdP#

As an ideal gas, the water vapor's molar volume would be given by

#barV = (RT)/P#

And so,

#dbarG = (RT)/PdP#

Integrating, we get the change in molar Gibbs' free energy to be:

#int dbarG = DeltabarG = RTln(P_2/P_1)#

where #P# is the vapor pressure at this temperature.

The result is:

#DeltabarG_"fus" = "0.082057 L"cdot"atm/mol"cdot"K" cdot "268.15 K" cdot ln("3.012 mm Hg"/"3.163 mm Hg")#

#= -"1.076 L"cdot"atm/mol"#

You should then take it upon yourself to convert to #"kJ/mol"#, and show that you get #color(blue)(-"0.1091 kJ/mol")#.

It should be slightly negative because this is freezing at a supercooled temperature, which is spontaneous.

Once you do convert to #"kJ/mol"#, show that since #DeltaG = DeltaH - TDeltaS# at constant temperature,

#color(blue)(DeltabarS_"fus") = (DeltabarH_"fus" - DeltabarG_"fus")/T = color(blue)("22.22 J/mol"cdot"K")#

You should recognize that at #0^@ "C"#, #DeltabarS_"fus"# would be #"21.82 J/mol"cdot"K"#.