5 Steps to a 5: AP Chemistry 2024 - Moore J.T., Langley R.H. 2023

STEP 4 Review the Knowledge You Need to Score High
15 Equilibrium

IN THIS CHAPTER

In this chapter, the following AP topics are covered:

7.1 Introduction to Equilibrium

7.2 Direction of Reversible Reactions

7.3 Reaction Quotient and Equilibrium Constant

7.4 Calculating the Equilibrium Constant

7.5 Magnitude of the Equilibrium Constant

7.6 Properties of the Equilibrium Constant

7.7 Calculating Equilibrium Concentrations

7.8 Representations of Equilibrium

7.9 Introduction to Le Châtelier’s Principle

7.10 Reaction Quotient and Le Châtelier’s Principle

7.11 Introduction to Solubility Equilibria

7.12 Common-Ion Effect

7.13 pH and Solubility

7.14 Free Energy of Dissolution

Summary: We’ve been discussing chemical reactions for several chapters. In the Kinetics chapter you saw how chemical reactions take place and some of the factors that affect the reactions’ speed. In this chapter, we will discuss another aspect of chemical reactions: equilibrium.

A few chemical reactions proceed to completion, using up one or more of the reactants and then stopping. However, most reactions behave in a different way. Consider the general reaction:

a A + b B → c C + d D

Reactants A and B are forming C and D. Then C and D start to react to form A and B:

c C + d D → a A + b B

These two reactions proceed until the two rates of reaction become equal. That is, the speed of production of C and D in the first reaction is equal to the speed of production of A and B in the second reaction. Since these two reactions are occurring simultaneously in the same container, the amounts of A, B, C, and D become constant. A chemical equilibrium has been reached, in which two exactly opposite reactions are occurring at the same place, at the same time, and with the same rates of reaction. When a system reaches the equilibrium state, the reactions do not stop. A and B are still reacting to form C and D; C and D are still reacting to form A and B. But because the reactions proceed at the same rate, the amounts of each chemical species are constant. This state is sometimes called a dynamic equilibrium state to emphasize the fact that the reactions are still occurring—it is a dynamic, not a static, state. An equilibrium state is indicated by a double arrow instead of a single arrow. For the reaction above it would be shown as:

a A + b B ⇆ c C + d D

It is important to remember that at equilibrium the concentrations of the chemical species are constant, not necessarily equal. There may be a lot of C and D and a little A and B, or vice versa. The concentrations are constant, unchanging, but not necessarily equal.

At any point during the preceding reaction, a relationship may be defined called the reaction quotient, Q. It has the following form:

The reaction quotient is a fraction. In the numerator is the product of the chemical species on the right-hand side of the equilibrium arrow, each raised to the power of that species’ coefficient in the balanced chemical equation. It is called the Q_c in this case because molar concentrations are being used. If this were a gas-phase reaction, gas pressures could be used, and it would become a Q_p.

Remember: products over reactants

Keywords and Equations

Equilibrium Expressions

The reactant quotient can be written at any point during the reaction, but the most useful point is when the reaction has reached equilibrium. At equilibrium, the reaction quotient becomes the equilibrium constant, K_c (or K_p if gas pressures are being used). Usually this equilibrium constant is expressed simply as a number without units. In addition, the concentrations of solids or pure liquids (not in solution) that appear in the equilibrium expression are assumed to be 1, since their concentrations do not change. (The use of concentrations or pressures is only an approximation; equilibrium constants are based upon something called activities. Activities are, by definition, unitless, which makes the equilibrium constants unitless.) (Note: the equilibrium constant is constant as long as the temperature is constant. However, if the temperature is only slightly changed, it may be assumed that the change in the constant is negligible.)

Consider the Haber process for producing ammonia:

N₂(g) + 3H₂(g) ⇆ 2NH₃(g)

The equilibrium constant expression would be written as:

If the partial pressures of the gases were used, then K_p would be written in the following form:

There is a relationship between K_c and K_p: K_p = K_c(RT)^Δn, where R is the ideal gas constant (0.08206 L atm mol^—1 K^—1) and Δn is the change in the number of moles of gas in the reaction. However, this topic will not appear on the AP Exam.

Magnitude of the Equilibrium Constant

The numerical value of the equilibrium constant can give an indication of the extent of the reaction after equilibrium has been reached. If the value of K is large, that means the numerator is much larger than the denominator and the reaction has produced a relatively large quantity of products (reaction lies far to the right). If K is small, then the numerator is much smaller than the denominator and not much product has been formed (reaction lies far to the left).

There is a wide range of values for the equilibrium constant. The magnitude of the equilibrium not only says something about the degree of reaction but also says something about how to solve a problem involving equilibrium.

Very large equilibrium constants mean that nearly everything to the left of the equilibrium arrow will be converted to the substances to the right of the arrow. Very small equilibrium constants mean that nearly everything to the right of the equilibrium arrow will be converted to the substances to the left of the arrow. Identifying the constants as very large or very small is a subjective judgment.

The following example will illustrate how a problem may or may not be simplified. If you have not covered ICE tables yet, you may wish to return to this later.

Given the following general equilibrium and equilibrium constant expression:

This also applies if a K_p was used. If the original concentrations of the reactant and products are expressed as [A], [C], and [D]. A table for this equilibrium would look like the following.

The ± will be + in some cases and — in others; it depends upon the original concentrations and the value of the equilibrium constant. The general solution will require a quadratic formula. However, a very small equilibrium constant means that [C] and/or [D] must be small. If the initial are equal to 0, then [C] ± cx ≈ cx and/or [D] ± dx ≈ dx, and [A] ± ax ≈ [A], it is no longer be necessary to do a quadratic.

