Birthday Borax and An Explanation of Crystal Nucleation

For my birthday a while back, a friend of mine gave me food coloring, a box of borax and spools of thread. I asked what they were for, and she said crystals.

Fig. 1: Borax crystal growing birthday gift (orig.)

Of course! I hadn’t ever seen people grow borax crystals before, only sugar or salt, so I looked up a tutorial on YouTube. The procedure was pretty standard: heat water up to near boiling, dissolve the borax, insert a pipe cleaner or thread and wait overnight for the solution to cool down and precipitate out crystals. But while the procedure is simple, the science behind nucleation is more complicated and pretty interesting.

In order for a crystal to form in solution, molecules of a substance must conglomerate to an adequate size to encourage the spontaneous coordination of more molecules. A cluster of this adequate size is called a "nucleus," and the process of its formation is called "nucleation." Clusters that are not large enough to be nuclei are called "embryos." The nucleation process can be described in terms of Gibbs free energy, which accounts for both enthalpy, related to the nucleus’ internal energy, and entropy. Gibbs free energy is given by the equation

                      1.       ΔG=ΔH-TΔS (H is enthalpy, T is temperature in Kelvins and S is entropy)

For a newborn crystal born homogenously (meaning suspended in a medium without contact with other surfaces), there are two main energy changes that are occurring. The first is a lowering of molecular energy due to the formation of attractions between coordinating substance molecules for reasons described in Why Cold Drinks "Sweat". This change in energy can be describe by the equation

                  2.       ΔG=VΔG_v (V is volume, ΔG_v is the change in internal energy per unit volume)

To find ΔG_v, we will assume that the crystal-solution system is cooled to slightly below the substance's melting point, T_m. At this small undercooling, ΔH and ΔS can be approximated as temperature independent [1]. But first, at T_m the difference in free energy ΔG between a substance’s solid and liquid forms is 0. Therefore,

                                                                      3.       ΔG_m=ΔH-T_mΔS=0

                                                                      4.       ΔS_m= ΔH/T_m

Now, assuming that ΔH and ΔS are temperature independent, we can use the same expression for entropy as used at T_m to find the value of ΔG_v. While we’re at it, let’s assume the crystal nucleus is spherical for simplicity’s sake. The following are therefore true:

          5.       ΔG_v=ΔH_fv-TΔS= ΔH_fv-T(ΔH_fv/T_m)= ΔH_fv(T_m-T)/ T_m= ΔH_fvΔT/T_m (from equations 1 and 4)

          6.       ΔG=VΔG_v=4/3πr³(ΔH_fvΔT/T_m) (from equations 2 and 5, V_sphere=4/3πr³)

The second energy change occurring is an increase in a newborn crystal’s energy because the surface of the crystal is disrupting bonding in the liquid medium around it. This change can be described as

          7.       ΔG=Aγ_s=4πr²γ_s (A is the nucleus’ surface area, γ_s is the surface free energy value
           characteristic of the medium-solid interface)

This expression is just the disruption free energy per unit area multiplied by the surface area of a sphere. Together, these two energy changes dictate nucleation. Putting the two expressions together, we get

          8.       ΔG_hom=VΔG_v+Aγ_s= 4/3πr³ΔH_fvΔT/T_m+4πr²γ_s (ΔG_hom indicates that this equation is for
           homogenous nucleation)

This equation is very useful and can be used to describe nucleation events from borax crystals precipitating to homogenous cloud formation. To tease a little more information out of equation 8, let’s see how ΔG_hom changes in response to a substance clump slowly growing in radius. This rate of change corresponds to the derivative of ΔG_hom with respect to radius r [2];

                                                             9.       d(ΔG_hom)/dr=4πr²ΔG_v+8πrγ_s

The point where ΔG_hom no longer changes with r corresponds to the peak on the following graph of equation 8 and its component parts, equations 2 and 7.

Fig. 2: Free energy of nucleation as radius increases (Materials Science and Engineering, an Introduction)

Equation 2 is shown as a decreasing curve because ΔH_f is negative for crystallization. This makes sense because as a material crystallizes, energy is lost to its surroundings, so ΔH_f must flow out of the system. To find the critical nucleation radius,

                                           10.   d(ΔG_hom)/dr=4πr²ΔG_v+8πrγ_s=0 (from equation 9)

                                           11.   r^*=2γ_s/ΔG_v

Plugging the critical radius into equation 8 and extracting the negative from ΔH_f to avoid confusion, we get

                                  12.   ΔG^*_hom=-4/3π(2γ_s/ΔG_v)³ΔG_v+4π(2γ_s/ΔG_v)²γ_s=16/3πγ_s³/ΔG_v²

This expression describes the magnitude of the energy change needed to get a molecular cluster up to the size of the critical radius r^*, a sort of activation energy [3].

However, homogenous nucleation is relatively difficult compared to heterogenous nucleation, as I’m sure you’ve probably heard. Heterogenous nucleation occurs when a nucleus forms on a surface, such as a dust particle for clouds or thread for nucleating borax crystals. Because of the complexity of finding heterogenous nucleation equations due to complex volume and surface area terms as well as extra surface energy considerations, I will not be posting these calculations. The calculations can be found in reference 3, but the summarized result is that the critical radius in heterogenous nucleation remains the same as with homogenous nucleation while the nucleation free energy lowers and makes the nucleation process more accessible. This is why minimizing dust or rough surfaces is important for growing larger crystals rather than clusters of small ones.

What has been described above is just a small bit of the complexity of crystal growing. Knowledge of nucleation rates and crystal growth is widely used in processes such as tuning metals to have properties fit for specific purposes or growing single-crystal silicon for computer processors, and I’m sure this information will come up again on later posts. But mild difficulty of math aside, it’s cool stuff, huh? As a treat for reading through this post, watch this fun Minute Earth video on the homogenous nucleation of clouds and see if you recognize some of the concepts we discussed.

I’m planning on growing some borax crystals soon, and when I do I’ll likely write an experiment post about it so be sure to come back and check that out. I’ve already had my first few days of classes and so far it seems that I will be able to continue posting once a week, likely on Sunday or Monday. As always, thanks for reading and I’ll be posting new stuff soon!