Junk_Enginerd
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What causes the delays in a HCl+xNO3+Cu reaction?
I've used this reaction to produce NO2. I'd like to understand it better, because it acts really odd sometimes.
I usually mix HCl and a nitrate, typically NaNO3, but I've used KNO3 as well, and then add copper. As far as I understand it, HCl can't dissolve Cu,
but CuO is fine, and the nitrate donates oxygen to make that happen, allowing CuO to be dissolved and releases NO2 in the process.
The thing I'm not getting is how it is incredibly slow to get going sometimes. The usual is that it takes about 10 minutes before it is going full
speed. At this pace 100-200 ml of liquid can dissolve maybe 15 grams of copper in a minute or two. Sometimes it refuses to get going. Sometimes it
proceeds at a blazing pace and consumes all the copper, only to refuse to start up again when new copper is added.
I just don't understand it. It doesn't seem to be primarily temperature that controls this behaviour, nor a limiting reagent. What's happening?
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Bedlasky
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This is not a good rout how to produce NO2. Your NO2 will be contaminated with NOCl and Cl2. Use KNO3 and H2SO4 mixture instead of it.
I never tried dissolve Cu in aqua regia, but from my experience with molybdenum, it reacts much faster than nitric acid alone due to complex
formation. Essentially this is what happens when you dissolve Cu in aqua reagia:
2HNO3 + Cu + 2H3O+ --> Cu2+ + 2NO2 + 4H2O
2HNO3 + 3Cu + 6H3O+ --> 3Cu2+ + 2NO + 10H2O
Cu2+ + 4Cl- <--> [CuCl4]2-
HNO3 + 3HCl --> NOCl + Cl2 + 2H2O
NOCl <--> NO + Cl2
NO + O2 --> NO2
NO2 <--> N2O4
[Edited on 5-7-2020 by Bedlasky]
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Junk_Enginerd
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Quote: Originally posted by Bedlasky  | This is not a good rout how to produce NO2. Your NO2 will be contaminated with NOCl and Cl2. Use KNO3 and H2SO4 mixture instead of it.
I never tried dissolve Cu in aqua regia, but from my experience with molybdenum, it reacts much faster than nitric acid alone due to complex
formation. Essentially this is what happens when you dissolve Cu in aqua reagia:
2HNO3 + Cu + 2H3O+ --> Cu2+ + 2NO2 + 4H2O
2HNO3 + 3Cu + 6H3O+ --> 3Cu2+ + 2NO + 10H2O
Cu2+ + 4Cl- <--> [CuCl4]2-
HNO3 + 3HCl --> NOCl + Cl2 + 2H2O
NOCl <--> NO + Cl2
NO + O2 --> NO2
NO2 <--> N2O4
[Edited on 5-7-2020 by Bedlasky] |
Hm. I wasn't aware. Well, I use NaHSO4 + NaNO3 now anyways.
I'm not sure I'm following. Which of these reactions might explain the delay and sudden boost in the reaction?
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AJKOER
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My advice, test with a small amount of your KNO3 of replacing the HCl with NaHSO4 and check the amount of NO2 generated.
Some understanding of the reaction mechanics would likely be helpful to explain my suggested precaution. To quote a source: "Fenton chemistry in
biology and medicine*" by Josef Prousek, to quote reaction (15) on page 2330: "For Fe(II) and Cu(I), this situation can be generally depicted as
follows [20,39],
Fe2+/Cu+ + HOX → Fe3+/Cu2+ + .OH + X- (15) where X = Cl, ONO, and SCN. "
Now, in the case of Cu (or Cu+) acted upon by HONO2, we have:
Cu/Cu(+) + HONO2 → Cu(+)/Cu(2+) + .OH + NO2-
The above fenton-type chemistry could be followed by:
.OH + NO2- = OH- + .NO2
with the creation of nitrogen dioxide from concentrated nitric acid as is reported.
Now, why is introducing the HSO4- or SO4(2-) a potential bad idea, in my opinion, because the sulfate is a known scavenger of the hydroxyl radical
(.OH) per the reaction:
.OH + HSO4- = H2O + .SO4-
with the introduction of the sulfate radical.
Now, one might think:
.SO4- + .NO2- = SO4(2-) + .NO2
However, I am not sure what direction is favored in the above radical equilibrium reaction. So, if you get little to no nitrogen dioxide gas
generation, you now have an understanding as to why.