Properties of the Equilibrium Constant

The general reaction quotient is written as:

If concentrations are used, this is a Q_c, and if pressures are used, this is a Q_p. (Do not mix concentrations and pressures.) If the values used in the expression are the equilibrium values, then Q becomes K.

Note: using concentrations or pressures is only an approximation. True reaction quotients and equilibrium constants require a concept from beyond AP Chemistry, notable activities. The activity of a substance is a unitless quantity, which may be approximated by the concentration or partial pressure.

If the above equilibrium were written as:

c C + d D ⇆ a A + b B

Then the reaction quotient would change to:

The relationship between Q and which should be apparent if one compares the two reaction quotient expressions. This relationship applies to equilibrium constants also. That is, if the equilibrium reaction is reversed, the reaction quotient or equilibrium constant for the reversed reaction is the reciprocal of the initial.

Consider the following two equilibria:

Equation (2) is simply equation (1) doubled. The equilibrium constant expressions for these two equilibria are (subscripts indicate to which equilibrium the constant applies):

Comparing the two equilibrium constant expressions shows that . This is a property of equilibria. Doubling a reaction squares the constant and halving an equation gives the square root of the original constant. Similarly, tripling the reaction cubes the constant, and so on. This applies to all equilibria.

Consider the following two equilibria and their sum:

Sum

The sum is determined by a procedure similar to a Hess’s law problem.

Let’s explore the relationship between the three equilibrium constants. What happens if we multiply K₁ by K₂? Let’s begin by writing the individual expressions:

This illustrates another general property of equilibria. The constant for the sum of any number of separate equilibria is equal to the product of the separate equilibrium constants.

These relationships apply to any K (both K_c and K_p) or Q (both Q_c and Q_p).

Calculating Equilibrium Concentrations

The following example will illustrate a common method that may be used to solve equilibrium problems:

Gaseous BrCl will decompose to the elements when heated. At 500 K, the equilibrium constant for this decomposition reaction is K_c = 33. What are the equilibrium concentrations of the three substances involved in this equilibrium if the initial concentration of BrCl(g) is 1.0 M?

The first step is to write the reaction and the equilibrium constant expression.

Sometimes one or both of these is given, which will save you some work.

The next step is to begin a table. The headings of the columns are the substances involved in the equilibrium. It will help to list them in the order they appear in the reaction.

Below each substance in a row labelled “Initial” are the initial values of the concentration for each substance (0 if none of the substance was given). This gives:

While an initial value may be 0, no equilibrium value will ever be 0 (it may be close). Therefore, for any substance that is 0, some must be produced, and the expense of any substances on the opposite of the equilibrium arrow. The values are normally assigned to be x, with every x multiplied by the coefficient from the equilibrium reaction. In this case, we get a “Change” row as shown here:

The final row of the table is the “Equilibrium” row, which is the sum of the Initial and Change rows. This gives the following complete table:

This table is sometimes called an “ICE” table from the first letter in the label of each of the three lines.

The next step is to enter the values from the Equilibrium line into the equilibrium constant:

Solving this equation for x will allow you to determine the equilibrium concentrations of the substances involved. Since the key concept for this section is the construction of an ICE table, we will not solve this problem for the concentrations at this point.

Let’s look at two variations on this problem and see what the ICE table will look like.

Variation 1

Instead of the initial being 1.0 M BrCl, the initial is 1.0 M Br₂ and 1.0 M in Cl₂.

Variation 2

Now use the initial BrCl = 1.0 M, and the initial Br₂ = 1.0 M.

Since the initial chlorine concentration was 0, the equilibrium must shift to the right to alleviate the problem of having a 0 concentration.

If none of the initial values was 0, then a Q calculation would be needed to predict which direction the equilibrium would shift. If it shifts to the right, substances on the left would have a negative change and substances on the right would have a positive change. If it shifts to the left, substances on the right would have a negative change and substances on the left would have a positive change. Finally, if Q_c = K_c, the system is already at equilibrium, and the concentrations given are the equilibrium concentrations.

If the problem dealt with a K_p instead of a K_c, the procedure would be the same except partial pressures would replace the concentrations.

Once you have an equilibrium expression with the appropriate values entered from the Equilibrium line of the ICE table, the problem becomes a math problem where it is necessary to solve for x. Some of these problems will be more of a challenge than others to solve. You will find that in many cases, there will be a simplifying assumption to make the solution easy to find. In general, the simplifying assumptions can be made if K is very large or very small.

Representations of Equilibrium

There are many ways to represent a system moving to equilibrium. Figure 15.1 shows two of these methods. Reaction A illustrates the reaction A(g) + B(g) → AB(g), from pure A + B to equilibrium. Reaction B illustrates the reaction AB(g) → A(g) + B(g), from pure AB to equilibrium. Together, these two reaction schemes show the same equilibrium being achieved by two methods. The equilibrium is A(g) + B(g) ⇆ AB(g).

Figure 15.1 Two reaction schemes (A and B) to represent the reaction A + B → AB reaching equilibrium. The dark circles represent atoms of A, and the open circles represent atoms of B. All species are in the gas phase.

You should prove for yourself that each of the reaction schemes do achieve the same equilibrium. In addition, you should calculate the equilibrium constant before going on to the solution.

Solution:

Begin by writing the equilibrium constant expression. Then enter the number of A, B, and AB you count in the box representing equilibrium.

The last box in each reaction scheme is at equilibrium. You may recognize this because the number of atoms/molecules has not changed from the preceding box.

Le Châtelier’s Principle

At a given temperature, a reaction will reach equilibrium with the production of a certain amount of product. If the equilibrium constant is small, that means that not much product will be formed. But is there anything that can be done to produce more? Yes, there is—through the application of Le Châtelier’s principle. Le Châtelier, a French scientist, discovered that if a chemical system at equilibrium is stressed (disturbed), it will reestablish equilibrium by shifting the reactions involved. This means that the amounts of the reactants and products will change, but the final ratio will remain the same. The equilibrium may be stressed in numerous ways: changes in concentration, pressure, and temperature. Many times the use of a catalyst is mentioned. However, a catalyst will have no effect on the equilibrium amounts because it affects both the forward and reverse reactions equally. It will, however, cause the reaction to reach equilibrium faster.

