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Author: Subject: Eliminating NO2 from HNO3
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[*] posted on 28-5-2002 at 11:48
Eliminating NO2 from HNO3


I have heard that urea, CO(NH2)2, removes NO2 from HNO3. I have also found out from my own experience that oxamide, (CONH2)2, removes NO2 from HNO3. I believe this is what is happening:

2CO(NH2)2 + 2NO2 --> 2CO + 3N2 + 4H2O
2(CONH2)2 + 2NO2 --> 4CO + 3N2 + 4H2O

I believe that H2O2 could also be used to prepare NO2-free HNO3. This method would convert the NO2 into HNO3, instead of simply removing the NO2.

2NO2 + H2O2 --> 2HNO3




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[*] posted on 28-5-2002 at 12:51


Something seems fishy with the H2O2 idea. After the H2O2 has given up an O to become H2O, why should it split its hydrogens among two molecules (the 2HNO3) and give up the remaining oxygen to form one of the HNO3? What I know about thermodynamics you could write on the back of a postage stamp but this just doesn't seem plausible to me. If you had a way of blasting all those pesky NO2s with hydroxyl radicals, that might work, but H2O2 is not two hydroxyl radicals that happen to be hanging out together. Of course I could be very wrong even with this last speculation... In any case, try the H2O2 method out and let us know how it worked.
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[*] posted on 29-5-2002 at 09:29


Would it be possible electrolysing the remaining water in the HNO3 ??

4H20 + electricity --> 4H2 + 2O2

2NO2 + 4H2 + 2O2 --> 2HNO3 + 3H2 + O2

If this could work (Al electrodes) it would be a way to remove NO2 and increase percentage at once !!

Can someone with a degree in chemistry give us some feedback ?
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[*] posted on 29-5-2002 at 18:13


Hi
It could be pissible but I would use platinum electrodes. I think that the Al would corrode.




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[*] posted on 30-5-2002 at 11:43


Everything is pissible exepct high voltage electric fences ;-)

And Al doesn't get corroded by the HNO3, but it probably will by the oxygen evolving from the anode (or is it the kathode ? I allways mix up ..)
So Pt would probably be best choice.
As long as it can withstand pure O2 and HNO3, it will do ...
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[*] posted on 31-5-2002 at 07:18


You have to remember, though, that if there is water in the HNO3, then some of the HNO3 will be ionised, as H3O+ and NO3-. Since these are charged, they will also be attracted to the electrodes. I guess the HNO3 would decompose:
2HNO3 --> H2 + 2NO2 + O2.
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[*] posted on 2-6-2002 at 14:48


I'm fairly certain these equations occur:

2NO2 + H2O --> HNO3 + HNO2
Nitrous acid is unstable, and so this occurs to a degree:
2HNO2 --> H2O + NO + NO2

However, if H2O2 is used instead of H2O:

2NO2 + H2O2 --> 2HNO3

No HNO2, therefore no NOx.

At least, that's my theory.




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[*] posted on 3-6-2002 at 08:58


This would be analogical to the method of reacting SO2 and H2O2 to form sulfuric acid. I don't see why it couldn't be possible.
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[*] posted on 7-6-2002 at 16:06


Purging NO2 from HNO3 with oxamide doesn't work. Details have been posted in the nitroxamide thread, in the Energetic Materials section. Here's a link for the lazy: http://www.sciencemadness.org/talk/post.php?action=reply&fid...



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mad.gif posted on 28-9-2002 at 07:09


You destroy a little part of highly concetrated HNO3 when you remove the
NOX with adding urea... :mad:
(HNO3 become diluted !)
The best method i know,
warm up the HNO3 careful to ~ 40 C
and bubble a slight stream of dry CO2 gas
through the HNO3 for 15-20 min.




all you need is C - H - O - N
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[*] posted on 28-9-2002 at 08:43


H2O2 did remove NO2 from yellow HNO3.

I'm guessing that the CO2 method of purging NO2 from HNO3 works by displacing the NO2 from solution... a good idea.




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[*] posted on 28-9-2002 at 11:21


I think,
it`s a very expensive method
to produce H2O and gas by
adding urea to HNO3 !:mad:
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thumbup.gif posted on 31-10-2003 at 06:49


"Something seems fishy with the H2O2 idea. After the H2O2 has given up an O to become H2O, why should it split its hydrogens among two molecules (the 2HNO3) and give up the remaining oxygen to form one of the HNO3? What I know about thermodynamics you could write on the back of a postage stamp but this just doesn't seem plausible to me. If you had a way of blasting all those pesky NO2s with hydroxyl radicals, that might work, but H2O2 is not two hydroxyl radicals that happen to be hanging out together. Of course I could be very wrong even with this last speculation... In any case, try the H2O2 method out and let us know how it worked."
There's nothing fishy with that. The O-O bond in H2O2 is the weakest of all, that's why it's such a atrong oxidizer. Sunlight can split it, and hey presto - two OH radicals. :D
On second thoughts, why am I writing this, if it works it doesn't need no theoretical backup. :)
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[*] posted on 31-10-2003 at 08:00


NOx doesnt seem to dissolve in HNO3 DCM extracts....



