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A proton obviously has no electronic energy, no vibrational and no rotational degrees of freedom. Therefore I think it is fair to assume, that $$H(\ce{H+})=\frac32\cdot \mathcal{R}\cdot T$$ holds for the translational degrees of freedom. I am looking for some quotable literature where they state that, or if that has not been stated anywhere, how to derive this.

I am doing quantum chemical calculations, i.e. gas phase. I define the proton affinities via $$\text{PA}(\ce{M})=-\left(H(\ce{H+M})-\bigr[H(\ce{M})+H(\ce{H+})\bigr]\right)$$ for the reaction $$\ce{H+ + M -> H+M}.$$ This is respecting the (quite common) sign convention to show the energy released upon binding a proton. I need a reliable source or at least a best-practice example(s) for $H(\ce{H+})$.

Martin - マーチン
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1 Answers1

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A quite complete procedure in the calculation of proton affinities from ab initio calculations is referenced in an article by H. Audier et al., Int. J. Mass Spectrom. 1996. It goes much further in the procedure than only on the calculation of the enthalpy of a proton (including corrections due to the basis set superposition error and so on), but confirms that you are right in assuming that the enthalpy of a proton is $3/2 RT$.

PLD
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  • Unfortunately they also only state my suspicion as a fact (no proof) - but at least now I know that I am not the only one thinking so... (sorry for the delay in accepting the answer). – Martin - マーチン May 07 '14 at 15:18