Gibbs Free Energy and Chemical Reactions
We have seen that the Gibbs free energy, G, seeks a minimum at constant T and p. The
question now is does this have anything to do with chemical reactions. We define the
change in Gibbs free energy for a chemical reaction as,
rGo  Gproducts  Greactants . (1)
It is understood in Equation 1 that both the reactants and the products are pure, isolated,
in their standard states, and at the same temperature and pressure.
When we first introduced heats of reaction we said that it is not feasible to tabulate the
H o for every possible reaction. Instead, we tabulate heats of formation for compounds
and use Hess' law to find the heat of a reaction involving those compounds.
In the same manner we define Gibbs free energies of formation for compounds as G o
for the reaction:
pure isolated elements in their standard states 
one mole of pure compound in its standard state.
For example, fG o for CO2(g) at 25o and 1 atm is defined to be the G for the reaction:
C(s, graphite) + 2 O2(g)  CO2(g).
Then, using Hess' law, the rG o for some arbitrary reactions, say
a A + b B  c C + d D,
(2)
is rG o = c fGC o + d fGD o  a fGA o  b fGB o. (3)
Rarely one might want to run a reaction at constant volume and, therefor, would need the
change in the Helmholtz free energy. We can obtain rA o using the same
approximations we used to obtain U from H. That is, from
G  A  pV
we see that,
A  G  ( pV ).
Using, as before, the approximation that the change in the pV product for liquids and
solids is small and the approximation that the gases are ideal, we obtain the equation,
r Ao  rGo  RT ngas . (4)
Processes at Constant Temperature
Knowing that G = H – TS, and that reactions are constant temperature process, we can
relate the Gibbs free energy of a reaction to the enthalpy of a reaction. That is,
G  H  TS ,
G  H  (TS ) (5a, b, c)
GT  H T  T ST .
We have written a subscript, T, in Equation 5c to indicate that the equation is only valid
at constant T, but that is unusual. Usually we see
G  H  T S , (6)
and it is left to the reader to understand that the equations is only valid at constant T.
The "Driving Force" of a Chemical Reaction
Recall that at constant T and p the Gibbs free energy seeks a minimum. That means that
we can use rG o to tell whether or not a reaction will proceed spontaneously as written:
If rG o > 0 then the reaction will not go as written (the reverse reactions will go)
if rG o < 0 then the reaction will go as written.
Sometimes people say that rG o is a measure of the "driving force" of a chemical
reaction. Although the use of the word "force" is probably not appropriate, the statement
conveys the correct idea that rG o tells us whether or not a given reaction will really run
spontaneously. (You can probably write and balance a large number of reactions, but just
because you can write and balance a reaction is no guarantee that it will run.)
We see from Equation 6 that the "driving force" of chemical reactions has two
components:
H is the drive toward stability. When H < 0 the products are more stable than the
reactants (and vice versa).
S is the drive toward disorder. When S > 0 the products are more disordered than the
reactants. (The negative sign in front of the TS shows that a positive S makes a
negative contribution to G which tends to drive the reaction in the forward direction.)
Note that increasing T increases the influence of S on the reaction "driving force."
For a chemical reaction H and S are independent of each other. that is, you can not
calculate one from the other. You can have situations where both are positive, both are
negative, or one is positive and the other negative. Notice that if H and S have the
same sign they are working against each other. You can make the entropy win by
increasing the temperature or you can make the enthalpy win by decreasing the
temperature.
We will show later that rG o is related to the equilibrium constant for the reaction ( as
you might expect).
WRS