Nucleation and growth: a new phase must first beat a surface-energy barrier

A new phase starts as a tiny seed, and tiny seeds are fragile

Intuition

Think of a soap bubble. Making a bubble costs effort because you have to create a skin. A big bubble is worth it, but a tiny one collapses straight back. A new crystal inside a cooling liquid is the same: the inside wants to be solid, but the new skin around it costs energy. Only a seed that gets big enough survives.

So far Lecture 23 and earlier looked at the thermodynamics of phase changes. Lecture 24 says many phase transitions are kinetically controlled. Nucleation and growth is the main mechanism of phase transformation in materials. Everyday cases on the slide: clouds and cloud seeding, bubbles forming on a finger in a glass of water, and a Diet Coke bottle foaming.

• Heterogeneous nucleation: the new phase appears on container walls, impurity particles and similar.
• Homogeneous nucleation: solid nuclei appear spontaneously inside the supercooled liquid.

The energy of a nucleus is a gain from the inside and a cost from the skin

Take a solid nucleus of volume $V$ and surface area $A$ inside a supercooled liquid. Going from all liquid to liquid plus a nucleus changes the Gibbs energy by

$\Delta G = G_f - G_i = -V\left(G_V^{L} - G_V^{S}\right) + A\gamma = -V\,\Delta G_V + A\gamma$

The first term is negative below $T_m$: the solid is more stable, so growing the inside lowers the energy. The second is the interfacial energy $\gamma$ times the area. It is always positive.

For a sphere of radius $r$, $V = \tfrac{4}{3}\pi r^3$ and $A = 4\pi r^2$:

$\Delta G = -\frac{4}{3}\pi r^3\,\Delta G_V + 4\pi r^2\gamma$

At small $r$ the $r^2$ term wins and $\Delta G$ rises. At large $r$ the $r^3$ term wins and $\Delta G$ falls. So there is a peak at a critical radius $r^*$, with height $\Delta G^*$, the barrier to nucleation.

The sketch is schematic and follows slide 10: the interfacial term rises as $r^2$, the volume term falls as $-r^3$, and their sum (heavy line) has a maximum $\Delta G^*$ at $r^*$.

Try it: worked example

Find the critical radius r* and the barrier DeltaG* from DeltaG = -(4/3) pi r^3 DeltaG_V + 4 pi r^2 gamma.

Show answer and steps

Answerr* = 2 gamma / DeltaG_V and DeltaG* = 16 pi gamma^3 / (3 DeltaG_V^2).

1. The top of the hump is where the slope is zero, so differentiate with respect to $r$.
2. The derivative of $-\tfrac{4}{3}\pi r^3\Delta G_V$ is $-4\pi r^2\Delta G_V$. The derivative of $4\pi r^2\gamma$ is $8\pi r\gamma$.
3. Set the sum to zero: $-4\pi r^2\Delta G_V + 8\pi r\gamma = 0$.
4. Divide by $4\pi r$: $-r\,\Delta G_V + 2\gamma = 0$, so $r^* = \dfrac{2\gamma}{\Delta G_V}$.
5. For the height, put $r^*$ back into $\Delta G$. Use $\Delta G_V = 2\gamma/r^*$ in the first term: $-\tfrac{4}{3}\pi r^{*3}\cdot\dfrac{2\gamma}{r^*} = -\tfrac{8}{3}\pi r^{*2}\gamma$.
6. Add the second term $4\pi r^{*2}\gamma$: $\Delta G^* = \left(4 - \tfrac{8}{3}\right)\pi r^{*2}\gamma = \tfrac{4}{3}\pi r^{*2}\gamma$.
7. Put in $r^* = 2\gamma/\Delta G_V$, so $r^{*2} = 4\gamma^2/\Delta G_V^2$: $\Delta G^* = \tfrac{4}{3}\pi\gamma\cdot\dfrac{4\gamma^2}{\Delta G_V^2} = \dfrac{16\pi\gamma^3}{3\,\Delta G_V^2}$.

Nuclei smaller than $r^*$ shrink. Nuclei larger than $r^*$ grow. This is behind Ostwald ripening (coarsening): over time small particles dissolve and large ones grow.

