Chapter 13 Chemical Equilibrium

13.1 Chemical Equilibrium

Consider the following reaction occurring in a closed container (so that no material can go in or out):

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 H₂ and the I₂ react to make HI.

However, this is not the case. The reverse chemical reaction is also taking place:

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 equilibriumThe point at which forward and reverse chemical reactions balance each other’s progress. (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:

The double arrow implies that the reaction is going in both directions. Note that the reaction must still be balanced.

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

is the same thing as representing the same equilibrium as

The reaction must be at equilibrium for this to be the case, however.

13.2 The Equilibrium Constant

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 actionThe relationship of the amounts of reactants and products at equilibrium.. It relates the amounts of reactants and products at equilibrium for a chemical reaction. For a general chemical reaction occurring in solution,

the equilibrium constantA numerical value that relates to the ratio of products and reactants at equilibrium., also known as K_eq, is defined by the following expression:

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:

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

(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:

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:

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.

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.

The K_eq was defined earlier in terms of concentrations. For gas-phase reactions, the K_eq can also be defined in terms of the partial pressures of the reactants and products, P_i. For the gas-phase reaction

the pressure-based equilibrium constant, K_P, is defined as follows:

where P_A is the partial pressure of substance A at equilibrium in atmospheres, and so forth. As with the concentration-based equilibrium constant, the units are omitted when substituting into the expression for K_P.

There is a simple relationship between K_eq (based on concentration units) and K_P (based on pressure units):

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.

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:

This chemical equation includes all three phases of matter. This kind of equilibrium is called a heterogeneous equilibriumAn equilibrium in which more than one phase of reactants or products is present. 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 gas-phase 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

because the two solids and one liquid would not appear in the expression.

13.3 Shifting Equilibria: Le Chatelier’s Principle

Once equilibrium is established, the reaction is over, right? Not exactly. An experimenter has some ability to affect the equilibrium.

Chemical equilibria can be shifted by changing the conditions that the system experiences. We say that we “stress” the equilibrium. When we stress the equilibrium, the chemical reaction is no longer at equilibrium, and the reaction starts to move back toward equilibrium in such a way as to decrease the stress. The formal statement is called Le Chatelier’s principleIf an equilibrium is stressed, then the reaction shifts to reduce the stress.: If an equilibrium is stressed, then the reaction shifts to reduce the stress.

There are several ways to stress an equilibrium. One way is to add or remove a product or a reactant in a chemical reaction at equilibrium. When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.

It is worth noting that when reactants or products are added or removed, the value of the K_eq does not change. The chemical reaction simply shifts, in a predictable fashion, to reestablish concentrations so that the K_eq expression reverts to the correct value.

How does an equilibrium react to a change in pressure? Pressure changes do not markedly affect the solid or liquid phases. However, pressure strongly impacts the gas phase. Le Chatelier’s principle implies that a pressure increase shifts an equilibrium to the side of the reaction with the fewer number of moles of gas, while a pressure decrease shifts an equilibrium to the side of the reaction with the greater number of moles of gas. If the number of moles of gas is the same on both sides of the reaction, pressure has no effect.

What is the effect of temperature changes on an equilibrium? It depends on whether the reaction is endothermic or exothermic. Recall that endothermic means that energy is absorbed by a chemical reaction, while exothermic means that energy is given off by the reaction. As such, energy can be thought of as a reactant or a product, respectively, of a reaction:

Because temperature is a measure of the energy of the system, increasing temperature can be thought of as adding energy. The reaction will react as if a reactant or a product is being added and will act accordingly by shifting to the other side. For example, if the temperature is increased for an endothermic reaction, essentially a reactant is being added, so the equilibrium shifts toward products. Decreasing the temperature is equivalent to decreasing a reactant (for endothermic reactions) or a product (for exothermic reactions), and the equilibrium shifts accordingly.

In the case of temperature, the value of the equilibrium has changed because the K_eq is dependent on temperature. That is why equilibria shift with changes in temperature.

