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

1 Answer
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".