Two examples of an equilibrium constant dealt with in other sections, namely,

for the reaction       cis-C₂H₂F₂ trans-C₂H₂F₂

and                      

for the reaction      N₂O₄ 2NO₂

are both particular examples of a more general law governing chemical equilibrium in gases. If we write an equation for a gaseous equilibrium in general in the form

      aA(g) + bB(g) cC(g) + dD(g)      (1)

then the equilibrium constant defined by the equation

            (2)

is found to be a constant quantity depending only on the temperature and the nature of the reaction. This general result is called the law of chemical equilibrium, or the law of mass action.

EXAMPLE 1 Write expressions for the equilibrium constant for the following reactions:

a) 2HI(g) H₂(g) + I₂(g)

b) N₂(g) + 3H₂(g) 2NH₃(g)

c) O₂(g) + 4HCl(g) 2H₂O(g) + 2Cl₂(g)

Solution

a)

b)

c)

EXAMPLE 2 A mixture containing equal concentrations of methane and steam is passed over a nickel catalyst at 1000 K. The emerging gas has the composition [CO] = 0.1027 mol dm^–3, [H₂] = 0.3080 mol dm^–3, and [CH₄] = [H₂O] = 0.8973 mol dm^–3. Assuming this mixture is at equilibrium, calculate the equilibrium constant K_c for the reaction

      CH₄(g) + H₂O(g) CO(g) + 3 H₂(g)

Solution The equilibrium constant is given by the following equation:

Note: The yield of H₂ at this temperature is quite poor. In the commercial production of H₂ from natural gas, the reaction is run at a somewhat higher temperature where the value K_c is larger.

As the above example shows, the equilibrium constant K_c is not always a dimensionless quantity. In general it has the units (mol dm^–3)^Δn, where Δn is the increase in the number of molecules in the equation. In the above case Δn = 2, since 4 molecules (3 H₂ and 1CO) have been produced from 2 molecules (CH₄ and H₂O). Only if Δn = 0, as is the case for the cis-trans isomerization considered above, is the equilibrium constant a dimensionless quantity.

We can also apply the equilibrium law to reactions which involve pure solids and pure liquids as well as gases. We find in such cases that as long as some solid or liquid is present, the actual amount does not affect the position of equilibrium. Accordingly, only the concentrations of gaseous species are included in the expression for the equilibrium constant. For example, the equilibrium constant for the reaction

      CaCO₃(s) CaO(s) + CO₂(g)      (3)

is given by the expression

      K_c = [CO₂]      (4)

in which only the concentration of the gas appears. Equation (4) suggests that if we heat CaCO₃ to a high temperature so that some of it decomposes, the concentration of CO₂ at equilibrium will depend only on the temperature and will not change if the ratio of amount of solid CaCO₃ to amount of solid CaO is altered. Experimentally this is what is observed.

EXAMPLE 3 Write expressions for the equilibrium constants for the following reactions:

a) C(s) + H₂O(g) CO(g) + H₂(g)

b) C(s) + CO₂(g) 2CO(g)

c) Fe₃O₄(s) + H₂(g) 3FeO(s) + H₂O(g)

Solution Since only gaseous species need be included, we obtain

a)

b)

c)

The equilibrium law can be shown experimentally to apply to dilute liquid solutions as well as to mixtures of gases, and the equilibrium-constant expression for a solution reaction can be obtained in the same way as for a gas-phase reaction. In solution only the concentrations of species in the liquid phase need be included. In some solution reactions, the solvent may be a reactant or product. Acetic acid, for example, reacts as follows when it dissolves in water:

      CH₃COOH + H₂O CH₃COO^– + H₃O⁺      (5)

As long as the solution is dilute, however, the concentration of the solvent is hardly affected by addition of solutes, even if they react with it. (The concentration of pure water may be calculated from the density:

Even if 0.1 mol dm^–3 of acetic acid were added, the concentration of water would be affected by much less than 1 percent.)

Because the concentration of solvent remains essentially constant, it is usually incorporated into the equilibrium constant. Following the usual rules, Eq. (5) would give

This can be rearranged to

      K_a = K_c × 55.5 mol dm^–3 =

Thus the concentration of water is conventionally included in the equilibrium constant K_a for a reaction in aqueous solution. Since it applies to a weak acid, K_a is called an acid constant. (The a stands for acid.) Other equilibrium constants which contain a constant concentration in this way are the base constant, K_b, for ionization of a weak base and the solubility product constant, K_sp, for dissolution of a slightly soluble compound.

EXAMPLE 4 Write out expressions for the equilibrium constants for the following ionic equilibria in dilute aqueous solution:

a) HF(aq) + H₂O F^–(aq) + H₃O⁺(aq)

b) H₂O + NH₃(aq) OH^–(aq) + NH₄⁺(aq)

c) H₂O + CO₃^2–(aq) HCO₃^–(aq) + OH^–(aq)

d) BaSO₄(s) Ba²⁺(aq) + SO₄^2–(aq)

Solution We leave out the concentration of H₂O in the first three examples and the concentration of solid BaSO₄ in the fourth.

a) K_a = K_c × [H₂O] =

b) K_b = K_c × [H₂O] =

c) K_b = K_c × [H₂O] =

d) K_sp = K_c × [BaSO₄] = [Ba²⁺][SO₄^2–]

EXAMPLE 5 Measurements of the conductivities of acetic acid solutions indicate that the fraction of acetic acid molecules converted to acetate and hydronium ions is

a) 0.0296 at a concentration of 0.020 00 mol dm^–3

b) 0.5385 at a concentration of 2.801 × 10^–5 mol dm^–3

Use these data to calculate the equilibrium constant for Eq. (5) at each concentration.

Solution Consider first 1 dm³ of solution a. This originally contained 0.02 mol CH₃COOH of which the fraction 0.0296 has ionized. Thus (1 – 0.0296) × 0.02 mol undissociated CH₃COOH is left, while 0.0296 × 0.02 mol H₃O⁺ and CH₃COO^– have been produced. In tabular form

Substituting into the expression for K_a gives

A similar calculation on the second solution yields

Note: The two values of the equilibrium constant are only in approximate agreement. In more concentrated solutions the agreement is worse. If the concentration is 1 mol dm^–3, for instance, K_a has the value 1.41 × 10^–5 mol dm^–3. This is the reason for our statement that the equilibrium law applies to dilute solutions.