Classical nucleation theory predicts that the growth of small nuclei is thermodynamically disfavoured, on account of the energy required to grow its surface. I am struggling to understand why it takes energy to grow a nuclei's surface.
I have done my best to expound my understanding of surface tension and classical nucleation theory below. My hope is that someone can identify a mistaken assumption or fault in my reasoning.
Consider a drop of water. Molecules on the surface of the droplet have fewer neighbours than molecules in the bulk of the liquid. This gives them a higher potential energy relative to their counterparts away from the surface.
We can assign a positive Gibbs free energy $G_s$ to the liquid's surface that is proportional to its surface area, with the constant of proportionality known as the surface tension. It therefore takes work to increase the drop's surface area through deformation.
However, it is important to note that water molecules on the surface still have less free energy compared to water in vapour outside of the drop. After all, a molecule on the surface has less potential energy due to cohesive forces with other molecules. $G_s$ is an opportunity cost: it's the free energy difference between the water droplet and the droplet if it were submerged entirely in water.
Classical nucleation theory says that growth of nuclei, such as water droplets in a supercooled vapour, is governed by two competing processes. The first is the free energy reduction from the vapour to liquid phase transition, which is proportional to the drop's volume. The second is the free energy increase from the formation of the drop's surface. The reasoning goes that, since the drop has surface tension, it must have a positive Gibbs free energy proportional to its surface area.
Since the surface area of a small drop grows faster than its volume, (homogeneous) nucleation is not thermodynamically favoured.
Where I'm Confused
My issue with the reasoning above is the claim that the formation of the drop's surface has a free energy cost. The preceding discussion on surface tension illustrates that the deformation of a drop of water with a fixed number of molecules increases its free energy. However, as far as I can tell, the addition of a molecule to its surface should still decrease the total free energy.
Phrased differently: I understand why it takes work to increase a drop's surface area through deformation. I don't understand why it takes work to increase a drop's surface area by adding molecules.