A catalystA substance that increases the speed of a reaction. is a substance that increases the speed of a reaction. Overall, a catalyst is not a reactant and is not used up, but it still affects how fast a reaction proceeds. However, a catalyst does not affect the extent or position of a reaction at equilibrium. It helps a reaction achieve equilibrium faster.

13.4 Calculating Equilibrium Constant Values

There are some circumstances in which, given some initial amounts and the K_eq, you will have to determine the concentrations of all species when equilibrium is achieved. Such calculations are not difficult to do, especially if a consistent approach is applied. We will consider such an approach here.

Suppose we have this simple equilibrium. Its associated K_eq is 4.0, and the initial concentration of each reactant is 1.0 M:

Because we have concentrations for the reactants but not the products, we presume that the reaction will proceed in the forward direction to make products. But by how much will it proceed? We don’t know, so let us assign it a variable. Let us assume that x M H₂ reacts as the reaction goes to equilibrium. This means that at equilibrium, we have (1.0 − x) M H₂ left over.

According to the balanced chemical equation, H₂ and Cl₂ react in a 1:1 ratio. How do we know that? The coefficients of these two species in the balanced chemical equation are 1 (unwritten, of course). This means that if x M H₂ reacts, x M Cl₂ reacts as well. If we start with 1.0 M Cl₂ at the beginning and we react x M, we have (1.0 − x) M Cl₂ left at equilibrium.

How much HCl is made? We start with zero, but we also see that 2 mol of HCl are made for every mole of H₂ (or Cl₂) that reacts (from the coefficients in the balanced chemical equation), so if we lose x M H₂, we gain 2x M HCl. So now we know the equilibrium concentrations of our species:

We can substitute these concentrations into the K_eq expression for this reaction and combine it with the known value of K_eq:

This is an equation in one variable, so we should be able to solve for the unknown value. This expression may look formidable, but first we can simplify the denominator and write it as a perfect square as well:

The fraction is a perfect square, as is the 4.0 on the right. So we can take the square root of both sides:

Now we rearrange and solve (be sure you can follow each step):

Now we have to remind ourselves what x is—the amount of H₂ and Cl₂ that reacted—and 2x is the equilibrium [HCl]. To determine the equilibrium concentrations, we need to go back and evaluate the expressions 1 − x and 2x to get the equilibrium concentrations of our species:

The units are assumed to be molarity. To check, we simply substitute these concentrations and verify that we get the numerical value of the K_eq, in this case 4.0:

We formalize this process by introducing the ICE chart, where ICE stands for initial, change, and equilibrium. The initial values go in the first row of the chart. The change values, usually algebraic expressions because we do not yet know their exact numerical values, go in the next row. However, the change values must be in the proper stoichiometric ratio as indicated by the balanced chemical equation. Finally, the equilibrium expressions in the last row are a combination of the initial value and the change value for each species. The expressions in the equilibrium row are substituted into the K_eq expression, which yields an algebraic equation that we try to solve.

The ICE chart for the above example would look like this:

Substituting the last row into the expression for the K_eq yields

which, of course, is the same expression we have already solved and yields the same answers for the equilibrium concentrations. The ICE chart is a more formalized way to do these types of problems. The + sign is included explicitly in the change row of the ICE chart to avoid any confusion.

Sometimes when an ICE chart is set up and the K_eq expression is constructed, a more complex algebraic equation will result. One of the more common equations has an x² term in it and is called a quadratic equation. There will be two values possible for the unknown x, and for a quadratic equation with the general formula ax² + bx + c = 0 (where a, b, and c are the coefficients of the quadratic equation), the two possible values are as follows:

One value of x is the + sign used in the numerator, and the other value of x is the − sign used in the numerator. In this case, one value of x typically makes no sense as an answer and can be discarded as physically impossible, leaving only one possible value and the resulting set of concentrations. Example 9 illustrates this.

13.5 Some Special Types of Equilibria

In one sense, all chemical equilibria are treated the same. However, there are several classes of reactions that are noteworthy because of either the identities of the reactants and products or the form of the K_eq expression.

