Talk:Dinitrogen tetroxide

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Di nitrogen tetroxide IS listed on the IUPAC site YOU list !

good inorganic IUPAC link: http://www.cofc.edu/~deavorj/101/nomenclature.html

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Appearance... Is it really brown? I seem to remember an experiment where 2 NO2 <->N2O4 via a change in pressure or temperature; and the glass vessel being a brown shade when in the NO2 form and clear when in the N2O4 form. It is this constant equilibrium between the two gases that make N2O4 appear brown, when really its the NO2 that is brown. see link http://chem-courses.ucsd.edu/Uglabs/Lecture/Demos//04.html Piyrwq 16:41, 15 August 2005 (UTC)

I added a note there. R6144 14:54, 26 August 2005 (UTC)

When frozen, N2O4 is a white solid (if pure)but as soon as it begins to melt(-11C) the drak brown NO2 molecule impart a progressive coloration to the liquid and when allowed to stand at room temperature the liquid is almost extremely dark brown (almost black). Those expericence wuth NTO learn to be able to tell the approximate temperture of the cooled liquid just by its colour.

Please verify the data for NO₂ and N₂O₄

The data for NO₂ and N₂O₄ are not very consistent in literature, probably because they always exist together in equilibrium. Verification of the data is recommended before use. R6144 14:54, 26 August 2005 (UTC)

Isn't it impossible to have accurate data on the individual bp/mp of no2/n2o4 since they are in eqm. anyway?

Molar mass

I'm almost cerain that molar mass of this compuond is not equal to 7.7 carbon atoms. --Krgeqewrjsif (talk) 09:57, 20 July 2008 (UTC)

The molecular weight of N₂O₄ is 92.011. I just checked it in ChemDraw.

Ben (talk) 16:13, 20 July 2008 (UTC)

Equilibrium Push

Disclaimer: I am *so* not a professional chemist, but from the article:

"Higher temperatures push the equilibrium towards nitrogen dioxide."

That doesn't seem right... why would higher pressures tilt the equilibrium toward twice as many molecules? In most cases that I am familiar with, increasing pressure on a gas favors the smaller number of molecules.

Riventree (talk) 03:13, 6 December 2011 (UTC)

$\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$

For the equilibrium:

N₂O₄ Template:Eqm 2 NO₂

ΔG can be calculated at any temperature, and it has been done [1] using known values for ΔH and ΔS. Where ΔG is negative, the reaction is spontaneous, whereas if ΔG is positive, the reaction is not. That's the quick answer. --Rifleman 82 (talk) 03:41, 6 December 2011 (UTC)

Har... well since my disclaimer was true, the above makes almost no sense to me, but appears sufficiently authoritative that I withdraw the question and leave it to the expert. Thank you, Rifleman 82.

Riventree (talk) 04:04, 6 December 2011 (UTC)

You can take a look at Gibbs free energy. --Rifleman 82 (talk) 04:46, 6 December 2011 (UTC)

hazard diamond

In the graphic, the yellow part of the diamond displays a zero, indicating 'un-reactive', despite that fact in the table above, the chemical is mentioned as reacting with water. Plus, since it IS a rocket fuel, one would expect it to be reactive.Mr Morden76 (talk) 21:33, 29 July 2013 (UTC)

Assessment comment

Template:Substituted comment Substituted at 13:27, 29 April 2016 (UTC)

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Safety issues

Nitrogen tetroxide is a very aggressive oxidant, and as such requires great handling precautions. I have read (but cannot confirm) that nitrogen tetroxide spilt on the skin will immediately and violently react with and oxidise subcutaneous fat, so violently that it can spontaneously burst into flame. 2001:8003:E40F:9601:645D:C915:B2CA:4B75 (talk) 23:13, 7 December 2023 (UTC)

If N₂O₄ is planar, how about C₂O₄²⁻?

The naive thought would be that the single N-N bond in N₂O₄ or the single C-C bond in C₂O₄²⁻ (oxalate ion) could both rotate. While the latter is true (the C-C bond in C₂O₄²⁻ can indeed rotate; see here), the N-N bond cannot rotate. I was wondering why. 2A04:CEC0:116F:1D5B:BCBB:17FB:3FBB:B50A (talk) 20:54, 14 June 2025 (UTC)