Good day!
I study hydrogenation reactions with organometallic catalyst. The software used is Orca.
If we calculate energy difference (see attached Figure with PES Scan final single point energy) or enthalpy change between 2 reactants and intermediate (~4 A), intermediate - reactants = -15 kJ/mol, so the intermediate is 15 kJ/mol lower on the figure. We understand this local minima at 4A as a stabilization of molecule with hydrogen bonds near the catalyst.
However, if we calculate Gibbs free energy change (which includes entropic contribution), the intermediate is 53 kJ/mol higher than reactants.
What energy difference we should consider if we want to speak about the energy barriers and rate-determining step (we compare multiple mechanisms)?
Reading the articles, I noticed the Gibbs free energy change is used everywhere. In my case, I'm embrassed by the fact that this hydrogen-bond formation is assisted with the big barrier 53 kJ/mol, and this reaction becomes the rate-determining step in the whole mechanism if we calculate Gibbs free energy energy barriers for all reactions.
Also, Orca (and me) usually watch total energy values from PES Scan to locate TS, and it is often useful. There is no TS for hydrogen bonding according to PES Scan (please, see Figure). How can it be: big Gibbs free energy barrier, and no TS? Please, shed some light on it.
Regards, Ivan.
Dear Bojidarka B. Ivanova , it's not a MD simulation. I have used relaxed PES Scan with DFT (triple-zeta + polarization basis and B3LYP functional).
Mr. Mitrichev,
I would recommend you to use comparative analysis between scan option, ADMP, BOMD and TSH approaches. Very frequently, using the latter methods we have found minima of PES (evidenced by frequency analysis) which have energies lower than those minima obtained using the approach shown in your posting.
In addiotion, very frequently there is needed a repetition of the computations starting from the minimum of the energy obtained on the base on the first computation.
Dear Ivan Mitrichev,
It seems that you have a barrier around 3 Angstroms and two intermediates at 2.5 and 4.5 Angstroms reading your PES scan.
Regarding the Gibbs free energies.
1. Did you consider in your calculations the weakly interacting reactants? Try run the scan starting with the intermediate and elongate along the desire coordinate to see where you end up.
2. How big is your system? Maybe you should account for low frequencies? Check for instance Grimme S. Chem. Eur. J. 2012, 18, 9955 and the python code to inlcude this quasi-RRHO correction with Funes-Ardoiz, I.; Paton, R. S. GoodVibes v2.0.2 DOI: https:/doi.org/10.5281/zenodo.1247565 3. Are you consider a solvent, charges, higher level of theory? This might change completely the picture.
4. Maybe you haven't find the transition state yet? ;-)
For that the code of Emilio Martinez-Nuñez might help you (http://forge.cesga.es/projects/tsscds).
Good luck ;-)
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D. G. P.
Hallo, Ivan!
Gibbs free energy calculations are of physical meaning only at PES minima (i.e. stable geometries, without any imaginary vibrations with negative frequencies). They are used to evaluate equilibrium constants only. Gibbs free energies are usually evaluated under assumption of harmonic approximation (i.e. quadratic potential) and in such case they are equal to Helmholtz energies. Anharmonic corrections might be used as well.
At the transition state geometry you obtain a single imaginary vibration and a standard Gibbs free energy calculation has no physical meaning (too far from the quadratic potential shape). If you would need the Gibbs free energy value at TS, some correction for the imaginary vibration is necessary.
Best wishes
mb
The so-called "entropy " doesn't exist at all.
During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.
The so-called entropy was such a concept that was derived by mistake in history.
It is well known that calculus has a definition,
any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.
Based on the definition of calculus, we know:
to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.
As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,
∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.
1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.
2) On the other hand, In fact, Q=f(P, V, T), then
∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).
We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.
that is, the so-called "entropy " doesn't exist at all.
Why did the wrong "entropy" appear ?
In summary , this was due to the following two reasons:
1) Physically, people didn't know Q=f(P, V, T).
2) Mathematically, people didn't know AΔB couldn‘t become AdB directely .
If people knew any one of them, the mistake of entropy would not happen in history.
Please read my paper and those answers of the questions related to my paper in my Projects.
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
What has entropy and enthalpy got to do with Entropic Intermediates as being considered in chemical reactions?
Off course I agree ENTROPY as being defined in Thermodynamics is a virtual reality which doesn't exists.
The starting point of thermodynamic logic is: 1) opposing perpetual motion. 2) The irreversibility of dynamics. Let's look at how to get Carnot efficiency = 1-T1/T2.
1.1 There are many kinds of type 2 perpetual motors, each of which is a natural phenomenon.
A) Machine A: against the irreversibility of thermodynamics (diffusion, heat conduction, friction, etc) - dynamics;
B) Machine B: Utilizing the Difference of Carnot Efficiency (Reversible Thermodynamics) - Thermodynamics
1.2 A and B belong to different disciplines. There is a parallel relationship between them and there is no logical mutual inevitability.
1.3. Logic of the Second Law of Thermodynamics:
Experience induction, deny that A machine==> B machine can not be manufactured==> All material Kano efficiency: 1-T1/T2.
1.4 The logic of the second law of thermodynamics violates the physical logic that A and B cannot be inferred from each other.
1.5 The second law of thermodynamics elevates the "irreversibility", which is in fact only a kinetic experience.
📷 Detailed discussion can be found in the following figure.
1. The second law of thermodynamics loses the ability of quantitative prediction.
2. Data processing method of law 2 of thermodynamics violates scientific discipline: modifying experimental data to satisfy theory
See the picture for details.
3. The second law of thermodynamics can only be called science if it reaches the same quantitative prediction ability as Newton's second law, otherwise it is pseudoscience.
4.Unfortunately, all the charts of enthalpy and entropy are organized in this way.
Comparison of New and Old Thermodynamics
1. Logic of the Second Law of Thermodynamics: Subjectivism, Logical Jump, Interdisciplinary Argumentation.
2. New thermodynamics pursues universality, two theoretical cornerstones:
2.1 Boltzmann formula: ro=A*exp(-Mgh/RT) - Isotope centrifugal separation experiments show that it is suitable for gases and liquids.
2.2. Hydrostatic equilibrium: applicable to gases and liquids.
3. The second and third sonic virial coefficients of R143a derived from the new thermodynamics are in agreement with the experimental results.
3.1. The third velocity Virial coefficient derived is in agreement with the experimental data, which shows that the theory is still correct when the critical density is reached.
4. See Appendix Pictures and Documents for details.
It is a classical case for adsorption as studied by DFT.
Adsorption has a low prefactor, low probability. Using both the Arrhenius and the Eyring equation, we have
k=A e-(Eact,ads/(RT)) = kB T / h e-(deltaGads/(RT))
Eact,ads = 0 (no barrier), G > 0 is high, and that's OK.
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