By a remarkable coincidence, I just learned what an azeotrope is in the last day or so.
By way of background, let's look at a solution of two types of molecules (A and B) where each component has a concentration-independent enthalpy of vaporization, and also that the entropy just depends on the concentration in the simple ideal way. That means that A is equally well stuck in a solution of A molecules or of B's. Then at some T, the vapor pressures of A and B, call them vp(A) and vp(B), just scale with their concentrations in the liquid, [A] and [B]. Unless those enthalpies happen to be equal, the vapor pressures are unequal and so boiling off a fraction of the liquid puts more of, say B, into the vapor, leaving more of A behind in the liquid. Repeating the process on the condensed vapor concentrates B some more.
Now let's say that the enthalpies do depend on the concentrations. (For simplicity, let's still say the entropies depend on concentration just like ideal solutes.) Let's look at the case where A and B stick better to each other than each to themselves. Say A is the one with the higher vapor pressure in the 50% solution. At low [A], A is surrounded by B and thus is really stuck, so it tends not to evaporate. So vp(A) fell faster than [A]. As you reduce [A] , vp(A)/[A] falls. Meanwhile vp(B)/[B] goes up because now B is surrounded mostly by B's, which it doesn't stick to as well as it sticks to A's. If the effect is big enough, the two ratios can cross, so that vp(A)/[A] = vp(B)/[B] at that particular [A]/[B]. Then the evaporated stuff has the same A/B ratio as the liquid left behind.
The case I've described above is called a "negative azeotrope". I believe that the key to it is a negative enthalpy of mixing. More generally, it would mean negative excess free-energy of mixing beyond the simple entropy of mixing term.
Wikipedia has a nice article on azeotropes, http://en.wikipedia.org/wiki/Azeotrope
, which also describes positive azeotropes. It looks like they are characterized by positive excess free-energy of mixing, i.e. the two components don't stick well to each other. Then in our example above (A with higher vapor pressure) when the solution is mostly A, B isn't well stuck and vp(B)/[B] can go up enough to match vp(A)/[A].
You need a big enough excess free-energy of mixing (positive or negative) to get an azeotrope. If it's too small, then the component with the higher vapor pressure in the pure form will always form a higher proportion of the vapor than of the liquid.
(published on 07/21/2011)