- #1
Le Luc
- 10
- 2
Hello. I wanted to see how much more efficient a car would be if hydrogen and oxygen (created through electrolysis of water) were injected into the engine alongside gasoline. I got lost when I found out that I would need 241.8 kJ/mol of water to electrolyze it (https://chemistry.stackexchange.com...ch-energy-is-needed-to-break-a-water-molecule), but I would get 572 kJ just from the combustion of 2H2+O2 (https://www.quora.com/How-much-energy-is-released-in-hydrogen-combustion-reaction).
So here are the steps I took:
1) 1 gallon water is 210 moles
2) 241.8 kJ to split 1 mole of water
3) 241.8*210 = 50778 kJ
4) 50778 kJ to split 210 moles
5) 2H2(g) + O2(g) --> 2H2(g) + 572kJ
6) 210/2 = 105 (because the formula produces 2 H2O)
7) 105*(6.02214179*10^23)*572 = 3.616898359*10^28 kJ
Sorry if the math is incorrect beyond explanation.
To clarify step 7, I divided the # of moles because there were 2 water molecules produced.
I think step 7 is incorrect.
So here are the steps I took:
1) 1 gallon water is 210 moles
2) 241.8 kJ to split 1 mole of water
3) 241.8*210 = 50778 kJ
4) 50778 kJ to split 210 moles
5) 2H2(g) + O2(g) --> 2H2(g) + 572kJ
6) 210/2 = 105 (because the formula produces 2 H2O)
7) 105*(6.02214179*10^23)*572 = 3.616898359*10^28 kJ
Sorry if the math is incorrect beyond explanation.
To clarify step 7, I divided the # of moles because there were 2 water molecules produced.
I think step 7 is incorrect.