2009/12/29 Jones Beene <[email protected]>: > -----Original Message----- > From: Michel Jullian > >> - but the 2 eV available >> from loading alone without deuterium (contrast that to about .5 eV if the >> hydrogen were burned in air) is a huge surprise - > > MJ: Jones, where did you get that .5 eV figure? I did the maths and found > about 1.5 eV instead, here is the Google calculator result; > > ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts > > > Michel, the half-eV figure is the common 'real world' estimate based on the > maximum average temperature of the resultant steam - but even so, it appears > you did not first deduct the dissociation energy of O2 and H2
Their formation enthalpy is zero, by convention > and then later > deduct the parasitic losses of NOx, peroxides etc. and the other losses that > are expected in actual practice, for combustion in air? Negligible > IOW there are lies, damn lies, and theoretical calculations ;) when trying > to go from 'paper numbers' to actual practice. Kitamura's numbers were > indicated to be actual practice (if they can be trusted) so it is fair to > contrast those numbers with that which would happen if one were to actually > burn H2 in air - and .5 eV is a fair estimate No (see below) > even if you discount the 80% > of air which is nearly inert. why would you not discount them??? > Since water can be split into H2 and O2 with 1.23 volts - does it stand to > reason that one could get 1.5 eV in return ? That was rhetorical; and of > course this one of nature's built-in cases of "systemic overunity" - This was not rhetorical at all actually, I hadn't made the connexion but yes, the combustion energy per D atom in eV should be, of course, exactly equal to the thermoneutral electrolysis voltage... and it is, as a matter of fact: the thermoneutral voltage for electrolysis of D2O is 1.54V, which confirms my 1.53V calculation. And BTW, it's 1.48V for H2O, not 1.23V. > ... except for the damn lie that it simply does not work out that way in > practice - but it does serve to contrast the large disparity of the "actual > with the calculated". > >> Did I get it wrong? > > Well, let's say that you got it partly right and mostly wrong Or rather, as it turns out, exactly right. Physics works, contrary to your suggestions :) Besides, you don't have to take my word, see http://en.wikipedia.org/wiki/Heat_of_combustion Hydrogen: 140 kJ/g, which is about 1.5eV per atom. The important result here is that the 2 eV you get by letting an hydrogen atom bond to the _surface_ of a Pd nanoparticle are comparable with the chemical energy you get by letting it bond to an oxygen atom (starting from molecular gas phase in both cases) Michel
