2009/12/28 Jones Beene <jone...@pacbell.net>: > - 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 -
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 294.6 kJ/mol is the energy released per mole of D2O formed (=minus the enthalpy of formation of D2O), which I divided by 2 (2 D per D2O) and by Avogadro's number and then converted to eV to find the burning energy in eV per D atom. Did I get it wrong? Michel
