Chapter 5 – Chemical Equilibrium

Introduction

The equilibrium constant was defined in Chapter 4, and calculations were done to establish the relationship between it and the standard free energy change of the reaction. However, the equilibrium constant is most valuable because it lets us predict the composition of an equilibrium mixture or to adjust the reactant concentrations so as to obtain an equilibrium mixture with the desired composition. These calculations are so important that they form the basis of this chapter and much of Chapters 6, 7, and 8, where the principles established in this chapter are applied to proton transfer, solubility, and electron transfer processes, respectively.

5.1 The Equilibrium Constant

Introduction

The thermodynamic equilibrium constant discussed in Chapter 4 is unitless because it is defined in terms of unitless activities. However, equilibrium constants can be defined in terms of partial pressures or molar concentrations. These equilibrium constants typically, but not always, have units. In this section, we introduce both types of equilibrium constants, show how to convert from one to the other, and demonstrate some of their properties.

Objectives

• •
  Determine the value of K for a reaction given the value of K for a related reaction that differs only by a multiple or in direction.
• •
  Determine the equilibrium constant of a reaction that is the sum of several other reactions.
• •
  Convert between K_p and K_c.

K_p Versus K_c

5.1-1. K_p and K_c Defined

An equilibrium constant in which partial pressures are used for activities is called K_p, while one in which molar concentrations are used for activites is called K_c. For a reaction that involves only gases, K_p equals K (the thermodynamic equilibrium constant that has no units) if the partial pressures are in atmospheres because the activity of a gas is numerically equal to its partial pressure in atmospheres. For a reaction that involves only solutes in a liquid solution, K_c equals K because the activity of a solute is numerically equal to its molar concentration. Consider the following equilibrium: The units of K_p would be determined as follows: which would have units of atm^–2. The exponent of the units equals the sum of exponents in the numerator minus the sum of exponents in the denominator or the number of moles of products minus the number of moles of reactants, which is Δn_g. This is true for any equilibrium, so we can state that for a gas phase equilibrium,

• •
  the units of K_p are atm^Δn_g
• •
  the units of K_c are M^Δn_g

For solution equilibria, the units of K_c are M^Δn, where Δn = the number of moles of solute products minus the number of moles of solute reactants.

5.1-2. Converting Between K_p and K_c

Since the thermodynamic K equals K_p for gas phase equilibria, K_p values are a common way to discuss their equilibria. However, there are times when it is more convenient to consider gas phase equilibria in terms of molar concentrations; i.e., in terms of K_c. In such cases, the K_p would have to be converted into a K_c. The conversion involves converting the partial pressures of the gases into molar concentrations, which is done with the ideal gas law. We first express the partial pressure of gas A in terms of its pressure as follows: where P_A is the partial pressure of A, n_A is the number of moles of gas A. n_A/V is the number of moles of A per liter of gas, which is [A], the molar concentration of A. Substitution of n_A/V = [A] into the above equation yields Equation 5.1, which relates the concentration of a gas to its partial pressure. Substitution of [A]RT for P_A into a K_p expression converts the expression from one of pressures to one of concentrations; i.e., it converts the K_p into a K_c. Equation 5.2 summarizes the relationship.

5.1-3. Converting with the Factor Label Method

If remembering which sign of Δn_g to use is a problem, then K_p ⇌ K_c conversions can be done as factor label conversions in which units of (atm^Δn_g) are converted into units of (M^Δn_g) using RT as the conversion factor. R, which is the ideal gas law constant, equals 0.08206 L·atm·K^–1·mol^–1. However, it can be made more useful for converting between atm and M by recognizing that L·mol^–1 = M^–1, so the units of R can be expressed as atm·K^–1·M^–1, so

• •
  RT has units of atm·M^–1 = atm/M.

Finally, note that when Δn_g = 0, K_c = K_p = K. Thus, using either molarities or partial pressures in the equilibrium constant expression of a reaction consisting only of gases yields the same value of K, which is unitless and numerically equal to the thermodynamic equilibrium constant, so long as the number of moles of gas phase products equals the number of moles of gas phase reactants.

