Why does it take energy to grow the surface of a drop?

Question

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.

Surface Tension

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.

Nucleation Theory

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.

Answer 1

(This is a model of a well-constructed question–my compliments.)

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.

This is true but incomplete. Recall that phase transitions occur because of an interplay between enthalpy $H$ and a temperature-mediated entropy term $TS$. Nature (contradictorily) prefers strong bonding but also many possibilities, and the temperature controls which aspect wins out (with the lower-entropy phase always seen at the lower temperature at equilibrium.)

The full driving force for a phase change is $V\Delta G$, with volume $V$ and free energy change $\Delta G$. Note that this term incorporates not only the additional drop volume but also the latent heat $L$ and undercooling/overcooling $\Delta T$ past the equilibrium phase transition temperature $T_\mathrm{t}$:

$$V\Delta G=V(\Delta H-T\Delta S)=VL-T\frac{L}{T_\mathrm{t}}=VL\frac{\Delta T}{T_\mathrm{t}},$$

where I've ignored any difference in the specific heat between the two phases (equivalent to taking only the first term of a Taylor-series expansion around $T_\mathrm{t}$ to linearize the response) and I've used the equality $\Delta G=\Delta H-T_\mathrm{t}\Delta S=0$ at $T_\mathrm{t}$ to obtain $\Delta S=\frac{\Delta H}{T_\mathrm{t}}=\frac{L}{T_\mathrm{t}}.$

So the energy benefit from a phase change is low at low temperature excursions $\Delta T$ past $T_\mathrm{t}$. However, the energy penalty from forming additional surface area $A$ remains constant: $\gamma A$, with interface energy $\gamma$. This is why the latter can be larger than the former. It's only when the sum of the two start to decrease that nucleation is widely expected (as discussed in this answer).

Therefore, I don't agree with this part of the question:

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.

The molecules can obtain a lower enthalpy from attaching to neighbors, but the loss in entropy from being constrained may make the $-T\Delta S$ term positive enough that the free energy is larger than that in the vapor. In this case, molecules that happen to collide to form a nucleate-like cluster (with a surface) will tend to disperse again as relatively unstable. The undercooling simply isn't enough for them to give up the high-entropy possibilities of free translation in the gas phase.

Does this help clear up the confusion?