h a l f b a k e r yMake mine a double.
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Any soft acid medium (even a neutral medium) contains a share of ionized molecules.
H20 -> H3(+)O + OH(-)
If, say, H3PO4 is kept in solution it can release 1, 2, or even 3 protons per molecule, which are not 'free', but hydrated as H3(+)O.
Good. Now, say the proton is the reducer agent. This way,
we prevent the use of any Carbon molecule as a fuel. And what about the oxydizing agent? Not O2 in the air, that's for sure. The most electronegative element, Fluorine, which is easily kept in plastic containers.
The exothermic reaction:
proton + fluoride -> HF + energy
Admittedly, the developed formula is:
2 H3(+)O + F2 -> 2 HF + 2 ENERGY
HF is not a very strong acid (pKa = 3,5). But, it presents abnormal boiling point (293 K), because of the hydrogen bonds HF molecules can build among them.
This would render dissociation in aqueous medium, so that F2 gets recycled:
2 HF + 2 H2O -> 2 H3(+)O + 1 F2 [gas]
The buffering effect (keeping pH almost constant) would be provided by H2PO4(-) and HPO4(2-) species, the cognate bases for phosphoric acid.
Then, the steam-machine power scheme could be easily adapted so that the energy from the reactor activates a turbine, and that's it!
I know it's halfbaked, but... can anyone see to some references?
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Erm, am I right in thinking that if you'd missed out the unnecessary steam/turbine steps, you would have just invented the car battery? |
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Hydrofloric acid is not something you want to have anything to do with period. There are much greater issues than just the strength of an acid. |
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"Deaths have been reported from concentrated acid burns to as little as 2.5% BSA." |
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This is not stuff you want to be anywhere near or involved with period. No questions asked. |
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Secondarily how exactly are you getting this solution to boil at 293K. You will not get that much heat out of the formation reaction. I think you just twiddled around in a chemistry book and found some nice sounding formulas that you dont understand. |
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Lets see you put some calculations together showing the number of jouls generated in the formation reaction the number of jouls required to boil the HF acid and seperated the H and F from one another. and exactly how much steam this would generate and then how much power that would represent. |
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I think you will find very quickly that the answer to many of these questions will be "not enough" |
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Yeah, I guess zet_tom is right. |
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For all other purposes, see below: |
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Enthalpy change per HF formed: -271'1 kJ/mol. |
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Depending on the acid you add at the beginning of the cycle, which is the real proton donor, well, pH DOES matter a lot.
If pH drops below HF's dissociation pH, we'll never get F2 back!! |
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As for boiling water, I guess a starter would do, wouldn't it? 373 K (100ºC) is well above HF's boiling temperature, so the real problems are at the beginning and the end of the cycle. |
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(H+ input, for example very low concentration H2SO4, would render around 5,4 MJ/kg; as compared to ethanol combustion's 28 MJ/kg) |
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C'mon, man, I'm not an idiot, if I use chemistry, I understand the concepts behind that. |
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Could you solve, say, SO4(2-) excess?
Would Ca(2+) drop as CaF2 or CaSO4????
Would Pb(2+) do better or worse than that?
Explain your answer.
I can also ask for the explosive mechanism of tri-nitro-toluene, then sit and watch you while you take your exam. |
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The proposal, anyway, is not to exhaust any CO2 at all, and the idea should stick to it were it to be baked.
What's the point in trying to bio-fuel every single motor / engine in the globe?
Not ecological concern, that much I take for granted. |
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how exactly were you planning to generate the required materials with which to power your car(assuming your no CO2 discharge goal that is) |
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It would have saved me some time if you'd
said this was intended as a means of
energy storage rather than energy
production. |
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So, the idea is to make acid and then react
it with a fluoride, yes? |
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[MaxwellBuchanan] got it right, acid + fluoride 'non-combustion'.
Sorry I wasn't straightforward before. |
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Made some calculations on efficiency:
It seems -540 kJ could be obtained per mol H2SO4.
It takes +2260 J to evaporate 1 gram of water. |
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Roughly 4'2 Molar concentrarion H2SO4 (239 grams) would be needed for 1 liter of water to boil. |
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Under those conditions, sulphuric acid would yield 30 MJ/kg, outperforming ethanol combustion's 28 MJ/kg).
Gave up on this one, though, because it would take real-world experiments to close the SO4(2-) cycle up -or whatever the anion involved. |
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Thanks for your comments, by the way. |
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