Insoluble Compounds

In Chapter 4 "Chemical Reactions and Equations", Section 4.2 "Types of Chemical Reactions: Single- and Double-Displacement Reactions", on chemical reactions, the concept of soluble and insoluble compounds was introduced. Solubility rules were presented that allow a person to predict whether certain simple ionic compounds will or will not dissolve.

Describing a substance as soluble or insoluble is a bit misleading because virtually all substances are soluble; they are just soluble to different extents. In particular for ionic compounds, what we typically describe as an insoluble compound can actually be ever so slightly soluble; an equilibrium is quickly established between the solid compound and the ions that do form in solution. Thus the hypothetical compound MX does in fact dissolve but only very slightly. That means we can write an equilibrium for it:

MX(s)⇄M+(aq) + X–(aq)

The equilibrium constant for a compound normally considered insoluble is called a solubility product constantThe equilibrium constant for a compound normally considered insoluble. and is labeled K_sp (with the subscript sp, meaning “solubility product”). Because the reactant is a solid, its concentration does not appear in the K_sp expression, so like K_w, expressions for K_sp do not have denominators. For example, the chemical equation and the expression for the K_sp for AgCl, normally considered insoluble, are as follows:

AgCl(s)⇄Ag+(aq) + Cl–(aq) Ksp=[Ag+][Cl–]

Table 13.2 "Solubility Product Constants for Slightly Soluble Ionic Compounds" lists some values of the K_sp for slightly soluble ionic compounds.

Table 13.2 Solubility Product Constants for Slightly Soluble Ionic Compounds

Compound	Ksp
BaSO4	1.1 × 10−10
Ca(OH)2	5.0 × 10−6
Ca3(PO4)2	2.1 × 10−33
Mg(OH)2	5.6 × 10−12
HgI2	2.9 × 10−29
AgCl	1.8 × 10−10
AgI	8.5 × 10−17
Ag2SO4	1.5 × 10−5

Example 13

Write the K_sp expression for Ca₃(PO₄)₂.

Solution

Recall that when an ionic compound dissolves, it separates into its individual ions. For Ca₃(PO₄)₂, the ionization reaction is as follows:

Ca3(PO4)2(s)⇄3Ca2+(aq) + 2PO43–(aq)

Hence the K_sp expression is

K_sp = [Ca²⁺]³[PO₄³⁻]²

Test Yourself

Write the K_sp expression Ag₂SO₄.

Answer

K_sp = [Ag⁺]²[SO₄²⁻]

Equilibrium problems involving the K_sp can also be done, and they are usually more straightforward than other equilibrium problems because there is no denominator in the K_sp expression. Care must be taken, however, in completing the ICE chart and evaluating exponential expressions.

Example 14

What are [Ag⁺] and [Cl⁻] in a saturated solution of AgCl? The K_sp of AgCl is 1.8 × 10⁻¹⁰.

Solution

The chemical equation for the dissolving of AgCl is

AgCl(s)⇄Ag+(aq) + Cl–(aq)

The K_sp expression is as follows:

K_sp = [Ag⁺][Cl⁻]

So the ICE chart for the equilibrium is as follows:

	AgCl(s)	⇄	Ag+(aq)	+	Cl−(aq)
I			0		0
C	−x		+x		+x
E			+x		+x

Notice that we have little in the column under AgCl except the stoichiometry of the change; we do not need to know its initial or equilibrium concentrations because its concentration does not appear in the K_sp expression. Substituting the equilibrium values into the expression:

(x)(x) = 1.8 × 10⁻¹⁰

Solving,

x² = 1.8 × 10⁻¹⁰ x = 1.3 × 10⁻⁵

Thus [Ag⁺] and [Cl⁻] are both 1.3 × 10⁻⁵ M.

Test Yourself

What are [Ba²⁺] and [SO₄²⁻] in a saturated solution of BaSO₄? The K_sp of BaSO₄ is 1.1 × 10⁻¹⁰.

Answer

1.0 × 10⁻⁵ M

Example 15

What are [Ca²⁺] and [PO₄³⁻] in a saturated solution of Ca₃(PO₄)₂? The K_sp of Ca₃(PO₄)₂ is 2.1 × 10⁻³³.