5.1-4. Thermodynamics and K Exercise

5.1-5. Exercise

Properties of the Equilibrium Constant

5.1-6. Properties

We will use the following two aqueous equilibria for our discussions of the properties of the equilibrium constant:

• •
  If the coefficients of a chemical equation are all multiplied by some number n, then the equilibrium constant of the resulting chemical equation equals the original equilibrium constant raised to the n^th power: K' = Kⁿ. For example, multiplying Equilibrium 1 by 3, we obtain

• •
  If the direction of the equation is reversed, then the equilibrium constant of the resulting equation is the reciprocal of the original equilibrium constant:
  K' = 1/K.
  For example, reversing Equilibrium 1, we obtain

• •
  If two chemical equations are added, then the equilibrium constant of the resulting equation is the product of the two original equilibrium constants. For example, adding Equilibria 1 and 2, we obtain A(aq) ⇌ C(aq), and the equilibrium constant is then determined as follows:

5.1-7. Exercise

5.1-8. Exercise

5.2 Le Châtelier's Principle

Introduction

In Chapter 4, the spontaneous direction of a reaction was predicted from known concentrations by using the fact that a reaction proceeds from left to right when

Q < K

and from right to left when

Q > K.

We also saw that a temperature rise causes the equilibrium constants of endothermic reactions to increase and the equilbirium constants of exothermic reactions to decrease. However, the same conclusions can be drawn without calculation by applying Le Châtelier's Principle. In this section, we apply this important principle to three types of disturbances:

• •
  changing an amount of a reactant or product
• •
  changing the volume
• •
  changing the temperature

Objectives

• •
  Explain how Le Châtelier's Principle is a summary of the relationship between Q and K of the reaction.
• •
  Use Le Châtelier's Principle and considerations of the reaction quotient and the equilibrium constant to predict the direction of the shift in equilibrium caused by stress placed on the equilibrium.

5.2-1. Statement of Le Châtelier's Principle

Le Châtelier's Principle

• •
  If a system at equilibrium is disturbed, then the system reacts so as to counteract the disturbance.

If a substance is added, then equilibrium shifts to remove it. If a substance is removed, then equilibrium shifts to replace it. If an equilibrium is heated, then it shifts so as to remove heat. If the volume is increased, then the equilibrium shifts to produce gas to fill the increased volume.

5.2-2. Changing the Amount of a Substance

Consider Figures 5.1a–c, which show the effect of changing one concentration in an equilibrium mixture of

A + B ⇌ C + D.

Figure 5.1a shows the original equilibrium in which [A] = [B] = 0.40 M and [C] = [D] = 0.80 M, so In Figure 5.1b, the concentration of B is increased to 0.800 M. Le Châtelier's Principle predicts that the equilibrium should shift so as to reduce the effect of the additional B by reducing its concentration. Alternatively, we determine that Q = (0.80)²/(0.40)(0.80) = 2.0 after the addition, so

Q < K,

and the reaction proceeds to the right, which has the effect of reducing [B] as predicted by Le Châtelier's Principle. Thus, B reacts with A to produce C and D. The changes in concentration resulting from the reaction cause Q to increase, which it does until it equals K and equlibrium is re-established. We use the symbol Δ for the changes in concentration; i.e., [A] and [B] each decrease by Δ, while [C] and [D] each increase by Δ. The magnitude of Δ is given by the arrows on the right side of the figure. We will show how to determine the value of Δ in the following section, but now we simply state that a

Δ = 0.105 M

is required to restore equilibrium, so the new concentrations would be the following:

• •
  [A] = 0.400 – Δ = 0.40 – 0.105 = 0.295 M
• •
  [B] = 0.800 – Δ = 0.80 – 0.105 = 0.695 M
• •
  [C] = [D] = 0.800 + Δ = 0.80 + 0.105 = 0.905 M

Note that we carry an extra decimal place in the concentrations to avoid rounding errors. Placing these concentrations back into the equation for Q, we obtain Thus,

Q = K

at these concentrations, so the system is back to equilibrium. In Figure 5.1c, the concentration of C is reduced to 0.20 M, so equilibrium should shift so as to replace some of the C; i.e., A and B must react in order to replace some of the removed C. The original conditions result in

Q = (0.20)(0.80)/(0.40)² = 1.0.

