Why is the change in free energy zero when substances are in their elemental state?

1 Answer
Truong-Son N.
Apr 10, 2018

Because you don't need to do anything to them to form them from their standard state.

The formation reaction is simply the reaction that starts with the elements in their standard state and forms #"1 mol"# of the product at that standard state.

If you wanted to form #"NH"_3(g)# at standard state, #25^@ "C"# and #"1 bar"#, you would write

#1/2"N"_2(g) + 3/2"H"_2(g) -> "NH"_3(g)#

since nitrogen element and hydrogen element are #"N"_2(g)# and #"H"_2(g)# respectively, at #25^@ "C"# and #"1 bar"#. If you simply want to form nitrogen gas at #25^@ "C"# and #"1 bar"#, then

#"N"_2(g) -> "N"_2(g)#

Clearly, nothing happened to go from #"N"_2(g)# to #"N"_2(g)#:

#DeltaG_(f,"N"_2(g))^@ -= DeltaG_(rxn)^@# for forming #"N"_2(g)# from #"N"_2(g)#

And the final and initial states are the same, so by definition of a state function, #DeltaG_(f,N_2(g))^@ = "0 kJ/mol"#.

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