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Alex Granovsky
gran@classic.chem.msu.su

Dear David,

how exactly were the numbers below obtained?

>free energy of HF (gaseous) : -100.426656 - 18.394/2625.5 = -100.433662 H

Could you comment on the -100.426656 - 18.394/2625.5  math?

Kind regards,
Alex Granovsky.



On Thu Feb 13 '14 6:33am, David G. wrote
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>Hi

>As a test to understand how Firefly does compute and consider solvent effects the free energy of HF diatomic molecule has been calculated in gaseous state, water, and benzene to compare the results.

>free energy of HF (gaseous) : -100.426656 - 18.394/2625.5 = -100.433662 H
>free energy of HF (water): -100.433893 H
>free energy of HF (benzene): -100.430009 H

>we expected a larger difference between the values in water and benzene since HF is a highly polar molecule. I have attached the input/output files and I appreciate any comment to help us how to improve the results if it looks unwise in this state. Many Thanks
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>Sincerely, David
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