Cooling deeper shrinks both the critical size and the barrier

How big is the driving force? $\Delta G_V = G_V^{L} - G_V^{S}$. At the melting temperature $T_m$ liquid and solid are in equilibrium, so

$\Delta G_V = \Delta H_V^{m} - T_m\,\Delta S_V^{m} = 0 \quad\Longrightarrow\quad \Delta S_V^{m} = \frac{\Delta H_V^{m}}{T_m}$

At a lower temperature $T^*$, assume $H$ and $S$ do not change with $T$ for a small undercooling $\Delta T = T_m - T^*$:

$\Delta G_V \approx \Delta H_V^{m} - T^*\frac{\Delta H_V^{m}}{T_m} = \frac{\Delta H_V^{m}\,\Delta T}{T_m}$

Put this into $r^*$ and $\Delta G^*$:

$r^* = \frac{2\gamma T_m}{\Delta H_V^{m}}\,\frac{1}{\Delta T} \qquad \Delta G^* = \frac{16\pi\gamma^3 T_m^2}{3\left(\Delta H_V^{m}\right)^2}\,\frac{1}{\Delta T^2}$

Exam detail

Both the nucleation radius and the barrier to nucleation decrease with increasing undercooling: $r^* \propto 1/\Delta T$ and $\Delta G^* \propto 1/\Delta T^2$.

The nucleation rate peaks at a sweet spot, and growth depends on diffusion

The rate of fluctuations big enough to cross the barrier $\Delta G^*$ follows the Arrhenius form.

$\dot{N} \sim v_d \exp\left(-\frac{\Delta G^*}{k_B T}\right) \sim \exp\left(-\frac{Q_d}{k_B T}\right)\exp\left(-\frac{A}{\Delta T^2}\right)$

$v_d$ is the attempt frequency for crossing the barrier. The first exponential is the probability of overcoming the barrier. $\exp(-Q_d/k_BT)$ is assumed to have the same kinetics as diffusion (atomic mobility), and $\exp(-A/\Delta T^2)$ is the $\Delta G^*$ term copied from the last slide.

Two things pull in opposite directions. Near $T_m$, $\Delta G^*$ is too high, so almost nothing nucleates. At low $T$, atomic mobility is too small, so atoms cannot rearrange. The product peaks in between, at the sweet spot (slide 15).

Growth comes after nucleation. It is generally diffusion controlled, because it needs long-range rearrangement of atoms, so it slows down as temperature falls. Diffusion-controlled growth often gives dendrites, because supersaturation is highest at the tips of the extremities. Snowflakes are the example.

The cooling rate decides the texture, from coarse crystals to glass

Slide 7 overlays the nucleation rate and the growth rate against temperature, with markers $T_a$ (small undercooling), $T_b$ and $T_c$ (deepest).

• Slow cooling: minor undercooling ($T_a$), rapid growth and slow nucleation, so fewer, coarse crystals.
• Rapid cooling: more undercooling ($T_b$), slower growth and rapid nucleation, so many fine crystals.
• Very rapid cooling: little if any nucleation or growth ($T_c$), producing a glass.

Examples from the lecture. Virgin coconut oil melts around 24 °C. Room temperature is just below that, so nucleation can be very slow and large grains can form. Igneous rocks are classified by texture (crystal size), set by how fast magma cools. Fine-grained rocks (rhyolite, basalt) form in volcanic eruptions with strong undercooling. Coarse-grained rocks (granite, gabbro) form when magma cools slowly underground. The same magma can give different rocks. The Nakaya diagram on slide 21 shows snowflake shapes depending on temperature and supersaturation: its boundaries are read from a plot, so check the slide before quoting them.

Spinodal decomposition is another mechanism. It occurs in mixtures where there is no barrier to nucleation, and its dynamics are driven solely by diffusion. At high $T$ a water-oil mixture is a single phase. Cool it suddenly to where oil-rich and water-rich phases coexist, and the molecules begin to cluster. The clusters grow and merge into large phases.

Compare

Nucleation and growth

There is a barrier. A seed must pass $r^*$ before it grows. The seeds appear at separate places.

Spinodal decomposition

No barrier. The whole mixture starts to separate at once, driven by diffusion alone.

Pitfall

Do not say more undercooling always means faster transformation. Deeper cooling lowers the barrier $\Delta G^*$, but also lowers atomic mobility. The nucleation rate peaks at a sweet spot and falls again at very low $T$.

Recall

Why do a fast cooled and a slowly cooled melt end up with different crystal sizes?

Fast cooling allows deep undercooling, where nucleation is fast and growth is slow, so there are many small crystals. Slow cooling stays near $T_m$, where nucleation is slow and growth is fast, so there are few large ones.