(1:1 ratio of oxygen to hydrogen)
sorry, I meant 2 hydrogen atoms to one oxygen atom
(may be broke/outdated!)
(1:1 ratio of oxygen to hydrogen)
sorry, I meant 2 hydrogen atoms to one oxygen atom
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Per the first link below, to cleave both the OH bonds in water would require 493.4 kJ/mol + 424.4 kJ/mol = 917.8 kJ/mol.
Per the second link below, to cleave an O=O bond requires 494 kJ/mol and to cleave an H-H bond requires 432 kJ/mol.
So, we take two H2 and one O2 molecules, cleave their bonds:
494 + 432 + 432 = 1358 kJ / mol O2
We then reassemble them into two H2O molecules: 2 * 917.8 kJ/mol = 1835.6 kJ / mol O2.
Thus, we generate 1836 – 1358 = 478 kJ / mol O2.
Unfortunately, if your 1:1 ratio is a molar ratio, then half of the oxygen doesn’t get used up. The other half, however, is available for the reaction. In 1 mol of gas particles we have 0.5 mol H2, 0.25 mol reacting O2, and 0.25 mol nonreacting O2. Assuming a 100% fuel consumption rate, you would then have 120 kJ / mol gas, one fourth of the 478 quoted above.
And one mole of an ideal gas at STP is something like 25L, so that would be something like 5 kJ/L.
Compare with petrol, which I think is something like 30 MJ/L.