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I'm reading this answer which refers to both red fuming nitric acid (RFNA) and white fuming nitric acid (WFNA).

Those two Wikipedia articles say:

Red fuming nitric acid (RFNA) is a storable oxidizer used as a rocket propellant. It consists of 84% nitric acid (HNO3), 13% dinitrogen tetroxide and 1–2% water. The color of red fuming nitric acid is due to the dinitrogen tetroxide, which breaks down partially to form nitrogen dioxide. The nitrogen dioxide dissolves until the liquid is saturated, and evaporates off into fumes with a suffocating odor. RFNA increases the flammability of combustible materials and is highly exothermic when reacting with water.

White fuming nitric acid (WFNA) is a storable liquid oxidizer used with kerosene and hydrazine rocket fuel. It consists of nearly pure nitric acid (HNO3). WFNA is commonly specified as containing no more than 2% water and less than 0.5% dissolved nitrogen dioxide or dinitrogen tetroxide.

WFNA as an oxidizer has somewhat less performance than red fuming nitric acid (RFNA) but is considerably safer (though extremely corrosive), as it has little to no dissolved nitrogen tetroxide, which is an extremely toxic and volatile chemical. If not inhibited, it will form nitrogen tetroxide on contact with most metals and some organic materials. WFNA can be converted from RFNA by simply leaving the RFNA out in low temperature for a couple of hours. (emphasis added)

I'm asking about that last sentence; why would red fuming nitric acid convert to white fuming nitric acid by leaving it out at low temperature?

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    $\begingroup$ The first two paragraphs you are quoting answer your question from the third one. By cooling down the solution with nitrogen dioxide, one shifts the equilibrium towards the formation of nitrogen tetroxide: $$\ce{\color{gray}{N2O4} <=>[^ $T$][v $T$] 2\color{brown}{NO2}}$$ $\endgroup$ – andselisk Aug 22 '17 at 15:26
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    $\begingroup$ @andselisk I'm having trouble understanding how RFNA is a "storable oxidizer" if the $\ce N_2 \ce O_4$ breaks down. Is it the partial pressure of $\ce N \ce O_2$ that keeps it stable, and letting the $\ce N \ce O_2$ escape leads to loss of $\ce N_2 \ce O_4$ by driving your reaction to the right, or is it the evaporation of the $\ce N_2 \ce O_4$ itself? $\endgroup$ – uhoh Aug 22 '17 at 15:38
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    $\begingroup$ "Breaks down" makes it sound like a permanent change, but it's reversible. When stored, they reach an equilibrium based on temperature and stay there. $\endgroup$ – Russell Borogove Aug 22 '17 at 15:47
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    $\begingroup$ Ehhm I'm not sure we are on the same page. Both RFNA and WFNA are oxidizers not because of the nature of nitrogen oxides dissolved, but because both are mainly $\ce{H\overset{+5}{N}O3}$. If you purge the system, say, when distilling nitric acid, then yes, the equilibrium between $\ce{NO2}$ and $\ce{HNO3}$ will shift, and at some point you get rid of dissolved gas (no more brown fumes). $\endgroup$ – andselisk Aug 22 '17 at 15:48
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    $\begingroup$ @andselisk the distinguishing feature between RFNA and WFNA seems to be the amount of dissolved $\ce N_2 \ce O_4$; ~13% and <0.5% respectively. When one "leav(es) the RFNA out in low temperature for a couple of hours" to make WFNA, what exactly is happening? This is the page that I and my question are on. If you can clearly explain what is happening, then please consider posting it as an answer. Thanks! $\endgroup$ – uhoh Aug 22 '17 at 15:57
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You can convert RFNA to WFNA by removing dissolved nitrogen oxides ($\ce{N2O4}$ and $\ce{NO2}$) from the RFNA.

Given the volatility of $\ce{N2O4}$ and $\ce{NO2}$ it might be possible to just let the gases evaporate from the solution at low temperature. However, since the boiling point of $\ce{N2O4}$ if about $21.69~\pu{°C}$, I do not think that this is a very practical method.

Another path that is theoretically possible is the reaction of residual water with excess $\ce{N2O4}$ and $\ce{O2}$ from air to $\ce{HNO3}$.

$$\ce{2 H2O + 2 N2O4 + O2 -> 4 HNO3}$$

Again, to me this seems not to be a very plausible way. Since the Wikipedia article does not give a source for its claim, we cannot decide which of the two paths, if at all, is the correct one.

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    $\begingroup$ Why does a room-temperature boiling point (21.69ºC for N2O4 and 21.2º C for NO2) make it impractical to "just let the gases evaporate"? It would seem the opposite to me. Nitric acid's boiling point is some 60 degrees higher. $\endgroup$ – Russell Borogove Aug 22 '17 at 20:24
  • $\begingroup$ I would assume that significant amounts of $\ce{N2O4}$ and $\ce{NO2}$ would stay in solution. Maybe bubbling nitrogen through it would help to evaporate, but this was not part of the question. $\endgroup$ – aventurin Aug 22 '17 at 20:56
  • $\begingroup$ Thank you for taking time to post an answer! I understand there will be a balance between $\ce N \ce O_2$ and $\ce N_2 \ce O_4$, but I'm wondering; does Henry's law tell us that if the partial pressure of both gases drops to zero above the liquid's surface, then the gases will continue to evaporate until their concentrations in solution also drops to zero? $\endgroup$ – uhoh Aug 23 '17 at 2:13
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$\ce{NO2}$ and $\ce{N2O4}$ mixture is an equilibrium mixture which depends on temperature. At low temp $\ce{NO2}$ dimerises more, so equilibrium will be more shifted to $\ce{N2O4}$ (colourless).

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