# Nucleation and Phase Equilibrium at the Melting Temperature

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...

• 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. – Jon Custer Nov 9 '16 at 17:00
• @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? – lightweaver Nov 9 '16 at 17:09
• 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. – Jon Custer Nov 9 '16 at 18:12
• @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 $t\rightarrow \infty$? – lightweaver Nov 10 '16 at 13:20
• @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. – lightweaver Nov 11 '16 at 16:19