One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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[*] posted on 31-10-2003 at 12:06
Dont mean to sound like an ass, but.....


You know, mny of these suppositions can be shooed away by a little literary searching. And linear equations of x + y --> z + a don't always work just because the reactantsand products merely balance out.

NO2 does in fact react with H2O2, that part of the reaction happy readily. The problem is that HNO3 also reacts with H2O2, and this leads to a messy mix if more than just a little H2O2 is used (read high concentration) with respect to HNO3.

I could probably give you some suggestions madscientist if I knew what you were attempting to ccomplish. It appears that you are attempting tocircumvent the issue of not enough oxygen in the reaction of NO2 + H2O to form nitric acid.

A question for you: have you considered what your measurable results will be? Very poor concentration of HNO3 in water, likelynot above 40%. If that's ok with you (distill it) then there you go.

In the reaction of NO2 and H2O, it is more convenient to describe what occurs if we explain what happens when N2O4 is subjected instead. I currentlyhave no drawing software, so I will rely on description of the molecule N2O4. Label the N atoms 1 and 2, and each N atom's oxygens a and b, resulting in N1 N2 O1a O1b O2a and O2b.

N2O4 is extremely weakly bonded at the N-N junction,much moreso than most other N-N single bonds. Personally, I am quite surprised the molecule exists at all at easily achievable temperatures, but it is nonetheless. This weak bond is largely due to positive polarization on both nitrogen atoms.

In NO2 we can write its structure as O=N(+)->O(-) meaning that the nitrogen is mostly covalently bonded using two orbitals to one oxygen atom (the left one here) and polar covalently to theother oxygen, with one orbital retaining covalent character and the other constrained to polarization. Of course which is which is not locked into place; they resonate. There is also resonance between which oxygen atoms are involved in the polarization. But for the large part, the nitrogen retains somepositive character.

This positive character weakens the N-N bond quite a chunk, which is why we see liquid and solid N2O4 but not gaseous, instead it deomposes to NO2. When N2O4 reacts with water, it is because it either reacts directly with solvated hydroxide ion, or more likely with the lone pair on an H2O oxygen, which then spits out a proton. The net result is the same, since for a hydroxide to exist for N2O4 to react with (in a neutral water solution) a proton must correspondingly exist elsewhere. The points of attack are effectively identical.

The oxygen of HO- attacks the positive character of N1, causing the electron density in this region to skyrocket temporarily. This increases the attraction between N1, now largely negative due to increased resonance, and N2(still positive) and the bond distance shrinks. The inductive effect is responsible for the next step, where the electron-pulling effect from N2's oxygens is carried all the way through the nitrogens to O1a and O1b, where one of them pushes an electron back to N1 as N1 extends an electron simultaneously to N2, breaking their bond and releasing the intermediate entity -O-N(+)-O- which is of course the nitrite ion. This ion now corresponds to the proton that belonged to our reactant hydroxide ion in our neutral water solution, which together account for the complete nitrous acid molecule, HNO2.

The action of O1a,b pushing an electron into the bond with N1 involves the conversion of the -O-N bond to a covalent O=N bond, which yields the structure of nitric acid, described in its unionized state, H-O-N(+)(=O)(->O-).

So N2O4 + H2O does yield equal amounts of nitric and nitrous acids; however this is only true in the beginning. Nitrous acid is a weak acid, and nitric acid is a strong acid. HNO3 in higher concentration (contrasted against H20, not HNO2) will protonate HNO2. At this point where the concentration is high enough for this first step to occur, it is likely to also be reaching the point at which introduction of water into the solution in a reversable process would start to favor water, thus HNO2 when protonated has a tendency to lose water in this situation, giving up N=->O+ where =-> represents a polar triple bond. Resonance dictates that the entity N(+)=O also exists.

A better way to destroy the HNO2 produced would be to oxidize it to HNO3 with air and catalytic Cu+2 ions. Cu(NO3)2 (maybe even produced in situ) would be a great catalyst for such a reaction. Cu+2 will oxidize the completely neutral nitrogen in H-O-N=O to reduce itself to Cu+1.

