0.23 Free energy and thermodynamic equilibrium  (Page 2/6)

Foundation

We have come a long way to reach this point in our studies, so we have a substantial foundation to build on. We know all the elements of the Atomic Molecular Theory, including the models for molecular structure and bonding, and we have developed the postulates of the Kinetic Molecular Theory. We have observed and defined phase transitions and phase equilibrium, and we have also observed equilibrium in a variety of reaction systems, including acids and bases. We will assume an understanding of the energetics of chemical reactions, including the idea of a “state function” and the concept of Hess’ Law.

In the previous study, we developed the expression ∆S-∆H/T>0 as a predictor of whether a process will be spontaneous. Chemists generally rewrite this inequality in a different and somewhat more convenient form by multiplying both sides by –T :

-T∆S+∆H<0

This is the same inequality, but take notice of the “less than” sign which replaced the “greater than” sign as a result of multiplying by – T.

Both terms in this inequality have units of energy, so ∆H - T∆S is often referred to as an energy. However, it is important to remember that this combination of terms arises from calculating the changes in entropy of the system and its surroundings, and the “less than” sign arises from the Second Law of Thermodynamics which deals with entropy and not energy. As such, chemists refer to this combination of terms as the “free energy” rather than the energy. Specifically, we define a new term as G = H – TS, which is also called the “Gibbs free energy.” If the temperature of a process is constant, then for that process ∆G = ∆H - T∆S. This means that

∆G<0 for any spontaneous process at constant T and P.

This might look and sound new and complicated, but this is just the Second Law of Thermodynamics in a succinct and convenient form. To see this, let’s recall that

∆S _universe = ∆S _system + ∆S _surroundings = ∆S - ∆H/T

If we multiply this equation by –T , we get

-T∆S _universe = -T∆S + ∆H = ∆G

These equations show that the change in the free energy is just the change in the entropy of the universe multiplied by –T . Since ∆ S _universe >0, then ∆G<0. As such, when you see G for “free energy,” you might find it convenient to think of it as the “negative entropy of the universe.” For that reason, during a spontaneous process, the free energy of the system always decreases. We will use this notation alternately with the entropy notation.

Observation 1: thermodynamic description of phase equilibrium

In the previous study, we carefully analyzed the freezing of water as an interesting example of a spontaneous process for which ∆Sº <0 but for which ∆S° - ∆H°/ T>0. The discussion above tells us that, when the freezing of water is at equilibrium, then ∆S° - ∆H°/T = 0. We should be able to solve this equation for T to predict the temperature at 1 atm when liquid water and solid water are in equilibrium. We can calculate ∆ Sº = Sº _liquid - Sº _solid = 22.0 J/mol·K, and ∆Hº = ∆Hº f (liquid) - ∆Hº f (solid) = 6.01 kJ/mol. When we solve for T we get T = 273 K, which is of course the melting point temperature of water. This calculation shows the power of our predictions from the Second Law of Thermodynamics. Not only can we determine what processes are spontaneous, but we also determine the conditions necessary for a process to come to equilibrium!