# Spontaneity and nature of attack of fluorine gas on aluminum

What is the nature of the reaction of attack of fluorine gas on aluminum metal? Is it spontaneous in nature?

I have studied reactions of halogens on aluminum but it had no information about the attack of fluorine and I would like to know about this reaction.

Aluminium has a tenacious oxide layer so the reaction of gaseous fluorine with aluminium metal is firstly the reaction of fluorine with aluminium oxide.

So finally $$\ce{F2}$$ will react with $$\ce{Al2O3}$$? Also will $$\ce{F2}$$ react with fresh $$\ce{Al}$$ metal surface.

• I have edited the question and made it more readable, getting a good idea on English sentences and writing would be great since it was very hard to read before. Your title should reflect the question that you want to ask. Hope you get an answer for this question and enjoy learning :D. – Safdar Faisal Apr 13 at 11:06
• Aluminium has a tenacious oxide layer so the reaction of gaseous fluorine with aluminium metal is firstly the reaction of fluorine with aluminium oxide. – Waylander Apr 13 at 11:27
• Use $\ce{} for chemical compounds – Safdar Faisal Apr 13 at 14:17 • What do you mean by "spontaneous"? If you mean "exergonic", I recommend to use that term. – Karl Apr 13 at 19:06 ## 1 Answer First and foremost! Do not do this reaction unless you are properly trained and have the appropriate safety equipment. Fluorine is one of the most dangerous substances out there, and one of the presumed products, $$\ce{OF2}$$, is also terrifying. If you can calculate the free energy change for a reaction, you will know whether it is spontaneous. To do this, you need to have more information than you currently provide. We need to know the products of the reaction in order to write a balanced chemical equation. Then we can look up appropriate thermodynamic values for each substance in the reaction in order to calculate the free energy change. Below is what I presume to be the balanced equation for the reaction of fluorine with aluminum oxide: $$\ce{Al2O3 + 6F2 -> 2AlF3 + 3OF2}$$ Using the NIST Chemistry Webbook, we can look up the enthalpies of formation and standard entropies of each substance so that we may calculate the enthalpy and entropy changes. $$\begin{array}{l|c|c|c|c|c|c} & \ce{Al2O3} & \ce{F2} & \ce{AlF3} & \ce{OF2} \\ \hline \Delta_f H^\circ\ \mathrm{(kJ/mol)} & -1675.7 & 0 & -1510.4 & 24.52 \\ S^\circ\ \mathrm{(J/K)} & 50.92 & 202.8 & 66.5 & 247.46 \\ \end{array}$$ We can use these data to calculate the free energy change of the reaction: $$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_rS^\circ$$ If the free energy change is negative at that temperature, the reaction is spontaneous. • But does it go kinetically? I can see it with aluminum hydroxide, but the anhydrous oxide may be hard to attack unless heated. – Oscar Lanzi Apr 13 at 15:00 • @Karl - I would like to believe that to be true everywhere in the world, but I cannot guarantee it. – Ben Norris Apr 13 at 20:31 • @nick012000 Electrolysis of a "bathtub" of fluorinated water seems to me unlikely to produce fluorine given fluorine will spontaneously react violently with water to produce oxygen (or ozone if there is excess fluorine which there surely won't be here) and HF. – abligh Apr 14 at 7:03 • What's your reasoning for the reaction you wrote down rather than$\ce{2Al2O3 + 6F2 -> 4 AlF3 + 3O2}\$? – Jan Apr 14 at 9:15
• @Jan Oxygen reacts with fluorine? Because everything reacts - if it hasn't already reacted - with fluorine? – Karl Apr 14 at 20:59