7.8 Equilibrium and the Equilibrium Constant
Introduction
Imagine you are stranded in a rowboat in the middle of the ocean. Suddenly, your boat springs a small leak, and you need to bail out water. You grab a bucket and begin to bail. After a few minutes, your efforts against the leak keep the water to only about half an inch, but any further bailing doesn’t change the water level; the leak brings in as much water as you bail out. You are at equilibrium. Two opposing processes have reached the same speed, and there is no more overall change in the process. Chemical reactions are like that as well. Most of them come to an equilibrium. The actual position of the equilibrium—whether it favors the reactants or the products—is characteristic of a chemical reaction; it is difficult to see just by looking at the balanced chemical equation. But chemistry has tools to help you understand the equilibrium of chemical reactions—the focus of our study in this chapter. So far in this text, when we present a chemical reaction, we have implicitly assumed that the reaction goes to completion. Indeed, our stoichiometric calculations were based on this; when we asked how much of a product is produced when so much of a reactant reacts, we are assuming that all of a reactant reacts. However, this is usually not the case; many reactions do not go to completion, and many chemists have to deal with that. In this chapter, we will study this phenomenon and see ways in which we can affect the extent of chemical reactions.
Learning Objectives
 Define chemical equilibrium.
 Recognize chemical equilibrium as a dynamic process.
Consider the following reaction occurring in a closed container (so that no material can go in or out):
H_{2} + I_{2} → 2HI
This is simply the reaction between elemental hydrogen and elemental iodine to make hydrogen iodide. The way the equation is written, we are led to believe that the reaction goes to completion, that all the H_{2} and the I_{2} react to make HI.
However, this is not the case. The reverse chemical reaction is also taking place:
2HI → H_{2} + I_{2}
It acts to undo what the first reaction does. Eventually, the reverse reaction proceeds so quickly that it matches the speed of the forward reaction. When that happens, any continued overall reaction stops: the reaction has reached chemical equilibrium (sometimes just spoken as equilibrium; plural equilibria), the point at which the forward and reverse processes balance each other’s progress.
Because two opposing processes are occurring at once, it is conventional to represent an equilibrium using a double arrow, like this:
H_{2 }+ I_{2 }⇄ 2HI
The double arrow implies that the reaction is going in both directions. Note that the reaction must still be balanced.
Example 7.12
Write the equilibrium equation that exists between calcium carbonate as a reactant and calcium oxide and carbon dioxide as products.
Solution
As this is an equilibrium situation, a double arrow is used. The equilibrium equation is written as follows:
CaCO_{3 }⇄ CaO + CO_{2}
Test Yourself
Write the equilibrium equation between elemental hydrogen and elemental oxygen as reactants and water as the product.
Answer
2H_{2 }+ O_{2} ⇄ 2H_{2}O
One thing to note about equilibrium is that the reactions do not stop; both the forward reaction and the reverse reaction continue to occur. They both occur at the same rate, so any overall change by one reaction is cancelled by the reverse reaction. We say that chemical equilibrium is dynamic, rather than static. Also, because both reactions are occurring simultaneously, the equilibrium can be written backward. For example, representing an equilibrium as
H_{2}+ I_{2 }⇄ 2HI
is the same thing as representing the same equilibrium as
2HI ⇄ H_{2 }+ I_{2}
The reaction must be at equilibrium for this to be the case, however.
Key Takeaways
 Chemical reactions eventually reach equilibrium, a point at which forward and reverse reactions balance each other’s progress.
 Chemical equilibria are dynamic: the chemical reactions are always occurring; they just cancel each other’s progress.
Exercises
Questions
 Define chemical equilibrium. Give an example.
 Explain what is meant when it is said that chemical equilibrium is dynamic.
 Write the equilibrium equation between elemental hydrogen and elemental chlorine as reactants and hydrochloric acid as the product.
 Write the equilibrium equation between iron(III) sulfate as the reactant and iron(III) oxide and sulfur trioxide as the products.
 Graphite and diamond are two forms of elemental carbon. Write the equilibrium equation between these two forms in two different ways.
 At 1,500 K, iodine molecules break apart into iodine atoms. Write the equilibrium equation between these two species in two different ways.