On concerns of NOCl creation, note per Wikipedia on NOCl to quote:
"In nitric acid, NOCl is readily oxidized into nitrogen dioxide."
Link: https://en.wikipedia.org/wiki/Nitrosyl_chloride
[Edited on 7-7-2020 by AJKOER]
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Bedlasky
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From my experience not all nitrosyl chloride is oxidized. When I dissovled molybdenum in aqua regia, solution was yellow due to dissolve NOCl. After
dilution NOCl hydrolyzed and solution became colourless.
White fuming nitric acid have better oxidation abilities than azeotropic and still - if you pour some of it on to a solid KCl, you obtain yellow gas
and yellow solution. If you add more acid gas turned more in to reddish, but still this isn't colour of pure NO2. See photos:
https://colourchem.wordpress.com/2020/02/24/chemical-propert...
| Quote: Originally posted by AJKOER |
However, I am not sure what direction is favored in the above radical equilibrium reaction. So, if you get little to no nitrogen dioxide gas
generation, you now have an understanding as to why.
[Edited on 7-7-2020 by AJKOER] |
There must be NO2 or NO formation. Nitric acid must be reduce in to something because it oxidized copper. Nitrous acid isn't stable. N2O, N2 and
NH4NO3 are produce only at low concentrations of HNO3. And NO is easily oxidized by air to NO2.
Btw. I forgot in my originals post that NOCl and Cl2 can also act as oxidizers to Cu.
[Edited on 7-7-2020 by Bedlasky]
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chornedsnorkack
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| Quote: Originally posted by Junk_Enginerd | I've used this reaction to produce NO2. I'd like to understand it better, because it acts really odd sometimes.
The thing I'm not getting is how it is incredibly slow to get going sometimes. The usual is that it takes about 10 minutes before it is going full
speed. At this pace 100-200 ml of liquid can dissolve maybe 15 grams of copper in a minute or two. Sometimes it refuses to get going. Sometimes it
proceeds at a blazing pace and consumes all the copper, only to refuse to start up again when new copper is added.
I just don't understand it. It doesn't seem to be primarily temperature that controls this behaviour, nor a limiting reagent. What's happening?
|
This sounds like an autocatalytic reaction. Some reactive intermediates formed by the reaction catalyze the reaction, and decompose when the reaction
stops so that the reaction again is slow when restarted.
The question is what those reactive catalysts are in this specific case, and how to produce those actively?
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Tsjerk
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| Quote: Originally posted by AJKOER | My advice, test with a small amount of your KNO3 of replacing the HCl with NaHSO4 and check the amount of NO2 generated.
Some understanding of the reaction mechanics would likely be helpful to explain my suggested precaution. To quote a source: "Fenton chemistry in
biology and medicine*" by Josef Prousek, to quote reaction (15) on page 2330: "For Fe(II) and Cu(I), this situation can be generally depicted as
follows [20,39],
Fe2+/Cu+ + HOX → Fe3+/Cu2+ + .OH + X- (15) where X = Cl, ONO, and SCN. "
Now, in the case of Cu (or Cu+) acted upon by HONO2, we have:
Cu/Cu(+) + HONO2 → Cu(+)/Cu(2+) + .OH + NO2-
The above fenton-type chemistry could be followed by:
.OH + NO2- = OH- + .NO2
with the creation of nitrogen dioxide from concentrated nitric acid as is reported.
Now, why is introducing the HSO4- or SO4(2-) a potential bad idea, in my opinion, because the sulfate is a known scavenger of the hydroxyl radical
(.OH) per the reaction:
.OH + HSO4- = H2O + .SO4-
with the introduction of the sulfate radical.
Now, one might think:
.SO4- + .NO2- = SO4(2-) + .NO2
However, I am not sure what direction is favored in the above radical equilibrium reaction. So, if you get little to no nitrogen dioxide gas
generation, you now have an understanding as to why.
On concerns of NOCl creation, note per Wikipedia on NOCl to quote:
"In nitric acid, NOCl is readily oxidized into nitrogen dioxide."
Link: https://en.wikipedia.org/wiki/Nitrosyl_chloride
[Edited on 7-7-2020 by AJKOER] |
AJKOER... There is no Fenton reaction going on here. There is absolutely no reason to think there is any hydroxyl radical present at any time. There
is a bunch of oxidizers present here, all perfectly capable of oxidizing copper.
