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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 '14 at 4:41
  • $\begingroup$ BarsMonster, I updated my answer, hopefully some helpful information $\endgroup$ – DavePhD Nov 15 '14 at 16:00
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The colorless area may be N2O4 which is in equilibrium with NO2. N2O4 is colorless and has a greater density and therefore is found at the bottom.

Dinitrogen Tetroxide

and green compound might be N2O3 which is in equilibrium with NO2 and NO

NO2-N2O4

Edit :
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

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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 '14 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 '14 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 '14 at 6:24
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Check if the green condensate is $\ce{N2O3}$

It has been reported to be blue-green.

See also http://pubs.rsc.org/en/Content/ArticlePDF/1914/CT/CT9140502310

In the present 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: http://nitrogen.atomistry.com/nitrogen_tetroxide.html

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 degrees 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.

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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 '14 at 22:05

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