Michel.
In the reaction 2 H-H + O-O ----> 2 H-O-H you are breaking three 498,000 Joule/mole
(5.17 eV Bonds) = 1.490E6 joules input to make four 498,000 Joule/mole (5.17 eV bonds)
= 1.99E6 joules for the 2 H-O-H molecules.
Hence, you should get 1.99E6 - 1.49E6 = 498,000 Joules free energy.
OTOH, 2 x 498,000 - 474,000 = 522,000 Joules, the higher calorimeter
value in your spreadsheet vs the 474,000 joule/mole dG Free Energy.
In Jones Beene's' Excellent post on "Water-based fuel for the ICE" this morning,
he points to the Anomalous Free Energy that comes from using a lot less than 1/4th the
energy (~ 1.0 - 2.5 eV or much less) to break the measured (5.17 eV) H-O-H
bonds with emphasis on restricting recombination to the 5.17 eV H-H or O-O bonds
in any electrolyzer if you want to maximize the combustion energy
(making H-O-H bonds) from H, O, and/or OH radicals in an ICE.
Reiterating using the Ellingham Diagrams for quick reference too.
