[Delfador]

If I have a simple molecule like H₂⁺, what's the right way to calculate its dissociation energy (making an estimate)?

The electron energy of the hydrogen atom is 13.6 eV, of H₂⁺ it's 30.0 eV, somehow subtracting two times the electron energy of hydrogen gives you the dissociation energy: 30.0 eV - 2*(13.6 eV) = 2.8 eV.

Why is this possible? Why subtracting two times?

[Mitch]

I'm not sure. But, since the DE to split H₂ is 4.5 at least the DE to split H₂⁺ makes sense that it is lower, almost by half.

[mjc123]

Where do you get your figures from? Subtracting twice the energy of the H atom seems to make no physical sense.

From Wikipedia (not infallible, I know) I get "The ionization energy of the hydrogen molecule is 15.603 eV. The dissociation energy of the ion is 1.8 eV." (Article on Dihydrogen cation). But is this right? From http://www.millsian.com/summarytables/SummaryTables022709S.pdf I find
H₂ bond energy 4.487 eV
H₂⁺ bond energy 2.651
H₂ ionisation energy 15.426
H₂⁺ ionisation energy 16.250
H₂ total energy 31.675

Your figure of 30 eV seems to be perhaps an approximation of the total energy of the two electrons of H₂. Subtracting from this twice the ionisation energy of the H atom should give you (by a Hess's law cycle) the dissociation energy of H₂.

By a similar cycle, DE(H₂⁺) = IE(H₂⁺) - IE(H).