Answers
 The situation when the forward and reverse chemical reactions occur, leading to no additional net change in the reaction position; H_{2}+ I_{2} ⇄ 2HI (answers will vary)
 H_{2} + Cl_{2} ⇄ 2HCl
 C (gra) ⇄ C (dia); C (dia) ⇄ C (gra)
The Equilibrium Constant
 Explain the importance of the equilibrium constant.
 Construct an equilibrium constant expression for a chemical reaction.
In the mid 1860s, Norwegian scientists C. M. Guldberg and P. Waage noted a peculiar relationship between the amounts of reactants and products in an equilibrium. No matter how many reactants they started with, a certain ratio of reactants and products was achieved at equilibrium. Today, we call this observation the law of mass action. It relates the amounts of reactants and products at equilibrium for a chemical reaction. For a general chemical reaction occurring in solution,
aA + bB ⇄ cC + dD
the equilibrium constant, also known as K_{eq}, is defined by the following expression:
[latex]K_{\text{eq}}=\dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}^a][\text{B}]^b}[/latex]
where [A] is the molar concentration of species A at equilibrium, and so forth. The coefficients a, b, c, and d in the chemical equation become exponents in the expression for K_{eq}. The K_{eq} is a characteristic numerical value for a given reaction at a given temperature; that is, each chemical reaction has its own characteristic K_{eq}. The concentration of each reactant and product in a chemical reaction at equilibrium is related; the concentrations cannot be random values, but they depend on each other. The numerator of the expression for K_{eq} has the concentrations of every product (however many products there are), while the denominator of the expression for K_{eq} has the concentrations of every reactant, leading to the common products over reactants definition for the K_{eq}.
Let us consider a simple example. Suppose we have this equilibrium:
A ⇄ B
There is one reactant, one product, and the coefficients on each are just 1 (assumed, not written). The K_{eq} expression for this equilibrium is
[latex]K_{\text{eq}}=\dfrac{[\text{B}]}{[\text{A}]}[/latex]
(Exponents of 1 on each concentration are understood.)
Suppose the numerical value of K_{eq} for this chemical reaction is 2.0. If [B] = 4.0 M, then [A] must equal 2.0 M so that the value of the fraction equals 2.0:
[latex]K_{\text{eq}}=\dfrac{[\text{B}]}{[\text{A}]}=\dfrac{4.0}{2.0}=2.0[/latex]
By convention, the units are understood to be M and are omitted from the K_{eq} expression. Suppose [B] were 6.0 M. For the K_{eq} value to remain constant (it is, after all, called the equilibrium constant), then [A] would have to be 3.0 M at equilibrium:
[latex]K_{\text{eq}}=\dfrac{[\text{B}]}{[\text{A}]}=\dfrac{6.0}{3.0}=2.0[/latex]
If [A] were not equal to 3.0 M, the reaction would not be at equilibrium, and a net reaction would occur until that ratio was indeed 2.0. At that point, the reaction is at equilibrium, and any net change would cease. (Recall, however, that the forward and reverse reactions do not stop because chemical equilibrium is dynamic.)
The issue is the same with more complex expressions for the K_{eq}; only the mathematics becomes more complex. Generally speaking, given a value for the K_{eq} and all but one concentration at equilibrium, the missing concentration can be calculated.
Example 7.13
Given the following reaction:
H_{2} + I_{2 }⇄ 2HI
If the equilibrium [HI] is 0.75 M and the equilibrium [H_{2}] is 0.20 M, what is the equilibrium [I_{2}] if the K_{eq} is 0.40?
Solution
We start by writing the K_{eq} expression. Using the products over reactants approach, the K_{eq} expression is as follows:
[latex]K_{\text{eq}}=\dfrac{[\text{HI}]^2}{[\ce{H2}][\ce{I2}]}[/latex]
Note that [HI] is squared because of the coefficient 2 in the balanced chemical equation. Substituting for the equilibrium [H_{2}] and [HI] and for the given value of K_{eq}:
[latex]0.40=\dfrac{(0.75)^2}{(0.20)[\ce{I2}]}[/latex]
To solve for [I_{2}], we have to do some algebraic rearrangement: divide the 0.40 into both sides of the equation and multiply both sides of the equation by [I_{2}]. This brings [I_{2}] into the numerator of the left side and the 0.40 into the denominator of the right side:
[latex][\ce{I2}]=\dfrac{(0.75)^2}{(0.20)(0.40)}[/latex]
Solving,
[latex][\ce{I2}]=7.0\text{ M}[/latex]
The concentration unit is assumed to be molarity. This value for [I_{2}] can be easily verified by substituting 0.75, 0.20, and 7.0 into the expression for K_{eq} and evaluating: you should get 0.40, the numerical value of K_{eq} (and you do).
