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It is known, that decomposition of Lead(II) nitrate is one of the ways of generating $\ce{NO2}$ for lab use.

I recently did this in order to acquire $\ce{NO2}$ in liquid form (ambient temperature was ~0 °C), and got quite strange results. First, liquid $\ce{NO2}$ had 2 visible layers, usual brown one at the top and transparent at the bottom:

enter image description here

Also, during condensation liquid initially had green color (I know that Graham condenser is not safe in this configuration - I wouldn't see this color change in conventional condenser setup).

enter image description here

Can anyone clarify why I got such variety of products instead of pure $\ce{NO2}$?

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    $\begingroup$ in case you want purer $\ce{NO2}$ without lead compound involved consider reaction $\ce{2HNO3 + NaNO2 \rightarrow 2NO2 + H2O + NaNO3}$ $\endgroup$
    – permeakra
    Nov 11, 2014 at 4:41
  • $\begingroup$ BarsMonster, I updated my answer, hopefully some helpful information $\endgroup$
    – DavePhD
    Nov 15, 2014 at 16:00

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Check if the green condensate is $\ce{N2O3}$. It has been reported to be blue-green.

In the presence of water vapor and oxygen, the nitrogen dioxide that you are trying to obtain is in equilibrium with nitric acid and nitrogen trioxide, as explained here.

As reported in the Journal of Industrial and Engineering Chemistry, volume 12, number 6, page 531-538 (1920), condensation of gaseous nitrogen dioxide in a spiral condensor at -5 °C in the presence of oxygen and water vapor resulted in liquid with distinct upper and lower layers, the upper layer being a concentrated nitric acid layer (essentially no water) and the lower layer being a diluted nitric acid layer (27 %) water.

So the reason you are getting side products is likely that water vapor and oxygen are also present and causing equilibrium reactions that yield nitric acid, nitrous acid and nitrogen trioxide.

Reference

  1. CCXIV.—The dissociation of gaseous nitrogen trioxide, Bernard Mouat Jones, J. Chem. Soc., Trans., 1914, 105, 2310-2322, DOI: 10.1039/CT9140502310
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  • $\begingroup$ That looks very close. Important difference though is that there should have been no water in the system, Lead (II) nitrate decomposition goes at 250C, when no humidity is left. So yes, there was plenty of oxygen but no water. $\endgroup$
    – BarsMonster
    Nov 15, 2014 at 22:05
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The colorless area may be $\ce{N2O4}$ which is in equilibrium with $\ce{NO2}$. $\ce{N2O4}$ is colorless and has a greater density and therefore is found at the bottom and green compound might be $\ce{N2O3}$ which is in equilibrium with $\ce{NO2}$ and $\ce{NO}$.

enter image description here

Edit :
Boiling point of $\ce{N2O4}$ is 21.69 °C, so the temperature at which you perform experiment is also critical. Moreover, there is high probability that $\ce{N2O4}$ might be present in liquid state too.

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    $\begingroup$ Hmm... Is N2O4 stable enough to form a separate layer (i.e. shouldn't NO2 form inside N2O4 layer and vice versa)? I had an impression that It should have been very hard to separate N2O4 and NO2 "gravitationally"... $\endgroup$
    – BarsMonster
    Nov 14, 2014 at 16:02
  • $\begingroup$ You cannot seperate them as they are in equilibrium as if you try to remove NO2 from here much of N2O4 will convert to NO2 to establish the equilibrium so you cannot seperate them.on basis of gravitationally as you are maintaining same pressure conditions $\endgroup$
    – DSinghvi
    Nov 15, 2014 at 6:19
  • $\begingroup$ ONE ore thing Boiling point 21.69 °C of N2O4 so the temperature at which you perform experiment is also critical . And there is high preobability that N2O4 might be present in liquid state too. $\endgroup$
    – DSinghvi
    Nov 15, 2014 at 6:24

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