Liquefying Silica

So much of the heavy lifting in the previous post about topographical heights involved calculating the liquefying energy needed for a silica molecule. I wanted to expand on that a bit in this note so I can provide myself closure about this topic that consumed me for a few hours.

All depends on the structure of the silica molecule. In the crystalline form, the SiO$_2$ lattice has each O atom bonded to two silicons and each Si atom bonded to four oxygens. For simplicity, we can consider this configuration to have zero entropy ($S=0$). In a liquid state, we can think of some fraction $(f_N)$ of the total molecules ($N$) being "disordered" by moving into holes, and then the entropy will be $S=k_B\ln{N\choose f_N}$. Another simplification ensues after we assume the liquid to be in the highest disordered state possible and the change in entropy per molecule will be $\Delta s\sim k_B\ln{2}$.

Figure 1. Typical subunit for low pressure silicon dioxide.

There are three steps in the process now, first we need to detach single SiO$_2$ molecules from the lattice, but to do that, we need to break bonds, or more accurately, overcome the binding energy of the bonds. Finally, we need to ask whether it is actually advantageous to completely detach every silica molecule from the lattice.

Each oxygen atom is bounded to one silicon atom in the molecule, but in the lattice, it is bounded to four silicon atoms. So we first need to break $3/4$ bonds. But each oxygen atom is part of two molecules, and so we're double counting. Therefore, we only need to break $3/8$ bonds.

Next, from high school physics, we remember that breaking a bond at the Bohr radius of $0.5$ Å is 13.6 eV. Silicon is on the second row of the periodic table and so we can assume a linear addition of an angstrom. And because two atoms form a bond, the binding energy of a single bond is $$E_{b}\sim Ry\cdot\frac{0.5\text{ Å}}{2\times1.5\text{ Å}}\sim2\text{ eV}.$$ But do we really want the binding energy? Breaking the lattice would mean converting the crystal to a gas, but we only want to achieve liquefaction. Assume that bond length changes by some $p$ % where $p<5$. Then we get the new binding energy $$E_{b}\sim Ry\cdot\frac{0.5\text{ Å}}{2\times1.5\text{ Å}}\cdot\left(\frac{0.0p}{1.0p}\right)\sim0.1\text{ eV}.$$