# For #A + B rightleftharpoons C#, the #DeltaG_f^@# are #"402.0 kJ/mol"#, #"387.7 kJ/mol"#, and #"500.8 kJ/mol"#, respectively. What is #DeltaG_(rxn)^@#? If both entropy and enthalpy changes are positive for this reaction at #25^@ "C"#, which one drives it?

##### 1 Answer

See the explanation below...

As with many other thermodynamic functions, they can be added together because they are extensive. This forms the basis for:

#DeltaG_(rxn)^@ = sum_(P) nu_P Delta_fG_(P)^@ - sum_(R) nu_R Delta_fG_(R)^@# ,where:

#Delta_fG^@# is thechange in Gibbs' free energy due to forming the substancefrom its elements in their standard states. This is the Gibbs' free energy of formation.#nu# is thestoichiometric coefficient.#P# and#R# stand for product and reactant, respectively.- Standard conditions here are defined to be
#"298.15 K"# and#"1 bar"# . Check your book to see if you still use#"1 atm"# .

This gives for

#1A + 1B rightleftharpoons 1C# ,

#DeltaG_(rxn)^@#

#= overbrace((1cdot402.0))^"Products" - overbrace((1cdot387.7 + 1cdot500.8))^"Reactants"# #"kJ/mol"#

#= ???#

You should have one decimal place.

Regardless of the magnitude of

But if **both positive**, you have

#DeltaG_(rxn)^@ = DeltaH_(rxn)^@ - TDeltaS_(rxn)^@#

#= (+) - (+)(+) = (+) - (+')#

That is, if

The spontaneous reaction is **driven by the remainder of the equation that is negative when one term goes to zero**, i.e.

#color(blue)(ul(DeltaG_(rxn)^@)) = (0) - (+)(+) color(blue)(ul(< 0))# when#DeltaH_(rxn)^@ = 0#

#color(red)(ul(DeltaG_(rxn)^@)) = (+) - (+)(0) color(red)(ul(> 0))# when#DeltaS_(rxn)^@ = 0#

One can see that if

So, the **entropy** drives the spontaneous (i.e. FORWARD) reaction at