Applications of Chemical Equilibrium

Chemical equilibria can be established in the gas phase and solution phase. All components present in the equilibrium are included in the equilibrium constant. However, for hetergeneous equilibria solid components will be assigned an activity of 1. Moreover, in solid-solid reactions the equilibrium constant is not defined.

If two phases of a substance coexist they are in equilibrium. The equilibrium will only be defined at the appropriate temperature and pressure defined by the coexistence curve. For example, the graphite ß à diamond equilibrium given in M&S (page 1083-1084) is an equilibrium between two phases of a substance. In fact, the sample problem calculates the pressure of phase transition at ambient temperature. As with any phase equilibrium the coexistence curve between graphite and diamond is defined by both temperature and pressure. Hence, we can calculate it for any pressure. Without explicitly making reference to the calculation this is an application of the Clapeyron equation.

Recall that the Clapeyron equation

defines the coexistence curve between two phases. We had previously considered only solid-liquid phase transitions. However, the Clapeyron equation also applies to solid-solid phase transitions and liquid-liquid phase transitions.

Protein Folding Example

We can consider protein folding as a special and interesting case of a phase transition. Under physiological conditions proteins are folded into well-defined structures. However, as the temperature increases above a folding or denaturation temperature, the protein will begin to unfold. Assuming for the moment that we can consider a reversible protein folding process between a folded and an unfolded state

F ßà U

we can examine the equilibrium constant.

Alternatively, we can express the equilibrium constant as the fraction folded, ff.

And, of course the equilibrium constant can be related to the free energy of unfolding DG_U, and therefore the enthalpy and entropy of unfolding.

Now solving for the fraction folded we have

We are interested in calculating the dependence of the folding transition on temperature. Note that in any phase transition DG = 0 at the transition temperature. So for a protein folding transition DG_U = 0 and T_U = DH_U/DS_U. The temperature dependence of the fraction folded is sigmoidal as shown in the curves plotted below.

Ligand-Receptor Binding

A charged ligand (L⁴⁺ with a charge of 4+) docks with a protein R (R^4- with a charge of 4-). The association constant is 10⁷. The initial concentration of R^4- and L⁴⁺ is 1 mM.

L⁴⁺ + R^4- ßà RL

a. What is the extent of association a?

b. What is the extent of association when the mean activity coefficient for L⁴⁺ and R^4- is included?

c. What is the extent of association in 1 mM NaCl?

Solution

The association constant is:

where a_j is the activity of each species j.

The activities are:

Where g_j is the activity coefficient and c_j is the molarity of the respective species. Given these relations we can express the association constant as:

For part a. we assume that the activity coefficients are all equal to one. Therefore, the activities are equal to the concentrations and K_a = K_c.

We can make a table:

Species	L^4+	R^4-	RL
Initial	10^-6	10^-6	0
Equilibrium	10^-6 - a	10^-6 - a	a

Substituting the equilibrium concentrations into the equilibrium constant we have:

Or

For K_c = 10⁷ we obtain a = 7.3 x 10^-7 M.

For part b. as assume that the activity coefficient for the neutral bound species RL is one and the activity coefficients of the charge species can be replaced by the mean activity coefficients according to the Debye-Huckel theory:

Thus,

To answer part b. we need to calculate the mean activity coefficient due to the ligand and receptor protein themselves:

Where I is the ionic strength. We assume that the molarity and molality of the species is equal here. c^o is the molarity in the standard state, i.e. c^o = 1 molar concentration. The charge z₊ and z_- correspond to the charges of the ligand and protein, respectively. We can calculate the ionic strength, I

We obtain the value of c+ and c- from our initial calculation of the extent of association, c₊ = c_- = 10^-6 - a = 2.7 x 10^-7 M.

Thus, I = 4.3 x 10^-6 and g = 0.96. For this value we obtain:

Substituting back into the expression for the dissociation

we calculate a = 7.2 x 10^-6.

The dissociation, which is (a/10^-6)100% changes by 1% (from 27% to 28%) by inclusion of the activity of the ligand and protein.

For part c. we consider the ionic strength in a 1 mM NaCl solution. Note that although we calculate the ionic strength based for NaCl, the charge numbers z⁺ and z^- in

remain 4⁺ and 4^- since these are the values of the species that are associating the in the equilibrium constant. In other words, the effect of adding salt will depend strongly on the nature of the charged species interacting in the equilibrium. Thus, addition of salt is a good experimental means of estimating the charges of species involved in binding equilibria.

For 1 mM NaCl, I = 10^-3 M. Therefore,

The dissociation increases to 44%.