Please check my math about this engine efficiency calculation

AI Thread Summary
The discussion centers on the efficiency of injecting hydrogen and oxygen, produced through electrolysis, into a gasoline engine. The original poster calculates the energy required for electrolysis and the energy released from combustion, initially expressing confusion over the math and efficiency. Key errors identified include misunderstanding the conservation of energy and unnecessary calculations involving Avogadro's number. It is clarified that the energy output from combustion should equal the energy input from electrolysis, with adjustments made for temperature affecting electrolysis efficiency. Ultimately, the conversation emphasizes the importance of accurate calculations and understanding the principles of energy conservation in chemical reactions.
Le Luc
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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.
 
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Le Luc said:
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.
It isn't clear to me what you are trying to do. You used the word "efficiency", but efficiency is expressed in percent and there are no percentages in your math. It almost looks like you are trying to calculate how much energy you get back by burning hydrogen and oxygen, which should be exactly how much you put into split it from water (per conservation of energy)...

...In step 5 you made the oxygen disappear...

But yeah, your big error is in step 7, in which you multiplied by Avagadro's number for no reason that I can see. You already had units of gallons, so there is no conversion needed.

Ultimately all you really did is do a really roundabout calculation of the heat of vaporization of water.
 
russ_watters said:
It isn't clear to me what you are trying to do. You used the word "efficiency", but efficiency is expressed in percent and there are no percentages in your math. It almost looks like you are trying to calculate how much energy you get back by burning hydrogen and oxygen, which should be exactly how much you put into split it from water (per conservation of energy)...

...In step 5 you made the oxygen disappear...

But yeah, your big error is in step 7, in which you multiplied by Avagadro's number for no reason that I can see. You already had units of gallons, so there is no conversion needed.

Ultimately all you really did is do a really roundabout calculation of the heat of vaporization of water.
Thank you very much. Since the energy I get out of combustion needs to be less than the energy I put in through electrolysis, would that mean that I would need to have more than 60060 kJ of input to get 60060 kJ of output, since step 7 without Avagadro's number is 105*572 = 60060 kJ?

In step 3, I got 50778 kJ to split 210 moles of water, so where would the missing 9282 kJ be? EDIT: I think I got the wrong number for energy required for electrolysis/mole. Now I got 285.8*210 = 60018 kJ and I see now that it is much closer to 60060 kJ. Is 60018 close enough to 60060?
 
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Le Luc said:
Thank you very much. Since the energy I get out of combustion needs to be less than the energy I put in through electrolysis...
Do you understand that that is impossible? The reactions are reversible - the energy is the same going forwards and backwards. This is demanded by conservation of energy.
...would that mean that I would need to have more than 60060 kJ of input to get 60060 kJ of output, since step 7 without Avagadro's number is 105*572 = 60060 kJ?
Pay attention to the state (liquid, gas) of the products and reactants. If the energy you get back isn't exactly the same as the energy you put in, it is probably because you started with one and ended with the other.
 
russ_watters said:
Do you understand that that is impossible? The reactions are reversible - the energy is the same going forwards and backwards. This is demanded by conservation of energy.

Oh. I said that because of electrical resistance. Sorry for not stating that.

Since temperature affects electrolysis, then my missing 42 kJ was probably from using the wrong temperature for water. Finally. I think I get it now.

Thank you very much.
 
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