Test Yourself
Given the following reaction:
H_{2} + I_{2 }⇄ 2HI
If the equilibrium [HI] is 0.060 M and the equilibrium [I_{2}] is 0.90 M, what is the equilibrium [H_{2}] if the K_{eq} is 0.40?
Answer
0.010 M
In some types of equilibrium problems, square roots, cube roots, or even higher roots need to be analyzed to determine a final answer. Make sure you know how to perform such operations on your calculator; if you do not know, ask your instructor for assistance.
Example 7.14
The following reaction is at equilibrium:
N_{2} + 3H_{2 }⇄ 2NH_{3}
The K_{eq} at a particular temperature is 13.7. If the equilibrium [N_{2}] is 1.88 M and the equilibrium [NH_{3}] is 6.62 M, what is the equilibrium [H_{2}]?
Solution
We start by writing the K_{eq} expression from the balanced chemical equation:
[latex]K_{\text{eq}}=\dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}[/latex]
Substituting for the known equilibrium concentrations and the K_{eq}, this becomes
[latex]13.7=\dfrac{(6.62)^2}{(1.88)[\ce{H2}]^3}[/latex]
Rearranging algebraically and then evaluating the numerical expression, we get
[latex][\ce{H2}]^3=\dfrac{(6.22)^2}{(1.88)(13.7)}=1.502112129[/latex]
To solve for [H_{2}], we need to take the cube root of the equation. Performing this operation, we get
[latex][\ce{H2}]=1.15\text{ M}[/latex]
You should verify that this is correct using your own calculator to confirm that you know how to do a cube root correctly.
Test Yourself
The following reaction is at equilibrium:
N_{2 }+ 3H_{2 }⇄ 2NH_{3}
The K_{eq} at a particular temperature is 13.7. If the equilibrium [N_{2}] is 0.055 M and the equilibrium [H_{2}] is 1.62 M, what is the equilibrium [NH_{3}]?
Answer
1.79 M
The K_{eq} was defined earlier in terms of concentrations. For gasphase reactions, the K_{eq} can also be defined in terms of the partial pressures of the reactants and products, P_{i}. For the gasphase reaction
aA(g) + bB(g) ⇄ cC(g) + dD(g)
the pressurebased equilibrium constant, K_{P}, is defined as follows:
[latex]K_{\text{P}}=\dfrac{P^c_{\text{C}}P^d_{\text{D}}}{P^a_{\text{A}}P^b_{\text{B}}}[/latex]
where P_{A} is the partial pressure of substance A at equilibrium in atmospheres, and so forth. As with the concentrationbased equilibrium constant, the units are omitted when substituting into the expression for K_{P}.
Example 7.15
What is the K_{P} for this reaction, given the equilibrium partial pressures of 0.664 atm for NO_{2} and 1.09 for N_{2}O_{4}?
2NO_{2}(g) ⇄ N_{2}O_{4}(g)
Solution
Write the K_{P} expression for this reaction:
[latex]K_{\text{P}}=\dfrac{P_{\ce{N2O4}}}{P^2_{\ce{NO2}}}[/latex]
Then substitute the equilibrium partial pressures into the expression and evaluate:
[latex]K_{\text{P}}=\dfrac{(1.09)}{(0.664)^2}=2.47[/latex]
Test Yourself
What is the K_{P} for this reaction, given the equilibrium partial pressures of 0.44 atm for H_{2}, 0.22 atm for Cl_{2}, and 2.98 atm for HCl?
H_{2} + Cl_{2} ⇄ 2HCl
Answer
91.7
There is a simple relationship between K_{eq} (based on concentration units) and K_{P} (based on pressure units):
[latex]K_{\text{P}}=K_{\text{eq}}\cdot (RT)^{\Delta n}[/latex]
where R is the ideal gas law constant (in units of L·atm/mol·K), T is the absolute temperature, and Δn is the change in the number of moles of gas in the balanced chemical equation, defined as n_{gas,prods} − n_{gas,rcts}.
