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

1 Answer
Truong-Son N.
Aug 12, 2017

Zero. What units are we in?

Pure, molecular oxygen, "O"_2(g), forms no free energy when it gets produced. Why?

Well, the standard change in Gibbs' free of formation, DeltaG_f^@, is defined relative to the elements in their elemental states... at 25^@ "C" and "1 atm", "O"_2(g) IS the elemental state of oxygen, i.e. the formation "reaction" is just

"O"_2(g) -> "O"_2(g)

Obviously, nothing happens. So, DeltaG_f^@ = "0 kJ/mol" for this "process".

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