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.
1 Answer
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
It should be slightly negative because this is freezing at a supercooled temperature, which is spontaneous.
Once you do convert to
#color(blue)(DeltabarS_"fus") = (DeltabarH_"fus" - DeltabarG_"fus")/T = color(blue)("22.22 J/mol"cdot"K")#
You should recognize that at