Gibbs free energy
At the freezing point, ice and water coexist. The tendency of the ice to melt is exactly counteracted by the tendency of water to freeze. The chemical potentials of the two phases are equal under these conditions:
µice = µpure waterwhere the chemical potentials are just the molar Gibbs free energies here.
What happens when you drop salt in the water? The dissolved salt will lower the chemical potential of the water without changing the chemical potential of the ice much at all. That means that water is more stable than ice now, and the ice will melt.
Why does water in a solution have a lower chemical potential than pure water? For an "ideal solution" (a solution in which all molecules interact in the same way) the chemical potential of water in the solution is given by
µsolution water = µpure water + RT ln xsolution waterwhere R is the gas law constant (8.314 J/mol K), T is the temperature, and xsolution water is the mole fraction of water in the solution. As salt is added to the water, the concentration of water in the solution goes down. That makes xsolution water less than one, and the natural log of a number less than one is negative. That makes the concentration correction negative, so the chemical potential of water will drop as more salt is added. Note that this has nothing to do with ordering of the water around the salt ions, because in an ideal solution salt-water interactions are assumed to be identical to water-water interactions. The decrease in chemical potential occurs because there is a lower concentration of water in the solution than in the pure liquid. Statistically, fewer water molecules escape a solution into the vapor phase or freeze out onto the solid phase. That's why salt lowers the chemical potential of water in the solution.
How can equilibrium be re-established by changing the temperature? To make the chemical potential of the water in the salt water equal to the chemical potential of the ice again, you'll have to lower the temperature. The freezing point depression equation T = - kf msolute can be derived from the defining equation for an ideal solution given above, without much trouble. Or you can just look at the graph at right, which shows how the chemical potential of the ice, pure water, and solution water change with temperature. Notice the following:
heat of fusion|
µice = µsolution waterAssume the solution is ideal. Then the defining equation for an ideal solution gives
µsolution water = µpure water + RT ln xsolution waterso the temperature T at the freezing point of the solution must satisfy
ln xsolution water = (µice - µpure water)/RTNow let's replace the chemical potentials with quantities that are easier to measure directly (or look up in the literature). Taking the derivative of both sides with respect to temperature at constant pressure,
d ln xsolution water/dT = -(µice - µpure water)/RT2 + ((µice/T)P + (µpure water/T)P)/RTSince µi = Hi - TSi, and (µi/T)p = -Si,
dln xsolution water/dT = - (Hice - Hpure water)/RT2 = Hfus/RT2where Hfus is the molar of fusion. Collecting all the temperature dependent terms on the right hand side and integrating from conditions in pure water to those in the solution,
|d ln x =||T
|ln xsolution water =||(||Hfus|
|- xsolute =||
|T = -||(||RT02|
Author: Fred Senese firstname.lastname@example.org
Copyright © 1997-2010 by Fred Senese
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Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101/solutions/faq/print-thermo-explanation-of-freezingpoint-depression.shtml