I am wondering if the Rubidium ($\ce{Rb}$) in gas form can react naturally with Sapphire crystal ($\ce{Al2O3}$)? The temperature is $500\ \mathrm{K}$ and the pressure is around $1\ \mathrm{Pa}$ to $10\ \mathrm{Pa}$.

One of the reaction that comes into my mind is $$\ce{4 Rb + Al2O3 -> 2 Rb2O + Al2O}$$

I was trying to calculate the Gibbs free energy of the reaction, but I could just find the Gibbs free energy for Rubidium and the Sapphire, which is $\Delta G_{\ce{Rb}}=53.1\ \mathrm{kJ/mol}$ and $\Delta G_{\ce{Al2O3}}=-1582.3\ \mathrm{kJ/mol}$. I can hardly find the Gibbs free energy for the products.

Is there any other reaction that can occur naturally?

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Aluminium%28I%29_oxide Al2O is unstable in such low temp. and rubidium is liquid. $\endgroup$
    – Mithoron
    Nov 10 '15 at 18:08
  • $\begingroup$ Why don't you just try reducing Al all the way to the metal? I mean, Al(I) is pretty unstable. $\endgroup$
    – orthocresol
    Nov 10 '15 at 18:18
  • $\begingroup$ @Mithoron, we are working on very low pressure, it is around $10^{-5}$ times lower than the atmospheric pressure (i.e. $P=0.7\ \mathrm{Pa}$). The vapor pressure of Rb is 1 Pa when the temperature is 434 K, so it is in the gas form. $\endgroup$
    – Firman
    Nov 10 '15 at 18:21

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