Changes in Concentration

If the equilibrium system is stressed by a change in concentration of one of the reactants or products, the equilibrium will react to remove that stress. If the concentration of a chemical species is decreased, the equilibrium will shift to produce more of it. In doing so, the concentration of chemical species on the other side of the reaction arrows will be decreased. If the concentration of a chemical species is increased, the equilibrium will shift to consume it, increasing the concentration of chemical species on the other side of the reaction arrows.

For example, again consider the Haber process:

N₂(g) + 3H₂(g) ⇆ 2NH₃(g)

If one increases the concentration of hydrogen gas, then the equilibrium shifts to the right to consume some of the added hydrogen. In doing so, the concentration of ammonia (NH₃) will increase, and the concentration of nitrogen gas will decrease. On the other hand, if the concentration of nitrogen gas was decreased, the equilibrium would shift to the left to form more, the concentration of ammonia would decrease, and the concentration of hydrogen would increase.

Again, remember that the concentrations may change, but the value of K_c or K_p would remain the same. Changes in concentration never change the value of the equilibrium constant.

Changes in Pressure

Changes in pressure are significant only if gases are involved. The pressure may be changed by changing the volume of the container or by changing the concentration of a gaseous species (although this is really a change in concentration and can be treated as a concentration effect, as above). If the container becomes smaller, the pressure increases because there is an increased number of collisions on the inside walls of the container. This stresses the equilibrium system, and it will shift to reduce the pressure. This can be accomplished by shifting the equilibrium toward the side of the equation that has the lesser number of moles of gas. If the container size is increased, the pressure decreases, and the equilibrium will shift to the side containing more moles of gas to increase the pressure. If the number of moles of gas is the same on both sides, changing the pressure will not affect the equilibrium.

Once again, consider the Haber reaction:

N₂(g) + 3H₂(g) ⇆ 2NH₃(g)

Note that there are 4 moles of gas (1 of nitrogen and 3 of hydrogen) on the left side and 2 moles on the right. If the container is made smaller, the pressure will increase, and the equilibrium will shift to the right because 4 moles would be converted to 2 moles. The concentrations of nitrogen and hydrogen gases would decrease, and the concentration of ammonia would increase. Changes in pressure never change the value of the equilibrium constant. It is important to use the partial pressures of the gases involved as a gas that is not part of the equilibrium does not count.

Remember: pressure effects are only important for gases.

Changes in Temperature

Changing the temperature changes the value of the equilibrium constant. It also changes the amount of heat in the system and can be treated as a concentration effect. To treat it this way, one must know which reaction, forward or reverse, is exothermic (releasing heat).

One last time, let’s consider the Haber reaction:

N₂(g) + 3H₂(g) ⇆ 2NH₃(g)

The formation of ammonia is exothermic (liberating heat), so the reaction could be written as:

N₂(g) + 3H₂(g) ⇆ 2NH₃(g) + heat

If the temperature of the reaction mixture were increased, the amount of heat would be increased, and the equilibrium would shift to the left to consume the added heat. In doing so, the concentration of nitrogen and hydrogen gases would increase, and the concentration of ammonia gas would decrease. If you were in the business of selling ammonia, you would probably want to operate at a reduced temperature, to shift the reaction to the right.

Remember, unlike concentration and pressure, changes in temperature will change the value of the equilibrium constant.

Consider the following equilibrium (endothermic as written), and predict what changes, if any, would occur if the following stresses were applied after equilibrium was established:

2 NaHCO₃(s) ⇆ Na₂CO₃(s) + H₂O(g)

a. Add H₂O(g)

b. Remove H₂O(g)

c. Add Na₂CO₃(s)

d. Increase T

e. Decrease V

f. Add a catalyst

Answers:

a. Left—The equilibrium shifts to remove some of the excess H₂O(g).

b. Right—The equilibrium shifts to replace some of the H₂O(g).

c. No change—Solids do not shift equilibria unless they are totally removed.

d. Right—Endothermic reactions shift to the right when heated.

e. Left—A decrease in volume, or an increase in pressure, will shift the equilibrium toward the side with less gas.

f. No change—Catalysts do not affect the position of an equilibrium.

Reaction Quotient and Le Châtelier’s Principle

If a system at equilibrium is disturbed, Le Châtelier’s principle governs what will happen. Predictions may also be made by comparing the value of the reaction quotient to the value of the equilibrium constant.

For this discussion, we will consider the following equilibrium:

N₂(g) + 3 H₂(g) ⇆ 2 NH₃(g)

Two reaction quotients may be written for this reaction:

If the system is in equilibrium, Q_c = K_c and Q_p = K_p. However, if the system is disturbed, one or more of the concentrations or pressures in these expression will be changed. Once the equilibrium has been changed, according to Le Châtelier’s principle, it will return. For example, a simple increase in the ammonia will result in a shift to the left (reactants). However, how will the reaction shift if nitrogen is increased and hydrogen is decreased? In this case, the new values should be entered into the reaction quotient expression. Once the values are entered, a reaction quotient may be calculated. There are three possible outcomes (for either Q and its respective K):

K > Q

K < Q

K = Q

If K is greater than Q, the reaction must shift to the right to increase Q until it equals K. If K is less than Q, the reaction must shift to the left to decrease Q until it equals K. If Q = K, the system is still in equilibrium, so no change will occur.

If the question includes something like “Is the system in equilibrium?” you most likely need to compare K and Q.