The response from the nitrous acid is to inductively spit out its proton yielding wht looks similar to a nitrite ion, but is actually an in-situ generated NO2 molecule, O=N(+)->O(-).

Oxygen easily oxidizes any Cu+1 present to Cu+2 when protons are present to form water.

For best results, I would assume passing a stream of not pure NO2 into water, but NO2 and air into a solution of water and Cu(NO3)2.

PrimoPyro
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sad.gif posted on 23-9-2017 at 03:57


The possible answer to removing and converting any excess NO2 to HNO3 is ozone. Source: https://books.google.com/books?id=FvbzrNYcvLsC&pg=PA151&...

Reactions gas phase:

.NO2 + O3 = .NO3 + O2

.NO3 + .NO2 = N2O5

N2O5 + H2O = 2 HNO3

Note, no gas phase reaction between O3 and HNO3, which is in line with the atmospheric research, see, for example, https://www.google.com/url?sa=t&source=web&rct=j&... , but in a HNO3/water mix, some reactive oxygen species and products may come into play (like .OH, .HO2, H2O2,...) with ozone.
------------------------------------------------------------------------------------

Another path, per the same source, introduce the hydroxyl radical:

.OH + .NO2 = HNO3

Or, add CO2 in the presence of the hydroxyl radical, creating the carbonate radical anion to further react with the nitrogen dioxide:

CO2 + H2O = H+ + HCO3-
HCO3- + .OH = H2O + .CO3-
.CO3- + .NO2 = CO2 + NO3-

To produce the required hydroxyl radicals, without introducing a problem contaminant, add say N2O to the acid. Then UV treatment in a sealed quartz vessel:

N2O + hv = N2 + .O-
.O- + H2O = .OH + OH- (reaction moves to the right except in highly alkaline conditions)

Other possible side reactions (not all desirable)::

HNO3 + hv = .OH + .NO2
.OH + HNO3 = H2O + .NO3
.....

So the ozone path may be best.
--------------------------------------------------

My last method is a takeoff of the dilutive H2O2 approach without being dilutive.

Try applying sonolysis to the aqueous nitric acid in the presence of oxygen. It is known that the sonolysis of oxygen rich water forms the superoxide radical anion (see http://pubs.acs.org/doi/full/10.1021/acs.chemrev.5b00407 and references cited therein) or, at low pH, .HO2. As an illustrative example of the use of sonolysis on the degradation of a compound, see, for example, http://pubs.acs.org/doi/abs/10.1021/es9502942.

My suspected path:

H2O --sonolysis--) .H + .OH
.H + O2 = .HO2 = H+ + .O2- (only at pH over 4.8)
.HO2 + .HO2 = H2O2 + O2
HO2- + .NO2 = .HO2 + NO2-
.HO2 + NO2- = OH- + .NO3

.OH + .NO2 = HNO3
.NO3 + .NO2 = N2O5
N2O5 + H2O = 2 HNO3

2 .NO2 + H2O = HNO2 + HNO3
H2O2 + HNO2 = H2O + HNO3
.......

Problem reactions:

HNO3 --sonolysis--) .OH + .NO2
.H (or, equivalently e(aq) + H+ ) + NO3- = .OH + NO2-

but the process is simply enough to give it a try, although my opinion stills places the ozone method as the better option.

[Edited on 23-9-2017 by AJKOER]
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[*] posted on 23-9-2017 at 05:59


I haven't checked the thermodynamics but a wild guess would be to use electrocatalysis with a nickel electrode:

N2O4 + 4OH- -> 2NO3- + H2O + 2e-
2Ni+ + 2NO3- +2e- -> 2NiNO3 2

Hmm the HNO3 would react first to make NiNO3 2 so it won't work like this. :D


[Edited on 23-9-2017 by gatosgr]




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[*] posted on 23-9-2017 at 11:13


Quote:
I haven't checked the thermodynamics but a wild guess would be to use electrocatalysis with a nickel electrode

Something somewhat less reactive (and pricier, I'm afraid) than nickel is required for those "fun" reaction conditions.


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[*] posted on 23-9-2017 at 13:56


Classic AJKOER: 14 year old thread, impossible sol'n to an easy problem.

The correct answer is to do... nothing!

http://en.wikipedia.org/wiki/White_fuming_nitric_acid

Quote:
WFNA can be converted from RFNA by simply leaving the RFNA out in low temperature for a couple of hours.


In order to make this less noxious, bubble nitrogen (or air) through the RFNA at 0-5C for a few hours using a carbonate sol'n to contain outgas.

Nitric acid cannot be 100% free of NO2 because it decomposes easily into NO2. But it can be mostly free of NO2, which makes it safer to work with. And for God's sake, have some respect for ozone. It's about 4x as toxic as NO2.

