Why iron reacts with steam to give Fe3O4 not Fe2O3? [duplicate]

Question

Why $\ce{Fe}$ with $\ce{H2O}$ gives $\ce{Fe3O4}$ not $\ce{Fe2O3}$?

Like, $\ce{Zn + H2O \rightarrow ZnO + H2}$, why $\ce{Fe}$ doesn't simply gives $\ce{Fe2O3}$ when it reacts with steam?

Answer 1

Quoted picture from Researchgate.net for Ellingham diagram for iron oxides and hydrogen:

On the diagram above, you can see equilibrium partial pressures (expressed as oxygen chemical potential} of oxygen for particular redox systems, depending on temperature.

Each system has a line $\mu(\ce{O2})=f(T)$. A system is able to oxidize systems with their respective lines below and vice versa.

We can conclude these consequences:

This reaction is thermodynamically favorable at all temperatures for the given range:

$$\ce{3 Fe2O3 + H2 -> 2 Fe3O4 + H2O}$$

This reaction is thermodynamically favorable below $\pu{900 K}$.

$$\ce{3 FeO + H2O-> Fe3O4 + H2}$$

The reaction is thermodynamically favorable below roughly $\pu{1700 K}$.

$$\ce{Fe + H2O -> FeO + H2}$$

It should be obvious that

• The reduction of water by iron stops at $\ce{Fe3O4}$.
• Above $\pu{900 K}$, it stops even at $\ce{FeO}$.
• Above $\pu{1700 K}$, hydrogen reduces $\ce{FeO}$ to iron.

Note that the respective temperatures are related to equilibrium states. The open system with the flow of steam reactant and removal of hydrogen somewhat shifts the reactions toward hydrogen as the product, so the effective decisive temperatures would be higher.

OTOH, for reduction of oxides by hydrogen, it is the opposite case and needed temperature would be lower.

I assume for $\ce{Fe + H2O <=> FeO + H2}$, for temperature near $\ce{1200 K}$ and above, the reaction would go in large extend forward if hot water steam goes over hot iron, or backward, if hydrogen goes over iron oxides.

Answer 2

Consider the Schikorr reaction, proceeding in two stages: "[A]naerobic oxidation of two Fe(II) into Fe(III) by the protons of water... [then] loss of two water molecules from the iron(II) and iron(III) hydroxides giving rise to... the formation of a thermodynamically more stable phase iron(II,III) oxide. Ma et al state, that crystalline $\ce{Fe3O4}$ (magnetite) is thermodynamically more stable than amorphous $\ce{Fe(OH)2}$.

BTW, the stability of magnetite, and its adherence to iron, is used to protect steel from rusting. "Blueing... creates a layer of magnetite over the metal to prevent rust."

See also this question.