# Heterogeneous and Multiple Equilibria

#### Learning Objective

• Calculate the equilibrium constant of a multiple-step reaction, given the equilibrium constant for each step

#### Key Points

• In heterogeneous equilibria, compounds in different phases react. However, the concentration of a pure solid or liquid per unit volume is always the same. As such, the activity (ideal concentration) of a solid or liquid is 1, and these phases have no effect on the equilibrium expression.
• In multiple equilibria, the equilibrium can be split into two or more steps. Both steps must be included in the equilibrium constant equation.
• The product of the equilibrium constants for each step in a reaction mechanism is equal to the equilibrium expression of the overall reaction.

#### Term

• heterogeneousHaving more than one phase (solid, liquid, gas) present in a system or process.

## Heterogeneous Equilibria

In heterogeneous equilibria, compounds in different phases react. For example, equilibrium could exist between solid and gaseous species, between liquid and aqueous species, etc.

## Example

The following equilibrium system involves both gas and solid phases:

$C(s)+{ CO }_{ 2 }(g)\rightleftharpoons 2CO(g)$

Therefore, the equilibrium expression for this reaction will be written as:

${ K }_{ eq }=\frac {[CO]^2}{[C][CO_2] }$

C(s) is omitted from the expression because it exists in the solid phase. The reason for this is because the concentration of a pure solid or a pure liquid is always the same; its "concentration" is really its density, which is uniform regardless of sample size. As a result, the activity, or ideal concentration, of a liquid or a solid is defined as 1. Since their activity is unity, and anything multiplied by 1 remains itself, solids and liquids have no effect whatsoever on the equilibrium expression. The above expression reduces to:

${ K }_{ eq }=\frac {[CO]^2}{(1)[CO_2] }=\frac {[CO]^2}{[CO_2] }$

## Multiple Equilibria

In multiple equilibria, the equilibrium can be split into two or more steps. Both steps must be included in the equilibrium constant equation.

## Example

Consider the case of a diprotic acid, such as sulfuric acid. Diprotic acids can be written as H2A. When dissolved in water, the mixture will contain H2A, HA-, and A2-. These equilibria can be split into two steps:

$H_2A \leftrightharpoons HA^- + H^+\quad\quad K_1 = \frac{[HA^-][H^+]}{[H_2A]}$

$HA^- \leftrightharpoons A^{2-} + H^+\quad\quad K_2 = \frac{[A^{2-}][H^+]}{[HA^-]}$

K1 and K2 are examples the equilibrium constants for each step. Next, we can write out the overall reaction equation, which is a sum of these two steps:

$H_2A\rightleftharpoons A^{2-}+2\;H^+\quad\quad K_{eq}=\frac{[A^{2-}][H^+]^2}{[H_2A]}$

Notice that the equilibrium expression for the overall reaction, Keq, is equal to the product of the equilibrium expressions for the two reaction steps. Thus, for a reaction involving two elementary steps:

$K_{eq}=K_1K_2$