8.5.1: Free Energies and Equilibrium Constants

Now we can calculate the equilibrium constant K_eq, assuming that we can measure or calculate the concentrations of reactants and products at equilibrium. All well and good, but is this simply an empirical measurement? It was certainly discovered empirically and has proven to be applicable to huge numbers of reactant systems. It just does not seem very satisfying to say this is the way things are without an explanation for why the equilibrium constant is constant. How does it relate to molecular structure? What determines the equilibrium constant? What is the driving force that moves a reaction towards equilibrium and then inhibits any further progress towards products?

You will remember (we hope) that it is the second law of thermodynamics that tells us about the probability of a process occurring. The criterion for a reaction proceeding is that the total entropy of the universe must increase. We also learned that we can substitute the Gibb’s free energy change (ΔG) for the entropy change of the universe, and that ΔG is much easier to relate to and calculate because it only pertains to the system. So it should not be a surprise to you that there is a relationship between the drive towards equilibrium and the Gibbs free energy change in a reaction. We have already seen that a large, negative Gibbs free energy change (from reactants to products) indicates that a process will occur (or be spontaneous, in thermodynamic terms¹⁶⁶), whereas a large, positive equilibrium constant means that the reaction mixture will contain mostly products at equilibrium.

Think about it this way: the position of equilibrium is where the maximum entropy change of the universe is found. On either side of this position, the entropy change is negative and therefore the reaction is unlikely. If we plot the extent of the reaction versus the dispersion of energy (in the universe) or the free energy, as shown in the graph, we can better see what is meant by this. At equilibrium, the system sits at the bottom of an energy well (or at least a local energy minimum) where a move in either direction will lead to an increase in Gibbs energy (and a corresponding decrease in entropy). Remember that even though at the macroscopic level the system seems to be at rest, at the molecular level reactions are still occurring. At equilibrium, the difference in Gibbs free energy, ΔG, between the reactants and products is zero. It bears repeating: the criterion for chemical equilibrium is that ΔG = 0 for the reactants ⇄ products reaction. This is also true for any phase change. For example, at 100^oC and 1 atmosphere pressure, the difference in free energy for H₂O_(g) and H₂O_(l) is zero. Because any system will naturally tend to this equilibrium condition, a system away from equilibrium can be harnessed to do work to drive some other non-favorable reaction or system away from equilibrium. On the other hand, a system at equilibrium cannot do work, as we will examine in greater detail.

The relationship between the standard free energy change and the equilibrium constant is given by the equation:

ΔG^o= –RTlnK

which can be converted into the equation

ln K_eq = – ΔG^o/RT or K_eq = e ^–ΔGo/RT.

As we saw earlier, the superscript ^o refers to thermodynamic quantities that are measured and calculated at standard states. In this case ΔG^o refers to 1 atmosphere pressure and 298 K and (critical for our present discussion) 1 M concentrations for both reactants and products. That is,: ΔG^o tells you about the free energy change if all the substances in the reacting system were mixed with initial concentrations of 1.0 M. It allows us to calculate equilibrium constants from tables of free energy values (see Chapter 9). Of course, this is a rather artificial situation and you might be tempted to think that ΔG^o is not very useful in the real world where initial concentrations of both reactants and products are rarely 1.0 M. But no, ΔG^o does tell us something useful: it tells us which way a reaction will proceed under these starting conditions. If we have a specific set of conditions, we can use ΔG^o to calculate the actual free energy change ΔG, where:

ΔG = ΔG^o + RT ln Q

In this equation, the variable Q is called the reaction quotient. It has the same form as K_eq ([products]/[reactants], except that the concentrations are not 1.0 M. Rather, they are the actual concentrations at the point in the reaction that we are interested in. The sign and magnitude of ΔG then will tell us which way the reaction will proceed and how far in that direction it will go.

The differences between Q and K_eq, ΔG, and ΔG^o are important to keep in mind. It is easy to get mixed up and apply them incorrectly. Q and ΔG relate to non-equilibrium systems whereas Keq and ΔG^o tell us about the equilibrium state itself. At equilibrium, Q = K_eq, and ΔG = 0, so that the equation ΔG = ΔG^o +RT ln Q becomes ΔG^o = – RT ln Keq. Note that K_eq and ΔG^o are constant for a given reaction at a given temperature, but Q and ΔG are not; their values vary according to the reaction conditions. In fact, by using Q and/or ΔG, we can predict how a system will behave under a specific condition as it moves towards the highest entropy state (to where ΔG=0).

References

¹⁶⁶ Once more it is important to note that in thermodynamic terms, reactions referred to as spontaneous (inappropriately, in our view) do not indicate the rate at which a reaction will happen, but rather whether it will ever happen. In fact some “Spontaneous” reactions either do not occur at all (wood in an atmosphere containing oxygen does not burn spontaneously) or occur quite slowly (iron rusting).