Matter Modeling Stack Exchange is a question and answer site for materials modelers and data scientists. It only takes a minute to sign up.
Sign up to join this community
This answer refers to a paper that calculates $\Delta H$ as $\Delta E_{\textrm{electronic}} + \Delta E_{\textrm{vibrational}}$.
However, what if I had enough computing power, and I wanted to get the exact value for $\Delta H$: Would I include a "rotational" term too? What about "torsional" or even fine-structure?
What are all the terms that would be included, if we had enough computing power and desired the accuracy?
At T=0K thermochemistry is kind to us.
At T = 0K $U = G = H$.
This is because H = U + PV, which for an ideal gas is H = U + RT, and T = 0K, and also, G = H-TS, and again, T = 0K so that cancels
The only direct calculation a QM program is doing, that I am aware of, is calculating contributions to the Energy. And it only does energy calculations at T=0K.
At T = 0K a QM calculation will give you $E_{\rm elec}$. It will also calculate the zero-point vibrational energy (ZPVE). You need to add this on. All other contributions to energy are zero. There are no rotations or translations at T=0K.
If you want to do thermochemistry at a different temperature, you need to calculate the $\Delta U$ between the energy at T=0K and the desired temperature, and also, as will be seen, calculate the entropy. I say "you", but I mean your QM program. In going to a different temperature you will need to consider
$\Delta E = \Delta E_{\rm trans} + \Delta E_{\rm rotation} + \Delta E_{\rm vib} +\Delta E_{\rm others}$.
The first three are calculated using statistical thermodynamics. Specifically, statistical thermodynamics for ideal gases, which are actually solvable. Professor Christopher Cramer has a great online course covering how to calculate thermodynamic properties from the partition function Cramer's Coursera course. This Gaussian white paper covers how Gaussian does thermochemistry Gaussian Thermochemistry White Paper
The take home is that you can calculate $\Delta E_{\rm trans}$, $\Delta E_{\rm rotation}$, $\Delta E_{\rm vib}$. A note on $\Delta E_{\rm others}$ this can contain many things, most notably nuclear contributions which we normally neglect, and excited state electronic energy, which is often neglected, but not necessarily wisely.
Now, also, from the same guiding principles of statistical thermodynamics we can calculate the entropy. So, we have in hand our $\Delta E$'s and our entropies, $S$.
Thermodynamics does the rest.
$H = U + PV$ and since this is an ideal gas, we know PV = nRT, so we thus have $\Delta H = \Delta U +nRT$. Typically $U$ refers to internal energy in a macroscopic sense, and we use $E$ to denote microscopic energy. I will use $E$ rather than $U$ for clarity.
We now have $\Delta H = \Delta E + nRT$ where R is the gas constant and n is number of moles, and T is temperature. Presumably you know n, can look up R, and had a T in mind when you did the QM computer experiment. So we have H now, at any temperature, for our single molecule, which, by fiat, is an ideal gas molecule.
The last thing usually of interest is the Gibbs free energy. It is certainly my favorite.
Again, thermodynamics has a relation which is G = H-TS. Remember, we calculated our $\Delta E$'s earlier and also our entropy S, and we know T, and we just calculated $\Delta H$, so we can plug and chug to get $\Delta G$ since we know $\Delta H$, T and S at this point.
it is useful to know that $\Delta E_{\rm trans} = \frac{3RT}{2}$ and, this applies to a proton (among other things)! often in proton affinity REACTIONS, a magical 5/2RT appears and people just accept it. it is because a proton does not have rotation, it is a single nuclei, nor does it vibrate, it only has a translational contrabution to the $\Delta E$ between two temperatures,presumably T=0K and your goal. I am not quite done. And this may be useful. Proton affinity is calculated as an Enthalpy - hence, the relevance to your question. So the proton affinity (enthalpy) must account for the $\Delta E_{\rm trans}$ of the proton is 3/2 RT, but Enthalpy is E + nRT, here n = 1, so the magic 5/2RT that appears is due to the 3/2RT from energy, and the RT to go from E to H. i.e., $\Delta H = 3/2RT + RT = 5/2RT$.
Probably overkill.
In this day and age, given the simplicity of a thermochemistry calculation, there should be no excuse for assuming $\Delta H = \Delta E$ at any temperature other than T = 0K.
