Dear Shilendra,

The best way to verify if the calculated value is reasonable or not is the experimental value, if available. If not, comparing your calculate value with values obtained by other computational methods proved to predict formation energy in high accuracy.

Hoping this will be helpful,

Rafik

Thank You Dr. Rafik Karaman for replying. I will try to find out experimental formation energy of my system.

 Hello Sharma!

It is not entirely clear to me what you want to calculate. Do you want to predict Standard formation enthalpy of your product?? regardless this,  In my opinion, your obtained value -301 eV (-790,000 kJ/mol) as a reaction energy is meaningless. This energy is equal to the energy released by explosion of 200 gr of TNT! Calculation of formation energy ( if it means standard enthalpy of formation) using DFT is not a straightforward task. However, I upload one of my articles in this field, declaring that we used some approximations and obtaining more accurate values needs more time-consuming calculations.

I hope it could be helpful!

Thank You Dr. Ali Ahmadi Peyghan for your help and suggestions, I will go through your reference.

You are right my question was related to standard formation enthalpy of my product. I calculated it through DFT by calculating total energy of my product- total energy of reactants. As enthaly H=U+PV and in DFT at optimize structure P=0 so H=U=Internal energy and it is equivalent to total energy in DFT at P=0.

So I calculated enthalpy of formation by taking 2*2*1 supercell in which my system contains 16 La, 16 V and 64 O as I mentioned in my question above. After that I replaced 1La atom by dopand Eu.

Since my original bulk structure 1*1*1 is 4 formula unit and I expanded it to 2*2*1 for doping suitability so now my supercell became 16 formula unit.

So I think I have to divide my enthalpy of formation by 16. After dividing 16 enthalpy of formation is coming out around -19.18 eV per formula unit

So I want to ask you I am doing right calculation and now this enthalpy of formation -19.18 eV/f.u is reasonable.

Dear  Shilendra (Answer to the private massage for  public discussion)

As I know, there are 3 ways to calculate the standard enthalpy of formation (∆Hf, 1 atom, and 298 K).

1) Computations based on the atomization energies

2) Computations based on the isodesmic reactions

3) Benson’s method (group additivity technique)

What we have used is the method 1. I explain here how this method can be used for calculation of ∆Hf of methane.

Assume a hypothetical reaction for CH4 formation as follows:

4 H (g) + C (g) ==> CH4 (g) + ∆Hr,    (1)      ∆Hf=?

It is clear that enthalpy change for this reaction (∆Hr) is not equal to ∆Hf because the atomic H and C is not in their standard states. Standard state at 1 atom for Hydrogen is H2 gas and for Carbon is solid graphite.

One can write

∆Hr = ∆Hf(CH4) - ∆Hf( C ) – 4 ∆Hf (H)   (2)

Calculation of ∆Hf (H) using DFT is possible from the below reaction

2H (g) == >H2 (g) + ∆Hr  (3)

Here, ∆Hf (H) = - ∆Hr which is a positive value

For calculating ∆Hr in eq. 3 you should do frequency analysis (for example: B3LYP 6-31g* freq=noraman temperature= 298 pressure= 1) for a single H atom and a H2 molecule therefore from the output

∆Hr  = H (H2) – 2H (H); all enthalpy should be zero point corrected (similar trend should be used for calculation of ∆Hr of Eq 1) .

In contrast to ∆Hf (H), for carbon atom calculation of ∆Hf is not straightforward task because the standard of carbon is solid graphite

To overcome this problem one can use available experimental ∆Hf of atoms in JANAF site. For consistency the ∆Hf (H) should be taken from this site. However, one can use the methods that already successfully tested and fitted by the experimental data. Self interaction error of DFT functionals in the calculating energy of atoms is a critical concern here.

Thank you Dr. Ali Ahmadi Peyghan for your help and long discussion.

My one doubt is remaining if we take ∆Hf from JANAF site then how will calculate automization energies D0. Other thing for Gibbs free energy of formation how will calculate entropy term.

Well,

1) In eq. 1, (- ∆Hr) is atomization energy (enthalpy)

2) Entropy terms are available in the output of the freq analysis (see the attachment)