Why does freezing point depression necessarily occur when adding a nonvolatile solute to a solvent?

We can consider the following equation for the chemical potential of a solution (closely related to the deviation of #DeltaG# from standard conditions) relative to a pure solvent standard state:

#mu_j = mu_j^"*" + ln chi_(j(l))#

where #mu_j# is the "chemical potential" of the solvent, #"*"# means pure, and #chi_(j(l))# is the mole fraction of the solvent in the solution phase.

[You may see this commonly being used in the context of liquids, but you CAN dope solids (such as semiconductors), which does create impurities in them and follows a similar principle.]

Melting point depression is the same as freezing point depression.

If you consider the following graph, we can explain this:

For starters, notice how the mole fraction #chi# (for any scenario) must be between #0# and #1#.

Since it starts at #1# for #100%# purity, it can only decrease. So, adding an impurity makes #chi_(j(l)) < 1#.

After adding an impurity, when #chi_(j(l)) < 1#, #ln chi_(j(l)) < 0#. Therefore,

#mu_j < mu_j^"*"#.

On the above graph, we have to consider the ratio #(Deltamu)/(DeltaT)#. That gives the slope of those diagonal lines corresponding to the solid, liquid, and gas.

To a first approximation, the slope remains essentially the same within the same condensed phase because it actually corresponds to the negative molar entropy of the phase:

#((delmu_j)/(delT))_(V, N_(i ne j)) = -barS = -S/n#,

and a solid has a pretty much constant entropy.

Now, if you specifically look at a liquid, when #mu_j# decreases, its diagonal line moves down.

In relative terms, you could then say that

• the solid's line moves up.
• the intersection IS the freezing point, and it moved left.

That means the freezing point was lowered.

(You could go through the same reasoning and determine that the boiling point actually increases when you dissolve something in a liquid.)