Chapter 17: Equilibrium 163

CHAPTER 17 Equilibrium

INTRODUCTION Because atoms and molecules are so tiny, it is hard to imagine what happens when they react and form new products. In this chapter you will learn what is necessary for a reaction to occur, why some reactions stop before all the reactants have been used up, and how to speed up a reaction. Learning how to control chemical reactions has led to many important applications such as new ways to keep food from spoiling.

CHAPTER DISCUSSION The collision model says that in order for molecules to react with each other, they must first collide. Increases in the temperature and concentration of reactants bring about more collisions, and the rate of reaction increases. The collision model explains many observations about reactions. Not all molecules that collide react. Colliding molecules must have a minimum amount of energy for a reaction to occur, and this is termed the activation energy. As the temperature increases, the molecules absorb more heat energy, move faster, and when they collide are more likely to meet the activation energy requirement. Therefore, as temperature increases, reaction rate increases. Some substances can cause the reaction rate to increase without increasing the temperature. These substances are called catalysts. Catalysts are useful because they increase the reaction rate without necessitating an increase in temperature or concentration. Many reactions do not continue until all of the reactants have been converted to products. This is because reactions are reversible. A reversible reaction is one where reactants form products, and products can also form reactants. Both the forward and reverse reactions eventually occur at the same rate, so the concentrations of the products and reactants do not change. Equilibrium is dynamic, though, because both reactions (forward and reverse) are always occurring, but the rates are equal. The reaction system must be in a closed container for this to occur. The results of measuring the concentrations of reactants and products for many reversible reactions led scientists to formulate the law of chemical equilibrium. This can be stated mathematically as: aA + bB € cC + dD [C]c [D] d K = [A]a [B] b The letter K is called the equilibrium constant, and the amounts of reactants and products are measured in molarity. The equilibrium constant remains the same as long as the temperature is held constant even though the initial concentrations may change. We can write the equilibrium expression for any equation as long as the equation is balanced. When we write an equilibrium expression for an equilibrium involving solids or pure liquid, we do not include the solid or pure liquid because the concentrations of these are constant.

Thus, the equilibrium expression for

2CO2 (g) € 2CO (g) + O2 (g) is

2 [CO] [O2 ] K = 2 [CO2 ] while the equilibrium expression for

Fe2O3 (s) + 3H2 (g) € 3H2O (g) + 2Fe (s) is

3 [H2 O] K = 3 [H2 ] Le Châtelier’s principle helps us predict what happens to systems at equilibrium when conditions are changed. Le Châtelier’s principle states that when a system at equilibrium is changed, the system will shift its equilibrium position in order to reduce the change. The most common changes are changes in concentration, volume, and temperature. When an equilibrium condition is changed by changing the concentration of a reactant or a product, the system will shift to counteract this change. For example, adding a reactant (or removing a product) will cause more product to be formed. Adding a product (or removing a reactant) will cause more reactant to be formed. Decreasing the volume of a reaction vessel at constant temperature and moles of gas causes an increase in pressure. The gas molecules are in a smaller volume, and they hit the walls more often. Le Châtelier’s principle predicts that the system will act to lower the pressure. How can pressure decrease? By having fewer molecules of gas. For example, consider the equation

2CO2 (g) € 2CO (g) + O2 (g)

If the volume of a reaction mixture with CO2, CO, and O2 is decreased, the system responds by shifting to the left, since three molecules react to form two molecules. K stays the same, but the concentrations change. Temperature causes the value of K to change, but we can still use Le Châtelier’s principle to predict the effect of a change in temperature. To do this, we treat heat just like a product or a reactant. If a reaction is endothermic (absorbs heat), an increase in temperature will shift the reaction to the right. If the reaction is exothermic (gives off heat), an increase in temperature will shift the reaction to the left. Decreasing the temperature has the opposite effect in each case. Ionic compounds dissolving in water can also reach equilibrium. The equilibrium constant for this type of reaction is called the solubility product and given as Ksp. For example, consider the following reaction

3+ − Al(OH)3 (s) € Al (aq) + 3OH (aq) The solubility product is given as

3+ − 3 Ksp = [Al ][OH ]

LEARNING REVIEW 1. Why does an increase in the concentration of reactant cause a reaction to speed up?