Solubility Equilibria

Many salts are soluble in water, but some are only slightly soluble. These salts, when placed in water, quickly reach their solubility limit and the ions establish an equilibrium system with the undissolved solid. For example, PbSO₄, when added to water, establishes the following equilibrium:

PbSO₄(s) ⇆ Pb²⁺(aq) + SO₄^2—(aq)

The equilibrium constant expression for systems of slightly soluble salts is called the solubility product constant, K_sp. It is the product of the ionic concentrations, each one raised to the power of the coefficient in the balanced chemical equation. It contains no denominator since the concentration of a solid is, by convention, 1 and does not appear in the equilibrium constant expressions. (Some textbooks will say that the concentrations of solids, liquids, and solvents are included in the equilibrium constant.) The K_sp expression for the PbSO₄ system would be:

K_sp = [Pb²⁺][SO₄^2—]

Never place any terms in the denominator of a K_sp expression.

For this salt, the numerical value of K_sp is 1.6 × 10^—8 at 25°C. Note that the Pb²⁺ and SO₄^2— ions are formed in equal amounts, so the right-hand side of the equation could be represented as [x]. If the numerical value of the solubility product constant is known, then the concentration of the ions can be determined. And if one of the ion concentrations can be determined, then K_sp can be calculated.

Here is an example of a K_sp problem. The K_sp of lead(II) fluoride in water is 3.6 × 10^—8. How many grams of lead(II) fluoride will dissolve in 0.500 L of water?

The mass (grams) dissolved in a solution is known as the solubility. The line immediately below the equilibrium equation is the Equilibrium line from the ICE table.

If a slightly soluble salt solution is at equilibrium and a solution containing one of the ions involved in the equilibrium is added, the solubility of the slightly soluble salt is decreased. For example, let’s again consider the PbSO₄ equilibrium:

PbSO₄(s) ⇆ Pb²⁺(aq) + SO₄^2—(aq)

Suppose a solution of Na₂SO₄ is added to this equilibrium system. The additional sulfate ion will disrupt the equilibrium, by Le Châtelier’s principle, and shift it to the left, decreasing the solubility. The same would be true if you tried to dissolve Na₂SO₄ in a solution of Na₂SO₄ instead of pure water—the solubility would be lower. This application of Le Châtelier’s principle to equilibrium systems of slightly soluble salts is called the common-ion effect (buffer solutions are another example of the common-ion effect). Calculations like the ones above involving finding concentrations and K_sp can still be done, but the concentration of the additional common ion will have to be inserted into the solubility product constant expression. Sometimes, if K_sp is very small and the common ion concentration is large, the concentration of the common ion can simply be approximated by the concentration of the ion added.

For example, calculate the silver ion concentration in each of the following solutions:

a.

b.

a.

b.

The molarity of the dissolved solid is known as the molar solubility. Knowing the value of the solubility product constant can also allow us to predict whether a precipitate will form if two solutions, each containing an ion component of a slightly soluble salt, are mixed. The ion-product, sometimes represented as Q (same form as the solubility product constant), is calculated taking into consideration the mixing of the volumes of the two solutions, and this ion-product is compared to K_sp. If it is greater than K_sp, precipitation will occur until the ion concentrations have been reduced to the solubility level.

If 10.0 mL of a 0.100 M CaCl₂ solution is added to 40.0 mL of a 0.0250 M Pb(NO₃)₂ solution, will PbCl₂ precipitate? K_sp for PbCl₂ = 1.7 × 10^—5.

To answer this question, the concentrations of the chloride ion and the lead(II) ion before precipitation must be used. These may be determined simply from M_dil = M_conV_con/V_dil. (You should recall this dilution formula from prior knowledge.) For both ions, V_dil = (10.0 + 40.0) = 50.0 mL. V_con for CaCl₂ is 10.0 mL and for Pb(NO₃)₂ it is 40.0 mL. The values of M_con are 2(0.100) = 0.200 M Cl^— and 0.0250 M Pb²⁺. It is now necessary to calculate M_dil for each ion. (The Ca²⁺ and NO₃^— ions are ignored because they are not part of the K_sp equilibrium.)

(Note: no volume conversions were necessary because all the volumes are in milliliters.)

Entering these values into the following relation produces:

Q = [Pb²⁺][Cl^—]² = (0.0200)(0.00400)² = 3.2 × 10^—7

Since Q is smaller than K_sp, no precipitation will occur.

Common-Ion Effect

Consider the following equilibrium:

At equilibrium [K⁺] = [IO₄^—] = 2.9 × 10^—2. Now assume that 1.0 mole of KNO₃ is added to 1.0 liter of a solution where the KIO₄ equilibrium has been established. The additional KNO₃ will generate . The added NO₃^—(aq) will not influence the established equilibrium as it does not appear in the equilibrium constant expression; however, the K⁺(aq) does appear, so it will influence the equilibrium. This changes the equilibrium constant expression to:

Solving for the concentrations gives [K⁺] = 1.0 M and [IO₄^—] = 8.3 × 10^—4 M. These values are different from the concentrations determined before the KNO₃ was added. Since K⁺(aq) was added, the change in K⁺(aq) would be obvious; however, the IO₄^—(aq) change is a different matter. This example illustrates the common-ion effect. The common-ion effect is the changing of an equilibrium by the addition of one of the equilibrium ions from an outside source. This is an application of Le Châtelier’s ionic equilibria.

The common-ion effect applies to all ionic equilibria. It will appear again in Chapter 16 in the discussion of buffer solutions.

pH and Solubility

You may wish to come back to this topic after Chapter 16.

