## Heterogeneous and Multiple Equilibria

Heterogeneous equilibria involve reactions with compounds in different phases; multiple equilibria involve reactions with two or more steps.

### Learning Objectives

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

### Key Takeaways

#### 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.

#### Key Terms

• heterogeneous: Having 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:

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

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

${ \text{K} }_{ \text{eq} }=\frac {[\text{CO}]^2}{[\text{C}][\text{CO}_2] }$

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:

${ \text{K} }_{ \text{eq} }=\frac {[\text{CO}]^2}{(1)[\text{CO}_2] }=\frac {[\text{CO}]^2}{[\text{CO}_2] }$

### 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:

$\text{H}_2\text{A} \leftrightharpoons \text{HA}^- + \text{H}^+\quad\quad \text{K}_1 = \frac{[\text{HA}^-][\text{H}^+]}{[\text{H}_2\text{A}]}$

Sulfuric acid: Sulfuric acid, the molecule pictured here, is an example of a diprotic acid.

$\text{HA}^- \leftrightharpoons \text{A}^{2-} + \text{H}^+\quad\quad \text{K}_2 = \frac{[\text{A}^{2-}][\text{H}^+]}{[\text{HA}^-]}$

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:

$\text{H}_2\text{A}\rightleftharpoons \text{A}^{2-}+2\;\text{H}^+\quad\quad \text{K}_{\text{eq}}=\frac{[\text{A}^{2-}][\text{H}^+]^2}{[\text{H}_2\text{A}]}$

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:

$\text{K}_{\text{eq}}=\text{K}_1\text{K}_2$

## Specialized Equilibrium Constants

Common reactions, such as the self-ionization of water, have specially named equilibrium constants.

### Learning Objectives

List the various special types of equilibrium constants

### Key Takeaways

#### Key Points

• The self-ionization of water is the dissociation of water into a proton and a hydroxide ion. The expression Kw is defined as the product of the concentration of hydrogen ions and the concentration of hydroxide ions.
• The acid dissociation constant, Ka, measures the relative strength of an acid.
• The base dissociation constant, Kb, measures the relative strength of a base.

#### Key Terms

• self-ionization: The process by which a water molecule donates a proton to a neighboring water molecule, yielding hydronium and hydroxide ions.
• dissociation: The process of breaking molecules apart into ions in solution.

Many reactions are so common or useful that they have their own special equilibrium constants.

### Self-Ionization of Water

The self-ionization, or autodissociation, of water is a reaction that occurs to a very small extent in neutral water. In this process, one molecule of water donates a proton to a neighboring water molecule, which yields hydronium and hydroxide ions.

Autodissociation of water: A water molecule protonates a neighboring water molecule, yielding hydronium and hydroxide ions.

The equilibrium expression for this reaction is written as follows:

$\text{K}_\text{W}=[\text{H}^+][\text{OH}^-]=1.0\times 10^{-14}$

Note that because water is a liquid, it is omitted from this equilibrium expression. The value of the dissociation constant of water, KW, is $1.0\times 10^{-14}$. This will come into use later, in a future discussion on acids and bases.

### Acid Dissociation Constant, Ka

An acid dissociation constant, Ka, is the equilibrium constant for the dissociation of an acid in aqueous solution. The general form of the balanced equation is:

$\text{HA}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq})+\text{A}^-(\text{aq})$

HA is a generic acid that dissociates by splitting into A, known as the conjugate base of the acid, and a hydrogen ion, or proton, H+. As described previously, hydrogen ions actually exist as solvated hydronium ions in aqueous solutions. The equilibrium expression is given as:

$\text{K}_\text{a}=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$

Recall that strong acids dissociate completely or almost completely into their ions. As such, strong acids will have large values of Ka that are greater than one, which indicates that the forward reaction of dissociation is strongly favored. Weak acids, on the other hand, will have small values of Ka that are less than one, indicating that the reverse reaction is strongly favored; weak acids dissociate only to a small extent. As such, Ka acts a relative indicator of acid strength.

Acetic acid dissociation: A ball-and-stick model of the dissociation of acetic acid to acetate. A water molecule is protonated to form a hydronium ion in the process. The acidic proton that is transferred from acetic acid to water is shown in green.

### Base Dissociation Constant, Kb

The base dissociation constant, Kb, is analogous to the acid dissociation constant. For a generalized reaction of a base, B, in water, we have:

$\text{B}(\text{aq})+\text{H}_2\text{O}(\text{l})\rightleftharpoons \text{BH}^+(\text{aq})+\text{OH}^-(\text{aq})$

The equilibrium expression, Kb, is given by:

$\text{K}_\text{b}=\frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$

As with the acid dissociation constant, large values of Kb are indicative of a stronger base, while small values of Kb are indicative of a weaker base.

## Reaction Quotients

The reaction quotient is a measure of the relative amounts of reactants and products during a chemical reaction at a given point in time.

### Learning Objectives

Calculate the reaction quotient, Q, and use it to predict whether a reaction will proceed in the forward or reverse direction

### Key Takeaways

#### Key Points

• Just like the equilibrium constant, Keq, the reaction quotient is a function of activities and/or concentrations of reactants and products.
• The reaction quotient can be used to determine in which direction a reaction will proceed. If Q = Keq, the reaction is at equilibrium. If Q < Keq, the reaction will move toward the products to reach equilibrium. If Q > Keq, the reaction will move toward the reactants in order to reach equilibrium.
• As the reaction proceeds, the species’ concentrations, and hence the reaction quotient, change. Eventually the concentrations become constant; at this point, the reaction is at equilibrium.