Note that this equation implies that if the number of moles of gas are the same in reactants and products, K_{eq} = K_{P}.
Example 7.16
What is the K_{P} at 25°C for this reaction if the K_{eq} is 4.2 × 10^{−2}?
N_{2}(g) + 3H_{2}(g) ⇄ 2NH_{3}(g)
Solution
Before we use the relevant equation, we need to do two things: convert the temperature to kelvins and determine Δn. Converting the temperature is easy:
[latex]T=25+273=298\text{ K}[/latex]
To determine the change in the number of moles of gas, take the number of moles of gaseous products and subtract the number of moles of gaseous reactants. There are 2 mol of gas as product and 4 mol of gas of reactant:
[latex]\Delta n=24=2\text{ mol}[/latex]
Note that Δn is negative. Now we can substitute into our equation, using R = 0.08205 L·atm/mol·K. The units are omitted for clarity:
[latex]K_{\text{P}}=(4.2\times 10^{2})(0.08205)(298)^{2}[/latex]
Solving,
[latex]K_{\text{P}}=7.0\times 10^{5}[/latex]
Test Yourself
What is the K_{P} at 25°C for this reaction if the K_{eq} is 98.3?
I_{2}(g) ⇄ 2I(g)
Answer
2.40 × 10^{3}
Finally, we recognize that many chemical reactions involve substances in the solid or liquid phases. For example, a particular chemical reaction is represented as follows:
2NaHCO_{3}(s) ⇄ Na_{2}CO_{3}(s) + CO_{2}(g) + H_{2}O(ℓ)
This chemical equation includes all three phases of matter. This kind of equilibrium is called a heterogeneous equilibrium because there is more than one phase present.
The rule for heterogeneous equilibria is as follows: Do not include the concentrations of pure solids and pure liquids in K_{eq} expressions. Only partial pressures for gasphase substances or concentrations in solutions are included in the expressions of equilibrium constants. As such, the equilibrium constant expression for this reaction would simply be
[latex]K_{\text{P}}=P_{\ce{CO2}}[/latex]
because the two solids and one liquid would not appear in the expression.
Key Takeaways
 Every chemical equilibrium can be characterized by an equilibrium constant, known as K_{eq}.
 The K_{eq} and K_{P} expressions are formulated as amounts of products divided by amounts of reactants; each amount (either a concentration or a pressure) is raised to the power of its coefficient in the balanced chemical equation.
 Solids and liquids do not appear in the expression for the equilibrium constant.
Exercises
Questions
 Define the law of mass action.
 What is an equilibrium constant for a chemical reaction? How is it constructed?
 Write the K_{eq} expression for each reaction.
 H_{2} + Cl_{2} ⇄ 2HCl
 NO + NO_{2} ⇄ N_{2}O_{3}
 Write the K_{eq} expression for each reaction.
 C_{2}H_{5}OH + NaI ⇄ C_{2}H_{5}I + NaOH
 PCl_{3} + Cl_{2} ⇄ PCl_{5}
 Write the K_{P} expression for each reaction.
 2H_{2}(g) + O_{2}(g) ⇄ 2H_{2}O(g)
 2H_{2}O_{2}(g) ⇄ 2H_{2}O(g) + O_{2}(g)
 Write the K_{P} expression for each reaction.
 CH_{4}(g) + 2O_{2}(g) ⇄ CO_{2}(g) + 2H_{2}O(g)
 CH_{4}(g) + 4Cl_{2}(g) ⇄ CCl_{4}(g) + 4HCl(g)
 The following reaction is at equilibrium:
PBr_{3} + Br_{2} ⇄ PBr_{5}
The equilibrium [Br_{2}] and [PBr_{5}] are 2.05 M and 0.55 M, respectively. If the K_{eq} is 1.65, what is the equilibrium [PBr_{3}]?
 The following reaction is at equilibrium:
CO + Cl_{2} ⇄ CoCl_{2}
The equilibrium [CO] and [Cl_{2}] are 0.088 M and 0.103 M, respectively. If the K_{eq} is 0.225, what is the equilibrium [COCl_{2}]?
