13.2 The Equilibrium Constant

In the mid 1860s, Norwegian scientists C. M. Guldberg and P. Waage noted a peculiar relationship between the amounts of reactants and products in an equilibrium. No matter how many reactants they started with, a certain ratio of reactants and products was achieved at equilibrium. Today, we call this observation the law of mass actionThe relationship of the amounts of reactants and products at equilibrium.. It relates the amounts of reactants and products at equilibrium for a chemical reaction. For a general chemical reaction occurring in solution,

$$a\text{A + }b\text{B}\rightleftarrows c\text{C + }d\text{D}$$

the equilibrium constantA numerical value that relates to the ratio of products and reactants at equilibrium., also known as K_eq, is defined by the following expression:

$$K_{\text{eq}}=\frac{{[\text{C]}}^c{[\text{D]}}^d}{{[\text{A]}}^a{[\text{B]}}^b}$$

where [A] is the molar concentration of species A at equilibrium, and so forth. The coefficients a, b, c, and d in the chemical equation become exponents in the expression for K_eq. The K_eq is a characteristic numerical value for a given reaction at a given temperature; that is, each chemical reaction has its own characteristic K_eq. The concentration of each reactant and product in a chemical reaction at equilibrium is related; the concentrations cannot be random values, but they depend on each other. The numerator of the expression for K_eq has the concentrations of every product (however many products there are), while the denominator of the expression for K_eq has the concentrations of every reactant, leading to the common products over reactants definition for the K_eq.

Let us consider a simple example. Suppose we have this equilibrium:

$$\text{A}\rightleftarrows \text{B}$$

There is one reactant, one product, and the coefficients on each are just 1 (assumed, not written). The K_eq expression for this equilibrium is

$$K_{\text{eq}}=\frac{[\text{B]}}{[\text{A]}}$$

(Exponents of 1 on each concentration are understood.) Suppose the numerical value of K_eq for this chemical reaction is 2.0. If [B] = 4.0 M, then [A] must equal 2.0 M so that the value of the fraction equals 2.0:

$$K_{\text{eq}}=\frac{[\text{B]}}{[\text{A]}}=\frac{4.0}{2.0}=2.0$$

By convention, the units are understood to be M and are omitted from the K_eq expression. Suppose [B] were 6.0 M. For the K_eq value to remain constant (it is, after all, called the equilibrium constant), then [A] would have to be 3.0 M at equilibrium:

$$K_{\text{eq}}=\frac{[\text{B]}}{[\text{A]}}=\frac{6.0}{3.0}=2.0$$

If [A] were not equal to 3.0 M, the reaction would not be at equilibrium, and a net reaction would occur until that ratio was indeed 2.0. At that point, the reaction is at equilibrium, and any net change would cease. (Recall, however, that the forward and reverse reactions do not stop because chemical equilibrium is dynamic.)

The issue is the same with more complex expressions for the K_eq; only the mathematics becomes more complex. Generally speaking, given a value for the K_eq and all but one concentration at equilibrium, the missing concentration can be calculated.

In some types of equilibrium problems, square roots, cube roots, or even higher roots need to be analyzed to determine a final answer. Make sure you know how to perform such operations on your calculator; if you do not know, ask your instructor for assistance.

The K_eq was defined earlier in terms of concentrations. For gas-phase reactions, the K_eq can also be defined in terms of the partial pressures of the reactants and products, P_i. For the gas-phase reaction

$$a\text{A(g) + }b\text{B(g)}\rightleftarrows c\text{C(g) + }d\text{D(g)}$$

the pressure-based equilibrium constant, K_P, is defined as follows:

$$K_{\text{P}}=\frac{P_{\text{C}}^c{P}_{\text{D}}^d}{P_{\text{A}}^a{P}_{\text{B}}^b}$$

where P_A is the partial pressure of substance A at equilibrium in atmospheres, and so forth. As with the concentration-based equilibrium constant, the units are omitted when substituting into the expression for K_P.

There is a simple relationship between K_eq (based on concentration units) and K_P (based on pressure units):

$$K_{\text{P}}=K_{\text{eq}}\cdot {(RT)}^{\Delta n}$$

where R is the ideal gas law constant (in units of L·atm/mol·K), T is the absolute temperature, and Δn is the change in the number of moles of gas in the balanced chemical equation, defined as n_gas,prods − n_gas,rcts. Note that this equation implies that if the number of moles of gas are the same in reactants and products, K_eq = K_P.

Finally, we recognize that many chemical reactions involve substances in the solid or liquid phases. For example, a particular chemical reaction is represented as follows:

$${\text{2NaHCO}}_{\text{3}}\text{(s)}\rightleftarrows {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{(s) + CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(}\ell \text{) }$$

This chemical equation includes all three phases of matter. This kind of equilibrium is called a heterogeneous equilibriumAn equilibrium in which more than one phase of reactants or products is present. because there is more than one phase present.

The rule for heterogeneous equilibria is as follows: Do not include the concentrations of pure solids and pure liquids in K_eq expressions. Only partial pressures for gas-phase substances or concentrations in solutions are included in the expressions of equilibrium constants. As such, the equilibrium constant expression for this reaction would simply be

$$K_{\text{P}}=P_{{\text{CO}}_2}$$

because the two solids and one liquid would not appear in the expression.