----- Original Message -----
Sent: Sunday, June 04, 2006 12:39
AM
Subject: dH vs dG (was Re: Free Radical
Chain Reactions)
Fred,
"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
