CCL: Formation Energy - Summary

Dear All,

Thanks for your response.
I am summarizing the replies.

Question :

Dear All,

I have three composition.


I would like to understand
which structures are stable and hence
decided to calculate formation (or Cohesive) energy.

Lets take the first two cases and 
calculate Formation energy by the below formulae.

Formation Energy of 1 (E1) = Total energy of A1-B3-C4 -(Energy of 1*A + Energy of 3*B + Energy of 4*C).
Formation Energy of 1 (E2) = Total energy of A2-B4-C5 -(Energy of 2*A + Energy of 4*B + Energy of 5*C).

Does the direct comparison (E1, E2) makes sense?
Can i conclude if E1>E2, then composition 1 is more stable.

I feel, i am missing something.
Your help would be appreciable.

Best regards,
Dear Vimal,

I would say that your approach is valid, under the condition that the A, B, and C and
sensible moieties. It reminds a bit of atomization energies.

However, I would say that if E1 > E2 (e.g.  +2  vs. -10) that 2 is more stable, not 1.


Prof Dr Marcel Swart, FRSC

ICREA Research Professor at
Institut de Química Computacional i Catàlisi (IQCC)
Univ. Girona (Spain)

COST Action CM1305 (ECOSTBio) chair
Girona Seminar 2016 organizer
IQCC director
Young Academy of Europe (Board) member


Dear Vimal:

Unfortunately, the concept of stability is not that straightforward. Thermodynamics does not provide the final answer. The stability of a molecule not only depends on its formation energy, but it depends also on the relative energies of its neighboring local minima on its potential energy surface and the energy barrier between the minimum and its immediate neighboring minima (kinetic stability). To begin, I would rather check the depth of minimum energy well by looking at frequencies. In the next step ab initio/MC/DFT- MD simulations can provide more insight toward the kinetic stability. Then of course we can define stability with respect to a particular reaction. Is your molecule stable when it is in the presence of air/water/light/etc.
You must be more precise about stability. 

I hope my general answer can help you to solve your problem.

Good luck,

Cina Foroutan-Nejad, PhD
CEITEC-Central European Institute of Technology
Kamenice 5/A4, Masaryk University, Brno, Czech Republic

Hi Vimal:

I am not exactly certain what you want to do but if you want to estimate is composition A1-B3-C4 is more stable than A2-B4-C5, you should calculate the energy of this "reaction" ...

A1-B3-C4 + A + B + C --> A2-B4-C5

Now you have a mass balanced equation. Assuming the energies of these components are negative, then if E(Right Hand Side) - E(Left Hand Side) is negative the LHS is more stable. This means A2-B4-C5 as a compound is more stable than A1-B3-C4 and an extra A, B, and C.

Joe Golab

Dear Jeya, how are you?
The comparison kind of makes sense, but to get the correct energies it is not enough to calculate the single points without the thermodynamic corrections (and using the correct stoichiometry in your equations).
Please, if you can, take a look at this link[1] (it is in Portuguese, but maybe Google Translator can deal with this issue for you).

All the best


Thanks and best regards,