You even explain why there is no need to think this reaction is going via the hydroxyl radical... if sulfate is such a good scavenger, then why does
this reaction run perfectly well with sulfuric acid/nitrate?
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AJKOER
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Per Tsjerk:
"You even explain why there is no need to think this reaction is going via the hydroxyl radical... if sulfate is such a good scavenger, then why does
this reaction run perfectly well with sulfuric acid/nitrate?"
OK, NOT my imagination that the reaction proceeds via a hydroxyl radical, it is ascribed to one Josef Prousek (and others), out of the Slovak
University of Technology in Bratislava, in their Department of Environmental Engineering, holding an MS, PhD.
Note, I did not say that a reaction would NOT proceed, I questioned to what extent, hence employing a small test amount and doing a rough visible
comparison on the quantity of NO2 generated.
Now, I may have misread the literature of whether the sulfate anion is a scavenger of the hydroxyl radical. Here is a quote from a source (https://www.sciencedirect.com/science/article/abs/pii/000398...):
"The sulfate anion radical is known to react with many compounds more commonly thought of as hydroxyl radical scavengers such as formate and ethanol.
Free radicals derived from these scavengers are trapped in systems where (bi)sulfite peroxidation has been inhibited by these scavengers."
But one could also read the above to claim that the NO3- is a scavenger of .OH (forming .NO2) which is removed by .SO3-, but the latter can still
engage in a reaction with NO3- to create .NO2 as I noted.
But alas, you do not apparently ascribe at all to any radical activity here, accept of course, to the formation of the stable free radical .NO2 itself
somehow.
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kmno4
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This behaviour of Cu in HNO3 solutions is known for ages.
To start the reaction at once, HNO2 must be present (but it does not mean that HNO2 itself is an active catalyst). Interesting - when urea is
present, no reaction occurs, ha. Just urea acts as HNO2 scavenger.
So, you can add some NaNO2 at the begining to the acid and reaction with Cu starts at once.
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Героям слава !
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woelen
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I noticed the same with adding KBr or NaBr to 60% HNO3. If you do this, no reaction occurs initially, and the reaction slowly starts or does not start
at all if it is cold, like in winter. If you add a pinch of NaNO2, then suddenly all of the bromide is oxidized very quickly, and copious amounts of
Br2 are produced (you can tell the difference with NO2 by different color and smell).
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Junk_Enginerd
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| Quote: Originally posted by kmno4 |
This behaviour of Cu in HNO3 solutions is known for ages.
To start the reaction at once, HNO2 must be present (but it does not mean that HNO2 itself is an active catalyst). Interesting - when urea is
present, no reaction occurs, ha. Just urea acts as HNO2 scavenger.
So, you can add some NaNO2 at the begining to the acid and reaction with Cu starts at once. |
Ahhh now this makes total sense. I've actually done this indirectly, come to think of it. Sometimes when it's been too slow to start, I've scraped
some of the dry left over stuff from earlier reactions and noted that it kickstarts the reaction like some sour dough starter culture lol. It's been a
little unpredictable, but I suppose that would depend on whether that particular reaction consumed all the NaNO2 or not.
Thanks a lot. I hate not understanding why something is happening, even if I can sometimes guess at a workaround.
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kmno4
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One more thing.
This example cleanly shows, how misleading chemical equations may be.
For example reaction given above:
2HNO3 + 3Cu + 6H3O+ --> 3Cu2+ + 2NO + 10H2O
This shows only stoichiometry of reaction, but strictly speaking, pure Cu and HNO3(aq) do not react. It is because of high activation
energy and/or steric factors, even though it is energetically allowed. Formation Cu2+ and NO3-/NO is only end effect, telling nothing about this
multistep and, as was said, autocatalytic reaction.
What is exact nature of Cu and "N-something" initial interaction ?
I do not know, but as far as I remember, there are (at least) several papers trying to solve the puzzle.
The (initial) need of HNO2 presence in various HNO3 oxidations is commonly known. Again - strictly speaking, in these cases there is no HNO3
oxidation.
HNO3 serves as "N-something" reoxidizer, but according to Hess's law (for example), total heat effect corresponds to a "simple" reaction:
Red + HNO3 -> Ox + NO (or N2O) + H2O
[Edited on 8-7-2020 by kmno4]
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