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# Chapter 15 Chemical Equilibrium

In Chapter 14 "Chemical Kinetics", we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibriumThe point at which the forward and reverse reaction rates become the same so that the net composition of the system no longer changes with time., the point at which the composition of the system no longer changes with time.

A smoggy sunset in Shenzhen, China. The reaction of O2 with N2 at high temperature in an internal combustion engine produces small amounts of NO, which reacts with atmospheric O2 to form NO2, an important component of smog. The reddish-brown color of NO2 is responsible for the characteristic color of smog, as shown in this true-color photo.

We introduced the concept of equilibrium in Chapter 11 "Liquids", where you learned that a liquid and a vapor are in equilibrium when the number of molecules evaporating from the surface of the liquid per unit time is the same as the number of molecules condensing from the vapor phase. Vapor pressure is an example of a physical equilibrium because only the physical form of the substance changes. Similarly, in Chapter 13 "Solutions", we discussed saturated solutions, another example of a physical equilibrium, in which the rate of dissolution of a solute is the same as the rate at which it crystallizes from solution.

In this chapter, we describe the methods chemists use to quantitatively describe the composition of chemical systems at equilibrium, and we discuss how factors such as temperature and pressure influence the equilibrium composition. As you study these concepts, you will also learn how urban smog forms and how reaction conditions can be altered to produce H2 rather than the combustion products CO2 and H2O from the methane in natural gas. You will discover how to control the composition of the gases emitted in automobile exhaust and how synthetic polymers such as the polyacrylonitrile used in sweaters and carpets are produced on an industrial scale.

## 15.1 The Concept of Chemical Equilibrium

### Learning Objective

1. To understand what is meant by chemical equilibrium.

Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (N2O4) to nitrogen dioxide (NO2). You may recall from Chapter 14 "Chemical Kinetics" that NO2 is responsible for the brown color we associate with smog. When a sealed tube containing solid N2O4 (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of NO2 appears (Figure 15.1 "The "). The reaction can be followed visually because the product (NO2) is colored, whereas the reactant (N2O4) is colorless:

Equation 15.1

$N 2 O(g) colorless ⇌ 2NO 2 (g) red-brown$

The double arrow indicates that both the forward and reverse reactions are occurring simultaneously; it is read “is in equilibrium with.”

Figure 15.1 The $N2O(g)⇌2NO2(g)$ System at Different Temperatures

(left) At dry ice temperature (−78.4°C), the system contains essentially pure solid N2O4, which is colorless. (center) As the system is warmed above the melting point of N2O4 (−9.3°C), the N2O4 melts and then evaporates, and some of the vapor dissociates to red-brown NO2. (right) Eventually the sample reaches room temperature, and a mixture of gaseous N2O4 and NO2 is present. The composition of the mixture and hence the color do not change further with time: the system has reached equilibrium at the new temperature.

Figure 15.2 "The Composition of N" shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of NO2 were zero, then it increases as the concentration of N2O4 decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no N2O4 but an initial NO2 concentration twice the initial concentration of N2O4 in part (a) in Figure 15.2 "The Composition of N", in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition, as shown in part (b) in Figure 15.2 "The Composition of N". Thus equilibrium can be approached from either direction in a chemical reaction.

Figure 15.2 The Composition of N2O4/NO2 Mixtures as a Function of Time at Room Temperature

(a) Initially, this idealized system contains 0.0500 M gaseous N2O4 and no gaseous NO2. The concentration of N2O4 decreases with time as the concentration of NO2 increases. (b) Initially, this system contains 0.1000 M NO2 and no N2O4. The concentration of NO2 decreases with time as the concentration of N2O4 increases. In both cases, the final concentrations of the substances are the same: [N2O4] = 0.0422 M and [NO2] = 0.0156 M at equilibrium.

Figure 15.3 "The Forward and Reverse Reaction Rates as a Function of Time for the " shows the forward and reverse reaction rates for a sample that initially contains pure NO2. Because the initial concentration of N2O4 is zero, the forward reaction rate (dissociation of N2O4) is initially zero as well. In contrast, the reverse reaction rate (dimerization of NO2) is initially very high (2.0 × 106 M/s), but it decreases rapidly as the concentration of NO2 decreases. (Recall from Chapter 14 "Chemical Kinetics" that the reaction rate of the dimerization reaction is expected to decrease rapidly because the reaction is second order in NO2: rate = kr[NO2]2, where kr is the rate constant for the reverse reaction shown in Equation 15.1.) As the concentration of N2O4 increases, the rate of dissociation of N2O4 increases—but more slowly than the dimerization of NO2—because the reaction is only first order in N2O4 (rate = kf[N2O4], where kf is the rate constant for the forward reaction in Equation 15.1). Eventually, the forward and reverse reaction rates become identical, kF = kr, and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.