#### Key Terms

• reaction quotient: A measure of the activities or concentrations of the chemical species involved in a chemical reaction at a given point in time.
• equilibrium constant: A numerical value derived from the ratio of the concentrations of products and reactants at equilibrium.
• equilibrium: The state of a reaction in which the rates of the forward and reverse reactions are the same.

The reaction quotient, Q, is a measure of the relative amounts of reactants and products during a chemical reaction at a given point in time. By comparing the value of Q to the equilibrium constant, Keq, for the reaction, we can determine whether the forward reaction or reverse reaction will be favored. Take the following generic equation:

$\text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD}$

The reaction quotient, Q, takes the following form:

$\text{Q} = \frac{[\text{C}]^{\text{c}}[\text{D}]^{\text{d}}}{[\text{A}]^{\text{a}}[\text{B}]^{\text{b}}}$

Note that the reaction quotient takes the exact same form as the equilibrium constant, and is a function of concentrations and/or activities of the reactants and products. The difference is that Q applies when the reaction is at non-equilibrium conditions, and therefore its value can vary. Just as for the equilibrium constant, the reaction quotient can be a function of activities or concentrations.

The reaction quotient can be used to determine whether a reaction under specified conditions will proceed spontaneously in the forward direction or in the reverse direction. Three properties can be derived from this definition of the reaction quotient:

1. If Q = Keq, the reaction is at equilibrium.
2. If Q < Keq, the reaction will move to the right (in the forward direction) in order to reach equilibrium.
3. If Q > Keq, the reaction will move to the left (in the reverse direction) in order to reach equilibrium.

Moving toward equilibrium: The ball in the initial state is indicative a reaction in which Q < K; in order to reach equilibrium conditions, the reaction proceeds forward.

As the reaction proceeds, assuming that there is no energy barrier, the species’ concentrations, and hence the reaction quotient, change. Eventually, the concentrations become constant; at this point, the reaction is at equilibrium. The equilibrium constant, Keq, can be expressed as follows:

${\text{K}}_{\text{eq}}=\lim_{\text{t}\to\infty}{\text{Q}(\text{t})}$

This expression shows that Q will eventually become equal to Keq, given an infinite amount of time. However, most reactions will generally reach equilibrium in a finite period of time.

## Expressing the Equilibrium Constant of a Gas in Terms of Pressure

For gas-phase reactions, the equilibrium constant can be expressed in terms of partial pressures, and is given the designation KP.

### Learning Objectives

Write the equilibrium expression, KP, in terms of the partial pressures of a gas-phase reaction

### Key Takeaways

#### Key Points

• According to the ideal gas equation, pressure is directly proportional to concentration, assuming volume and temperature are constant.
• Since pressure is directly proportional to concentration, we can write our equilibrium expression for a gas-phase reaction in terms of the partial pressures of each gas. This special equilibrium constant is known as KP.
• KP takes the exact same form as KC. To avoid confusion between the two, do not use brackets ([ ]) when expressing partial pressures.

#### Key Terms

• partial pressure: The pressure that one component of a mixture of gases contributes to the total pressure.
• equilibrium: The state of a reaction in which the rates of the forward and reverse reactions are the same.

### Equilibrium Constants for Gases

Up to this point, we have been discussing equilibrium constants in terms of concentration. For gas-specific reactions, however, we can also express the equilibrium constant in terms of the partial pressures of the gases involved. Take the general gas-phase reaction:

$\text{aA}(\text{g})+\text{bB}(\text{g})\rightleftharpoons \text{cC}(\text{g})+\text{dD}(\text{g})$

Our equilibrium constant in terms of partial pressures, designated KP, is given as:

$\text{K}_\text{P}=\frac{\text{P}^\text{c}_\text{C}\text{P}^\text{d}_\text{D}}{\text{P}^\text{a}_\text{A}\text{P}^\text{b}_\text{B}}$

Note that this expression is extremely similar to KC, the equilibrium expression written in terms of concentrations. In order to prevent confusion, do not use brackets ([ ]), when writing KP expressions.

### KP and the Ideal Gas Law

The reason we are allowed to write a K expression in terms of partial pressures for gases can be found by looking at the ideal gas law. Recall that the ideal gas law is given by:

$\text{PV}=\text{nRT}$

Re-writing this expression in terms of P, we have:

$\text{P}=\frac{\text{n}}{\text{V}}\text{RT}$

Note that in order for K to be constant, temperature must be constant as well. Therefore, the term RT is a constant in the above expression. As for n/V ( moles per unit volume) this is simply a measure of concentration. Pressure is directly proportional to concentration, so we are justified in our use of KP.

Lastly, there is a very important equation that relates KP and KC. It is given as follows:

$\text{K}_{\text{p}} = \text{K}_{\text{c}}(\text{RT}) ^{ \Delta \text{n}}$

In this expression, $\Delta \text{n}$ is a measure of the change in number of moles of gas in the reaction. For instance, if a reaction produces three moles of gas, and consumes two moles of gas, then $\Delta \text{n}=(3-2)=1$.

Liquefied gas: Inside this tank, propane is compressed into a liquid, which is in equilibrium with its gaseous headspace. The internal pressure of the gaseous propane is a function of temperature.