2. a. Which letter represents the energy of the products of a reaction? b. Which letter represents the energy of the reactants?

c. Which letter represents the activation energy (Ea) of a reaction? d. Which letter represents the catalyzed reaction pathway? 3. What is a catalyst, and how does it work? 4. A beaker of liquid water in a sealed container is allowed to reach equilibrium vapor pressure. What happens to the concentration of water vapor in the beaker from the time the water is placed in the beaker until equilibrium is reached? 5. Which of the following statements about equilibrium are true? a. After equilibrium is established, the rate of the forward reaction is greater than the rate of the reverse reaction. b. Before equilibrium is reached, the concentration of products increases as time passes. c. Before equilibrium is reached, the concentration of reactants increases as time passes. 6. Write equilibrium constant expressions for each of the reactions below.

a. I2(g) + Cl2(g) € 2ICl(g)

b. 2NO2(g) + 7H2(g) € 2NH3(g) + 4H2O(g)

c. 4NH3(g) + 5O2(g) € 4NO(g) + 6H2O(g) 7. Write the equilibrium expression, and calculate the equilibrium constant, K, for the reaction below at each set of equilibrium concentrations.

− + C2H4O2 € C2H3O2 + H Experiment Equilibrium Concentrations Eq. Expression K

+ − [C2H4O2] [H ] [C2H3O2 ] I 1.0 M 0.0042 M 0.0042 M II 0.50 M 0.0030 M 0.0030 M III 2.0 M 0.0060 M 0.0060 M

8. Why do we not include the concentration of solids or pure liquids in the equilibrium expression? 9. Write the equilibrium constant expression for each of the reactions below.

2+ − a. Pb(OH)2(s) € Pb (aq) + 2OH (aq)

b. 2Sb(s) + 3Cl2(g) € 2SbCl3(g)

c. Fe2O3(s) + 3H2(g) € 3H2O(g) + 2Fe(s) 10. What would happen to the position of the equilibrium when the following changes are made to the equilibrium reaction below?

2SO2(g) + O2(g) € 2SO3(g)

a. SO2 is removed from the reaction vessel.

b. SO3 is added to the reaction vessel. c. Oxygen is removed from the reaction vessel. 11. What will happen to the position of the equilibrium when the following changes are made to the reaction below?

2HgO(s) € Hg(l) + O2(g) a. Solid HgO(s) is added to the reaction vessel. b. The pressure in the reaction vessel is increased by lowering the volume. 12. When the volume of the following mixture of gases is increased, what will be the effect on the equilibrium position?

4HCl(g) + O2(g) € 2H2O(g) + 2Cl2(g) 13. Predict the effect of decreasing the volume of the container on the position of each equilibrium below.

a. SiF4(g) + 2H2O(g) € SiO2(s) + 4HF(g)

b. 2H2(g) + 2NO(g) € 2H2O(g) + N2(g)

c. C(s) + H2O(g) € CO(g) + H2(g) 14. Predict the effect of increasing the temperature on the position of each equilibrium below.

a. H2(g) + Cl2(g) € 2HCl(g) + heat exothermic

b. 2NH3(g) + heat € N2(g) + 3H2(g) endothermic

c. CO2(g) + H2(g) + heat € CO(g) + H2O(g) endothermic

15. If the equilibrium constant for the reaction below is 51.47, the concentration of HI is 0.50 M, and

the concentration of H2 is 0.069 M, what is the concentration of I2?

H2(g) + I2(g) € 2HI(g)

16. Write the balanced equation describing the dissolution of the solids below. Then write the Ksp expression.

a. Ca3(PO4)2 b. FeS

c. Al(OH)3

−4 17. The solubility of PbSO4(s) is 1.3 × 10 mol/L at 25C. What is the Ksp of PbSO4(s)?

