Answer
The equilibrium constant for that reaction is equal to $8.0 \times 10^{9}$.
Work Step by Step
1. Calculate the concentrations in mol per liter:
$Fe(s):$ Solid. So we don't consider it on the expression
$O_2(g): \frac{1.0\times 10^{-3}mol}{2.0L}= 5.0 \times 10^{-4} mol/L$
$Fe_2O_3(s) $: Solid. So we don't consider it on the expression.
2. Write the equilibrium constant expression:
** Since the coefficient for $O_2$ is "3", that will be its concentration exponent.
$K = \frac{[Products]}{[Reactants]} = \frac{1}{[O_2]^3} \times \frac{1}{(5.0 \times 10^{-4})^3} = 8.0 \times 10^{9}$
