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## How can freezing point depression be explained in terms of free energies?
**I wish I could relate solutes and freezing-point depression to the Gibbs Free energy equation. It seems that the process can be explained as an entropy-driven one, but in practical terms (ordering of the water around salt ions, etc.) I don't see how the delta-S for the ice to liquid transition should increase. Can you help me out or point me to a reference? Thanks.** Brett Skinner (Teacher) 7/12/99-
**Vocabulary**chemical potential Gibbs free energy ideal solution mole fraction
*(This is a bit above the general chemistry level. General chemistry students should look at Why does salt melt ice? for a more accessible explanation of freezing point depression.)*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: µ where the chemical potentials are just the molar Gibbs free energies here._{ice}= µ_{pure water}**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 byµ where R is the gas law constant (8.314 J/mol K), T is the temperature, and x_{solution water}= µ_{pure water}+ RT ln x_{solution water}_{solution 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 x_{solution 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 = - k_{f}m_{solute}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:- The chemical potential for all three substances goes up as temperature goes down. That's because the slope of the curve is equal to minus the substance's partial molar entropy (just the molar entropy, in the case of the pure ice and pure water). The entropies are positive, so the slopes of these curves must be negative.
- Liquid forms have higher molar entropies than solid forms of the same substance (usually), so the liquid curves are steeper. (If the curve for a gas were placed on this graph, do you think it would be steeper or shallower than the liquid curves?)
- The solution water curve is shifted below the curve for pure water because the RT ln x
_{solution water}term is less than zero. - When the ice curve intersects with a liquid curve, the chemical potentials are equal and the phases are in equilibrium. The intersections correspond to freezing points. On the graph, T
_{0}is the freezing point of the pure water. - The lower chemical potential of solution water shifts the point of intersection to the left, so the freezing point of the solution is lower than the freezing point of the pure solvent. T is the freezing point of the solution.
**Vocabulary**heat of fusion
### Appendix: Derivation of the freezing point depression equationThe condition for equilibrium is between the solution and the ice isµ Assume the solution is ideal. Then the defining equation for an ideal solution gives_{ice}= µ_{solution water}µ so the temperature T at the freezing point of the solution must satisfy_{solution water}= µ_{pure water}+ RT ln x_{solution water}ln x Now 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,_{solution water}= (µ_{ice}- µ_{pure water})/RTd ln x Since µ_{solution water}/dT = -(µ_{ice}- µ_{pure water})/RT^{2}+ ((µ_{ice}/T)_{P}+ (µ_{pure water}/T)_{P})/RT_{i}= H_{i}- TS_{i}, and (µ_{i}/T)_{p}= -S_{i},dln x where H_{solution water}/dT = - (H_{ice}- H_{pure water})/RT^{2}= H_{fus}/RT^{2}_{fus}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 T_{0}( H _{fus} RT^{2}) dT where Tln x _{solution water}=( H _{fus} R) ( T TT_{0}) _{0}is the freezing point of pure water. If we can assume that the solution is very dilute, the equation becomes much simpler. ln x_{solution water}equals ln (1 - x_{solute}), which is approximately equal to -x_{solute}in a dilute solution.In a dilute solution, the freezing point depression will be small, so T_{0}is approximately equal to T. Then or- x _{solute}=H _{fus} R( T T_{0}^{2}) Notice that the terms in parentheses depend on the solvent, not on the solute. That would suggest that equal concentrations of ions or molecules cause equal freezing point depressions, regardless of what the solute is. Experimental results agree with this prediction, so long as the concentrations are low.T = - ( RT _{0}^{2} H_{fus}) x _{solute}
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