Acids and bases can alter the solubility of an ionic salt. Acids will alter the solubility of salts of weak acids, and bases will alter the solubility of salts of weak bases. The solubility of salts of strong acids with strong bases are not altered. Examples are BaCO₃ (salt of a weak acid), K_sp = 5.1 × 10^—9, (NH₄)₂PtCl₆ (salt of a weak base), K_sp = 9 × 10^—6, BaSO₄ (salt of a strong acid and a strong base), K_sp = 1.1 × 10^—10. (Both acids and bases will affect the pH of a solution.) This is related to Le Châtelier’s principle.

Let’s use AgCN (K_sp = 1.4 × 10^—16) as an example of a salt of a weak acid. In this case, the weak acid is HCN (K_a = 6.2 × 10^—10). The equilibrium expressions for these two reactions are:

Reversing the K_a equilibrium requires the reciprocal of the equilibrium constant (this procedure may be applied to any equilibrium, as both situations are occurring simultaneously).

If solid AgCN is added to an HCN solution, or HCN is added to a solution where the AgCN has been established, then one equilibrium is being added to the other. In this example, we will add the K_sp to the K_rev:

Summing these two equilibria and calculating a new constant gives:

The new K for the dissolution of AgCN is much larger than the original K_sp, which means the AgCN is more soluble in the acid solution.

In most cases, you will not be required to do this type of calculation; you will only need to be able to predict the change in solubility. To make a correct prediction, you will need to recognize what type of salt you are considering.

Free Energy of Dissolution

Solubility constants, like all equilibrium constants, may be related to the Gibbs free energy by the following relationship:

ΔG = —RT ln K

For soluble materials, ΔG is negative (thermodynamically favorable), and for “insoluble” materials, ΔG is positive (thermodynamically unfavorable).

The Gibbs free energy may be defined by the relationship ΔG = ΔH — T ΔS. The enthalpy, ΔH, relates to the breaking and forming of interactions between the solvent molecules and the solute ions. The magnitude of these interactions depends upon the number and strength of the intermolecular forces. Breaking these interactions leads to a positive contribution, and forming new interactions leads to a negative contribution. To dissolve an ionic solid, it is necessary to break the ionic bonds in the solid. To make room for the ions in the solution, some of the interactions between the solvent molecules must be broken. New interactions (ion—dipole forces) can then form.

The entropy, ΔS, relates to the breakdown of the ordered crystalline structure of the solid (a positive contribution) and the ordering of the solvent molecules around the ions (ion—dipole forces), which leads to a negative contribution. The combination of these contributions yields the entropy of solution.

The enthalpy of solution and the entropy of solution lead to the Gibbs free energy of solution, the value, and more importantly the sign, of which dictates how soluble a substance is in a particular solvent.

Other Equilibria

Other types of equilibria can be treated in much the same way as the ones discussed above. For example, there is an equilibrium constant associated with the formation of complex ions. This equilibrium constant is called the formation constant, K_f. Zn(H₂O)₄²⁺ reacts with cyanide ion to form the Zn(CN)₄^2— complex ion according to the following equation:

The K_f of Zn(CN)₄^2—(aq) is 4.2 × 10¹⁹, indicating that the equilibrium lies to the right. The actual equilibrium process is more complicated than shown.

Experiments

Equilibrium experiments such as 10, 11, and 13 in Chapter 20, Experimental Investigations, directly or indirectly involve filling a table like the following:

The initial amounts—concentrations or pressures—are normally zero for the products, and a measured or calculated value for the reactants. Once equilibrium has been established, the amount of at least one of the substances is determined. Based on the change in this one substance and the stoichiometry, the amounts of the other materials may be calculated (not measured).

Measurements may include the pressure, the mass (to be converted to moles), the volume (to be used in calculations), and the pH (to be converted into either the hydrogen ion or hydroxide ion concentration). Some experiments measure the color intensity (with a spectrophotometer), which may be converted to a concentration.

Do not make the mistake of “measuring” a change. Changes are never measured; they are always calculated.

Common Mistakes to Avoid

1. Be sure to check the units and significant figures of your final answer.

2. When writing equilibrium constant expressions, use products over reactants. Each con- centration is raised to the power of the coefficient in the balanced chemical equation.

3. Remember, in working Le Châtelier problems, pressure effects are important only for gases that are involved in the equilibrium.

Review Questions

Use these questions to review the content of this chapter and practice for the AP Chemistry Exam. First are 20 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry Exam. These questions also include ones designed to help you review prior knowledge. Following those is a long free-response question like the ones in Section II of the exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.

Multiple-Choice Questions

Answer the following questions in 30 minutes. You may use the periodic table and the equation sheet at the back of this book.

1. A student wishes to reduce the zinc ion concentration in a saturated zinc iodate solution to 1 × 10^—6 M. How many moles of solid KIO₃ must be added to 1.00 L of solution? [Zn(IO₃)₂ = 4 × 10^—6 at 25°C]

(A) 1 mole

(B) 0.5 mole

(C) 2 moles

(D) 4 moles

2. At constant temperature, a change in volume will NOT affect the moles of substances present in which of the following?

(A) H₂(g) + I₂(g) ⇆ 2 HI(g)

(B) CO(g) + Cl₂(g) ⇆ COCl₂(g)

(C) PCl₅(g) ⇆ PCl₃(g) + Cl₂(g)

(D) N₂(g) + 3 H₂(g) ⇆ 2 NH₃(g)

3. The equilibrium constant for the hydrolysis of C₂O₄^2— is best represented by which of the following?

(A)

(B)

(C)

(D)

4. C(s) + H₂O(g) ⇆ CO(g) + H₂(g) endothermic

An equilibrium mixture of the reactants is placed in a sealed container at 150°C. The amount of the products may be increased by which of the following changes?

