Why is it hard to reduce titanium dioxide?

Titanium is expensive, although it is not rare and titanium dioxide is a ubiquitous component, it's part of such mundane articles as toothpaste, paper or plastics.

The reason for that is that the processes that extract titanium from titanium dioxide (which occurs naturally and just needs some inexpensive purification) are quite costly. The established industrial method for this is the Kroll process. It works by first chlorinating the titanium to $TiCl_4$ and then moving the chlorine over to some $Mg$, yielding $MgCl_2$ and pure $Ti$. Then follows a costly remelting process under vacuum to further purify the metal.

Recycling titanium scrap is even harder, and removing the other components from the alloys is not entirely possible, so recycling only yields titanium based alloys, not the pure stuff (which is not necessarily a problem since it's very rarely used in its pure form).

What specific chemical properties of titanium make it so hard to purify? For example, the chlorine process at the beginning of the Kroll process takes place with coke in a fluidized bed reactor, very similar to the reduction of $Fe$ in a blast furnace, but if the carbon takes care of the reduction, why is that not sufficient? Why is the chlorine route needed?

Research into titanium production dates back to the late 1880s, and the Kroll process, patented in 1940, made it feasible for mass production, but remains costly and has not been improved upon since, despite much effort. How would a chemist look for suitable reactants during such research?

• Titanium dioxide reacts directly with carbon when heated to form Titanium Carbide (see Wikipedia page on Titanium - History). Maybe a fluidized bed reactor could prevent that reaction in some way. Mar 19 '13 at 4:02
• @kaliaden, it appears the chlorine (rather than the reactor) does, grabbing the titanium and leaving the carbon to pair with the oxygen, lest it will spend the rest of the reaction alone. I'm looking for the actual scientific explanation for that. Mar 19 '13 at 10:33

Reasons why Titanium is hard to extract and purify:

1. The +4 oxidation state of $Ti$ is the most stable (due to its noble gas configuration) and so reducing $TiO_2$ to $Ti$ is quite difficult.

2. $Ti$ has chemical properties which are similar to its neighbouring elements (Vanadium, Zirconium, Chromium) and the presence of compounds containing these elements in its ores makes purification even more difficult. For example: Vanadium and Zirconium also form volatile halides ($ZrCl_4$)and oxyhalides ($VOCl_3$) which have to be separated from $TiCl_4$ by fractional distillation in Kroll's Process.

I have tried to explain why the chlorine route is taken using only thermodynamics (I'm not absolutely sure about this but these reasons are what I could come up with).

Consider these reactions which could take place in the fluidized bed reactor:

1. $\ce{TiO_2(s) + 2Cl_2(g) + 2C(s) -> TiCl_4(g) + 2CO(g)}$

2. $\ce{Ti(s) + C(s) -> TiC(s)}$

Comparing the Gibbs' Energies of these reactions can tell us which one is spontaneous since if the change in Gibbs' Energy is negative, the reaction will be spontaneous.
And the equation for change in Gibbs' Energy is:

$\Delta{G} = \Delta{H} - T\Delta{S}$

At normal temperatures, reaction 2 would occur more readily than reaction 1 since $TiC$ is extremely stable (reflected by its very high melting and boiling points) while $TiCl_4$ is unstable (it hydrolyses immediately on contact with air or oxygen to form $TiO_2$ and $HCl$). And so it is almost impossible to reduce Ti using C at low temperatures.

At higher temperatures (Kroll's Process is carried out at around $1000^oC$), $T\Delta{S}>>\Delta{H}$ so we can neglect $\Delta{H}$.

For the first reaction, since all the products are gaseous (while some of the reactants are solid), $\Delta{S}$ is positive. Therefore $\Delta{G}$ will be negative and the reaction is spontaneous and will occur easily.

For the second reaction, since the number of molecules is reduced, $\Delta{S}$ is negative. Therefore $\Delta{G}$ will be positive and the reaction becomes unfavourable.

Notes:

1. For the states of the reactants and products, I compared the melting and boiling points of the reactants and products with $1000^oC$ which is the temperature at which Kroll's process is carried out.

2. Actually, from what I read, there are many possible products for the first reaction. depending on the reaction conditions, a mixture containing $CO$, $CO_2$ and $COCl_3$ can be formed. But all these products are still gaseous so it doesn't affect the explanation.