Convex hull and compositional phase diagram
This note is originally planned for the condensed coffee on Sept. 29, 2017 in UNLV, but is subjected to undergo a few revisions. If you have any questions, please contact Qiang Zhu at [email protected].
From time to time, we have found numerous literature presenting convex hull like phase diagram.
The formation energy for each structure is determined with respect to the most stable structure of the pure elements. To determine the ground states of a system one needs to find, as a function of composition, the ordered compounds that have an energy lower than any other structure or any linear combination of structures that gives the proper composition. This set of ground state structures forms a convex hull, as all other structures have an energy that falls above the set of tie lines that connects the energy of the ground states.
Thermodynamically, the convex hull represents the Gibbs free energy of the compounds at zero temperature.
The above statement looks a bit awkward. Here is just some simple math. Suppose we have a binary system, say A-B. For elemental A and B, let say
- A has an enthalpy of -5 eV/atom,
- B has -3 eV/atom.
- If we find a compound with stoichiometry of AB, its enthalpy is -5 eV/atom. Will it be stable?
A simple way to do it is to calculate the following reaction A + B -> AB
- A: -5 eV/atom,
- B: -3 eV/atom.
- AB: -10 eV/2atom
The formation enthalpy of AB is -1 eV/atom. Yes, it is stable!
Mathematically, we can create the following plot. Let's make a scatter plot of enthalpy versus composition for each structure.
If we draw a line from A to B, it defines the boundary whether the decomposition to A+B is favorable or not. To make it more convenient, we can also make the line of AB horizontal. In this case, the y-axis exactly has the meaning of formation enthalpy! So, we shall talk about the formation enthalpy relative to A+B from now on.
- A: 0 eV/atom,
- B: 0 eV/atom.
- AB: -1 eV/atom.
Now, let's try to add more compounds. Suppose we have another compound AB2, which has the formation enthalpy of -0.5 eV/atom, will AB2 be stable?
AB2 is definitely stable towards the decomposition to A+B. However, if we calculate another pathway of AB+B, it actually unfavorable (above the decomposition line of AB+B). Thus it is unstable.
If we tried to add another compound A2B with a formation enthalpy of -0.8 eV/atom. What will happen?
Bazinga! A2B is below all possible decomposition lines. Therefore, it is a 100% stable points so far.
Therefore, the only stability criterion is that the new compounds need to be stable towards all possible decomposition pathways between any two stable substance (in this case, they are A+B, A+AB, AB+B).
Mathematically, it simply means that all stable structures in such formation enthalpy vs composition diagram will form a convex hull.
You might ask that we don't we just calculate the formation energy one by one and show the stability in a more straight way. Yes, it is much easier to understand if we only have a few structures under consideration. Say, if we only consider A, AB and B above.
However, it is very likely that we need to consider much more structures with different compositions. Therefore, the convex hull representation is more advantageous for the are following reasons:
- 1, it gives a compact view of stability if we have many structures to check simultaneously
- 2, it is mathematically easy to deal with.
I hope you know the meaning of convex hull so far. As you can see, it is very simple with only some iterative calculations on the formation enthalpy with different decompositions paths.
Such tricks recently become very popular in high pressure sciences and perhaps some other fields as well, mainly driven by the advances in computational crystal structure predictions and high-throughput calculations. Thanks to these methods, nowadays one could purely use the computer to predict a lot of possible structures for any compounds at any arbitrary conditions. Therefore, it becomes really necessary to use convex hull to present the stability of massive structures.
However, one has to be cautious with the interpretation of his/her results.
- It is important to note that all compounds present on the convex hull are stable. One shouldn't say compound A is more stable than B because it has more negative formation enthalpy relative to the elemental substances.
This is really a common mistake. Such claims have been found in many papers Remember in our thermodynamics class, we always use the Gibbs free energy to construct the phase diagram. Since the compositional space is involved in determining the stability. We are now dealing with the grand canonical ensemble, in which G=H-TS+uN. For simplicity, let's skip TS term since we are focusing on zero temperature. If you check the convex hull diagram, you will immediately find that we only use H , where does uN term come to play? It is actually the slope of each line segment in the convex hull. Yes, the left and right slopes for each point gives the information about the stability range as a function of chemical potential. Of course, one needs to to careful math since the slopes have changed when we make the horizontal line between the elemental substance. However, you could intuitively conclude that under the extremes, the compounds with more concentrations of those rich chemical species will be more favorable. So chemical stability also depends on the initial compositional condition. Convex hull does nothing else but split the compositional space according to the chemical potential. For instance, if the systems contains A4B composition, the most favorable thermodynamic equilibrium would be A+A2B. In this case, AB is not favorable at all, even though it appears to be the "ground state" by the formation enthalpy as shown in many literature!
-
Most convex hull construction are based on DFT calculations at 0 K. The reliability of results crucially depends on the choice of the reference energy(enthalpy) for the elemental substance. If one wants to check the stability of compounds like nitrides, oxides, s/he has to be cautious about the calculation of reference energy on the gas phases. If convex hull is made at 0 K, it might be dangerous to use it speculate the chemical stability at high temperature.
-
One shouldn't simply use the convex hull to exclude the presence of metastable structure. It only tells you the information of thermodynamics at 0 K. The metastable structures might be favored due to kinetically reasons.
So far, we only talked about the binary systems. In fact, such schemes could be easily extended to multi-component systems as well. The philosophy is still to check all possible decomposition paths among all possible known stable compounds. The only problem is that such convex hull would be no longer convenient to visualize in high dimensional space. But the results should be reliable since the mathematical foundation is obviously quite solid as shown above.
Nevertheless, we can still do good visualization for ternary systems. Here are some nice pictures.
- A: 0 eV/atom,
- B: 0 eV/atom.
- C: 0 eV/atom.
- ABC: -0.2 eV/atom
- AB2C3: -0.4 eV/atom
- ABC2 : -0.6 eV/atom
- A3B2C: -0.5 eV/atom
Which compositions will be stable? What is the convex hull?
All the pictures were made by the scripts shown in my github CMS repo.