I put you raw copy so check before debate.
I simplify the writing by turning the electronic charges:
code: Select all
2H --> H2 + 437,6 kJ/mol
O --> 1/2 O2 + 248,4 kJ/mol
These 2 first enthalpies would be to check I do not know where I found them!
code: Select all
H2 + 1/2 O2 --> H2O + 242,7 kJ/mol (celle là, tout le monde connait)
So if we start from atoms to get water here are the enthalpies released:
code: Select all
2H + O --> H2 + 1/2 O2 (+ 437,6 + 248,4 kJ/mol) --> H2O (+ 242,7 kJ/mol)
Total of the energy released by this reaction of "combustion" of the atomic elements: 437,6 +248,4 + 242,7 = 928,7 kJ / mol
You know that obviously thinks about doping with water.
So 1 L of fractional water at 100% in a motor (under the effect of heat of combustion) in H ^ + and O ^ 2- would give:
928,7 * 1000 / 18 = 51 600 kJ / L (because molar mass of water = 18 g / L and 1 L of water = 1000 g) and it is more than oil! It is about 1L of water = 1,3L / oil
If we make the same reasoning but starting from the H2 dihydrogen we have: 120 000 kJ / kg H2 (121 300 according to the enthalpy)
1 L of water = 2 / 18 * 1000 = 111,1 g of H2 is 120 000 * 0.111 = 13 330 kJ / L of water.
We see that it is much less than if we started with atomic elements!
Conclusion: 1 L of fractional water with 100% in di-hydrogen would thus contain the equivalent of 0.37 L of oil it is 3.5 times less than passing through the atomic states.
In other words: atomic hydrogen is 3.5 times more energy than diatomic hydrogen.
Question: do we arrive at atomic states in an engine? This is likely at some high T ° so load and since the reduction in consumption is related to the (high) load engine ... there is surely a rational explanation track (one more) to doping at the water.
In short: reasoning to check and track to follow ...