(A) decreasing the volume of the container

(B) raising the temperature of the container and increasing the volume of the container

(C) lowering the temperature of the container

(D) adding 1 mole of C(s) to the container

5. CH₄(g) + CO₂(g) ⇆ 2 CO(g) + 2 H₂(g)

A 1.00-L flask is filled with 0.30 mole of CH₄ and 0.40 mole of CO₂ and allowed to come to equilibrium. At equilibrium, there are 0.20 mole of CO in the flask. What is the value of K_c, the equilibrium constant, for the reaction?

(A) 1.2

(B) 0.027

(C) 0.30

(D) 0.060

6. 2 NO₂(g) ⇆ 2 NO(g) + O₂(g)

The above materials were sealed in a flask and allowed to come to equilibrium at a certain temperature. A small quantity of O₂(g) was added to the flask, and the mixture was allowed to return to equilibrium at the same temperature. Which of the following has increased over its original equilibrium value?

(A) the quantity of NO₂(g) present

(B) the quantity of NO(g) present

(C) the equilibrium constant, K

(D) the rate of the reaction

7. 2 CH₄(g) + O₂(g) ⇆ 2 CO(g) + 4 H₂(g) ΔH < 0

To increase the value of the equilibrium constant, K, which of the following changes must be made to the above?

(A) increase the temperature

(B) increase the volume

(C) decrease the temperature

(D) add CO(g)

8. At constant temperature, a change in the volume of the system will NOT affect the moles of the substances present in which of the following?

(A) C(s) + H₂O(g) ⇆ H₂(g) + CO(g)

(B) 3 O₂(g) ⇆ 2 O₃(g)

(C) Xe(g) + 2 F₂(g) ⇆ XeF₄(g)

(D) 6 CO₂(g) + 6 H₂O(l) ⇆ 6 O₂(g) + C₆H₁₂O₆(aq)

9. A chemistry student adds some dilute ammonia solution to some insoluble silver chloride. The solid dissolves. Which of the following is the correct net ionic equation for the reaction?

(A) AgCl(s) + 2 NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl^—(aq)

(B) AgCl(s) + 2 NH₄⁺(aq) → [Ag(NH₄)₂]³⁺(aq) + Cl^—(aq)

(C) AgCl(s) + NH₄⁺(aq) → Ag⁺(aq) + NH₄Cl(aq)

(D) AgCl(s) + NH₃(aq) → Ag⁺(aq) + NH₃Cl(aq)

10. Which of the following contains only equilibria that are normally always homogeneous?

(A) K_a, K_b, and K_sp

(B) K_a, K_sp, and K_f

(C) K_a, K_b, and K_f

(D) K_sp, K_b, and K_f

11. For which of the following equilibriums will K_c = K_p?

(A) H₂(g) + I₂(g) ⇆ 2 HI(g)

(B) CO(g) + Cl₂(g) ⇆ COCl₂(g)

(C) PCl₅(g) ⇆ PCl₃(g) + Cl₂(g)

(D) N₂(g) + 3 H₂(g) ⇆ 2 NH₃(g)

12. CaCO₃(s) ⇆ CaO(s) + CO₂(g)

Which of the following is the correct K_p expression for the above reaction?

(A)

(B)

(C)

(D)

13. Which of the following has the lower molar solubility in water CaCO₃ or SrF₂?

(A) CaCO₃

(B) SrF₂

(C) Both are nearly equal.

(D) There is insufficient information.

14. Altering the pH of a solution of which of the following will change the solubility of that substance?

(A) PbSO₄(s)K_sp = 1.6 × 10⁻⁸

(B) PbF₂(s)K_sp = 3.6 × 10⁻⁸

(C) PbCl₂(s)K_sp = 1.7 × 10⁻⁵

(D) PbI₂(s)K_sp = 7.9 × 10⁻⁹

15. Potassium perchlorate, KClO₄, has K_sp = 1.1 × 10^—2. Adding which of the following to a saturated solution of KClO₄ will decrease its solubility.

(A) HCl(aq)

(B) HClO₄(aq)

(C) NaOH(aq)

(D) Ca(C₂H₃O₂)₂(aq)

16. The K_sp for silver nitrite, AgNO₂(s), is 1.6 × 10^—4. A sample containing 1.0 × 10^-3 moles of Ca(NO₂)₂ and 8.0 × 10^—3 moles of AgNO₃ is added to a 1 liter volumetric flask and distilled water is added to dilute to the mark on the flask and both solids dissolve completely. Will AgNO₂ precipitate.

(A) Yes

(B) No

(C) Impossible to determine

(D) Insufficient information

17. A petroleum chemists is investigating the equilibrium:

For one experiment, she finds that at equilibrium the reaction mixture has the following composition:

The container had a volume of 1.00 L and was at 298 K. What is the approximate value of K_p?

(A) 0.038

(B) 19

(C) 26

(D) 0.053

18. In determining the solubility of a substance, a substance with a large positive ΔG of dissolution is likely to be

(A) Soluble

(B) Slightly soluble

(C) Insoluble

(D) Not predictable

19. In an aqueous solution of H₂S, there are two equilibria directly related to the H₂S. These equilibria are:

One way to simplify problems dealing with these two equilibria is:

(A) Add the two equations and add the constants.

(B) Ignore the first equilibrium.

(C) Both equation must always be dealt with separately.

(D) Add the two equations and multiply the constants.

20. An inorganic chemist was investigating the following equilibrium in aqueous solution:

In one experiment he found that the equilibrium concentrations of the substances in the equilibrium were:

What is the value of K_f?

(A) 1.0 × 10⁴

(B) 1.0 × 10^—4

(C) 3.0 × 10⁷

(D) 1.0 × 10^—7

Answers and Explanations

1. C—The solubility-product constant expression is K_sp = [Zn²⁺][IO₃^—]² = 4 × 10. This may be rearranged to . Inserting the desired zinc ion concentration gives [IO₃^—]² = . Taking the square root of each side leaves a desired IO₃^— concentration of 2 M. Two moles of KIO₃ must be added to 1.00 L of solution to produce this concentration. Since you can estimate the answer, no calculator is needed.

