The road is downhill:

Free energy and equilibrium

This means, of course, that if the total free energy G of a mixture of reactants and products goes through a minimum value as the composition changes, then all net change will cease— the reaction system will be in a state of chemical equilibrium. You will recall that the relative concentrations of reactants and products in the equilibrium state is expressed by the equilibrium constant. In this lesson we will examine the relation between the Gibbs free energy change for a reaction and the equilibrium constant.

Recall that a homogeneous reaction is one in which everything takes place in a single gas or liquid phase.

To keep things as simple as possible, we will consider a homogeneous chemical reaction of the form

A + B→ C + D

in which all components are gases at the temperature of interest. If the sum of the standard free energies of the products is less than that of the reactants, ΔG° for the reaction will be negative and the reaction will proceed to the right. But how far? If the reactants are completely transformed into products, the equilibrium constant would be infinity. The equilibrium constants we actually observe all have finite values, implying that even if the products have a lower free energy than the reactants, some of the latter will always remain when the process comes to equilibrium.

Although these relations are strictly correct only for perfect gases, we will see later that equations of similar form can be applied to many liquid solutions by substituting concentrations for pressures.

In order to understand how equilibrium constants relate to ΔG° values, assume that all of the reactants are gases, so that the free energy of gas A, for example, is given at all times by

G_A = G_A° + RT ln P_A      (5-1)

(This relation was developed in the previous lesson.) The free energy change for the reaction is sum of the free energies of the products, minus that of the reactants:

ΔG = G_C + G_D – G_A – G_B      (5-2)

Using Eq. 5-1 to expand each term on the right, we have

ΔG = (G°_C + RT ln P_C) + (G°_D + RT ln P_D) – (G°_B + RT ln P_B) – (G°_A + RT ln P_A) (5-3)

We can now express the G° terms collectively as ΔG°, and combine the logarithmic pressure terms into a single fraction

     (5-4)

which is more conveniently expressed in terms of the reaction quotient Q_P

ΔG = ΔG° + RT ln Q      (5-5) Must know this!

The free energy G is a quantity that becomes more negative during the course of any natural process. Thus as a chemical reaction takes place, G only falls and will never become more positive. Eventually a point is reached where any further transformation of reactants into products would cause G to increase. At this point G is at a minimum (see the plot below), and no further net change can take place; the reaction is at equilibrium.

Example 1: the dissociation of dinitrogen tetroxide

The gas-phase reaction N₂O₄ → 2 NO₂ furnishes a simple example of the free energy relationships in a homogeneous reaction. The free energy of 1 mole of N₂O₄ (1) is smaller than that of 2 moles of NO₂ (2) by 5.3 kJ; thus ΔG° = +5.3 kJ for the complete transformation of reactants into products. The straight diagonal line shows the free energy of all possible compositions if the two gases were prevented from mixing. The red curved line show the free energy of the actual reaction mixture. This passes through a minimum at (3) where 0.814 mol of N₂O₄ are in equilibrium with 0.372 mol of NO₂. The difference (4) corresponds to the free energy of mixing of reactants and products which always results in an equilibrium mixture whose free energy is lower than that of either pure reactants or pure products. Thus some amount of reaction will occur even if ΔG° for the process is positive.

In contrast to ΔG° which is a constant for a given reaction, ΔG varies continuously as the composition changes, finally reaching zero at equilibrium. ΔG is the “distance” (in free energy) from the equilibrium state of a given reaction.

Thus for the limiting cases of pure N₂O₄ or NO₂ (as far from the equilibrium state as the system can be!),
Q = [NO₂]²/[N₂O₄] = ±∞, making the logarithm in Eq 5-5, and thus the value of ΔG, approach the same asymptotic limits (1) or (2). As the reaction proceeds in the appropriate direction ΔG approaches zero; once there (3), the system is at its equilibrium composition and no further net change will occur.

Example 2: Isomerization of butane

 

Why reactions lead to mixtures of reactants and products

We are now in a position to answer the question posed earlier: if ΔG° for a reaction is negative, meaning that the free energies of the products are more negative than those of the reactants, why will some of the latter remain after equilibrium is reached? The answer is that no matter how low the free energy of the products, the free energy of the system can be reduced even more by allowing some of the products to be "contaminated" (i.e., diluted) by some reactants. Owing to the entropy associated with mixing of reactants and products, no homogeneous reaction will be 100% complete. An interesting corollary of this is that any reaction for which a balanced chemical equation can be written can in principle take place to some extent, however minute that might be.

 

Now let us return to Eq. 5-5 which we reproduce here:

ΔG = ΔG° + RT ln Q_p      (5-5)

As the reaction approaches equilibrium, ΔG becomes less negative and finally reaches zero. At equilibrium ΔG = 0 and Q = K, so we can write

ΔG° = –RT ln K_p (5-6) Must know this!

in which K_p, the equilibrium constant expressed in pressure units, is the special value of Q that corresponds to the equlibrium composition.

