How to find the balanced reaction equation for the decomposition of potassium dichromate in non aqueous conditions?

I can't find a method to balance a reaction: the heating of potassium dichromate, which forms potassium dichromate (III), dichromate oxide (III) and oxygen. I can't add water products as $\ce{H+}$, $\ce{{}^{-}OH}$ because we are not in solution. So, how can I balance it?

Who oxidises in this redox? The oxygen inside the salt? The oxygen of $\ce{H2O}$ which is in the air?

$$\ce {4K2Cr2O7 -> 4K2CrO4 + 2Cr2O3 + 3O2}$$
As for what is oxidized and reduced; the chromium is reduced from chromium(VI) to chromium(III) in going from $\ce{K2Cr2O7}$ to $\ce{Cr2O3}$ and some of the oxygen is oxidized from −2 in the dichromate to 0 in the oxygen gas.
• Small nitpick, the oxidation state in potassium chromate is not three, but six: $$\ce{4\overset{+I}{K}_2\overset{+VI}{Cr}_2\overset{-II}{O}_7 -> 4\overset{+I}{K}_2\overset{+VI}{Cr}\overset{-II}{O}_4 + 2\overset{+III}{Cr}_2\overset{-II}{O}_3 + 3\overset{\pm0}{O}_2}$$ – Martin - マーチン Nov 17 '14 at 2:51