2. A—When dealing with gaseous equilibriums, volume changes are important when there is a difference in the total number of moles of gas on opposite sides of the equilibrium arrow. All the answers, except A, have differing numbers of moles of gas on opposite sides of the equilibrium arrow.

3. C—The hydrolysis of any ion begins with the interaction of that ion with water. Thus, both the ion and water must be on the left side of equilibrium arrow and hence in the denominator of the equilibrium constant expression (water, like all solvents, will be left out of the expression). The oxalate ion is the conjugate base of a weak acid. As a base, it will produce OH^— in solution along with the conjugate acid, HC₂O₄^—, of the base. The equilibrium reaction is C₂O₄^2—(aq) + H₂O(l) ⇆ OH^—(aq) + HC₂O₄^—(aq).

4. B—The addition or removal of some solid, as long as some remains present, will not change the equilibrium. An increase in volume will cause the equilibrium to shift toward the side with more moles of gas (right). Raising the temperature of an endothermic process will shift the equilibrium to the right. Any shift to the right will increase the amounts of the products.

5. B—Using the following table:

The presence of 0.20 mole of CO (0.20 M) at equilibrium means that 2x = 0.20 and that x = 0.10. Using this value for x, the bottom line of the table becomes

Entering the equilibrium values into the equilibrium expression gives As usual, you do not need to do an exact calculation as estimation will work.

6. A—The addition of a product will cause the equilibrium to shift to the left. The amounts of all the reactants will increase, and the amounts of all the products will decrease (the O₂ will not go below its earlier equilibrium value since excess was added). The value of K is constant unless the temperature is changed. The rates of the forward and reverse reactions are equal at equilibrium.

7. C—The only way to change the value of K is to change the temperature. For an exothermic process (ΔH < 0), K increases with a decrease in temperature.

8. D—If there are equal numbers of moles of gas on each side of the equilibrium arrow, then volume or pressure changes will not affect the equilibrium. The presence of solids, liquids, or aqueous phases does not make any difference as long as some of the phase is present.

9. A—Aqueous ammonia contains primarily NH₃, which eliminates B and C. NH₃Cl does not exist, which eliminates D. The reaction produces the silver-ammonia complex, [Ag(NH₃)₂]⁺.

10. A—A K is never homogeneous; all the others are always homogeneous.

11. A—K_c = K_p occurs whenever there are equal moles of gaseous molecules on each side of the equilibrium arrow.

12. D—This is a heterogeneous equilibrium; therefore, the solids (CaCO₃ and CaO) will not appear in the equilibrium expression. B is the K_c expression, not the K_p expression.

13. A—Lower molar solubility means which one will have the lower concentration (M ) in solution. We can start by writing the equilibrium equations and the equilibrium constant expressions:

Replacing the concentrations with x values (if this is not obvious, create and ICE table):

If the stoichiometries were the same (both x² or both 4 x³), the smaller K_sp is the less soluble; however since the stoichiometries are not the same, it is necessary to solve for x.

To save time, note the following: Both values are 10^—9 and the square root of this value (≈ 10^—4.5) will be smaller than the cube root (≈ 10^—3). Thus, CaCO₃ is less soluble.

14. A—The lead(II) is a constant; therefore, it cannot be a factor. The anions are F^—, SO₄^2—, Cl^—, and I^—. Since changing the pH is what causes the change, we will examine the parent acid of these anions. The parent acids are HF, H₂SO₄, HCl, and HI. All these acids except HF are strong acids, so the solubility of the fluoride salt, PbF₂, is pH dependent.

15. B—The equilibrium reaction equation is: KClO₄(s) ⇋ K⁺(aq) + ClO₄⁻(aq).

Any substance that increases the K⁺ or ClO₄^— from this equilibrium will decrease the solubility. The ions are from KOH and HClO₄, which are a strong base and a strong acid, respectively. Since the parents are both strong, pH will not be a factor; thus, adding an acid of a base will not alter the solubility. The calcium acetate, Ca(C₂H₃O₂)₂(aq), has nothing in common with the KClO₄ equilibrium. The strong acid, HClO₄, will decrease the solubility not because it is an acid, but because it supplies ClO₄^— ions to the solution. This supplied ClO₄^— will decrease the solubility of KClO₄ by the common-ion effect.

16. B—The question is asking if something will happen; in general, this means that a reaction quotient, Q, will need to be calculated. The first step in this calculation will be to determine the concentrations of the Ag⁺ and NO₂^— ion (the remaining ions are spectator ions).

The volume of the solution is 1.00 L (a volumetric flask is very accurate, so a 1-liter flask holds 1.00 L of solution—some volumetric flasks are even more accurate). Therefore, the concentrations of the ions are the moles just determined divided by 1.00 L to give: 8.0 × 10^—3M Ag⁺ = [Ag⁺] and 2.0 × 10^—3 M NO₂^— = [NO₂^—]

The equilibrium reaction and the reaction quotient relationship for AgNO₂ are

Since Q < K_sp no precipitate will form.

17. A—First, set up the K_p equilibrium constant expression: . Hopefully, you did not make the error of setting up a Entering the values from the partial pressure line of the data table gives: (you can round to simplify the final calculation and pick the nearest answer). The value of 26 is the inverse of the correct answer, which means that you wrote the K_p expression upside-down. The value of 19 means that you used the K_c expression instead of the correct K_p. The value of 0.053 is the inverse of the K_c answer, which means that you wrote the K_c expression upside down.

