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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.
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    $\begingroup$ $\ce{Ti + 2I2 <=> TiI4}$ 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. $\endgroup$ – Buttonwood Sep 15 '20 at 18:44
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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.

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

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