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I was going through the properties of titanium when a certain thing caught my eye: It was the reaction of burning of titanium in nitrogen. I was astonished to read it as I knew that neither is nitrogen a supporter of combustion nor does it burn itself. But then how can titanium burn in nitrogen? Moreover it is one of the few elements to do so.

It was written that titanium burns in pure nitrogen.

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    $\begingroup$ Titanium, and many (most!) other transition metals, form very stable nitrides. The trick is that the N2 molecule is actually pretty tough to break up, so only the nitrides with the most negative free energies of formation can manage to 'burn' in nitrogen. With ammonia it is a different story... $\endgroup$ – Jon Custer Nov 24 '15 at 14:33
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    $\begingroup$ @Abhi Titanium can burn in nitrogen for the same reasons as anything burning in anything else: it is a fiercely exothermic reaction which is not prohibited by kinetics. Does not it surprise you that water can burn in fluorine? $\endgroup$ – Ivan Neretin Nov 24 '15 at 15:30
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    $\begingroup$ From 2019, spacex.com/news/2019/07/15/… SpaceX claims that a "reaction between titanium and nitrogen tetraoxide at high pressure was not expected." Not pure Nitrogen, but relevant. $\endgroup$ – Criggie 2 days ago
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There are two questions one has to answer when asking ‘will $\ce{X}$ react with $\ce{Y}$?’

  • Thermodynamics

    This basically asks ‘will the products of a hypothetical reaction be more stable than the reactants?’, i.e. whether the entire process will release energy. A quick web search tells me that $\Delta_\mathrm{f} H^0 (\ce{TiN}) = - 337.65~\mathrm{\frac{kJ}{mol}}$ and therefore the reaction is thermodynamically possible.

  • Kinetics

    This is asking ‘can I overcome the activation energy required for said hypothetical reaction to occur?’ And this is where it gets complicated.

The process of titanium burning can be split into the following sub-equations:

$$\ce{2 Ti + N2 -> 2 TiN} \\ ~\\ \ce{Ti(s) -> Ti(g)}\\ \ce{Ti(g) -> Ti^3+ (g) + 3 e-}\\ \ce{N2(g) -> 2 N(g)}\\ \ce{N(g) + 3 e- -> N^3- (g)}\\ \ce{Ti^3+ (g) + N^3- (g) -> TiN(g)}\\ \ce{TiN(g) -> TiN(s)}$$

Broken down in this way, we should immediately realise:

  • The first step is generally okay-ish but may require some activation. (We don’t need to make the entire material gaseous in one step; an atom a time is enough.)

  • The second step requires energy (no surprises there).

  • The third step is extremely hard to do, since the $\ce{N#N}$ is so strong.

  • The fourth step also requires energy since nitrogen has an endothermic electron affinity.

  • The fifth and sixth steps are the ones where huge amounts of energy are liberated; especially the sixth one where lattice enthalpy is liberated.

Now if we consider a block of titanium, as DavePhD mentioned, the the first step is already rather hard to do. The block’s crystal lattice is rather stable as is so we cannot easily extract atoms from there. However, the larger the surface to volume ratio gets the easier this can happen, especially since we have multiple ‘attacking’ points at the same time. Thus, solid metal blocks are usually hard to burn, while metal powder often burns just by putting it in pure oxygen. (Iron wool doesn’t even need to be a powder, the surface to volume ratio is small enough already to oxidise in pure oxygen at room temperature.)

As Jon Custer mentioned in a comment and as I mentioned earlier, the cleavage of dinitrogen molecules is also extremely difficult. It will take a lot of activation energy to do so. This means, that a lot of energy must be gained in the formation of the ion lattice so that at least the dinitrogen cleavage can be overcome and combustion can continue.

Thus, the kinetics part of the question can be answered with: ‘it can, but.’

The question remains how to supply activation energy. Since we are most likely dealing with powder, it could be too slow to ignite titanium in air and then immediately immerse in nitrogen. Since nitrogen cleaving is so hard, many other ignition ideas that rely on, say, a match’s flame won’t work (the match is too weak). Maybe the activation energy is ‘low’ enough that heating to a few hundred degres will suffice. Unfortunately, I do not know the experimental details.

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A block of titanium won't burn even in air, let alone nitrogen.

However, if the surface area to volume ratio is high enough (for example titanium powder) it can burn in air and even nitrogen (forming titanium nitride).

See Principles of Fire Protection Chemistry and Physics, 3rd edition, by Raymond Friedman, at page 155.

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