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I'm trying to review how to calculate bond dissociation energy for C-C in ethane, a very simple calculation (or should be).

I get the following Hartree energies for SCF 6-31G(p) for methane, ethane, and H2:

ch4_spe            -40.194639920746,
ethane             -79.228125042573,
h2                 -1.126456057886,

I would think that to get the C-C bond dissociation energy, I would simply do (Perl code)

$energies{ethane} - $energies{ch4_spe} * 2 - $energies{h2}

correct?

The reason that I'm asking is that this give an answer of 2.2876 Hartree using NWChem, which converts to 6006 kJ/mol, which is off by about a factor of 17 from the correct value of about 346 kJ/mol. I know that SCF isn't that accurate, but it shouldn't be this far off. I optimized the bond lengths using Avogadro.

This is the correct method to calculate the C-C bond energy?

con
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  • Hint: The balanced "imaginary reaction" you are considering is
    H2 + H3C-CH3 -> 2 * CH4 This said, how should you calculate the variation in energy?     –  Jan 05 '20 at 23:14
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    I don't see how this reaction equation gives you the carbon-carbon bond dissociation energy; you'll get a mixture of $\ce{H-H}$, $\ce{C-C}$ breaking with $\ce{C-H}$ forming. For the BDE I would consider $\ce{H3C-CH3 -> 2 .CH3}$. (One order of magnitude is not that bad for HF bond energies though. You also might want to consider optimising the structure at the level you're using to calculate the energy.) – Martin - マーチン Jan 05 '20 at 23:38
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    I was about to write exacly what Martin wrote. I just tried the calculation, and you can get a very good result by considering the reaction written by Martin. Remember that you should use UHF, with a basis set which includes polarization and diffuse functions when you deal with radicals (I did it with 6-31++G**). For the geometry of the radical, just split the optimized ethane molecule in half and do not optimize it, just run an energy calculation. See: http://przyrbwn.icm.edu.pl/APP/PDF/127/a127z4p004.pdf –  Jan 05 '20 at 23:47
  • @The_Vinz and Martin, thanks, unrestricted HF on the doublet state gives a dissociation energy of 289 kJ/mol, which is much much more reasonable to the empirical 346 kJ/mol. If you write your comment down as an answer, I can accept it – con Jan 06 '20 at 01:28
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    You should be able to further refine it: with the method I wrote about I obtained an energy of 351.28 kJ/mol, which is very close to the experimental value :) Remember that if you are comparing calculations on different molecules, you should apply the same method and basis set to the whole set of molecules –  Jan 06 '20 at 01:43
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    I would like to encourage you to write a self-answer. I further recommend some reading about the UHF/R(O)HF methodology (cc @The_Vinz): 1. U- or RO-method for Singlet-Triplet Gap? 2. Calculating singlet-triplet gap and open-shell singlet character Even though the title suggests something else, they can be applied to open-shell in general. – Martin - マーチン Jan 06 '20 at 11:36

1 Answers1

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This "experiment" is about determining bond dissociation energy.

The general way to solve this isn't like calculating enthalpies of reaction, as I initially thought, but rather through calculating the energy of two CH3 radicals, and comparing it with ethane. As another user pointed out, this should be done with unrestricted Hartree Fock and with diffuse basis sets.

Thus, energy(ethane) - 2*energy(methyl radical) is the solution.

con
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