dH vs dG (was Re: Free Radical Chain Reactions)

[Michel Jullian]

Fred,   http://antoine.frostburg.edu/chem/senese/101/thermo/ :   "estimating H from bond enthalpies strategy: imagine reaction as a) dissociation of reactants into atoms, b) recombination of atoms into products. Add enthalpies for all product bonds Add enthalpies for all reactant bonds H is approximately the difference between the product and reactant bond enthalpies limitations procedure doesn't account for molecular attractions/repulsions, so doesn't work well for liquid/solid phase reactions bonds interact with each other within molecules, so bond enthalpies really aren't additive " so you see, the 498 kJ/mol you have calculated for 2 H-H  + O=O ---->  2 H-O-H from your bond energy values (exact values from link above yield (463 * 4) - ((436 * 2) + 502) = 478 ) is really an _approximation for dH_, not for dG.   I will grant you that in the case of the present reaction 2H2(g)+O2(g)->2H2O(l) the 478 kJ/mole calculated seem much closer to dG (474 kJ/mole) than to dH (572 kJ/mole), but this is principally because the (_intra_molecular) bond energies method disregards the _inter_molecular attractions at play in liquid water (first limitation listed above), which happen to have an energy of 44kJ/mole (the water vaporization energy at STP ).   If you add the 2*44=88kJ/mole intermolecular bonds energies for the 2*H2O in the products, you get 478+88=566kJ/mole which is now much closer to actual dH=572 kJ/mole.   The bonds method works much better when all products and reactants are gases, e.g. the same reaction where the water produced is gaseous instead of liquid: O2(g) + 2 H2(g) -> 2 H2O(g) + 483.636 kJ/mol (exothermic) Spontaneous at 700°C. Equilibrium at about 5170°C. Molar masses and thermodynamic properties Enthalpy change kJ/mol Entropy change J/K/mol Gibbs Free Energy change kJ/mol Sources: c.f. bottom of spreadsheet -483.64 -88.86 -397.18   here as you can see the bonds energies balance 478kJ/mole (same as in previous reaction of course) is pretty close to the actual dH value 484 kJ/mole, much closer than it is to the dG value 397kJ/mole.   Let me know if you were still not convinced Fred, I have one more argument in reserve.   Michel   ----- Original Message ----- From: Frederick Sparber To: vortex-l Sent: Saturday, June 03, 2006 9:58 PM Subject: Re: Free Radical Chain Reactions 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. http://www.chem.mtu.edu/skkawatr/Ellingham.pdf