# Reaction of iron with steam

While studying about reversible reactions, my professor stated that the following reaction is a reversible reaction:

$$\ce{Fe + H2O <=> Fe3O4 + H2}$$

She also stated that the product $\ce{Fe3O4}$ is formed by a mixture of $\ce{FeO}$ and $\ce{Fe2O3}$. However, there was no mention of the type of reaction or the reason due to which the mixture of $\ce{FeO}$ and $\ce{Fe2O3}$ resulted in the formation of $\ce{Fe3O4}$.

So, my questions are as follows:

1. What type of a reaction is it? (I think it's a displacement reaction but from what I know, there are ions present in a displacement reaction).

2. Why have we combined $\ce{FeO}$ and $\ce{Fe2O3}$ resulting in the formation of $\ce{Fe3O4}$? (According to me, mixtures and compounds are two different things and that we can't combine them.)

• We didn't combine anything. Nature did. – Ivan Neretin Apr 6 '16 at 11:07
• Our professor stated that the product is a mixture (excluding hydrogen gas). And mixtures do not form compounds. Then how were the two combined? – Parth Apr 6 '16 at 11:09
• Your professor is wrong, or you misunderstood him. $\ce{Fe3O4}$ is a compound and not a mixture. – Ivan Neretin Apr 6 '16 at 11:17
• Single displacement reactions do involve redox -- the element on the reactant side ends up in a combined state in the products and, therefore, must undergo a change of oxidation state. As to whether or not Fe3O4 is formed directly, that is doubtful since it would require an activated complex of three iron atoms and four water molecules -- very unlikely. I would guess that there are a number of steps involving various iron oxide and hydroxide intermediates. – user28106 Apr 6 '16 at 14:35
• Regarding whether or not Fe3O4 is a compound, it surely is as stated by MaxW and Ivan Neretin. However, the formula FeO⋅Fe2O3 can be misleading in that it might seem to suggest that Fe3O4 is made of FeO and Fe2O3 "molecules" in chemical combination. Actually, Fe3O4 has a crystal structure (an extended geometrically regular array) of Fe(III), Fe(II) and oxide ions in which the oxide ions exist in a cubic close packed pattern with the Fe(II) ions located at half of the octahedral sites and the Fe(III) ions distributed evenly between octrahedral and tetrahedral sites. – user28106 Apr 6 '16 at 15:00

I would describe it first and foremost as a redox reaction. Iron is oxidised and hydrogen is reduced; the half equations would be: $$\ce{2Fe -> Fe^2+ + Fe^3+ + 5e-}$$ $$\ce{H2O + 2e- -> H2 + O^2-}$$
Why have we combined $\ce{FeO}$ and $\ce{Fe2O3}$ resulting in the formation of $\ce{Fe3O4}$?
This is a good observation. $\ce{Fe3O4}$ contains both iron(II) and iron(III) ions and is sometimes written as $\ce{FeO \cdot Fe2O3}$. The dot in the middle signifies that there is some form of bonding between the two compounds (another common example of this is water of crystallisation). $\ce{Fe3O4}$ is a compound and not a mixture because it does not consist of two separate $\ce{FeO}$ and $\ce{Fe2O3}$ phases but rather it is a single crystal structure containing $\ce{Fe^2+}$, $\ce{Fe^3+}$ and $\ce{O^2-}$ ions.
$\ce{Fe3O4}$ has a cubic inverse spinel structure which consists of a cubic close packed array of oxide ions where all of the $\ce{Fe^2+}$ ions occupy half of the octahedral sites and the $\ce{Fe^3+}$ are split evenly across the remaining octahedral sites and the tetrahedral sites.