This equation is one of the most important in chemistry because it relates the equilibrium composition of a chemical reaction system to measurable physical properties of the reactants and products. If you know the entropies and the enthalpies of formation of a set of substances, you can predict the equilibrium constant of any reaction involving these substances without the need to know anything about the mechanism of the reaction.

Instead of writing Eq. 5-6 in terms of K_p, we can use any of the other forms of the equilibrium constant such as K_c (concentrations), K_x (mole fractions), K_n (numbers of moles), etc. Remember, however, that for ionic solutions especially, only the K_a, in which activities are used, will be strictly valid.

It is often useful to solve Eq. 5-6 for the equilibrium constant, yielding

(5-7)

This relation is most conveniently plotted against the logarithm of K as shown at the right, where it can be represented as a straight line that passes through the point (0,0).

 

 

We have already discussed how changing the temperature will increase or decrease the tendency for a process to take place, depending on the sign of ΔS°. This relation can be developed formally by differentiating the relation

ΔG ° = ΔH ° – TΔS° (5-8)

with respect to the temperature:

(5-9)

... so the sign of the entropy change determines whether the reaction becomes more or less allowed as the temperature increases.

We often want to know how a change in the temperature will affect the value of an equilibrium constant whose value is known at some fixed temperature. Suppose that the equilibrium constant has the value K₁ at temperature T₁ and we wish to estimate K₂ at temperature T₂. Expanding Eq. 5-7 in terms of ΔH° and ΔS°, we obtain

–RT₁ ln K₁ = ΔH ° – T₁ ΔS° and –RT₂ ln K₂ = ΔH ° – T₂ ΔS°

Dividing both sides by RT and subtracting, we obtain

(5-10)

Which is most conveniently expressed as the ratio

(5-11)

This is an extremely important relationship, but not just because of its use in calculating the temperature dependence of an equilibrium constant. Even more important is its application in the “reverse” direction to experimentally determine ΔH° from two values of the equilibrium constant measured at different temperatures. Direct calorimetric determinations of heats of reaction are not easy to make; relatively few chemists have the equipment and experience required for this rather exacting task. Measurement of an equilibrium constant is generally much easier, and often well within the capabilities of anyone who has had an introductory Chemistry course. Once the value of ΔH° is determined it can be combined with the Gibbs free energy change (from a single observation of K, through Eq. 5-7) to allow ΔS° to be calculated through Eq. 5-9.

You should now understand that for homogeneous reactions (those that take place entirely in the gas phase or in solution) the equilibrium composition will never be 100% products, no matter how much lower their free energy relative to the reactants. As was summarized in the N₂O₄-dissociation example shown above, this is due to "dilution" of the products by the reactants. In heterogeneous reactions (those which involve more than one phase) this dilution, and the effects that flow from it, may not be possible.

A particularly simple but important type of a heterogeneous process is phase change. Consider, for example, an equilibrium mixture of ice and liquid water. The concentration of H₂O in each phase is dependent only on the density of the phase; there is no way that ice can be “diluted” with water, or vice versa. This means that at all temperatures other than the freezing point, the lowest free energy state will be that corresponding to pure ice or pure liquid. Only at the freezing point, where the free energies of water and ice are identical, can both phases coexist, and they may do so in any proportion.

If the standard free energy of the first reaction is positive but that of the second reaction is sufficiently negative, then for the overall process will be negative and we say that the first reaction is “driven” by the second one. This, of course, is just another way of describing an effect that you already know as the Le Châtelier principle: the removal of substance B by the second reaction causes the equilibrium of the first to “shift to the right”. Similarly, the equilibrium constant of the overall reaction is the product of the equilibrium constants of the two steps.

Some very important practical applications of coupled reactions are discussed in Parts 2 and 3 of the next lesson.

(You are expected to be able to define and explain the significance of terms identified in green type.)

• As a homogeneous chemical reaction proceeds, the free energies of the reactants become more negative and those of the products more positive as the composition of the system changes.
• The total free energy of the system (reactants + products) always becomes more negative as the reaction proceeds. Eventually it reaches a minimum value at a system composition that defines the equilibrium composition of the system, after which time no further net change will occur ( Fig. 13.)
• The equilibrium constant for the reaction is determined the standard free energy change:
  ΔG° = RT ln K_p
• The sign of the temperature dependence of the equilibrium constant is governed by the sign of ΔH°. This is the basis of the Le Châtelier Principle.
• The free energies of solid and liquid components are constants that do not change with composition. Thus in heterogeneous reactions such as phase changes, the total free energy does not pass through a minimum and when the system is not at equilibrium only all-products or all-reactants will be stable.
• Two reactions are coupled when the product of one reaction is consumed in the other. If ΔG° for the first reaction is positive, the overall process can still be spontaneous if ΔG° for the second reaction is sufficiently negative— in which case the second reaction is said to "drive" the first reaction.

 

 

© 2003, 2007 by Stephen Lower - Simon Fraser University - Burnaby/Vancouver Canada

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