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".

Related questions
See all questions in Gibbs Free Energy
Impact of this question
8629 views around the world
You can reuse this answer
Creative Commons License