Heterogeneous Equilibria

6.5 Heterogeneous Equilibria

Equilibrium in a system that contains more than one phase is called heterogeneous equilibrium. Until now, we have mostly looked at reactions where everything is in the same phase — all gases, or all in aqueous solution. But many important chemical equilibria involve solids, liquids, and gases together.

A simple everyday example is the equilibrium between liquid water and water vapour in a closed container. Two phases are present — the liquid and the gas — and they are in dynamic equilibrium:

Another example is the equilibrium between a solid salt and its saturated solution. Consider calcium hydroxide:

Here, the solid phase coexists with the aqueous ions. Both are examples of heterogeneous equilibria.

The Key Simplification: Pure Solids and Liquids Have Constant Concentration

Heterogeneous equilibria very often involve pure solids or pure liquids. This fact allows us to simplify the equilibrium expression dramatically.

Why? The molar concentration of a pure solid or a pure liquid is constant. It does not depend on how much of the substance is present — it depends only on its density and molar mass, which are fixed at a given temperature. In other words, for a substance X:

• $[\text{X}(s)]$ and $[\text{X}(l)]$ are constant, regardless of the amount of X present.
• In contrast, $[\text{X}(g)]$ and $[\text{X}(aq)]$ do vary as the amount of X in a given volume changes.

Because these concentrations are constant, they can be absorbed into the equilibrium constant. This means that pure solids and pure liquids do not appear in the equilibrium expression.

The Classic Example: Thermal Decomposition of Calcium Carbonate

This is the most important and frequently cited example of heterogeneous equilibrium. When calcium carbonate is heated, it decomposes reversibly:

If we naively wrote the equilibrium constant based on the stoichiometric equation, we would get:

But both $[\text{CaCO}_3(s)]$ and $[\text{CaO}(s)]$ are constants — they are pure solids. So we can rearrange:

The left-hand side is a product of constants, which is itself a constant. We call this new constant $K'_c$:

For the equilibrium constant in terms of partial pressures, the same logic applies. Since the solids do not exert a partial pressure, we get:

This is a remarkable result: at a given temperature, the concentration (or partial pressure) of CO$_2$ in equilibrium with CaO(s) and CaCO$_3$(s) is fixed. It does not depend on how much solid is present, as long as some of each solid is there.

Experimentally, at 1100 K, the pressure of CO$_2$ in equilibrium with CaCO$_3$(s) and CaO(s) is found to be $2.0 \times 10^5$ Pa. Therefore, the equilibrium constant at 1100 K is:

(The division by $10^5$ Pa is because standard state pressure is 1 bar = $10^5$ Pa, making the constant dimensionless — more on this shortly.)

Another Example: Purification of Nickel

The Mond process for purifying nickel involves the formation and decomposition of nickel carbonyl:

Nickel is a pure solid, so it does not appear in the equilibrium expression:

Again, the solid nickel must be present for equilibrium to exist, but its concentration is constant and is omitted.

The General Rule for Heterogeneous Equilibria

Consider another example:

Here, Ag$_2$O is a pure solid and H$_2$O is a pure liquid. Both are omitted. The equilibrium constant is:

Units of Equilibrium Constants

The value of $K_c$ is calculated by substituting concentrations in mol/L. The value of $K_p$ is calculated by substituting partial pressures in Pa, kPa, bar, or atm. This means that equilibrium constants often carry units — unless the sum of exponents in the numerator equals the sum of exponents in the denominator.

For example:

• For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, both $K_c$ and $K_p$ have no units (the exponents cancel).
• For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$, $K_c$ has units of mol/L and $K_p$ has units of bar (or atm, depending on the unit used).

Making Equilibrium Constants Dimensionless

To avoid the confusion of varying units, equilibrium constants can be expressed as dimensionless quantities by specifying a standard state for each reactant and product.

• For a pure gas, the standard state is 1 bar. A pressure of 4 bar, when expressed relative to the standard state, becomes $4 \text{ bar} / 1 \text{ bar} = 4$, which is a pure number.
• For a solute in solution, the standard state ($c^\ominus$) is 1 molar solution. All concentrations are measured relative to this: $[\text{X}] / (1 \text{ mol/L})$, giving a dimensionless number.

The numerical value of the equilibrium constant depends on the standard state chosen. So $K_p$ and $K_c$ for the same reaction may have different numerical values because they use different standard states (1 bar vs. 1 mol/L), but both are dimensionless. …