18. C—Dissolution is related to ΔG by the relationship: ΔG = -RT ln K. If you remember this equation, fine; however, if you do not, it is in the equation information supplied with the exam. Note that the negative sign in the equation means that there is a sign change and a positive ΔG leads to a negative exponent for K. A negative exponent indicates low solubility, and the larger the positive value of ΔG, the lower the value of K. A small positive value of ΔG leads to a slightly soluble substance, and a large positive value indicates an insoluble substance. Soluble materials have a negative ΔG.

19. D—One common way of dealing with multiple related equilibria is to add the equilibrium reactions, which requires that the equilibrium constants be multiplied, NOT added. In a series of related equilibria, the first one is probably the largest, and the largest K should not be ignored.

20. A—As is the case with most equilibrium calculations it is useful begin by writing the correct equilibrium constant expression (K). For a K_f reaction, such as this one, the expression is:

The next step is to enter the values from the table into the equilibrium expression:

The answer 1.0 × 10^—4 comes from setting the equilibrium constant expression up incorrectly (inverted). The answer 3.0 × 10⁷ comes from incorrectly including the solvent (H₂O) in the equilibrium constant expression. The answer 1.0 × 10^—7 comes from incorrectly including the solvent (H₂O) in the inverse of the equilibrium constant expression.

Free-Response Question

You have 20 minutes to answer the following long question. You may use a calculator and the tables in the back of the book.

Question

An aqueous solution is prepared that is initially 0.100 M in CdI₄^2—. After equilibrium is established, the solution is found to be 0.013 M in Cd²⁺. The products of the equilibrium are Cd²⁺(aq) and I^—(aq).

(a) Derive the expression for the dissociation equilibrium constant, K_d, for the equilibrium, and determine the value of the constant.

(b) What will be the cadmium ion concentration arising when 0.400 mol of KI is added to 1.00 L of the solution in part (a)?

(c) A solution is prepared by mixing 0.500 L of the solution from part (b) and 0.500 L of 2.0 × 10^—5 M NaOH. Will cadmium hydroxide, Cd(OH)₂, precipitate? The K_sp for cadmium hydroxide is 2.2 × 10^—14.

(d)When the initial solution is heated, the cadmium ion concentration increases. Is the equilibrium an exothermic or an endothermic process? Explain how you arrived at your conclusion.

(e)Complete the Lewis structure below for the cyanate ion and determine the formal charge for each atom. If resonance is possible, you only need to show one resonance form.

Answer and Explanation

(a) The equilibrium is: [CdI₄]^2—(aq) ⇆ Cd²⁺(aq) + 4 I^—(aq).

Give yourself 1 point for this expression (the “d” subscript is unnecessary).

Using the following table:

The value of [Cd²⁺] is given (= 0.013), and this is x. This changes the last line of the table to:

Entering these values into the K_d expression gives .

Give yourself 1 point for this answer. You can also get 1 point if you correctly put your values into the wrong K_d equation.

(b) The table in part (a) changes to the following:

(Since the initial [Cd²⁺] = 0, the equilibrium must shift to the right, which also changes the [I^—].)

Give yourself 1 point for the correct setup and 1 point for the correct answer. If you got the wrong value for K in part (a), you can still get one or both points for using the value correctly in this calculation.

(c) The K_sp equilibrium is: Cd(OH)₂(s) ⇆ Cd²⁺(aq) + 2 OH^—(aq)

The dilution reduces both the Cd²⁺ (4.3 × 10^—6 M) and OH^— (2.0 × 10^—5 M) concentrations by a factor of 2. This gives:

The reaction quotient is Q = [Cd²⁺][OH^—]² = (2.2 × 10^—6) (1.0 × 10^—5)² = 2.2 × 10^—16. This value is less than the K_sp, so no precipitate will form. (We have seen students lose points by incorrectly saying that 2.2 × 10^—14 < 2.2 × 10^—16, so be careful.)

Give yourself 1 point for the correct setup and 1 point for the correct answer. If you got the wrong value for K in part (a), you can still get 1 or both points for using the value correctly in this calculation.

(d) Since the cadmium ion concentration increases, the equilibrium must shift to the right. Endothermic processes shift to the right when they are heated. This is in accordance with Le Châtelier’s principle.

Give yourself 1 point for endothermic. Give yourself 1 point for correctly mentioning Le Châtelier’s principle.

(e) There are three acceptable resonance forms. It is only necessary for you to draw one of them correctly. For an atom, the formal charge = valence electrons — nonbonding electrons — ½ (bonding electrons).

Each of the resonance forms is shown below with the formal charge for each atom calculated:

The total is a check in case of a calculation error. The total must equal the charge on the ion (0 for a molecule).

You get 1 point for a correct Lewis structure and 1 point for a correct set of formal charges. You will get no more points for drawing more than one Lewis structure. If you draw more than one Lewis structure, you may not get the point for this part if the additional structure is incorrect.

Total your points. There are 10 points possible. Subtract 1 point if any numerical answer has an incorrect number of significant figures.

Rapid Review

A chemical equilibrium is established when two exactly opposite reactions occur in the same container at the same time and with the same rates of reaction. At equilibrium, the concentrations of the chemical species become constant, but not necessarily equal.

For the reaction a A + b B ⇆ c C + d D, the equilibrium constant expression would be: Know how to apply this equation.

Le Châtelier’s principle says that if an equilibrium system is stressed, it will reestablish equilibrium by shifting the reactions involved. A change in concentration of a species will cause the equilibrium to shift to reverse that change. A change in pressure or temperature will cause the equilibrium to shift to reverse that change.

The solubility product constant, K_sp, is the equilibrium constant expression for sparingly soluble salts. It is the product of the ionic concentration of the ions, each raised to the power of the coefficient of the balanced chemical equation.

Know how to apply ion-products and K_sp values to predict precipitation.

Formation constants describe complex ion equilibria.