Dinitrogen tetroxide decomposes to nitrogen dioxide at room temperature, but rockets that use it are usually said to use N₂O₄ and not NO₂ - why?

Question

To the best of my understanding, both dinitrogen tetroxide and nitrogen dioxide are usable and perform quite similarly as oxidizers, but rockets that use either are almost always said to use dinitrogen tetroxide specifically, even when they are stored above the temperature where the equilibrium between the two compounds would cause it to convert to NO₂.

Is "dinitrogen tetroxide" here just used as a shorthand for "either N₂O₄ or NO₂, whatever it converts to at the storage temperature", or is there some trick (slightly cooling it down before launch?) used to ensure it's converted to N₂O₄?

If it is N₂O₄ specifically, and it is deliberately prevented from decomposing to NO₂, why is this done when the two seemingly perform so similarly?

Answer 1

Nitrogen dioxide exists in an equilibrium with dinitrogen tetroxide according to the reaction:

$$\require{mhchem}\ce{N_2O_4(g) <=> 2NO_2(g)}$$

In short, as you correctly noted, increasing temperature shifts this equilibrium to the right, while decreasing temperature shifts it to the left. However, this equilibrium is not dependent exclusively on temperature -- it depends on pressure as well; the higher the pressure is, the more this equilibrium shifts to the left side. Since oxidizer tanks are naturally highly pressurized in order to liquify their contents in the first place, the formation of $\ce{N_2O_4}$ is much more favoured than it would be in the same temperature under standard atmospheric pressure.

One more important detail is that the mentioned equilibrium applies exclusively to the gaseous phase, while the majority of the tank's contents is going to be in liquid phase (except the marginal case of nearly-empty tanks). In liquid phase, the formation of $\ce{N_2O_4}$ is favoured much more than it is in gaseous phase^[1].

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More detailed explanation:

From the equilibrium formula, we can derive the formula for the equilibrium constant:

$$K_c = \ce{\frac{[NO_2]^2}{[N_2O_4]}}$$

where $\ce{[NO_2]}$ is the molar concentration of nitrogen dioxide, and $\ce{[N_2O_4]}$ is, analogously, the molar concentration of dinitrogen tetroxide.

The value of said constant at around room temperature $T = 298 K (25 °C)$ is $K_c \approx 0.006$.

Important thing to note there is, shifting the equilibrium to the left leaves us with less moles of gas, and shifting it to the right with more moles of gas. Since we consider a system consisting of a rigid-walled tank, it is reasonable to assume that the volume stays constant; if we assume the temperature to stay constant as well, the value that is going to be variable is pressure. Therefore, increasing pressure favours the formation of $\ce{N_2O_4}$, and decreasing pressure favours the formation of $\ce{NO_2}$.

However, that is not the full picture. The oxidizer tank's contents are going to be mainly liquified oxidizer, with only a small fraction of the oxidizer existing in the gaseous phase above the liquid; thus, there is going to be another set of equilibria:

$$\ce{N_2O_4(l) <=> N_2O_4(g)}$$

$$\ce{NO_2(l) <=> NO_2(g)}$$

In both cases, the equilibria are dependent of both the temperature and pressure; higher temperatures and lower pressures favour shifting to gas formation, while lower temperatures and higher pressures favour liquid formation.

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^{[1] I'm sure that's true, but still looking for a reference. Will include it later.}