Figure 15.3 The Forward and Reverse Reaction Rates as a Function of Time for the $N2O4(g)⇌2NO2(g)$ System Shown in Part (b) in Figure 15.2 "The Composition of N"

The rate of dimerization of NO2 (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of N2O4 is zero, the rate of the dissociation reaction (forward reaction) at t = 0 is also zero. As the dimerization reaction proceeds, the N2O4 concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of N2O4 and NO2 no longer change.

### Note the Pattern

At equilibrium, the forward reaction rate is equal to the reverse reaction rate.

### Example 1

The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation $2A⇌B,$ where the blue circles are A and the purple ovals are B. Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium?

Given: three reaction systems

Asked for: relative time to reach chemical equilibrium

Strategy:

Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium.

Solution:

In systems 1 and 3, the concentration of A decreases from t0 through t2 but is the same at both t2 and t3. Thus systems 1 and 3 are at equilibrium by t3. In system 2, the concentrations of A and B are still changing between t2 and t3, so system 2 may not yet have reached equilibrium by t3. Thus system 2 took the longest to reach chemical equilibrium.

Exercise

In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium?

### Summary

Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached.

### Key Takeaway

• At equilibrium, the forward and reverse reactions of a system proceed at equal rates.

### Conceptual Problems

1. What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the concentrations or amounts of the reactants and the products?

2. Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example.

3. Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of NaCl in water. What is occurring on a microscopic level? What is happening on a macroscopic level?

4. Which of these systems exists in a state of chemical equilibrium?

1. oxygen and hemoglobin in the human circulatory system
2. iodine crystals in an open beaker
3. the combustion of wood
4. the amount of 14C in a decomposing organism

1. Both forward and reverse reactions occur but at the same rate. Na+ and Cl ions continuously leave the surface of an NaCl crystal to enter solution, while at the same time Na+ and Cl ions in solution precipitate on the surface of the crystal.

## 15.2 The Equilibrium Constant

### Learning Objectives

1. To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
2. To write an equilibrium constant expression for any reaction.

Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the system described in Equation 15.1, the decomposition of N2O4 to NO2. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:

Equation 15.2

forward rate = kf[N2O4]

Equation 15.3

reverse rate = kr[NO2]2

At equilibrium, the forward rate equals the reverse rate:

Equation 15.4

kf[N2O4] = kr[NO2]2

so

Equation 15.5

$k f k r = [ NO 2 ] 2 [ N 2 O 4 ]$

The ratio of the rate constants gives us a new constant, the equilibrium constant (K)The ratio of the rate constants for the forward reaction and the reverse reaction; that is, $K=kf/kr.$ It is also the equilibrium constant calculated from solution concentrations: $K=[C]c[D]d/[A]a[B]b$ for the general reaction $aA+bB⇌cC+dD,$ in which each component is in solution., which is defined as follows:

Equation 15.6

$K = k f k r$

Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions.

### Note the Pattern

The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.

Table 15.1 "Initial and Equilibrium Concentrations for " lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation 15.1. At equilibrium the magnitude of the quantity [NO2]2/[N2O4] is essentially the same for all five experiments. In fact, no matter what the initial concentrations of NO2 and N2O4 are, at equilibrium the quantity [NO2]2/[N2O4] will always be 6.53 ± 0.03 × 10−3 at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations.

Table 15.1 Initial and Equilibrium Concentrations for $NO2/N204$ Mixtures at 25°C

Initial Concentrations Concentrations at Equilibrium
Experiment [N2O4] (M) [NO2] (M) [N2O4] (M) [NO2] (M) K = [NO2]2/[N2O4]
1 0.0500 0.0000 0.0417 0.0165 6.54 × 10−3
2 0.0000 0.1000 0.0417 0.0165 6.54 × 10−3
3 0.0750 0.0000 0.0647 0.0206 6.56 × 10−3
4 0.0000 0.0750 0.0304 0.0141 6.54 × 10−3
5 0.0250 0.0750 0.0532 0.0186 6.50 × 10−3

## Developing an Equilibrium Constant Expression

In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form

Equation 15.7

$a A + b B ⇌ c C + d D$

where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass actionFor the general balanced chemical equation $aA+bB⇌cC+dD,$ the equilibrium constant expression is $K=[C]c[D]d/[A]a[B]b.$ and can be stated as follows:

Equation 15.8

$K = [C] c [D] d [A] a [B] b$

where K is the equilibrium constant for the reaction. Equation 15.7 is called the equilibrium equationFor the general balanced chemical equation $aA+bB⇌cC+dD,$ the equilibrium constant expression is $K=[C]c[D]d/[A]a[B]b.$, and the right side of Equation 15.8 is called the equilibrium constant expressionFor a balanced chemical equation, the ratio is $[C]c[D]d/[A]a[B]b$ for the general reaction $aA+bB⇌cC+dD.$. The relationship shown in Equation 15.8 is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism.

The equilibrium constant can vary over a wi