−45 18. Copper(II) sulfide has a Ksp of 8.0 × 10 . What is the solubility of CuS(s) in water at 25C? ANSWERS TO LEARNING REVIEW 1. The collision model says for reactions to occur the reactants must first collide with each other. Higher concentrations of reactants cause the reaction rate to speed up because the number of collisions increases as the concentration of reactants increases. 2. a. Letter b represents the energy of the products. In this part of the graph, the reactants have achieved the activation energy, and products have formed. b. Letter a represents the energy of the reactants. The average energy of the reactants is lower than the activation energy. c. Letter c represents the activation energy, the minimum amount of energy needed for a reaction to occur. d. Letter e represents the catalyzed reaction pathway. A catalyst lowers the activation energy for a reaction. 3. A catalyst is a substance added to the reaction that causes the reaction to speed up. Catalysts are not used up during the reaction. They are not reactants and do not form products. A catalyst works by providing a new path for the reaction, which lowers the activation energy for that reaction. 4. When water is first sealed in the beaker, the level of water in the beaker decreases as water from the beaker enters the vapor phase. After some period of time, the level of water stops decreasing and stays at the same level. At this point a balance occurs between the processes of evaporation and condensation. The system is at equilibrium, and the concentration of water vapor does not change. 5. a. This statement is false. At equilibrium the rate of forward reaction equals the rate of the reverse reaction. b. This statement is true. When reactants are mixed, they continue to react to form product. The amount of product increases until equilibrium is reached. c. This statement is false. The concentration of reactants decreases until equilibrium is reached, at which point the concentration of reactants remains constant. 6. When writing equilibrium expressions, coefficients become powers. Products appear in the numerator and reactants in the denominator. [ICl]2 a. K = [I2 ][Cl 2 ]

4 2 [H2 O] [NH 3 ] b. K = 2 7 [NO2 ] [H 2 ]

4 6 [NO] [H2 O] c. K = 4 5 [NH3 ] [O 2 ]

7. Experiment Equilibrium Concentrations Eq. Expression K

+ − [C2H4O2] [H ] [C2H3O2 ] [H+ ][C H O- ] I 1.0M 0.0042M 0.0042M K = 2 3 2 1.8 × 10−5 [C2 H 4 O 2 ] [H+ ][C H O- ] II 0.50M 0.0030M 0.0030M K = 2 3 2 1.8 × 10−5 [C2 H 4 O 2 ] [H+ ][C H O- ] III 2.0M 0.0060M 0.0060M K = 2 3 2 1.8 × 10−5 [C2 H 4 O 2 ]

8. The concentrations of pure solids and liquids are constant and do not change. They are not shown in the equilibrium expression. 9.

2+- 2 a. Ksp = [Pb ][OH ]

2 [SbCl3 ] b. K = 3 [Cl2 ]

3 [H2 O] c. K = 3 [H2 ] 10.

a. The equilibrium will shift to the left and begin producing SO2.

b. The equilibrium will shift to the left and begin consuming SO3.

c. The equilibrium will shift to the left and begin producing O2. 11. a. Pure solids and liquids have no effect on the equilibrium, so adding or removing HgO(s) or Hg(l) will not affect the position of the equilibrium. b. The equilibrium will shift toward the left to reduce the pressure inside the system by consuming oxygen gas. 12. On the left side there are five gaseous molecules, and on the right there are four. When the volume is increased, the pressure decreases, and the equilibrium will shift toward the side that increases the pressure, to the left. 13. Decreasing the volume causes an increase in pressure. The equilibrium shifts in the direction to relieve the pressure, or toward the side with fewer numbers of gaseous molecules. a. The equilibrium would shift toward the left because the left has three gaseous molecules while the right has four gaseous molecules. b. The equilibrium would not shift in either direction because the number of gaseous molecules on the right and the left is the same. c. The equilibrium would shift toward the left because the left has one gaseous molecule and the right has two gaseous molecules. Solid carbon on the left does not affect the equilibrium.

