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Let's say we have the following reaction:

$$\ce{RH -> R- + H+}$$

By modelling it using quantum chemical calculations in the gas phase I would make calculations for both $\ce{RH}$ and $\ce{R-}$, but $\ce{H+}$ has no electrons.

On the other hand, $\ce{RH}$ and $\ce{R-}$ are isoelectronic and zero energy is assigned for infinitely separated particles. I have a gut feeling that the energetics of $\ce{RH}$ and $\ce{R-}$ can't be compared directly in this case, but I can't think of a good argument for this. So, can I compare them? Why or why not?

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    $\begingroup$ I actually read something somewhere that discussed this, but I can't remember now what the argument/approach is. Glad to see it as a question here, thank you for posting! $\endgroup$
    – hBy2Py
    Feb 10 '17 at 2:43
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    $\begingroup$ only relative energies for balanced chemical reactions make sense $\endgroup$
    – Jan Jensen
    Feb 10 '17 at 8:08
  • $\begingroup$ But if a proton has zero electronic energy, isn't the equation balanced? (I understand that when one considers thermal effects the proton will acquire translational energy.) $\endgroup$ Feb 10 '17 at 9:36
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    $\begingroup$ Ah OK, in that sense. Yes, the energy difference is the electronic contribution to the proton affinity. $\endgroup$
    – Jan Jensen
    Feb 10 '17 at 13:18
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    $\begingroup$ A little bit too theoretical question: while technically it is true what you say, proton as a lone species hardly exist in chemical environment. In other words, charged specii, esp very small ones's energy is strongly influenced by the interaction with environment and it generally doesn't make any sense to compare neutral and charged entities' energy in vacuum. $\endgroup$
    – Greg
    Feb 10 '17 at 14:08
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I think they can and should be compared. Consider the $\ce{H_2^+}$ molecular ion. When dissociating this correctly (that is, including symmetry breaking), one should eventually end up with $\ce{H}+\ce{H^+}$. One can then extract the dissociation energy of the bond from the calculations. This is equivalent to the proton affinity of atomic $\ce{H}$. By applying this logic to larger species, one reaches the conclusion, that yes, $\ce{H+}$ by definition has zero energy and should enter calculations of relative energies as such.

As an aside, the calculation of negatively charged species in the gas-phase can lead to erroneous results, but any argument made for $\ce{A-/AH}$ will certainly hold for $\ce{B/BH^+}$.

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