The Equilibrium Constant

The relationship between forward and reverse reactions in dynamic equilibrium can be expressed mathematically in what is known an equilibrium expression, or K_eq expression. Most often, this expression is written in terms of the concentrations of the various reactants and products, and is given by:

$K_{eq}=\frac{[C]^{m}[D]^{n}}{[A]^{a}[B]^{b}}$

Species in brackets represent the concentrations of products, which are always in the numerator, and reactants, which are always in the denominator. Each of the concentrations is raised to a power equal to the stoichiometric coefficient for each species.

Derivation of the Equilibrium Expression from the Law of Mass Action

Consider the following general, reversible reaction:

$aA+bB\rightleftharpoons mC+nD$

Assuming this reaction is an elementary step, we can write the rate laws for both the forward and reverse reactions:

$\text{rate}_{\text{forward}}=k_1[A]^a[B]^b$

$\text{rate}_{\text{reverse}}=k_2[C]^m[D]^n$

However, we know that the forward and reverse reaction rates are equal in equilibrium:

$k_1[A]^a[B]^b=k_2[C]^m[D]^n$

Rearranging this equation and separating the rate constants from the concentration terms, we get:

$\frac{k_1}{k_2}=\frac{[C]^m[D]^n}{[A]^a[B]^b}$

Notice that the left side of the equation is the quotient of two constants, which is simply another constant. We simplify and write this constant as K_eq:

$\frac{k_1}{k_2}=K_{C}=\frac{[C]^{m}[D]^{n}}{[A]^{a}[B]^{b}}$

Keep in mind that the only species that should be included in the K_eq expression are reactants and products that exist as gases or are in aqueous solution. Reactants and products in the solid and liquid phases, even if they are involved in the reaction, are not included in the K_eq expression, as these species have activities of 1.

"Activity" is a term in physical chemistry used to describe a substance's ideal concentration. The activity for solids and liquids is 1, so they essentially have a constant concentration of 1, and thereby have no effect on the K_eq expression. As such, they are omitted.

Example

Write the K_eq expression for the following reaction:

$H_2O(g)+C(s)\rightleftharpoons H_2(g)+CO(g)$

The expression would be written as:

$K_{eq}=\frac{[H_2][CO]}{[C][H_2O]}=\frac{[H_2][CO]}{1\times[H_2O]}=\frac{[H_2][CO]}{[H_2O]}$

Note that because it is a solid, the activity of C(s) is 1, and it is omitted from the final K expression.

Predicting the Direction of a Reaction From the Value of K_eq

When looking at the K_eq expression, we should notice that it is essentially a ratio relating the concentrations of products to the concentrations of reactants at equilibrium. If we know the value of K_eq, we can draw some conclusions about the thermodynamics of the forward and reverse reactions. These conclusions are summarized as follows:

• A K_eq value of << 1 is indicative that the reverse reaction is highly favored over the forward reaction, and the concentrations of reactants are much higher than those of the products at equilibrium.
• A K_eq value $\approx$ 1 is indicative that the forward and reverse reactions are about equally favorable, for the ratio of concentrations of reactants and products is close to unity.
• A K_eq >>1 is indicative that the forward reaction is highly favored over the reverse reaction, and at equilibrium, the concentrations of the products are much greater than those of the reactants.