4
$\begingroup$

I have been told by a metallurgist that the reason a factory oxidizes the iron oxides in ilmenite before reducing them down to pure iron, is that the the products of this have a more spacious crystal structure, thus allowing the $\ce{CO}$ gas (which they use to reduce the iron) to diffuse more wholly and deeply into the crystal structure of the products. Now, what are the products? According to [1, p. 555], the reactions are as follows:

$$ \begin{align} \ce{4 FeTiO3 + O2 &-> 2 Fe2O3 + 4 TiO2} \tag{R1} \\ \ce{6 FeTiO3 + O2 &-> 2 Fe3O4 + 6 TiO2} \tag{R2} \\ \ce{4 FeTiO3 + O2 &-> 2 Fe2TiO5 + 2 TiO2} \tag{R3} \end{align}$$

My post can be one of two question. A general one, which if too hard/lengthy, means my post defers to the specific case I am most interested in.

General question: How do I find out which of two given crystal structures any given gas would most wholly and deeply diffuse into?

Specific question: Would CO gas diffuse more wholly and deeply into the above reactions' products than it would their solid reactants?

Reference

  1. Tathavadkar, V and Kari, Chandrakala and Rao, S Mohan (2006) Oxidation of Indian Ilmenite: Thermodynamics and Kinetics Considerations. In: Proceedings of the International Seminar on Mineral Processing Technology and Indo-Korean Workshop on Resource Recycling (MPT-2006), March 8-10, 2006, NML, Chennai. http://eprints.nmlindia.org/6302 (PDF)
$\endgroup$
3
  • 1
    $\begingroup$ Sounds like an urban legend to me. $\endgroup$ May 7, 2023 at 19:44
  • 1
    $\begingroup$ Besides, I must say I admire your perseverance at making the formula look right. If not for one slanted O, I would never suspect it was made without the tools we use around here. $\endgroup$ May 7, 2023 at 19:53
  • $\begingroup$ @IvanNeretin Ah, one slipped past me! Thanks for the edit. Also, I didn't know you use a different display code here. I will acquaint myself with mhchem. $\endgroup$
    – user110391
    May 7, 2023 at 20:10

2 Answers 2

8
$\begingroup$

The process of heating a crystal, transforming into a different chemical, would destroy the crystal structure, making it into many small crystals or an amorphous solid. Inevitably, this makes the substance porous, having nothing to do with the defunct crystal structure. Consider what would happen to the selenite crystals in Cueva de los cristales if the temperature varied or humidity dropped.

That said, there are some interesting materials that are surprisingly permeable to gases:

$\endgroup$
1
  • $\begingroup$ Interesting. Are you saying that compounds within the same phases transforming into distinct compounds leads to a very large number of small phases, thus causing the bulk material to be quite porous? That makes a lot of sense, but makes me wonder why the metallurgist was talking about the crystal structure. $\endgroup$
    – user110391
    May 7, 2023 at 22:26
1
$\begingroup$

I do not think your general question has any answer. It is oversimplification to state that the greater distance between atoms in crystal lattice makes the crystal to be more permeable by gases. It can be a contributing factor, but most likely not the only factor. Specific interactions between gas atoms/molecules and constituents of the lattice (atoms, ions, molecules) will be of great importance here. So, what is permeable for, say, oxygen can be almost non-permeable for carbon monoxide. The crystal degradation correctly described in another answer because of the incommensurability of lattices of the matrix phase and the nascent one results in formation of polycrystalline or amorphic phases. Also, considerable number of pores can be created. All that can result in forming much more permeable material. This is what the metallurgist could have meant (if this is the case then he was not exactly accurate while choosing terms to convey his idea, but the idea itself still makes sense).

As for specific question, the answer will depend on the conditions of the process. Should there be reasons to believe it to be an equilibrium process, you could calculate which one of the listed reactions leads to the largest energy gain (see any thermochemistry textbook). But I doubt that is the case. There are a number of factors in non-equilibrium processes that can favour this or that reaction. It will depend on specific values of thermodynamic and kinetic parameters. If you are really interested in this, I suggest you check the papers describing experiments.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.