I am considering layered oxides $\ce{LiCoO2}$. Previous papers suggested that vdW interaction should be added to get lattice parameters closer to experimental data. I would like to dope other elements to Co sites of layered $\ce{LiCoO2}$ and calculate the formation energy of the doped structure. In the formula for estimating formation energy, I need to know the chemical potentials of dopant elements. For example,

$$\textrm{formation energy} = E(\ce{LiCo}_{1-y}\ce{Na_yO_2})-[E(\ce{LiCoO2})+yE(\ce{Na})-yE(\ce{Co})]\tag{1}$$

The idea of adding vdW to estimate chemical potential seems to be strange, but I still wonder if I should add vdW interaction to calculate the chemical potentials of dopant elements? I found that when I add vdW interaction to calculated Na chemical potential, there is a big gap compared to results without adding vdW. For example, $E(\ce{Na}_{\textrm{vdwD3}})$ = -0.385 eV, E(Na_without vdW)= -1.283eV (using same calculation parameters such as ENCUT, K-POINTS in VASP).

Adding to the complexity is the difference in listed chemical potentials based on the phase of the compounds. For example, the chemical potential $\mu$ of Na is listed as both 0 (solid) and 0.5 kG (liquid): https://www.job-stiftung.de/index.php?data-collection

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    $\begingroup$ +1 but I'd like you to take a look at my edit and finish fixing the formatting issues that I started fixing. It would be appreciated! Can you tell us which papers you're referring to in the 2nd sentence of your question's body? Also, you got a big difference between using vdW and not using vdW for $\ce{Na}$, but what's the experimental value (i.e. which one is more accurate)? $\endgroup$ May 11, 2021 at 20:00
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    $\begingroup$ @NikeDattani. Please take a look at the following paper for vdw adding while calculating LiCoO2. pubs.acs.org/doi/10.1021/acs.jpcc.5b06240. I checked chemical potential on this website job-stiftung.de/index.php?data-collection and I got the value of 0 in solid, but 0.5kG in liquid. $\endgroup$
    – Binh Thien
    May 12, 2021 at 1:51

1 Answer 1

Short version

If the $\text{LiCoO}_2$ is computed with van der Waals included, so should everything else.

Long version

This seems more of a thermodynamics question than a simulations question given the comment from May 12 '21. What is set to zero by definition is the Gibbs free energy of formation at standard temperature and pressure (1 atm, 298.15 K) for pure elements in their natural state. This explains the difference between the cited values for the solid and the liquid (comment from May 12 '21): the difference is the enthalpy of melting. More generally, the chemical potential of the various species must be taken into account appropriately to compute voltages (which are, strictly speaking, temperature-dependent) and for example defect formation energies. See for example:

What is computed in VASP energy minimization is the electronic potential energy of the atoms, relative to the same atoms being infinitely far apart as elemental gas. This is equal to the Gibbs free energy only at effectively zero temperature, plus the vaporization. They are different by the entropy term times the finite temperature. The DFT code does not know about us humans setting a reference state for Gibbs free energies, so everything has to be built up from the energies at 0K plus chemical potentials.

[Someone correct me if I'm wrong:] The zero-point energy of vibrations is not included, and sometimes that is non-negligible, see for example: https://pubs.acs.org/doi/10.1021/acs.chemmater.0c03442

Because of this, the energies computed for pure elements (e.g. the Na and the Co) do not have to be zero. All energies will change with the functional that is used. For consistency purposes, it is necessary to compute all parts of the reaction with the same level of theory and functional. If the $\text{LiCoO}_2$ is computed with van der Waals included, so should everything else.


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