14. For exothermic reactions, treat heat energy as a product of the reaction. If heat energy is added to the system by raising the temperature, then the equilibrium shifts to the left to consume the heat energy. For endothermic reactions, treat heat energy as a reactant. If heat energy is added to the system by raising the temperature, then the equilibrium shifts to the right to consume the heat energy. a. This is an exothermic reaction. Raising the temperature would shift the equilibrium to the left to consume the energy. b. This is an endothermic reaction. Raising the temperature would shift the equilibrium to the right to consume excess energy. c. This is an endothermic reaction. Raising the temperature would shift the equilibrium to the right to consume the energy. 15. First write the equilibrium expression for this reaction. [HI]2 K = [H2 ][I 2 ]

2 Rearrange this equation to isolate I2 on one side. Divide both sides by [HI] . K [HI]2 1 2 = 2 [HI] [HI] [H2 ][I 2 ] K 1 2 = [HI] [H2 ][I 2 ]

Now, multiply both sides by [H2].

K[H2 ] 1 2 = [HI] [I2 ] Take the inverse of both sides. [HI]2 [I2 ] = K[H2 ]

Substitute values into the equation and find [I2]. [0.25 mol/L]2 [I ] = 2 51.47[0.069 mol/L]

[I2] = 0.070 mol/L

16. The Ksp expression does not include the concentration of the solid salt, only the ions in solution.

2+ 3 3− 2 a. Ksp = [Ca ] [PO4 ]

2+ 2− b. Ksp = [Fe ][S ]

3+ − 3 c. Ksp = [A1 ][OH ]

17. This problem asks us to calculate the Ksp for PbSO4 given the solubility of PbSO4. When PbSO4 dissolves in water, lead ions and sulfate ions are released into the aqueous environment.

2+ 2− PbSO4(s) € Pb (aq) + SO4 (aq)

The Ksp expression for the reaction is

2+ 2− Ksp = [Pb ][SO4 ]

2+ 2− We need to know the molar concentrations of Pb and SO4 to find the Ksp. We know that the −4 −4 solubility of PbSO4 is 1.3 × 10 mol/L. This means that 1.3 × 10 mol PbSO4 dissolves per liter of −4 −4 2+ −4 2− solution. Each 1.3 × 10 mol PbSO4 produces 1.3 × 10 mol Pb and 1.3 × 10 mol SO4 . The 2+ −4 2− −4 concentration of Pb is 1.3 × 10 mol/L, and the concentration of SO4 is 1.3 × 10 mol/L. We can use these concentrations to calculate the Ksp.

2+ 2− Ksp = [Pb ][SO4 ]

−4 −4 Ksp = (1.3 × 10 mol/L)(1.3 × 10 mol/L)

−8 2 2 Ksp = 1.7 × 10 mol /L

−8 The units for Ksp are usually omitted, so the Ksp would be reported as 1.7 × 10 .

18. This problem asks us to find the solubility of copper(II) sulfide. We are given the Ksp, which is 8.0 × 10−45. When CuS dissolves in water, each mole of CuS that dissolves produces one mole of Cu2+ and one mole of S2−. CuS(s) € Cu2+(aq) + S2− (aq)

The Ksp expression for the dissolution of CuS in water is

2+ 2− Ksp = [Cu ][S ] mol mol mol x of CuS(s) dissociates into x Cu2+(aq) and x S2−(aq). L L L mol mol At equilibrium [Cu2+] = x and [S2−] = x . L L We can substitute these values into the equilibrium expression.

−45 2+ 2− 2 Ksp = 8.0 × 10 = [Cu ][S ] = (x)(x) = x We can say that x2 = 8.0 × 10−45

x = 8.0 10-45 = 8.9 × 10−23 mol/L The solubility of CuS is 8.9 × 10−23 mol/L.