When doing an experiment in the laboratory with copper, dilute nitric acid and finally hydrochloric acid, the first reaction between copper and dilute nitric acid should end in nitric monoxide, cupric nitrate and water:

$$\ce{3 Cu(s) + 8 HNO3(aq) -> 3 Cu(NO3)2(aq) + 4 NO(g) + 2 H2O(l)},\tag{R1}$$

but somehow nitric dioxide, a reddish-brownish gas appears with a poignant smell, develops from this type of reaction:

$$\ce{Cu(s) + 4 HNO3(aq) -> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)}.\tag{R2}$$

There was 5 mL of nitric acid and as we added nitric acid we quickly covered it with a watch glass. Then we put it in a sand bath instead of heating it normally with a Bunsen burner. Unfortunately, I cannot as it was one of the experiments that we just wanted to know what was going to happen and write down our observations and conclusion.


2 Answers 2


There are two ways to interpret an observation nitrogen dioxide.

  1. Reactions between a metal and nitric acid rarely proceed by a single path to generate a specific nitric acid reduction product; in all likelihood there will be a combination of multiple reactions and thus a mixture of products. With copper you could get both $\ce{NO}$ and $\ce{NO2}$. And with a more powerful reducing metal like magnesium even more products could get into the mix, including ammonium nitrate.

  2. You may have conducted the reaction in the presence of air. $\ce{NO}$ is readily oxidized by the air to make $\ce{NO2}$ "after the fact". This is in fact part of the natural production of nitric acid in thunderstorms. Further complicating this is the fact that nitrogen dioxide in turn reacts with water, forming nitric oxide and regenerating some of the nitric acid (this is also indicated in the above link). To get the true product mix out of the nitric acid you needed to use an environment free of air and unintended water.


The cited reaction of Copper metal and dilute Nitric acid has long known to produce ${NO}$ gas, although the precise reaction mechanism remains complex.

The mechanics, in my opinion, likely involves a REDOX pathway with, for example, ${NO2}$ as a radical product in the case of concentrated Nitric acid proceeding, in this case, somewhat simply as follows:

$\ce{Cu + 2 HNO3 -> Cu++ + 2 OH- + 2 .NO2}$

and, with more acid, Copper nitrate, also a cited product, with the net reaction now being completely in agreement with noted Reaction R2 above.

Perhaps, less complex is to quote Wikipedia on the mechanics of the interaction between Nitric oxide and oxygen:

When exposed to oxygen, nitric oxide converts into nitrogen dioxide:

$\ce{2 .NO + O2 → 2 .NO2}$

This conversion has been speculated as occurring via the ONOONO intermediate.

Now, I actually agree with the speculation that ${N2O4}$ is likely of assistance here, but how?

To answer, I refer to another educational source, which again repeats, albeit, noting the improbability of the termolecular pathway, quoting:

An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:

$\ce{2 .NO + O2 → 2 .NO2}$

However, of import on the speculated pathway, to continue quoting:

An elementary reaction is at equilibrium when it proceeds in both the forward and reverse directions at equal rates. Consider the dimerization of ${NO}$ to termolecular, with ${k_1}$ used to represent the rate constant of the forward reaction and ${k_{-1}}$ used to represent the rate constant of the reverse reaction:

$\ce{.NO + .NO ⇋ N2O2}$

So, my opinion as to the speculated pathway that avoids the unlikely, but widely promulgated termolecular reaction (which may instead involve ${N2O4}$) is as follows:

$\ce{.NO + .NO ⇋ N2O2}$ (fast equilibrium)

$\ce{N2O2 + O2 -> N2O4 (slow)}$ Source here

$\ce{N2O4 ⇋ .NO2 + .NO2}$

where the net reaction is in accord with the more widely claimed single step reaction:

$\ce{2 .NO + O2 → 2 .NO2}$

implying, I would argue, only apparently speculatively (as there is a confirming source) that there is no direct action of oxygen on the ${NO}$ radical absence water (in complete contravention to the widely promulgated reaction above), but relatedly per a cited reference, instead on its possible self-reaction non-radical product ${N2O2}$, which then forms the non-radical intermediate ${N2O4}$, being probabilistically more likely the precise path.

Those doubting this comment are welcomed to review the associated biochemistry of the ${NO}$ radical in the human body as, for example, presented in this 2015 article Breathing new life into nitric oxide signaling: A brief overview of the interplay between oxygen and nitric oxide☆.

  • 2
    $\begingroup$ It is absolutely impossible to produce any $\ce{OH-}$ ion out of a reaction consuming an acid like $\ce{HNO3}$. In case this would happen, the $\ce{OH-}$ ion would be immediately destroyed according to : $$\ce{OH- + HNO3 -> H2O + NO3^-}$$ $\endgroup$
    – Maurice
    Dec 7, 2020 at 12:58
  • $\begingroup$ @Maurice: Actually what it implies, is that an excess of HNO3 is favorable in moving the cited net reaction to the right! $\endgroup$
    – AJKOER
    Dec 7, 2020 at 14:22
  • $\begingroup$ Supporting my comment to Maurice, interesting readings here chemistry.stackexchange.com/questions/17067/… which was already a cited reference by Nilay Ghosh above. $\endgroup$
    – AJKOER
    Dec 7, 2020 at 14:48
  • $\begingroup$ Continuing, reading a prior question's answer link just provided an interesting reply by Oscar Lanzi, which accounts for NO in dilute conditions via a secondary reaction involving NO2 with excess water. $\endgroup$
    – AJKOER
    Dec 7, 2020 at 14:59

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