Q < K,

so again the reaction proceeds to the right and Q increases again until

Q = K

and equilibrium is re-established. Again the extent of reaction, Δ, is represented by the arrows on the right side of the figure. In this case,

Δ = 0.125 M,

which produces the following concentrations:

• •
  [A] = [B] = 0.400 – Δ = 0.400 – 0.125 = 0.275 M
• •
  [C] = 0.200 + Δ = 0.200 + 0.125 = 0.325 M
• •
  [D] = 0.800 + Δ = 0.800 + 0.125 = 0.925 M

Q = K,

so the new concentrations are indeed equilibrium concentrations.

5.2-3. Changing the Volume

Changing the volume changes the pressures of the gases, and Le Châtelier's Principle states that an equilibrium mixture reacts to counteract the change. At constant temperature, the only way to change the pressure is to change the number of moles of gas. Consequently, equilibrium mixtures react so as to increase the number of moles of gas when the volume of the mixture is increased or decrease the number of moles of gas when the volume is decreased. Changing the volume changes both the pressures and the concentrations of the gases, which changes Q. The change in Q depends upon Δn_g in such a way that the same conclusions are drawn from Q and K considerations as with Le Châtelier's Principle.

5.2-4. Changing the Temperature

Changing the temperature at constant pressure changes K not Q, and Le Châtelier's Principle summarizes the effect that temperature has on the equilibrium constant. Heat can be viewed as a product of an exothermic reaction and as a reactant in an endothermic reaction. Heating an exothermic reaction is the same as adding a product, which shifts the reaction toward the reactant side (←). The shift increases the equilibrium concentrations of the reactants and decreases the equilibrium concentrations of the products. The result is that the value of K is decreased in agreement with the conclusion reached using the equation

ln(K₂) = ln(K₁) +

ΔH°

R

1

T1

−

1

T2

.

5.2-5. Exercises

5.3 Using the Equilibrium Constant

Introduction

The stoichiometry problems of Chapter 1 all assumed that the limiting reactant disappeared completely during reaction. However, this assumes that the equilibrium constant is very large. In this section, we address problems in which the equilibrium constant is not very large, so the limiting reactant does not disappear completely. We will again use a reaction table as introduced in Chapter 1. The reaction table consists of three lines: the initial, change, and final. In equilibrium problems, the unknown can lie on any one of these lines or it can require filling in the table to determine the value of the equilibrium constant. In this section we show how to solve equilibrium problems.

Objectives

• •
  Calculate the concentrations of one substance in an equilibrium mixture from the other concentrations and the equilibrium constant.
• •
  Determine the equilibrium constant for a reaction given the initial amounts and one equilibrium amount.
• •
  Determine the equilibrium composition from the initial composition and the equilibrium constant.
• •
  Calculate the amount of one reactant required to react with a given amount of another reactant to produce a given amount of product.
• •
  Determine the extent of the change caused by the addition of known amounts of reactants or products to an equilibrium mixture, given K and the initial equilibrium concentrations.
• •
  Calculate the amount of one substance that would have to be added to an equilibrium to change the concentration of another substance in the equilibrium by a given amount.

Solving for One Concentration when K and the Other Concentrations are Known

5.3-1. Determining an Unknown Concentration

We begin with the easiest of equilibrium problems: determining an unknown concentration when all other concentrations and the equilibrium constant are known. Solving this type of problem involves setting up the equilibrium constant expression, solving it for the unknown, and substituting the known values into the resulting expression. The next two exercises demonstrate the process.

