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Using e.g. the VWN5 functional, what is a simple way to calculate the energy of a single proton? Of course, it is not possible to perform a calculation of a proton in isolation, as it bears no electrons. So I first attempted to use the experimentally determined value (see e.g. M. Dewar and K. Dieter, J. Am. Chem. Soc. 1986, 108, 8075.), but this leads to highly erratic proton affinities.

Could it be that I need to determine the proton affinity for any given functional + basis set (e.g. VWN5, 6-31+G** in my case)? What would be an effective way to get a good rough estimate of the energy of a proton?

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closed as unclear what you're asking by Mithoron, airhuff, Tyberius, Jan, bon Oct 22 '17 at 7:43

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    $\begingroup$ What "energy of single" proton? It can have any kinetic energy you can imagine. $\endgroup$ – Mithoron Oct 21 '17 at 19:53
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    $\begingroup$ A good rough estimate of the energy of a proton is $0$. $\endgroup$ – Ivan Neretin Oct 21 '17 at 20:50
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    $\begingroup$ Energy of a proton? That doesn't make any sense. Referenced to what? $\endgroup$ – Zhe Oct 21 '17 at 22:03
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    $\begingroup$ I do not have access to the paper referenced. If you wish to calculate gasphase proton affinities (which would be simple energies), you assign the energy zero for the proton. Thus, two calculations and you're done. If you look for pKb values, you need a solvation treatment and then the functional/basis set come into play. $\endgroup$ – TAR86 Oct 21 '17 at 22:33
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    $\begingroup$ I think looking at a single proton takes it out of the realm of chemistry and into physics. $\endgroup$ – Tyberius Oct 22 '17 at 2:32
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An isolated proton will have energy given by the Schrodinger equation (just like any molecular system):

$$ -\frac{\hbar^2}{2\mu}\nabla^2\psi + V\psi = E\psi $$

($\mu$ is reduced mass; $\hbar$ is $h/2\pi$; $V$ is potential energy; $\psi$ is the wavefunction)

The potential of this single-proton system is zero; the kinetic energy will depend on the momentum of the particle.

If you're using a generic quantum-chemistry software, the Born-Oppenheimer approximation is probably being applied, i.e. the proton ($\ce{H+}$) will be fixed in space with zero momentum. Thus its total energy will be zero, unless you deactivate BO-approximation or introduce another particle.

It's not typical to consider the energy of an isolated proton for obvious reasons (try finding one in nature!). Nonetheless it is calculable if the wavefunction is known.

Another way to consider its kinetic energy is by the classical equation $K = \frac{1}{2} mv^2$; if you consider an approximation of the lonely proton's speed as roughly that of atoms in liquid water, 1 angstrom per picosecond, you obtain ~ $5.03$ kJ/mol. This is similar to the thermal energy available at room temperature, $k_B T$, ~ 2.48 kJ/mol.

You're better off calculating energy of $\ce{H+}$ by getting energy for a molecule, and subtracting the energy of its deprotonated form. In that way you can evaluate proton affinity for a given molecule.

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    $\begingroup$ I think I have figured it out now: Dewars method (see above reference) is based on AM1 which gives different energy values than ab initio methods. Therefore, he has to consider the heat of formation of a single proton, which is less stable than hydrogen and hence corresponds to a positive heat of formation. However, ab initio methods yield the electronic energy, which is obviously zero for a proton. Therefore, PA = Edeprot - Eprot. $\endgroup$ – logical x 2 Nov 1 '17 at 15:32
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    $\begingroup$ don't forget the charge will change for the deprotonated form, and you have to indicate the charge of the system in QM software to compute the wavefunction/energy/etc. $\endgroup$ – khaverim Nov 1 '17 at 19:35
  • $\begingroup$ Thanks I know this much ; ) Anyway, multiplicity wouldn't add up for the wrong charge meaning it will fail if the charge state is wrong making it pretty "idiot-safe" at least in the ORCA implementation... $\endgroup$ – logical x 2 Nov 1 '17 at 19:51

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