Solution

This is similar to Example 14, but the ICE chart is much different because of the number of ions formed.

	Ca3(PO4)2(s)	⇄	3Ca2+(aq)	+	2PO43−(aq)
I			0		0
C	−x		+3x		+2x
E			+3x		+2x

For every unit of Ca₃(PO₄)₂ that dissolves, three Ca²⁺ ions and two PO₄³⁻ ions are formed. The expression for the K_sp is also different:

K_sp = [Ca²⁺]³[PO₄³⁻]² = 2.1 × 10⁻³³

Now when we substitute the unknown concentrations into the expression, we get

(3x)³(2x)² = 2.1 × 10⁻³³

When we raise each expression inside parentheses to the proper power, remember that the power affects everything inside the parentheses, including the number. So

(27x³)(4x²) = 2.1 × 10⁻³³

Simplifying,

108x⁵ = 2.1 × 10⁻³³

Dividing both sides of the equation by 108, we get

x⁵ = 1.9 × 10⁻³⁵

Now we take the fifth root of both sides of the equation (be sure you know how to do this on your calculator):

x = 1.1 × 10⁻⁷

We are not done yet. We still need to determine the concentrations of the ions. According to the ICE chart, [Ca²⁺] is 3x, not x. So

[Ca²⁺] = 3x = 3 × 1.1 × 10⁻⁷ = 3.3 × 10⁻⁷ M

[PO₄³⁻] is 2x, so

[PO₄³⁻] = 2x = 2 × 1.1 × 10⁻⁷ = 2.2 × 10⁻⁷ M

Test Yourself

What are [Mg²⁺] and [OH⁻] in a saturated solution of Mg(OH)₂? The K_sp of Mg(OH)₂ is 5.6 × 10⁻¹².

Answer

[Mg²⁺] = 1.1 × 10⁻⁴ M; [OH⁻] = 2.2 × 10⁻⁴ M

Food and Drink App: Solids in Your Wine Bottle

People who drink wine from bottles (as opposed to boxes) will occasionally notice some insoluble materials in the wine, either crusting the bottle, stuck to the cork, or suspended in the liquid wine itself. The accompanying figure shows a cork encrusted with colored crystals. What are these crystals?

The red crystals on the top of the wine cork are from insoluble compounds that are not soluble in the wine.

Source: Photo courtesy of Paul A. Hernandez, http://www.flickr.com/photos/paul_a_hernandez/2940862302/.

One of the acids in wine is tartaric acid (H₂C₄H₄O₆). Like the other acids in wine (citric and malic acids, among others), tartaric acid imparts a slight tartness to the wine. Tartaric acid is rather soluble in H₂O, dissolving over 130 g of the acid in only 100 g of H₂O. However, the potassium salt of singly ionized tartaric acid, potassium hydrogen tartrate (KHC₄H₄O₆; also known as potassium bitartrate and better known in the kitchen as cream of tartar), has a solubility of only 6 g per 100 g of H₂O. Thus, over time, wine stored at cool temperatures will slowly precipitate potassium hydrogen tartrate. The crystals precipitate in the wine or grow on the insides of the wine bottle and, if the bottle is stored on its side, on the bottom of the cork. The color of the crystals comes from pigments in the wine; pure potassium hydrogen tartrate is clear in its crystalline form, but in powder form it is white.

The crystals are harmless to ingest; indeed, cream of tartar is used as an ingredient in cooking. However, most wine drinkers don’t like to chew their wine, so if tartrate crystals are present in a wine, the wine is usually filtered or decanted to remove the crystals. Tartrate crystals are almost exclusively in red wines; white and rose wines do not have as much tartaric acid in them.

Key Takeaway

• Equilibrium constants exist for certain groups of equilibria, such as weak acids, weak bases, the autoionization of water, and slightly soluble salts.

Exercises

1. Explain the difference between the K_eq and the K_sp.

2. Explain the difference between the K_a and the K_b.

3. Write the balanced chemical equation that represents the equilibrium between HF(aq) as reactants and H⁺(aq) and F⁻(aq) as products.

