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I used to think that at the melting point, the solid and liquid phases are in equilibrium. That’s because the Gibbs free energy change for the transformation is 0, as is characteristic of equilibria.

However, I recently learnt about nucleation in phase transformations, and I’m very confused. My textbook tells me that below the melting temperature the Gibbs free energy change is not 0 and this energy can go into increasing nuclei surfaces and surmounting the nucleation activation energy barrier. It goes on to say that at the melting temperature, no such energy is liberated (since ∆G = 0) and the nucleation activation energy barrier cannot be achieved. Thus, no nucleation occurs at the melting temperature.

If no nucleation occurs, then how can the solid and liquid phases be in equilibrium? I know about supercooling and that the principle behind it is that it takes time for such an equilibrium to be achieved. However, if nucleation simply doesn’t occur at the melting temperature, then equilibrium will never be achieved...

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  • $\begingroup$ Nucleation is a kinetics issue, not a thermodynamics issue. Note that nucleation of the melt is very easy (it is very very hard to superheat a solid), while nucleation of the solid can be difficult (supercooling by 100s of K is not that difficult under the right conditions). But, again, that is a kinetics issue. Wait a very long time at the melting point, and it will eventually melt/solidify. $\endgroup$ – Jon Custer Nov 9 '16 at 17:00
  • $\begingroup$ @JonCuster Can you further explain the kinetics issue by addressing what I mentioned about activation energy? Where is the activation energy supplied from if there’s no Gibbs free energy released? $\endgroup$ – lightweaver Nov 9 '16 at 17:09
  • $\begingroup$ The activation energy required for, e.g. nucleation, has little to do with the release of Gibbs free energy. It is the energy of a transitional state required configurationally to get from A to B. Again, kinetics vs thermodynamics. $\endgroup$ – Jon Custer Nov 9 '16 at 18:12
  • $\begingroup$ @JonCuster I see. Going back to your first comment, my lecture notes tell me that at the melting temperature, nucleation rate is 0. Yet you say that “wait a very long time at the melting point and it will eventually melt/solidify.” How long are you talking about? In the limit $t\rightarrow \infty$? $\endgroup$ – lightweaver Nov 10 '16 at 13:20
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    $\begingroup$ @JonCuster Perhaps I’ve made myself unclear by referring to the “melting point” thus far, but I’m actually more interested in the freezing process. $\endgroup$ – lightweaver Nov 11 '16 at 16:19
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You probably know that water can be supercooled down to well below -30°C. And it has probably been suggested (if not outright stated) that at those temperatures, the water is in a "metastable" liquid state and that "eventually" it will spontaneously solidify - even if "eventually" may require trillions of years. Since thermodynamics is empirical, you could wonder how that has been empirically observed...but let's not go there. You've been taught that ΔG is a measure of the spontaneity of a process at constant temperature and pressure and at equilibrium. (How a "process" happens when the system is at equilibrium (steady-state) is often ignored. It's fair to say that nothing "happens" at equilibrium, rather when we say "equilibrium process" we mean "a process occurring sufficiently close to equilibrium". But I think I'm getting off topic, again. In considering nucleation, it is vital you distinguish between homogeneous and heterogeneous nucleation. This is the crux of the issue. Water is almost never without contaminates which provide nucleation sites, even if those sites are only on the surface of the container holding the liquid (getting the water this pure is an enormously difficult feat, and how you create a liquid phase without a container I leave to your imagination (hint:rain)). So, when reading about nucleation, you need to distinguish between the situations where there is preexisting centers (contaminants) or whether the water is assumed to be absolutely pure and centers must form spontaneously. Perhaps to answer your question (it would have been most useful if you had copy/pasted the exact text from your book). Once centers (seeds?) exist, the continued growth is pretty much identical whether their origin was homo- or hetero-. Obviously there is no equilibrium between a dust particle (solid) and liquid water. So how can nuclei exist at the MP? Obvious, right? They're hetero-, no formation is required. (Here I avoid the issue of whether FP is (or is not) identical to MP). Wikipedia has a short article on Classical Nucleation Theory which may help. Consider a solid. At its surface, the atoms (or molecules) are different than in the bulk (for most solids, bulk properties aren't reached for top 2 or 3 (or more) layers) but just consider an ideal crystal. Obviously, the surface has fewer bonds and hence has higher energy (is less stable). We call that surface tension (or exactly equivalently, surface energy). The classical theory states it's always positive. This is the energy which needs to be "overcome" to get nuclei, and it's proportional to the surface area. You should note here that our discussion has gone down to the microscopic level, where classical thermodynamic quantities become less and less useful. In other words, the macroscopic classical world thermodynamics needs to be treated carefully here. Statistical thermodynamics is most suitable transition from the classical to the quantum world. These types of processes can be studied using either, but fundamentally the world behaves according to the rules of (statistical) quantum mechanics when you're concerned with atoms depositing on or leaving a surface. The perspective of Classical Thermodynamics is of limited use. (In this case, I think it is probably just as useful, if not more useful, than the quantum mechanical approach. But you're unlikely to get many "fundamental" insights using it.) I'll point out one other apparent flaw in the "activation energy" approach: if the energy barrier is symmetric, then at its peak there should be no preferential direction: the odds of the system moving towards the solid or towards the liquid should be exactly the same (i.e. 50:50), yet the Classical Nucleation Theory requires a fudge factor for this probability - it is an indication that there is more going on here than this approach can handle, that the nucleation activation energy is, at best an approximation. Meaning don't take it too seriously; unless it can be derived from the quantum mechanics, it has no fundamental basis. I don't know much about this area of chemistry. I don't know whether or not it can be derived from the QM, but given that the Classical Nucleation Theory article in wikipedia is speaking about radius as if it is continuous, then I have my doubts. In some systems you begin to see bulk properties at N=100 to N=1000 (N=number of particles), so you'd have to be above this threshold for "classical" behavior to emerge.

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