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How does nitrogen tetraoxide dissociate into nitrosyl ion and nitrate ion?

Is it similar to the auto-dissociation of $\ce{SO2}$ into thionyl ion and sulfite ion by acceptance and donation of oxide ion?

I couldn't find a mechanism for the auto-dissociation of nitrogen tetraoxide.

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    $\begingroup$ Do you have any evidence that $$\ce{N2O4 <=> NO+ + NO3-}$$ is actually happening and under which conditions? Maybe then one would be able to also find mechanistic evidence. I would actually be surprised when this would be more feasible than $$\ce{N2O4 <=>2NO2}.$$ $\endgroup$ Commented Jan 24, 2017 at 12:59
  • $\begingroup$ Are you interested in the gas-phase dissociation? Liquid-phase dissociation? Dissociation in aqueous media? $\endgroup$
    – hBy2Py
    Commented Jan 24, 2017 at 18:31
  • $\begingroup$ Maybe you confused it with N2O5 which is comprised of nitronium and nitrate ion? $\endgroup$
    – EJC
    Commented Jan 24, 2017 at 22:09
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/58564/… $\endgroup$ Commented Jan 25, 2017 at 3:53

1 Answer 1

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Dinitrogen tetroxide $\ce{N2O4}$ is a colourless liquid composed of planar dimers of nitrogen dioxide $\ce{NO2}$.$^{[1]\ \text{[a]}\ [2]}$

Nitrogen dioxide dimerisation. Chemsketch and reference [1]

As Martin noted and is to be expected from the structure of $\ce{N2O4}$, the process

$$\ce{N2O4 <=> NO+ + NO3-}$$

is not very likely to happen. Autoionisation of $\ce{NO2}$ is more common$^{[1]}$.

$$\ce{NO2 -> NO2+ + e^-\\ NO2 + e^- -> NO2-}$$

Ionisation energy of $\ce{NO2}$ is $9.8\ \mathrm{eV}$ while its electron affinity equals $1.62\ \mathrm{eV}$.$^{[1]}$ Technically you could look at this as heterolytic bond dissociation $\ce{N2O4}$ but that is probably not what is happening. This autoionisation is, in a way, realised for aqueous solutions$^{[2]}$:

$$\ce{\underset{electron transfer}{NO2 + NO2} + H2O -> \underset{really just HNO3}{NO2OH} + HNO2},$$ $$\ce{\underset{electron transfer}{NO2 + NO2} + 2KOH -> KNO3 + KNO2 + H2O}.$$

Table 1. Melting and poiling points of $\ce{NO2}$ and $\ce{N2O4}$. Note the "equality" of melting points.$^{[2]\ [3]}$ $$\begin{array}{|c|c|c|} \hline \mathbf{Molecule} & \mathbf{mp\ (\deg \mathrm{C})} & \mathbf{bp\ (\deg \mathrm{C})}\\ \hline \ce{NO2} & -11.2 & 21 \\ \ce{N2O4} & -11.2 & 22 \\ \hline \end{array}$$

Table 2. Percentage of $\ce{NO2}$ and $\ce{N2O4}$ at various temperatures (pressure $1 \ \mathrm{atm}$).$^{[2]}$ $$\begin{array}{|c|c|c|c|} \hline \mathbf{Molecule} & \text{below}\ \mathrm{mp}\ (\mathrm{\%}) & \mathbf{40\ \deg \mathrm{C}\ (\mathrm{\%})} & \text{over}\mathbf{\ 150\deg \mathrm{C}\ (\mathrm{\%})}\\ \hline \ce{NO2} & 0 & 31 & 100 \\ \ce{N2O4} & 100 & 69 & 0 \\ \hline \end{array}$$


$\text{[a]}$ To the best of my knowledge, the enthalpy $-55\ \mathrm{kJ}$ applies at $20 \deg \mathrm{C}$ and $1\ \mathrm{atm}$ for both components in the gas phase. NIST gives $-57.12\ \mathrm{kJ}$ at $25 \deg \mathrm{C}$ and $1\ \mathrm{bar}$. See $\ce{N2O4}$ and $\ce{NO2}$.


$[1]$ N. N. Ahmetov. Anorgaaniline keemia. (1974) (pages 375, 376)

$[2]$ H. Karik, Kalle Truus. Elementide keemia. (2003) (pages 424, 426)

$[3]$ Wikipedia:

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