Of course there is energy stored in those gases John, exactly the same amount
that was put in their dissociation. Here is a transcript of what I posted on
August 16, 2006 to our restricted list of CMNS researchers on the subject (it
was for H2O electrolysis mind you, figures are slightly different for D2O). I
authorized some of the top researchers there to quote my derivation (fairly
basic stuff I thought) in their papers or books so I guess I can authorize
myself to post it here:
-----------------------------
Hi M****** and E*,
I did post a calculation, that was in a July 2 post "**************". Here is a
revised version:
For water dissociation the enthalpy tables (e.g. those in my calculator) give:
H2O(l) -> 0.5 O2(g) + H2(g) - 285.83 kJ/mol (endothermic) [P.S. MJ 2007-03-14
I should have written more generally "endoenergetic"]
The delta H ~286 kJ per mole is of course Avogadro's number (6.02*10^23) times
the required electric energy per dissociated molecule, which is therefore:
E= 286*10^3/6.02*10^23 = 4.75*10^-19 J/molecule
We also know that we need to circulate 2 electrons per molecule, so if V is the
part of the electrolysis voltage used to dissociate the molecule and e is the
electron charge 1.6*10^-19 C, another expression for the required electric
energy (work) for the molecule dissociation is: E=V*2e. Thus:
V=E/2e = 4.75*10-19/3.2*10^-19 = 4.75/3.2 = 1.48V
Please note that the -285.83 kJ/mol reaction enthalpy this calculation is based
on is for STP conditions 1atm and 25°C, so for different conditions e.g. 1atm
and 100°C the appropriate reaction enthalpy must be used and will yield a
different thermo-neutral voltage value.
Michel
--------------------------------
So if what we are electro-lyzing (hyphen as a hint for Edmund) is H2O, and if
both the evolved H2 and O2 and the initial H2O _are at 1 atm and 25°C_, the
energy stored in those gases is simply 1.48V times the total charge circulated
through the cell (integral of current over time).
Let me know if this makes sense to you John.
> A REWARD for the effort, in advance, even if no one helps with my
> questions:
What's the reward BTW? ;-)
Michel
----- Original Message -----
From: "john herman" <[EMAIL PROTECTED]>
To: "vortex-l" <[email protected]>; <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Wednesday, March 14, 2007 8:54 AM
Subject: [Vo]:
...
> [3] I offer a note here about Cold Fusion:
>
> In advance Isit corrected if someone can show a paper or other
> that
> addresses the note below....
>
> (A) In almost every paper I have read and conferences and
> dedicated presentations I have attended that are directed toward the ideas
> of CF....
> There is one big "hole" which will-would-can change the figures of all
> of the energy
> figures:
> (B) CF is an electrochemical experiment. Aqueous....hence H2 and
> O2 are
> liberated, generated, result of electrolysis of water or how some ever
> (C) But no one seems to give a measure of the volume of these
> gases....and what
> energy would be resulting if the set up collected the gases as
> separate gas
> flow or stream outputs AND
> i) burned them and measured the heat
> ii) combined them with a catalyst set up INTENDED to make
> and measured the same
> iii) combined the gases in a fuel cell and measured the
> electrical output
>
> -------------> And then offer figures of the energy of the use-chemical-fuel
> cell of these
> gases which are the result of nearly Any CF cell I have come across.
>
> Yes.... I have read and talked to presenters and sometime the
> gases are vented, recombined often with some proprietary or not defined
> catalyst .... but when I ask...
> "How much gas and what energy it would contribute ...if combusted or fuel
> cell or
> thermal by other means ...I get no answer...or "not part of the calculation"
> or
> ?I do not know? or"itdoesnomatter"
>
> Note Bene: If the gases are produced...the energy they represent
> Does
> make a difference in the total intermediate andor end result.
>
> Any comment or help on these gasses?
>
> Thanks to all,
> JHS
