# What is the Gibbs' free energy of formation for "O"_2(g)?

Aug 12, 2017

Zero. What units are we in?

Pure, molecular oxygen, ${\text{O}}_{2} \left(g\right)$, forms no free energy when it gets produced. Why?

Well, the standard change in Gibbs' free of formation, $\Delta {G}_{f}^{\circ}$, is defined relative to the elements in their elemental states... at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, ${\text{O}}_{2} \left(g\right)$ IS the elemental state of oxygen, i.e. the formation "reaction" is just

${\text{O"_2(g) -> "O}}_{2} \left(g\right)$

Obviously, nothing happens. So, $\Delta {G}_{f}^{\circ} = \text{0 kJ/mol}$ for this "process".