Spontaneity and nature of attack of fluorine gas on aluminum
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:
$$\ce{Al2O3 + 6F2 -&gt; 2AlF3 + 3OF2}$$
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.
$$\begin{array}{l|c|c|c|c|c|c}
&amp; \ce{Al2O3} &amp; \ce{F2} &amp; \ce{AlF3} &amp; \ce{OF2}  \\ \hline
\Delta_f H^\circ\ \mathrm{(kJ/mol)} &amp; -1675.7 &amp; 0 &amp; -1510.4 &amp; 24.52 \\
S^\circ\ \mathrm{(J/K)} &amp; 50.92 &amp; 202.8 &amp; 66.5 &amp; 247.46 \\
\end{array}$$
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.