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,
(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 ΔH 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 the Gibbs free energy of formation for a compound as ΔG o for the reaction:
pure isolated elements in their standard states →
one mole of pure compound in its standard state.
For example, ΔfG o for CO2(g) at 25o and 1 atm is defined to be the ΔG for the reaction:
C(s, graphite) + 2 O2(g) → CO2(g).Then, using Hess' law, the ΔrG o for some arbitrary reaction, say
a A + b B → c C + d D, (2)is
ΔrG o = c ΔfGC o + d ΔfGD o − a ΔfGA o − b ΔfGB o. (3)Rarely one might want to run a reaction at constant volume and, therefore, would need the change in the Helmholtz free energy. We can obtain ΔrAo using the same approximations we used to obtain ΔU from ΔH. That is, from
we see that,
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,
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,
(5a, b, c)We have written a subscript, T, in Equation 5c to indicate that the equation is only valid at constant T, but it is unusual to include the T. Usually we see
(6)and it is left to the reader to understand that the equation 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 ΔrG o to tell whether or not a reaction will proceed spontaneously as written:
If ΔrG o > 0 then the reaction will not go as written (the reverse reaction will go)Sometimes people say that ΔrG 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 ΔrG 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.)
if ΔrG o < 0 then the reaction will go as written.
We see from Equation 6 that the "driving force" of a chemical reaction has two components:
ΔH is the drive toward stability. When ΔH < 0 the products are more stable than the reactants (and vice versa).For a chemical reaction ΔH and ΔS 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 ΔH and ΔS 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.
ΔS is the drive toward disorder. When ΔS > 0 the products are more disordered than the reactants.(The negative sign in front of the TΔ S shows that a positive ΔS makes a negative contribution to ΔG which tends to drive the reaction in the forward direction.)
Note that increasing T increases the influence of ΔS on the reaction "driving force."
We will show later that ΔrG
o is related to the equilibrium constant for the reaction ( as you
From here you can:
Copyright 2004, W. R. Salzman
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