# Why does the titanium go to the tungsten filament in the crystal bar process?

I've tried to read a report by the US atomic energy commission [1], but couldn't really find the answer, or maybe I missed it.

Why is the gaseous titanium deposited on the tungsten wire/filament during the van Arkel–de Boer process?

### Reference

1. Petersen, A. W. Preparation of Metallic Titanium by Film Boiling (Thesis); UCRL-2523, 4393936; United States, 1954. DOI: 10.2172/4393936.
• $\ce{Ti + 2I2 <=> TiI4}$ is an endothermic reaction. In the chamber, temperatures are high enough that there is $\ce{TiI4}$, but the tungsten filament is so hot that $\ce{TiI4}$ decomposes again. Or, to cite the Dutch Wikipedia: forward reaction at $\pu{600 ^\circ{}C}$, backward reaction at $\pu{1200 ^\circ{}C}$ for a $\Delta{}H = \pu{−427 kJ/mol}$. A bit like a halogen lamps, if you still have at your bicycle. – Buttonwood Sep 15 '20 at 18:44

## 1 Answer

By "gaseous titanium", do you mean "titanium tetraiodide ($$\ce{TiI4}$$) vapor"?

Van-Arkel De-Boer process is a purification process of titanium and zirconium. Basically, what happens is that the impure metal, let's say titanium is heated in iodine environment at a temperature of $$\ce{250 ^\circ C}$$ to form volatile titanium tetraiodide ($$\ce{TiI4}$$) vapor. The impurities are left behind, as they do not react with iodine. This vapor is then passed over a hot tungsten filament at $$\ce{1400 ^\circ C}$$ for which the vapor gets decomposed and pure titanium is deposited on the filament and is removed. The iodine is reused. The overall reaction is:

$$\ce{Ti _{(impure)} + 2I2 ->[250 ^\circ C] TiI4 ->[1400 ^\circ C][W filament] Ti_{(pure)} + 2I2}$$

The thermodynamics of this reaction is discussed in Buttonwood's comment.

• Except that Buttonwood stated it upside down (although the point is clear, that the dissociation is endothermic) – Buck Thorn Sep 16 '20 at 11:22