In reply to Mike Carrell's message of Thu, 23 Oct 2008 15:48:33 -0400: Hi, [snip] >There is something much simpler. NaH is formed by reactions given from NaOH >coating of the R-Ni and heating. At some point the NaH decomposes, releasing >Na and H atoms in close proximity, whereby Na++ then catalyses the H >producing H[1/3]. There are aspects of this which puzzle me. [snip] According to Randy, the NaH decomposes directly in Na+++ + H[1/3] + 3e- .
Na++ is not a catalyst. (The ionization energy is 71.641 eV). In going from H[1] to H[1/3] the H requires an energy hole of 54.4. eV. This is the sum of the first and second ionization energies of Na (5.1391 eV & 47.286 eV resp.) and the energy required to break NaH into atoms (about 1.98 eV). IOW the molecule can decompose directly into the final products, and in so doing provides its own "energy hole". This is probably why it is so effective (the coupling is all internal within the molecule). BTW the whole hydrino reaction actually produces 108.8 eV, so the difference between the total energy released and the "energy hole" (54.4 eV) will likely be released as additional kinetic energy IMO. Regards, Robin van Spaandonk <[EMAIL PROTECTED]>