4. Write the balanced chemical equation that represents the equilibrium between CaF₂(s) as reactants and Ca²⁺(aq) and F⁻(aq) as products.

5. Assuming that all species are dissolved in solution, write the K_eq expression for the chemical equation in Exercise 3.

6. Noting the phase labels, write the K_sp expression for the chemical equation in Exercise 4.

7. Determine the concentrations of all species in the ionization of 0.100 M HClO₂ in H₂O. The K_a for HClO₂ is 1.1 × 10⁻².

8. Determine the concentrations of all species in the ionization of 0.0800 M HCN in H₂O. The K_a for HCN is 6.2 × 10⁻¹⁰.

9. Determine the pH of a 1.00 M solution of HNO₂. The K_a for HNO₂ is 5.6 × 10⁻⁴.

10. Determine the pH of a 3.35 M solution of HC₂H₃O₂. The K_a for HC₂H₃O₂ is 1.8 × 10⁻⁵.

11. Write the chemical equations and K_a expressions for the stepwise dissociation of H₃PO₄.

12. Write the chemical equations and K_a expressions for the stepwise dissociation of H₃C₆H₅O₇.

13. If the K_a for HNO₂ is 5.6 × 10⁻⁴, what is the K_b for NO₂⁻(aq)?

14. If the K_a for HCN is 6.2 × 10⁻¹⁰, what is the K_b for CN⁻(aq)?

15. What is [OH⁻] in a solution whose [H⁺] is 3.23 × 10⁻⁶ M?

16. What is [OH⁻] in a solution whose [H⁺] is 9.44 × 10⁻¹¹ M?

17. What is [H⁺] in a solution whose [OH⁻] is 2.09 × 10⁻² M?

18. What is [H⁺] in a solution whose [OH⁻] is 4.07 × 10⁻⁷ M?

19. Write the balanced chemical equation and the K_sp expression for the slight solubility of Mg(OH)₂(s).

20. Write the balanced chemical equation and the K_sp expression for the slight solubility of Fe₂(SO₄)₃(s).

21. What are [Sr²⁺] and [SO₄²⁻] in a saturated solution of SrSO₄(s)? The K_sp of SrSO₄(s) is 3.8 × 10⁻⁴.

22. What are [Ba²⁺] and [F⁻] in a saturated solution of BaF₂(s)? The K_sp of BaF₂(s) is 1.8 × 10⁻⁷.

23. What are [Ca²⁺] and [OH⁻] in a saturated solution of Ca(OH)₂(s)? The K_sp of Ca(OH)₂(s) is 5.0 × 10⁻⁶.

24. What are [Pb²⁺] and [I⁻] in a saturated solution of PbI₂? The K_sp for PbI₂ is 9.8 × 10⁻⁹.

Answers

1. The K_sp is a special type of the K_eq and applies to compounds that are only slightly soluble.

3. HF(aq)⇄H+(aq)+F–(aq)

5. Keq=[H+][F–][HF]

7. [HClO₂] = 0.0719 M; [H⁺] = [ClO₂⁻] = 0.0281 M

9. 1.63

11. H3PO4(aq)⇄H+(aq) + H2PO4–(aq); Ka=[H+][H2PO4–][H3PO4]

    H2PO4–(aq)⇄H+(aq) + HPO42–(aq); Ka=[H+][HPO42–][H2PO4–]

    HPO42–(aq)⇄H+(aq) + PO43–(aq); Ka=[H+][PO43–][HPO42–]

13. 1.8 × 10⁻¹¹

15. 3.10 × 10⁻⁹ M

17. 4.78 × 10⁻¹³ M

19. MgOH2(s)⇄Mg2+(aq) + 2OH–(aq); K_sp = [Mg²⁺][OH⁻]²

21. [Sr²⁺] = [SO₄²⁻] = 1.9 × 10⁻² M

23. [Ca²⁺] = 0.011 M; [OH⁻] = 0.022 M