 The following reaction is at equilibrium:
CH_{4} + 2Cl_{2} ⇄ CH_{2}Cl_{2} + 2HCl
If [CH_{4}] is 0.250 M, [Cl_{2}] is 0.150 M, and [CH_{2}Cl_{2}] is 0.175 M at equilibrium, what is [HCl] at equilibrium if the K_{eq} is 2.30?
 The following reaction is at equilibrium:
4HBr + O_{2} ⇄ 2H_{2}O + 2Br_{2}
If [HBr] is 0.100 M, [O_{2}] is 0.250 M, and [H_{2}O] is 0.0500 M at equilibrium, what is [Br_{2}] at equilibrium if the K_{eq} is 0.770?
 Write the K_{P} expression for the following gasphase reaction:
4NO_{2}(g) + O_{2}(g) ⇄ 2N_{2}O_{5}(g)
 Write the K_{P} expression for the following gasphase reaction:
ClO(g) + O_{3}(g) ⇄ ClO_{2}(g) + O_{2}(g)
 What is the equilibrium partial pressure of COBr_{2} if the equilibrium partial pressures of CO and Br_{2} are 0.666 atm and 0.235 atm and the K_{P} for this equilibrium is 4.08?
CO(g) + Br_{2}(g) ⇄ COBr_{2}(g)
 What is the equilibrium partial pressure of O_{3} if the equilibrium partial pressure of O_{2} is 0.0044 atm and K_{P} for this equilibrium is 0.00755?
3O_{2}(g) ⇄ 2O_{3}(g)
 Calculate the K_{P} for this reaction at 298 K if the K_{eq} = 1.76 × 10^{−3}.
3O_{2}(g) ⇄ 2O_{3}(g)
 Calculate the K_{P} for this reaction at 310 K if the K_{eq} = 6.22 × 10^{3}.
4NO_{2}(g) + O_{2}(g) ⇄ 2N_{2}O_{5}(g)
 Calculate the K_{eq} for this reaction if the K_{P} = 5.205 × 10^{−3} at 660°C.
CO(g) + F_{2}(g) ⇄ COF_{2}(g)
 Calculate the K_{eq} for this reaction if the K_{P} = 78.3 at 100°C.
4HCl(g) + O_{2}(g) ⇄ 2H_{2}O(g) + 2Cl_{2}(g)
 Write the correct K_{eq} expression for this reaction.
NaOH(aq) + HCl(aq) ⇄ NaCl(aq) + H_{2}O(ℓ)
 Write the correct K_{eq} expression for this reaction.
AgNO_{3}(aq) + NaCl(aq) ⇄ AgCl(s) + NaNO_{3}(aq)
 Write the correct K_{P} expression for this reaction.
CaCO_{3}(s) ⇄ CaO(s) + CO_{2}(g)
 Write the correct K_{P} expression for this reaction.
C_{2}H_{2}(g) + 2I_{2}(s) ⇄ C_{2}H_{2}I_{4}(g)
Answers
 The relationship between the concentrations of reactants and products of a chemical reaction at equilibrium

 [latex]K_{\text{eq}}=\dfrac{[\ce{HCl}]^2}{[\ce{H2}][\ce{Cl2}]}[/latex]
 [latex]K_{\text{eq}}=\dfrac{[\ce{N2O3}]}{[\ce{NO}][\ce{NO2}]}[/latex]

 [latex]K_{\text{P}}=\dfrac{P^2_{\ce{H2O}}}{P^2_{\ce{H2}}P_{\ce{O2}}}[/latex]
 [latex]K_{\text{P}}=\dfrac{P^2_{\ce{H2O}}P_{\ce{O2}}}{P^2_{\ce{H2O2}}}[/latex]
 0.163 M
 0.272 M
 [latex]K_{\text{P}}=\dfrac{P^2_{\ce{N2O5}}}{P^4_{\ce{NO2}}P_{\ce{O2}}}[/latex]
 0.639 atm
 7.20 × 10^{−5}
 K_{eq} = 3.98 × 10^{−1}
 [latex]K_{\text{eq}}=\dfrac{[\ce{NaCl}]}{[\ce{NaOH}][\ce{HCl}]}[/latex]
 [latex]K_{\text{P}} = P_{\ce{CO2}}[/latex]