5.3-2. Exercise

5.3-3. Exercise

Using Reaction Tables in Equilibrium Problems

5.3-4. Filling in the Reaction Table

The amount of a substance present in an equilibrium mixture is determined from the concentrations of all reactants and products, not just a single limiting reactant, and the reaction table is again the best way to track the changes. However, the label of the last line is changed from "final" to "eq" to emphasize that equilibrium is established. The Δ line represents the amount of each substance that must form or react in order to change from initial to equilibrium concentrations. When making entries into it, remember the following:

• •
  It is the only line to which stoichiometry applies. If one entry on the Δ line is known, then all other entries can be determined by stoichiometry.
• •
  If the initial system must shift to the right (→,
  Q < K),
  the concentrations of everything on the right (product) side of the reaction must increase (Δ > 0), while all concentrations on the left side must decrease (Δ < 0).
• •
  If the shift is to the left (←,
  Q > K),
  the concentrations of everything on the left side must increase
  (Δ > 0),
  while the concentrations of everything on the right side must decrease (Δ < 0).

5.3-5. Types of Equilibrium Problems

We will be doing equilibrium problems in most of the remaining chapters as we examine acid-base, solubility, and redox equilibria. However, there are only three different kinds of equilibrium problems:

• 1
  Determine K from the equilibrium concentrations. If K is the unknown, then the reaction table can contain no unknowns.
• 2
  Determine the equilibrium concentrations from K and a set of initial concentrations. The unknown is in the Δ line and therefore also in the equilibrium line. We will use x as our unknown identifier.
• 3
  Determine the amount of substance that must be added (removed) to (from) a known amount of another reactant to produce a desired equilibrium concentration. The unknown is in the initial line.

You should realize that the methods set forward in the remainder of this chapter will be used repeatedly in subsequent chapters. We examine each of these types in the following exercises.

Determining K

5.3-6. Method

In order to determine K, you must be given sufficient information to determine the equilibrium concentrations. This information can be given directly, but many times only the initial concentrations and the final concentration of just one of the substances are given. Setting up a reaction table is the easiest way to establish the other concentrations. In these cases, use the following method:

5.3-7. K_c for a Gas Phase Reaction

5.3-8. Aqueous Reaction

Determining Equilibrium Amounts

5.3-9. Method

If you are asked for an equilibrium concentration or pressure, you must be given the value of K and the initial conditions. The unknown (x) is in the reaction table. To find equilibrium concentrations or pressures given the equilibrium constant and the initial concentrations, use the following method.

5.3-10. Hints to Simplify the Problems

K_c depends upon the equilibrium concentrations, which depend upon the volume of the container and the number of moles of each species present. However, if Δn_g = 0, the volumes in the numerator and denominator cancel, so K_c depends only upon the number of moles. In other words, equilibrium problems on gas phase reactions in which
Δn_g = 0 can be done in moles rather than molarities. However, if Δn_g is not zero, you must convert to molarities in order to use a K_c. Solving an equilibrium expression can be very difficult because many expressions contain high order terms. However, the problems in this course have been constructed to simplify the algebra, so you can focus on the equilibrium. Many of the problems involve squares and can be solved with the quadratic equation, but most of these can be simplified because the equilibrium expression is a perfect square. In these cases, take the square root of both sides of the equation before solving for x.

5.3-11. Equilibrium Moles

5.3-12. Determining Equilibrium Pressures

5.3-13. Re-establishing Equilibrium After an Addition

Determining the Amount to Add or Remove

5.3-14. Method

Another important calculation involves determining how much of one substance to add or remove in order to get a desired amount of one of the substances. In these cases, the unknown lies in the initial line. Sometimes, you add or remove the substance from an equilibrium mixture, but the equilibrium mixture cannot be the initial line because nothing would happen if it were at equilibrium. However, adding or removing a substance from an equilibrium line makes it an initial line. Remember that you will have only one unknown in each problem, so there are no other unknowns if one is already in the initial line.

5.3-15. Percent Yield

5.3-16. Adding Reactants to Shift Equilibrium to Desired Position

Putting it Together

5.3-17. Mixing an Acid and a Base

5.4 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.

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