Why Iron Rusts but Does Not React with Nitrogen

Air is roughly 21% oxygen and 78% nitrogen. Iron sitting in this mixture reacts readily with the oxygen — that's rust — but it does essentially nothing with the nitrogen, even though nitrogen is four times more abundant. Why does one gas win and not the other?

What Decides Whether a Reaction Happens on Its Own

A process runs spontaneously, with no external push, when the total entropy of the universe increases. Packed into a single quantity at constant temperature and pressure, this condition becomes the Gibbs free energy:

$$\Delta G = \Delta H - T\Delta S$$

\(\Delta H\) is the heat released or absorbed, \(T\) is the absolute temperature, and \(\Delta S\) is the entropy change of the system. A reaction is spontaneous when \(\Delta G < 0\).

The sign of each term tells you the story before you even compute a number:

Iron and Oxygen: \(4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3\)

Here \(\Delta H\) is strongly negative — this reaction is highly exothermic, which is why rusting steel wool can visibly warm up and why thermite-style Fe/O reactions are violent. \(\Delta S\) for the system is also negative: three moles of free, disordered \(\text{O}_2\) gas are locked into a rigid solid lattice. Gas turning into solid is always an entropy loss for the system.

So both terms are working against the simple "always spontaneous" case — \(\Delta H\) is favorable, but \(\Delta S\) is not. The reaction proceeds anyway because the \(\Delta H\) term is enormous. If you draw the system boundary around just the iron and oxygen, the system's own entropy drops as the gas disappears into ordered solid. But the heat that reaction releases doesn't vanish — it pours into the surroundings, where it sets molecules jostling more vigorously. That's an entropy gain in the surroundings, and because the reaction is so exothermic, that gain is large enough to outweigh the system's loss.

$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0$$

The system becomes more ordered. The universe as a whole does not. \(\Delta G\) ends up negative because \(|\Delta H|\) is large enough to dominate the \(-T\Delta S\) term — rust forms.

Iron and Nitrogen: \(2\text{Fe} + \text{N}_2 \rightarrow 2\text{FeN}\)

The entropy side of the story is identical in shape: gas is again being converted into an ordered solid, so \(\Delta S\) for the system is negative here too.

The difference is on the enthalpy side. \(\text{N}_2\) is held together by a nitrogen triple bond, one of the strongest bonds in chemistry. Breaking it costs a large amount of energy up front, and the iron–nitrogen bonds that form afterward don't release nearly enough energy to pay that cost back. The net \(\Delta H\) ends up small, and for iron nitride formation under ordinary conditions it isn't negative enough — in practice the reaction is not favorable at room temperature.

Now the same tug-of-war runs the other way. The heat released into the surroundings is too small to generate enough entropy gain there to outweigh the system's entropy loss from gas-to-solid ordering:

$$\Delta S_{surroundings} \not> |\Delta S_{system}|$$

\(\Delta G\) stays positive. The nitrogen triple bond is simply too strong for iron to casually break apart at room temperature, so no reaction occurs — despite nitrogen making up nearly four-fifths of the air iron sits in.

The Core Difference

Both reactions pay the same entropy penalty — gas locked into solid. What separates them is the size of the enthalpy term. Iron oxidation releases enough heat to flood the surroundings with entropy and overwhelm the system's ordering. Iron nitridation doesn't, because the nitrogen triple bond refuses to give up its energy that easily. Same entropy story, different enthalpy budget — and that's the whole reason rust forms in air while iron nitride does not.