[Edited on 23-9-2017 by clearly_not_atara]
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[*] posted on 23-9-2017 at 23:15


Quote: Originally posted by clearly_not_atara  
Classic AJKOER: 14 year old thread, impossible sol'n to an easy problem.

The correct answer is to do... nothing!

http://en.wikipedia.org/wiki/White_fuming_nitric_acid

Quote:
WFNA can be converted from RFNA by simply leaving the RFNA out in low temperature for a couple of hours.


In order to make this less noxious, bubble nitrogen (or air) through the RFNA at 0-5C for a few hours using a carbonate sol'n to contain outgas.

Nitric acid cannot be 100% free of NO2 because it decomposes easily into NO2. But it can be mostly free of NO2, which makes it safer to work with.

Solids have difficulties incorporating impurities. And tend to be integer molecular compositions.
HNO3 melts at -41 degrees
An eutectic solution of 91 % HNO3 and 9 % H2O melts at -60 degrees.
I do not know the specific melting properties of ternary system N2O4-H2O-HNO3. But I expect that near 100 % HNO3, crystals of solid HNO3 will not incorporate N2O4 either.
So, starting from 98 % nitric acid with modest impurities of H2O and N2O4, freeze it slowly from -41 to -60 degrees. Separate the HNO3 crystals from the mother liquor. Store HNO3 as a solid below its melting point of -41 degrees. If needed as a liquid, thaw shortly before use.
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[*] posted on 24-9-2017 at 04:01


Clearly_not_atara:

I will make it simple for you.

My analysis suggests if one stores the fuming decomposing solution of HNO3 under an atmosphere of ozone, you may no longer have a cloud of fumes (as the NO2 is consumed by the O3).

That is a possible answer (but I agree not at all practical for nearly everyone) to this thread's question on how to visibly eliminate NO2 from HNO3 (at least until all the O3 is consumed, and not possibly regenerated in situ from the created O2 from, say, a pulsing electric arc, like that would be crazy cool). The ideal of storing it below -41 C is also a solution, and requires extraordinary cooling, which is more practical if you has access to a very expensive refrigeration unit (I am cold to this idea).

Doing nothing may be the proper action, but it is not an answer that removes NO2 and preserves the HNO3. Actually, with the ozone approach, or is it the oxygen/electric arc method, there may be even stronger HNO3 as the decomposition reaction has been reversed.

So if one had to transport a large amount of nitric acid, which may be exposed to heating (from the sun or a fire condition), and one could reduce the danger of leaking NO2, while further preserving the quality of the HNO3, inexpensively no less, that method would be of what value?

[Edited on 24-9-2017 by AJKOER]
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[*] posted on 24-9-2017 at 15:07


Quote: Originally posted by hissingnoise  
Quote:
I haven't checked the thermodynamics but a wild guess would be to use electrocatalysis with a nickel electrode

Something somewhat less reactive (and pricier, I'm afraid) than nickel is required for those "fun" reaction conditions.


I'm not so eager to try electrolysis of HNO3 and vacuum distilling NO2 but thanks for the read at 2 am :D:(:P

They have superfreezers in biology labs, putting HNO3 in there will kill all the cultures, plus I doubt it will purify the HNO3, there is a chance it will form hydration shell with NO2.

It's not hydration shell it's forming I don't remember how they call it.

[Edited on 24-9-2017 by gatosgr]




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[*] posted on 25-9-2017 at 03:01


Quote:
I'm not so eager to try electrolysis of HNO3 and vacuum distilling NO2 but thanks for the read at 2 am :D:(:P

Distilling NO2 is necessary in continuous operation, obviously ─ in a home batch reactor you'd just need a supply of RFNA as top-up feedstock.


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[*] posted on 25-9-2017 at 03:16


If your goal is to convert NO2 to HNO3, oxidants H2O2 and O3 are unreliable.
Unreliable because they are both unstable to dismutation:
2H2O2->2H2O+O2
2O3->3O2
You may get lucky and oxidize NO2, but you may instead just have NO2 catalyze the dismutation (or worse).
Cannot happen with O2 itself.
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[*] posted on 25-9-2017 at 06:14


Quote:
If your goal is to convert NO2 to HNO3, oxidants H2O2 and O3 are unreliable.
Unreliable because they are both unstable to dismutation:
2H2O2->2H2O+O2
2O3->3O2

"Dismutation" is more descriptive of reactions in biology, I think, but ozone readily oxidises NO2 to N2O5 in the gas-phase and in solution, making it the preferred oxidiser ─ its high cost of production is a serious drawback, however...


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