Question

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.

Answer 1

For the supercooled water,

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

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

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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"`.