Determining melting point with equilibrium

Question

A well known problem in thermodynamics is the determination of boiling points from the Van’t Hoff equation using the liquid-vapor phase equilibrium. When taking the temperature dependence of the enthalpy of vaporization and entropy of vaporization into account, it is possible to quite accurately predict the boiling temperature as the point at which the liquid’s vapor pressure matches the ambient pressure (assuming the system is open so that the external pressure is fixed, of course).

My question is whether a similar procedure can be done for the solid-liquid phase transition. Unfortunately, the most relevant equilibrium expression would be quite useless since there are just two condensed phases on either side of the reaction. My thought is that the solid and liquid phases each have a characteristic vapor pressure that changes with temperature, and at the point of the phase transition the two vapor pressures must be equal. This is because of Le Chatelier’s principle that if the vapor pressure over the solid exceeded the vapor pressure the liquid phase would have at that temperature, the equilibrium would be shifted such that the liquid would form. I did some quick and dirty calculations using thermochemical data for water, and this method got the melting point wrong by about 7 K without taking the temperature dependence of enthalpy and entropy into account, which seems like a plausible amount of error. Is this line of reasoning correct/is there a simpler line of reasoning that would allow one to calculate the melting transition temperature?

EDIT: Some details on my rough calculations. The chemical reactions that were considered were simply, \begin{align} &\text{H$_2$O$_{(s)}$} \leftrightarrow \text{H$_2$O$_{(g)}$} &&K_{sub} = P_{\text{H$_2$O}} \\ &\text{H$_2$O$_{(l)}$} \leftrightarrow \text{H$_2$O$_{(g)}$} &&K_{vap} = P_{\text{H$_2$O}} \end{align} Thus, as the temperature varies, we can use the definition of the equilibrium constant or the Van't Hoff equation to write, \begin{align} P_{sub} &= e^{-\frac{\Delta G_{sub}^\circ}{RT}} \\ P_{vap} &= e^{-\frac{\Delta G_{vap}^\circ}{RT}} \end{align} where the relevant $\Delta G$ value is obtained from literature values of the enthalpy and entropy for the two phases (i.e. the CRC Handbook). By approximating the enthalpy and entropy for the phase transition as fixed (a somewhat poor approximation), the calculated temperature at which the vapor pressures of the solid and liquid forms would be equal is at $T = 280$ K, which is only 7 K above the proper melting temperature of 273 K under a standard pressure of $P^\circ = 1$ atm. Since the vapor pressure of the solid grows faster than that of the liquid past this temperature, Le Chatelier's principle dictates that the equilibrium would shift for the liquid equilibrium to generate the liquid species until we ran out of the solid, which is exactly what we observe at a melting transition.

Answer 1

Your pressure argument is not the whole picture. The general argument is that phase coexistence happens whenever the chemical potential of each phases are equal. This comes from studying the Gibbs potential $G$ (free enthalpy). Assuming you are imposing pressure and temperature, this is the relevant potential, equilibrium being characterised by being a minimum of $G$.

Say you have two phases, $1,2$, using the definition of the chemical potential: $$ \mu_i = \frac{\partial G}{\partial n_i} $$ the free enthalpy change of transitioning $dn$ elements from phase $1$ to $2$ is: $$ dG = (\mu_2-\mu_1)dn $$ Thus, if $\mu_2<\mu_1$ and $dn>0$, $dG<0$ so we are closer to equilibrium and conversely. In general, the mixture will be dominated by the phase of lowest chemical potential. The exception being when both are equal, in which case to have phase coexistence, i.e. a transition line in the $p,T$ diagram. Equivalently, at fixed $p$, gives you the melting point.

Usually, you define vapour pressure as the pressure corresponding to the given temperature in the coexistence line with vapour. It therefore comes from the definition that you can find the boiling point from it. If I understand correctly, you are looking at the solid-gas line which gives a corresponding "solid" vapour pressure. However, they have little to do with the melting point, which corresponds to the solid liquid line. Plus, you cannot equate the two since for example in the case of water they have distinct domains of definition. The liquid water vapour is defined for temperatures higher than the triple point and the solid vapour pressure is defined for temperatures below the triple point. Actually, it looks like you are rather describing a way of calculating the triple point. I also don't see how you impose the ambient pressure, the melting point depends on it after all. Perhaps you could detail your calculations a bit more.

The analogous reasoning would be to look at the liquid-solid curve which gives a liquid-solid pressure as a function of temperature. The melting point would be where the ambient pressure matches this liquid-solid pressure.

Now it all depends on what data you have at your disposition. If you have the liquid-solid pressure, you can call it a day. Otherwise, if you have access to entropies and enthalpies, you can calculate the difference in chemical potential and find the melting point.

Hope this helps.

Answer 2

It turns out that the approach outlined above is a classic way of quantitatively predicting the solid-liquid phase transition. It is explained in Zumdahl’s chemistry textbook explicitly.