Gibbs Free Energy

You will learn how to combine energy and entropy driving forces into a single new state function, the Gibbs free energy. You will then learn how to apply it to different circumstances.

Synopsis

Because most processes involve a change in energy and entropy, it would be useful to have a state function that combines both driving forces: energy striving for a minimum, and entropy striving for a maximum. In most processes the energy does not reach the absolute minimum value that it seeks, and the entropy does not reach the absolute maximum value that it seeks. A compromise is reached so that the process attains the lowest energy that it can concomitant with the highest energy that it can attain. This new function, the Gibbs free energy, allows us to so combine the two driving forces. This energy is called "free" because it is the net energy available to carry out some useful process, other than expansion, relating to the system being studied.

We can obtain the equation for the Gibbs free energy by starting with the entropy change for the universe:

S^o_universe = S^o_{surroundingss} + S^o_system = -H^o_system/T + S^o_system

Then if we multiply by -T, we have

-TS^o_universe = H^o_system - TS^o_system

Gibbs defined

-TS^o_universe = G^o_system

so that we finally have

G^o_system = H^o_system - TS^o_system

You see that this function combines the energy and entropy effects, and it is only related to the system. Because -TS^o_universe = G^o_system, if G^o_system is negative then S^o_universe would be positive which informs us that the process would spontaneously occur in the direction used to calculate the delta values. We than have the following:

• If G^o_system > 0 the process will spontaneously occur in the opposite direction used to calculate G^o_system.
• If G^o_system < 0 the process will spontaneously occur in the same direction used to calculate G^o_system
• If G^o_system = 0, the reaction is at equilibrium.

We then can calculate the H^o and the S^o for a chemical reaction assuming a preferred direction (usually toward the reactants) and from these values calculate G^o_system. Then we can use the above guide to determine the direction of spontaneity of the reaction.

We can also use the same basis to define a G^o_f as we did to define H^o_f. G^o_f is the change in Gibbs free energy when one mol of a compound is formed from the elements when the reactants and the products are in their standard state of aggregation at standard temperature and pressure. Table 20.3 contains some values of G^o_f for some compounds and many more values can be found in the Handbook of Chemistry and Physics. Using these standard Gibbs free energies of formation, we can then calculate the G^o_reaction for a chemical reaction precisely as was done to calculate the H^o_reaction:

G^o_reaction = G^o_f(products) - G^o_f(reactants)

Now let's look at the Gibbs free energy expression a little closer. The defining relationship is

G^o_system = H^o_system- TS^o_system

and from this expression we see that G^o_system can be negative by three different scenarios:

• The reaction can be exothermic, H^o_system negative, and S^o_system positive.
• The reaction can be exothermic and S^o_system can be negative but with \[delta]H^o_system\ > \T[delta]S^o_system\ so that the overall result is negative [\x\ means the absolute value of x].
• The reaction can be endothermic, H^o_system positive, and S^o_systempositive.

The second scenario requires that H^o_system be very negative so that the overall result will be negative and is usually termed an enthalpy-driven reaction.

The third scenario requires a positive value for S^o_system so that \TG^o_system\ > \H^o_system\ and is usually referred to as an entropy-driven reaction. One of two ways that this inequality can occur is if the entropy change is huge, and the another way is that the entropy change is not large enough, but the inequality is driven by the temperature being very large. Many endothermic reactions are carried out at very high temperatures so that this inequality will occur and the reaction is spontaneous to the product side of the reaction.

Some reactions are not spontaneous toward the reactants, but there is not a more convenient way to make the product. The reaction is not product favored because the G^o_reaction is positive. If we can couple this reaction with one whose G^o_reaction is more negative than the first one is positive, then the sum of the two reactions will have a negative G^o_reaction. An example of such a set of coupled reactions is the following:

TiO₂(s) + 2 Cl₂(g) --> TiCl₄(°) + O₂(g) G^o₂₉₈ = 152 kJ/mol C(s) + O₂(g) --> CO₂(g) G^o₂₉₈ = -394 kJ/mol

The overall reaction has G^o₂₉₈ = -242 kJ/mol and thus is product favored for the production of TiCl₄. Your text illustrates other reactions using this coupling of reactions to produce very useful products.

The usual driving force for the mixing of two or more substances is entropy. For an ideal solution (Chapter 14) the interaction energies between the molecules do not change and hence entropy is the only driving force for this mixing.

Review Question

1. Calculate S^o_system, S^o_surroundings, and S^o_universe for the following reaction

   NO(g) + O₃(g) --> NO₂(g) + O₂(g)

   Molecule	So(J/molK)	Hof(kJ/mol)	Hof(kJ/mol)
   NO(g)	211	90	87
   O3(g)	239	143	163
   NO2(g)	240	34	52
   O2(g)	205	0	0

2. Is the above reaction entropy-driven or enthalpy-driven?
3. Calculate G^o_system.

Discussion Question for Bulletin Board Activities

Find a book in the library that discusses the coupled reaction concept above and share another example of such a coupled reaction to the class on WebCT.

Web Author: Dr. Leon L. Combs
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