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?

Question

Serenity Allen · Accepted 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:

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 #DeltaG_(rxn)^@# at #25^@ "C"# and standard pressure, as long as it is negative, your reaction is spontaneous AT #25^@ "C"# and standard pressure. It is important to recognize that #DeltaG^@ ne DeltaG# in general.

But if #DeltaH_(rxn)^@# and #DeltaS_(rxn)^@# are both positive, you have

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

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

That is, if #DeltaS_(rxn)^@# is large or the temperature is high, the reaction is spontaneous.

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 #DeltaS_(rxn)^@ = 0#, the influence of the entropy is no longer there, and the reaction is no longer spontaneous.

So, the entropy drives the spontaneous (i.e. FORWARD) reaction at #25^@ "C"# and standard pressure.

Serenity Allen · Answer

undefined ΔG_(rxn)^@ = ΔG_f^@(C) - [ΔG_f^@(A) + ΔG_f^@(B)]

ΔG_(rxn)^@ = (500.8 kJ/mol) - [(402.0 kJ/mol) + (387.7 kJ/mol)]

ΔG_(rxn)^@ ≈ -288.9 kJ/mol

Positive enthalpy drives the reaction.