Homogeneous vs Heterogeneous Equilibria
In a homogeneous equilibrium all substances are in the same phase, whereas in a heterogeneous equilibrium there is more than one phase; in the K expression, the concentrations of pure solids and pure liquids are omitted because they are constant. In brief: for CaCO3(s) ⇌ CaO(s) + CO2(g), the expression is simply Kp = P(CO2), and the amount of solid does not change it as long as some of each solid is present.
Many important equilibria involve more than one phase: the decomposition of limestone, the dissolving of a sparingly soluble salt or the vapor over a liquid. The rules for writing the equilibrium expression differ slightly for these cases. This article explains the distinction, gives the rules for writing K, shows examples in a table and explains why the concentration of a pure solid does not appear.
What are homogeneous and heterogeneous equilibria?
- Homogeneous equilibrium: all reactants and products are in the same phase, for example all gases (N2 + 3H2 ⇌ 2NH3) or all dissolved in solution (CH3COOH ⇌ H+ + CH3COO−).
- Heterogeneous equilibrium: the substances are in two or more phases, for example a solid and a gas (CaCO3(s) ⇌ CaO(s) + CO2(g)) or a solid and a solution (AgCl(s) ⇌ Ag+(aq) + Cl−(aq)).
What is the rule for pure solids and liquids?
The concentrations (or activities) of pure solids and pure liquids are not included in the equilibrium constant expression. Only gases (as partial pressures or concentrations) and solutes (as concentrations) appear. The amount of a pure solid or liquid may change in the reaction, but the amount is not relevant to the equilibrium constant as long as some of it is present.
Why are solids left out?
The “concentration” of a pure solid is its density divided by its molar mass, and it depends only on the temperature, not on how much solid is present. For CaCO3, with a density of 2.71 g/cm3 and a molar mass of 100.09 g/mol, the concentration is 2.71 / 100.09 = 0.0271 mol/cm3, or about 27 mol/L, whether you have a gram or a kilogram of it. Since it is a constant, it is absorbed into the equilibrium constant. In the thermodynamic definition, a pure solid in its standard state has an activity of 1. The same holds for pure liquids.
The consequence is that a solid must be present for the equilibrium to exist, but its amount does not change the position of the equilibrium. Adding more solid CaCO3 to a vessel at equilibrium does not change the pressure of CO2.
Examples of equilibrium expressions
| Reaction | Type | Expression |
|---|---|---|
| N2(g) + 3H2(g) ⇌ 2NH3(g) | Homogeneous (gas) | Kc = [NH3]2 / ([N2][H2]3) |
| CaCO3(s) ⇌ CaO(s) + CO2(g) | Heterogeneous | Kc = [CO2]; Kp = P(CO2) |
| C(s) + CO2(g) ⇌ 2CO(g) | Heterogeneous | Kp = P(CO)2 / P(CO2) |
| H2O(l) ⇌ H2O(g) | Heterogeneous | Kp = P(H2O), the vapor pressure |
| Fe3O4(s) + 4H2(g) ⇌ 3Fe(s) + 4H2O(g) | Heterogeneous | Kc = [H2O]4 / [H2]4 |
| AgCl(s) ⇌ Ag+(aq) + Cl−(aq) | Heterogeneous | Ksp = [Ag+][Cl−] |
Worked examples
Example 1: calcium carbonate decomposition
At about 898 °C the equilibrium pressure of CO2 over CaCO3 and CaO is 1 atm, so Kp ≈ 1 at that temperature. At about 800 °C the equilibrium pressure is only about 0.2 atm, so Kp ≈ 0.2. This is the reason why lime kilns operate at 900 °C or higher, and the CO2 is carried away, so that the reaction continues. In a closed vessel at a given temperature, the pressure of CO2 is fixed, whatever the amounts of solids.
Example 2: carbon and carbon dioxide
For C(s) + CO2(g) ⇌ 2CO(g), the equilibrium pressures at a particular temperature are P(CO) = 0.60 atm and P(CO2) = 0.30 atm. Kp = (0.60)2 / 0.30 = 0.36 / 0.30 = 1.2. This reaction is important in blast furnaces for the production of iron.
Example 3: Q for a solid–gas system
For Fe3O4(s) + 4H2(g) ⇌ 3Fe(s) + 4H2O(g), suppose Kc = 0.10. A mixture contains [H2] = 2.0 M and [H2O] = 1.0 M. Q = [H2O]4 / [H2]4 = (1.0 / 2.0)4 = 0.0625. Since Q < K, the reaction proceeds forward, producing more iron and water vapor.
Example 4: vapor pressure as an equilibrium constant
For H2O(l) ⇌ H2O(g), Kp = P(H2O). At 25 °C this is 23.8 mmHg, or 0.0313 atm, so Kp = 0.0313. The vapor pressure of water depends only on temperature, not on how much liquid is in the container.
What about the solvent?
In dilute aqueous solutions, water is the solvent and is present in great excess, with a concentration of about 55.5 mol/L (1000 g / 18.015 g/mol). Its concentration hardly changes in a reaction, so it is treated as a constant and does not appear in the expression. For this reason, the equilibrium constant of the ionization of a weak base, NH3 + H2O ⇌ NH4+ + OH−, is written Kb = [NH4+][OH−] / [NH3], without [H2O]. When water is a reactant or product in a reaction in a non-aqueous medium, or is present in a small quantity, it may need to be included.
Where does the rule come from?
Guldberg and Waage, in their law of mass action of 1864, recognized that the “active mass” of a solid in contact with a solution or gas does not change with the amount of solid. The thermodynamic treatment by Gibbs in the 1870s, with chemical potentials and activities, put it on a firm basis: the activity of a pure condensed phase at a given pressure and temperature is constant, and by convention it is taken to be 1 under standard conditions.
Which mistakes are common?
- Including solids and pure liquids: they are left out of K and Q.
- Forgetting that some solid must be present: without it, the system is not at equilibrium.
- Counting the solid in Δn for Kp and Kc: only gases count.
- Thinking that more solid shifts the equilibrium: it does not.
- Leaving out a liquid that is not pure: mixtures of liquids, such as ethanol in water, need their concentrations.