1. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu How to Use This Presentation To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.” To advance through the presentation, click the right-arrow key or the space bar. From the resources slide, click on any resource to see a presentation for that resource. From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. You may exit the slide show at any time by pressing the Esc key.

2. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter Presentation Transparencies Bellringer Standardized Test PrepVisual Concepts Sample Problems Resources

3. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Table of Contents Chapter 14 Chemical Equilibrium Section 1 Reversible Reactions and Equilibrium Section 2 Systems at Equilibrium Section 3 Equilibrium Systems and Stress

4. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Bellringer Describe what reversible means. Find a synonym for reversible. Section 1 Reversible Reactions and Equilibrium Chapter 14

5. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Contrast reactions that go to completion with reversible ones. Describe chemical equilibrium. Give examples of chemical equilibria that involve complex ions. Section 1 Reversible Reactions and Equilibrium Chapter 14

6. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions If enough oxygen gas is provided for the following reaction, almost all of the sulfur will react: S 8 (s) + 8O 2 (g) → 8SO 2 (g) Reactions such as this one, in which almost all of the reactants react, are called completion reactions. In other reactions, called reversible reactions, the products can re-form reactants. Section 1 Reversible Reactions and Equilibrium Chapter 14

7. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium One reversible reaction occurs when you mix solutions of calcium chloride and sodium sulfate. CaCl 2 (aq) + Na 2 SO 4 (aq) → CaSO 4 (s) + 2NaCl(aq) The net ionic equation best describes what happens. Section 1 Reversible Reactions and Equilibrium Chapter 14

8. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium, continued Solid calcium sulfate, the product, can break down to make calcium ions and sulfate ions in a reaction that is the reverse of the previous one. Section 1 Reversible Reactions and Equilibrium Chapter 14 Use arrows that point in opposite directions when writing a chemical equation for a reversible reaction.

9. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium, continued The reactions occur at the same rate after the initial mixing of CaCl 2 and Na 2 SO 4. The amounts of the products and reactants do not change. Chemical equilibrium is a state of balance in which the rate of a forward reaction equals the rate of the reverse reaction and the concentrations of products and reactants remain unchanged. Section 1 Reversible Reactions and Equilibrium Chapter 14

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium The reaction of hydrogen, H 2, and iodine, I 2, to form hydrogen iodide, HI, reaches chemical equilibrium. Section 1 Reversible Reactions and Equilibrium Chapter 14 Only a very small fraction of the collisions between H 2 and I 2 result in the formation of HI. H 2 (g) + I 2 (g) → 2HI(g)

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued After some time, the concentration of HI goes up. As a result, fewer collisions occur between H 2 and I 2 molecules, and the rate of the forward reaction drops. Similarly, in the beginning, few HI molecules exist in the system, so they rarely collide with each other. Section 1 Reversible Reactions and Equilibrium Chapter 14

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued As more HI molecules are made, they collide more often and form H 2 and I 2 by the reverse reaction. 2HI(g) → H 2 (g) + I 2 (g) The greater the number of HI molecules that form, the more often the reverse reaction occurs. Section 1 Reversible Reactions and Equilibrium Chapter 14

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Rate Comparison for H 2 (g) + I 2 (g)  2HI(g) Section 1 Reversible Reactions and Equilibrium Chapter 14

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued When the forward rate and the reverse rate are equal, the system is at chemical equilibrium. If you repeated this experiment at the same temperature, starting with a similar amount of pure HI instead of the H 2 and I 2, the reaction would reach chemical equilibrium again and produce the same concentrations of each substance. Section 1 Reversible Reactions and Equilibrium Chapter 14

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Equilibrium Visual Concepts Chapter 14

16. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Equilibria Are Dynamic If you drop a ball into a bowl, it will bounce. When the ball comes to a stop it has reached static equilibrium, a state in which nothing changes. Chemical equilibrium is different from static equilibrium because it is dynamic. In a dynamic equilibrium, there is no net change in the system. Two opposite changes occur at the same time. Section 1 Reversible Reactions and Equilibrium Chapter 14

17. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Equilibria Are Dynamic, continued In equilibrium, an atom may change from being part of the products to part of the reactants many times. But the overall concentrations of products and reactants stay the same. For chemical equilibrium to be maintained, the rates of the forward and reverse reactions must be equal. Arrows of equal length also show equilibrium. Section 1 Reversible Reactions and Equilibrium Chapter 14

18. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chemical Equilibria Are Dynamic, continued In some cases, the equilibrium has a higher concentration of products than reactants. This type of equilibrium is also shown by using two arrows. Section 1 Reversible Reactions and Equilibrium Chapter 14 The forward reaction has a longer arrow to show that the products are favored.

19. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria Even when systems are not in equilibrium, they are continuously changing in order to reach equilibrium. For example, combustion produces carbon dioxide, CO 2, and poisonous carbon monoxide, CO. After combustion, a reversible reaction produces soot. Section 1 Reversible Reactions and Equilibrium Chapter 14 This reaction of gases and a solid will reach chemical equilibrium. Equilibria can involve any state of matter, including aqueous solutions.

20. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria, continued Equilibria Involving Complex Ions Complex ion, or coordination compound, is the name given to any metal atom or ion that is bonded to more than one atom or molecule. Some ions have a metal ion surrounded by ligands, molecules or anions that readily bond to metal ions. Complex ions may be positively charged cations or negatively charged anions. Section 1 Reversible Reactions and Equilibrium Chapter 14

21. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued In this complex ion, [Cu(NH 3 ) 4 ] 2+, ammonia molecules bond to the central copper(II) ion. Section 1 Reversible Reactions and Equilibrium Chapter 14

22. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued Complex ions formed from transition metals are often deeply colored. Section 1 Reversible Reactions and Equilibrium Chapter 14

23. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued The charge on a complex ion is a sum of the charges on the species from which the complex ion forms. For example, when the cobalt ion, Co 2+, bonds with four Cl − ligands, the total charge is (+2) + 4(−1) = −2. Metal ions and ligands can form complexes that have no charge. These are not complex ions. Complex ions often form in systems that reach equilibrium. Section 1 Reversible Reactions and Equilibrium Chapter 14

24. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued Consider zinc nitrate dissolving in water: Section 1 Reversible Reactions and Equilibrium Chapter 14 In the absence of other ligands, water molecules bond with zinc ions. So, this reaction can be written: If another ligand, such as CN −, is added, the new system will again reach chemical equilibrium.

25. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued Both water molecules and cyanide ions “compete” to bond with zinc ions, as shown in the equation below. Section 1 Reversible Reactions and Equilibrium Chapter 14 All of these ions are colorless, so you cannot see which complex ion has the greater concentration.

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued In the chemical equilibrium of nickel ions, ammonia, and water, the complex ions have different colors. You can see which ion has the greater concentration. Section 1 Reversible Reactions and Equilibrium Chapter 14 green blue-violet The starting concentration of NH 3 will determine which one will have the greater concentration.

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Systems at Equilibrium Bellringer Make a list of numbers that are “constants” under constant conditions. An example is the speed of light. What do these constants have in common? Answer: Each constant is always the same number for a certain and constant set of conditions. Chapter 14

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives Write K eq expressions for reactions in equilibrium, and perform calculations with them. Write K sp expressions for the solubility of slightly soluble salts, and perform calculations with them. Section 2 Systems at Equilibrium Chapter 14

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq Limestone caverns form as rainwater, slightly acidified by H 3 O +, dissolves calcium carbonate. The reverse reaction also takes place, depositing calcium carbonate and forming stalactites and stalagmites. Section 2 Systems at Equilibrium Chapter 14 When the rates of the forward and reverse reactions become equal, the reaction reaches chemical equilibrium.

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq, continued There is a mathematical relationship between product and reactant concentrations at equilibrium. For limestone reacting with acidified water at 25°C: Section 2 Systems at Equilibrium Chapter 14 K eq is the equilibrium constant of the reaction. K eq for a reaction is unitless, applies only to systems in equilibrium, and depends on temperature and must be found experimentally or from tables.

31. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Equilibrium Constant Visual Concepts Chapter 14

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq, continued Determining K eq for Reactions at Chemical Equilibrium 1.Write a balanced chemical equation. Make sure that the reaction is at equilibrium before you write a chemical equation. 2.Write an equilibrium expression. Section 2 Systems at Equilibrium Chapter 14 To write the expression, place the product concentrations in the numerator and the reactant concentrations in the denominator.

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq, continued Determining K eq for Reactions at Chemical Equilibrium, continued The concentration of any solid or a pure liquid that takes part in the reaction is left out. For a reaction occurring in aqueous solution, water is omitted. 3.Complete the equilibrium expression. Section 2 Systems at Equilibrium Chapter 14 Finally, raise each substance’s concentration to the power equal to the substance’s coefficient in the balanced chemical equation.

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Equilibrium Constant Visual Concepts Chapter 14

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating K eq from Concentrations of Reactants and Products Sample Problem A An aqueous solution of carbonic acid reacts to reach equilibrium as described below. Chapter 14 Section 2 Systems at Equilibrium The solution contains the following solution concentrations: carbonic acid, 3.3 × 10 −2 mol/L; bicarbonate ion, 1.19 × 10 −4 mol/L; and hydronium ion, 1.19 × 10 −4 mol/L. Determine the K eq.

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating K eq from Concentrations of Reactants and Products Sample Problem A Solution Chapter 14 Section 2 Systems at Equilibrium Substitute the concentrations into the expression. For this reaction, the equilibrium constant expression is

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq, continued K eq Shows If the Reaction Is Favorable When K eq is large, the numerator of the equilibrium constant expression is larger than the denominator. Thus, the concentrations of the products will usually be greater than those of the reactants. In other words, when a reaction that has a large K eq reaches equilibrium, there will be mostly products. Reactions in which more products form than reactants form are said to be “favorable.” Section 2 Systems at Equilibrium Chapter 14

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq, continued K eq Shows If the Reaction Is Favorable, continued The synthesis of ammonia is very favorable at 25°C and has a large K eq value. Section 2 Systems at Equilibrium Chapter 14

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq, continued K eq Shows If the Reaction Is Favorable, continued However, the reaction of oxygen and nitrogen to give nitrogen monoxide is not favorable at 25°C. Section 2 Systems at Equilibrium Chapter 14

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu When K eq is small, the denominator of the equilibrium constant expression is larger than the numerator. The larger denominator shows that the concentrations of reactants at chemical equilibrium may be greater than those of products. A reaction that has larger concentrations of reactants than concentrations of products is an “unfavorable” reaction. Section 2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, K eq, continued K eq Shows If the Reaction Is Favorable, continued

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Equilibrium Constant, K eq, continued Keq Shows If the Reaction Is Favorable, continued These pie charts show the relative amounts of reactants and products for three K eq values of a reaction. Section 2 Systems at Equilibrium Chapter 14

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Concentrations of Products from K eq and Concentrations of Reactants Sample Problem B K eq for the equilibrium below is 1.8 × 10 −5 at a temperature of 25°C. Calculate when [NH 3 ] = 6.82 × 10 −3. Chapter 14 Section 2 Systems at Equilibrium

43. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Concentrations of Products from K eq and Concentrations of Reactants, continued Sample Problem B Solution The equilibrium expression is Chapter 14 Section 2 Systems at Equilibrium and OH − ions are produced in equal numbers, so So, the numerator can be written as x 2.

44. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem B Solution, continued K eq and [NH 3 ] are known and can be put into the expression. Calculating Concentrations of Products from K eq and Concentrations of Reactants, continued Chapter 14 Section 2 Systems at Equilibrium x 2 = (1.8  10 −5 )  (6.82  10 −3 ) = 1.2 × 10 −7

45. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Concentrations of Products from K eq and Concentrations of Reactants, continued Sample Problem B Solution, continued Take the square root of x 2. Chapter 14 Section 2 Systems at Equilibrium

46. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Solubility Product Constant, K sp The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water. Solubilities can be expressed in moles of solute per liter of solution (mol/L or M). For example, the solubility of calcium fluoride in water is 3.4 × 10 −4 mol/L. So, 0.00034 mol of CaF 2 will dissolve in 1 L of water to give a saturated solution. If you try to dissolve 0.00100 mol of CaF 2 in 1 L of water, 0.00066 mol of CaF 2 will remain undissolved. Section 2 Systems at Equilibrium Chapter 14

47. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solution Equilibrium Visual Concepts Chapter 14

48. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Solubility Product Constant, K sp, continued Calcium fluoride is one of a large class of salts that are said to be slightly soluble in water. The ions in solution and any solid salt are at equilibrium. Section 2 Systems at Equilibrium Chapter 14 Solids are not a part of equilibrium constant expressions, so K eq for this reaction is the product of [Ca 2+ ] and [F − ] 2, which is equal to a constant.

49. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Solubility Product Constant, K sp, continued Equilibrium constants for the dissolution of slightly soluble salts are called solubility product constants, K sp, and have no units. The K sp for calcium fluoride at 25°C is 1.6  10 −10. K sp = [Ca 2+ ][F − ] 2 = 1.6  10 −10 This relationship is true whenever calcium ions and fluoride ions are in equilibrium with calcium fluoride, not just when the salt dissolves. Section 2 Systems at Equilibrium Chapter 14

50. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Solubility Product Constant, K sp, continued For example, if you mix solutions of calcium nitrate and sodium fluoride, calcium fluoride precipitates. The net ionic equation for this precipitation is the reverse of the dissolution. Section 2 Systems at Equilibrium Chapter 14 This equation is the same equilibrium. So, the K sp for the dissolution of CaF 2 in this system is the same and is 1.6 × 10 −10.

51. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility Product Constant Visual Concepts Chapter 14

52. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Solubility Product Constant Visual Concepts Chapter 14

53. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Solubility Product Constant, K sp, continued Section 2 Systems at Equilibrium Chapter 14

54. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1.Write a balanced chemical equation. Solubility product is only for salts that have low solubility. Soluble salts do not have K sp values. Make sure that the reaction is at equilibrium. Equations are always written so that the solid salt is the reactant and the ions are products. Section 2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, K sp, continued Determining K sp for Reactions at Chemical Equilibrium

55. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 2.Write a solubility product expression. Write the product of the ion concentrations. Concentrations of solids or liquids are omitted. 3.Complete the solubility product expression. Section 2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, K sp, continued Determining K sp for Reactions at Chemical Equilibrium Raise each concentration to a power equal to the substance’s coefficient in the balanced chemical equation.

56. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Using Solubility Product Constants Visual Concepts Chapter 14

57. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating K sp from Solubility Sample Problem C Most parts of the oceans are nearly saturated with CaF 2.The mineral fluorite, CaF 2, may precipitate when ocean water evaporates. A saturated solution of CaF 2 at 25°C has a solubility of 3.4 × 10 −4 M. Calculate the solubility product constant for CaF 2. Chapter 14 Section 2 Systems at Equilibrium

58. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu [CaF 2 ] = 3.4  10 –4, [F − ] = 2[Ca 2+ ] K sp = [Ca 2+ ][F − ] 2 Because 3.4 × 10 −4 mol CaF 2 dissolves in each liter of solution, you know from the balanced equation that every liter of solution will contain 3.4 × 10 −4 mol Ca 2+ and 6.8 × 10 −4 mol F −. Thus, the K sp is: [Ca 2+ ][F − ] 2 = (3.4  10 −4 )(6.8  10 −4 ) 2 = 1.6 × 10 −10 Calculating K sp from Solubility, continued Sample Problem C Solution Chapter 14 Section 2 Systems at Equilibrium

59. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Calculating Ionic Concentrations Using K sp, Sample Problem D Copper(I) chloride has a solubility product constant of 1.2 × 10 −6 and dissolves according to the equation below. Calculate the solubility of this salt in ocean water in which the [Cl − ] = 0.55. Chapter 14 Section 2 Systems at Equilibrium

60. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem D Solution The product of [Cu + ][Cl − ] must equal K sp = 1.2 × 10 −6. [Cl − ] = 0.55 K sp = [Cu + ][Cl − ] = 1.2 × 10 −6 Calculating Ionic Concentrations Using K sp, continued Chapter 14 Section 2 Systems at Equilibrium This is the solubility of copper(I) chloride because the dissolution of 1 mol of CuCl produces 1 mol of Cu +. Therefore, the solubility of CuCl is 2.2 × 10 −6 mol/L.

61. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Though slightly soluble hydroxides are not salts, they have solubility product constants. Magnesium hydroxide is an example. Section 2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, K sp, continued Using K sp to Make Magnesium [Mg 2+ ][OH − ] 2 = K sp = 1.8 × 10 −11 This equilibrium is the basis for obtaining magnesium.

62. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The table at right lists the most abundant ions in ocean water and their concentrations. Mg 2+ is the third most abundant ion in the ocean. Section 2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, K sp, continued Using K sp to Make Magnesium, continued

63. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu To get magnesium, calcium hydroxide is added to sea water, which raises the hydroxide ion concentration to a large value so that [Mg 2+ ][OH − ] 2 would be greater than 1.8 × 10 −11. Magnesium hydroxide precipitates. Magnesium hydroxide is treated with hydrochloric acid to make magnesium chloride, MgCl 2. Section 2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, K sp, continued Using K sp to Make Magnesium, continued

64. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Finally, magnesium is obtained by the electrolysis of MgCl 2 in the molten state. One cubic meter of sea water yields 1 kg of magnesium metal. Because of magnesium’s low density and rigidity, alloys of magnesium are used when light weight and strength are needed. Section 2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, K sp, continued Using K sp to Make Magnesium, continued

65. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 Equilibrium Systems and Stress Bellringer List examples of everyday adjustments that are made to relieve stress on a system. Chapter 14

66. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Objectives State Le Châtelier’s principle. Apply Le Châtelier’s principle to determine whether the forward or reverse reaction is favored when a stress such as concentration, temperature, or pressure is applied to an equilibrium system. Discuss the common-ion effect in the context of Le Châtelier’s principle. Discuss practical uses of Le Châtelier’s principle. Section 3 Equilibrium Systems and Stress Chapter 14

67. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle Stress is another word for something that causes a change in a system at equilibrium. Chemical equilibrium can be disturbed by a stress, but the system soon reaches a new equilibrium. Le Châtelier’s principle states that when a system at equilibrium is disturbed, the system adjusts in a way to reduce the change. Section 3 Equilibrium Systems and Stress Chapter 14

68. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Chemical equilibria respond to three kinds of stress: changes in the concentrations of reactants or products changes in temperature changes in pressure When a stress is first applied to a system, equilibrium is disturbed and the rates of the forward and backward reactions are no longer equal. Section 3 Equilibrium Systems and Stress Chapter 14

69. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued The system responds to the stress by forming more products or by forming more reactants. A new chemical equilibrium is reached when enough reactants or products form. At this point, the rates of the forward and backward reactions are equal again. Section 3 Equilibrium Systems and Stress Chapter 14

70. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Le Châtelier's Principle Chapter 14

71. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition If you increase a reactant’s concentration, the system will respond to decrease the concentration of the reactant by changing some of it into product. Therefore, the rate of the forward reaction must be greater than the rate of the reverse reaction. The equilibrium is said to shift right, and the reactant concentration drops until the reaction reaches equilibrium. Section 3 Equilibrium Systems and Stress Chapter 14

72. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued In a reaction of two colored complex ions: Section 3 Equilibrium Systems and Stress Chapter 14 When the reaction mixture in a beaker is pale blue, we know that chemical equilibrium favors the formation of reactants. pale blue blue-purple

73. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued If ammonia is added, the system responds by forming more product and the solution becomes blue-purple. Section 3 Equilibrium Systems and Stress Chapter 14

74. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued The equilibrium below occurs in a bottle of soda. Section 3 Equilibrium Systems and Stress Chapter 14 After you uncap the bottle, the dissolved carbon dioxide leaves the solution and enters the air. The forward reaction rate of this system will increase to produce more CO 2.

75. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued This increase in the rate of the forward reaction decreases the concentration of H 3 O +. As a result, the drink gets “flat.” If you could increase the concentration of CO 2 in the bottle, the reverse reaction rate would increase, and would increase. Section 3 Equilibrium Systems and Stress Chapter 14

76. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems The effect of temperature on the gas-phase equilibrium of nitrogen dioxide, NO 2, and dinitrogen tetroxide, N 2 O 4, can be seen because of the difference in color of NO 2 and N 2 O 4. The intense brown NO 2 gas is the pollution that is responsible for the colored haze that you sometimes see on smoggy days. Section 3 Equilibrium Systems and Stress Chapter 14

77. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued Recall that endothermic reactions absorb energy and have positive ∆H values. Exothermic reactions release energy and have negative ∆H values. The forward reaction is an exothermic process, as the equation below shows. 2NO 2 (g) → N 2 O 4 (g)∆H = −55.3 kJ Section 3 Equilibrium Systems and Stress Chapter 14

78. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Temperature Changes Affect an Equilibrium System Section 3 Equilibrium Systems and Stress Chapter 14

79. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued Now suppose that you heat the flask to 100°C. The mixture becomes dark brown because the reverse reaction rate increased to remove some of the energy that you added to the system. The equilibrium shifts to the left, toward the formation of NO 2. Because this reaction is endothermic, the temperature of the flask drops as energy is absorbed. Section 3 Equilibrium Systems and Stress Chapter 14

80. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued This equilibrium shift is true for all exothermic forward reactions: Increasing the temperature of an equilibrium mixture usually leads to a shift in favor of the reactants. The opposite statement is true for endothermic forward reactions: Increasing the temperature of an equilibrium mixture usually leads to a shift in favor of the products. Section 3 Equilibrium Systems and Stress Chapter 14

81. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued The following is an endothermic reaction involving two colored cobalt complex ions: Section 3 Equilibrium Systems and Stress Chapter 14 The forward reaction is endothermic, so the forward reaction is favored at high temperatures. The reverse reaction is favored at low temperatures.

82. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued Equilibrium changes with changes in temperature they affect the value of equilibrium constants. Consider K eq for the ammonia synthesis equilibrium: Section 3 Equilibrium Systems and Stress Chapter 14 The forward reaction is exothermic (∆H = −91.8 kJ), so the equilibrium constant decreases a lot as temperature increases.

83. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Visual Concepts Factors Affecting Equilibrium Chapter 14

84. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium Changes in pressure can affect gases in equilibrium. The NO 2 and N 2 O 4 equilibrium can show the effect of a pressure stress on a chemical equilibrium. 2NO 2 (g) → N 2 O 4 (g) When the gas mixture has a larger concentration of N 2 O 4 than of NO 2, it has a pale color. Section 3 Equilibrium Systems and Stress Chapter 14

85. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium When the gas is suddenly compressed to about half its former volume, the pressure doubles. Section 3 Equilibrium Systems and Stress Chapter 14

86. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium, continued Before the system has time to adjust to the pressure stress, the concentration of each gas doubles, which results in a darker color. Le Châtelier’s principle predicts that the system will adjust in an attempt to reduce the pressure. According to the equation, 2 mol of NO 2 produce 1 mol of N 2 O 4. Section 3 Equilibrium Systems and Stress Chapter 14

87. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium, continued At constant volume and temperature, pressure is proportional to the number of moles. So, the pressure reduces when there are fewer moles of gas. Thus, the equilibrium shifts to the right, and more N 2 O 4 is produced. Section 3 Equilibrium Systems and Stress Chapter 14

88. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium, continued In an equilibrium, a pressure increase favors the reaction that produces fewer gas molecules, which for the following equilibrium is the reverse reaction. Section 3 Equilibrium Systems and Stress Chapter 14 When there is no change in the number of molecules, a change in pressure will not affect equilibrium.

89. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Equilibrium Visual Concepts Chapter 14

90. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Common-Ion Effect The solubility of CuCl in water is 1.1  10 −3 mol/L. The solubility of CuCl in sea water is 2.2  10 −6 mol/L CuCl is 500 times less soluble in sea water. The dramatic reduction in solubility of CuCl shows Le Châtelier’s principle and is described below. Section 3 Equilibrium Systems and Stress Chapter 14 [Cu + ][Cl − ] = K sp = 1.2 × 10 −6

91. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Common-Ion Effect, continued If you add chloride-rich ocean water to a saturated solution of copper(I) chloride, [Cl − ] increases. However, the K sp, or [Cl − ]  [Cu + ], remains constant. [Cu + ] must decrease. This decrease can occur only by the precipitation of the CuCl salt. The ion Cl − is the common-ion in this case. The reduction of the solubility of a salt in the solution due to the addition of a common ion is called the common-ion effect. Section 3 Equilibrium Systems and Stress Chapter 14

92. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Common–Ion Effect Visual Concepts Chapter 14

93. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Common-Ion Effect, continued Doctors use solutions of barium sulfate, BaSO 4, to diagnose problems in the digestive tract. BaSO 4 is in equilibrium with a small concentration of Ba 2+ (aq), a poison. To reduce the Ba 2+ concentration to a safe level, a common ion is added. Doctors add sodium sulfate, Na 2 SO 4, into the BaSO 4 solution that a patient must swallow. Section 3 Equilibrium Systems and Stress Chapter 14

94. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu The Common-Ion Effect, continued Another example of the common-ion effect emerges when solutions of potassium sulfate and calcium sulfate are mixed and either solution is saturated. Immediately after the two solutions are mixed, the product of the is greater than the K sp of calcium sulfate. Section 3 Equilibrium Systems and Stress Chapter 14 So, CaSO 4 precipitates to establish equilibrium:

95. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Practical Uses of Le Châtelier’s Principle The chemical industry makes use of Le Châtelier’s principle in the synthesis of ammonia by the Haber Process. High pressure is used to drive the following equilibrium to the right. Section 3 Equilibrium Systems and Stress Chapter 14 The forward reaction converts 4 mol of gas into 2 mol of another gas, so it is favored at high pressures.

96. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Practical Uses of Le Châtelier’s Principle, continued The ammonia synthesis is an exothermic reaction, so the forward reaction is favored at low temperatures. 0°CK eq = 6.5 × 10 8 250°CK eq = 52 500°CK eq = 5.8 × 10 −2 The Haber Process is operated at temperatures of 500°C even though the K eq is small at that temperature. The reaction proceeds too slowly at lower temperatures. Section 3 Equilibrium Systems and Stress Chapter 14

97. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1.In which of these reactions is the formation of the products favored by an increase in pressure? A.2O 3 (g)  3O 2 (g) B.C(s) + O 2 (g)  CO 2 (g) C.2NO(g) + O 2 (g)  2NO 2 (g) D. Standardized Test Preparation Understanding Concepts Chapter 14

98. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1.In which of these reactions is the formation of the products favored by an increase in pressure? A.2O 3 (g)  3O 2 (g) B.C(s) + O 2 (g)  CO 2 (g) C.2NO(g) + O 2 (g)  2NO 2 (g) D. Standardized Test Preparation Understanding Concepts Chapter 14

99. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 2.What is the effect of an increase in temperature on an exothermic reaction at equilibrium? F.It has no effect on the equilibrium. G.It shifts the equilibrium in favor of the forward reaction. H.It shifts the equilibrium in favor of the reverse reaction. I.It shifts the equilibrium in favor of both the forward and reverse reactions. Standardized Test Preparation Understanding Concepts Chapter 14

100. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 2.What is the effect of an increase in temperature on an exothermic reaction at equilibrium? F.It has no effect on the equilibrium. G.It shifts the equilibrium in favor of the forward reaction. H.It shifts the equilibrium in favor of the reverse reaction. I.It shifts the equilibrium in favor of both the forward and reverse reactions. Standardized Test Preparation Understanding Concepts Chapter 14

101. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3.Which of the following properties of a reactant is included in a K eq equation? A.charge B.concentration C.mass D.volume Standardized Test Preparation Understanding Concepts Chapter 14

102. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3.Which of the following properties of a reactant is included in a K eq equation? A.charge B.concentration C.mass D.volume Standardized Test Preparation Understanding Concepts Chapter 14

103. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 4.Explain how the common ion effect is involved in the addition of NaIO 3 to a solution of Cd(IO 3 ) 2, which is much less soluble than NaIO 3. Standardized Test Preparation Understanding Concepts Chapter 14

104. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 4.Explain how the common ion effect is involved in the addition of NaIO 3 to a solution of Cd(IO 3 ) 2, which is much less soluble than NaIO 3. Answer: Some of the Cd(IO 3 ) 2 would precipitate out of the solution, reducing the cadmium ion concentration. Standardized Test Preparation Understanding Concepts Chapter 14

105. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 5.Explain how pressure can be used to maximize the production of carbon dioxide in the reaction Standardized Test Preparation Understanding Concepts Chapter 14

106. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 5.Explain how pressure can be used to maximize the production of carbon dioxide in the reaction Answer: Increasing the pressure will cause the reaction to favor the production of carbon dioxide because there are three gas molecules in the reactants and only two in the products. Standardized Test Preparation Understanding Concepts Chapter 14

107. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Read the passage below. Then answer the questions. Coral reefs are made by tiny organisms known as polyps. They attach themselves permanently in one place and survive by eating tiny marine animals that swim past. The polyps secrete calcium carbonate to make their shells or skeletons. When the polyps die, the calcium carbonate structures remain and accumulate over time to form a reef. This reef building is possible because calcium carbonate is only slightly soluble in water. Standardized Test Preparation Reading Skills Chapter 14

108. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 6.What is the K sp expression for calcium carbonate? F. G. H. I. Standardized Test Preparation Reading Skills Chapter 14

109. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 6.What is the K sp expression for calcium carbonate? F. G. H. I. Standardized Test Preparation Reading Skills Chapter 14

110. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 7.What is the most likely source of calcium used by the polyps to build their shells? A.calcium atoms in solution in sea water B.calcium ions in solution in sea water C.calcium particles that reach the water in acid rain D.calcium containing rocks gathered from the ocean floor Reading Skills Chapter 14 Standardized Test Preparation

111. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 7.What is the most likely source of calcium used by the polyps to build their shells? A.calcium atoms in solution in sea water B.calcium ions in solution in sea water C.calcium particles that reach the water in acid rain D.calcium containing rocks gathered from the ocean floor Reading Skills Chapter 14 Standardized Test Preparation

112. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 8.The K sp for calcium carbonate is 2.8  10 -9 and the K sp value for calcium sulfate is 9.1  10 -6. If coral polyps secreted calcium sulfate rather than calcium carbonate, how would this affect the formation of the coral reef? Standardized Test Preparation Reading Skills Chapter 14

113. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 8.The K sp for calcium carbonate is 2.8  10 -9 and the K sp value for calcium sulfate is 9.1  10 -6. If coral polyps secreted calcium sulfate rather than calcium carbonate, how would this affect the formation of the coral reef? Answer: A reef would not be built as quickly if polyps secreted calcium sulfate, because calcium sulfate has a higher solubility in water. Standardized Test Preparation Chapter 14 Reading Skills

114. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Use the information from the table below to answer questions 9 through 12. Standardized Test Preparation Interpreting Graphics Chapter 14

115. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 9.What is the concentration of magnesium carbonate in a saturated aqueous solution? F.0.0000068 M G.0.0026 M H.0.84 M I.1.31 M Standardized Test Preparation Interpreting Graphics Chapter 14

116. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 9.What is the concentration of magnesium carbonate in a saturated aqueous solution? F.0.0000068 M G.0.0026 M H.0.84 M I.1.31 M Standardized Test Preparation Interpreting Graphics Chapter 14

117. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 10.If a saturated solution of silver carbonate is mixed with a saturated solution of zinc sulfide, a precipitate forms. What compound precipitates? A.Ag 2 CO 3 B.Ag 2 S C.ZnCO 3 D.ZnS Standardized Test Preparation Interpreting Graphics Chapter 14

118. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 10.If a saturated solution of silver carbonate is mixed with a saturated solution of zinc sulfide, a precipitate forms. What compound precipitates? A.Ag 2 CO 3 B.Ag 2 S C.ZnCO 3 D.ZnS Interpreting Graphics Chapter 14 Standardized Test Preparation

119. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 11.Calculate the concentration of S 2– ions in a saturated solution of FeS that contains 0.010 M Fe 2+ ions. Standardized Test Preparation Interpreting Graphics Chapter 14

120. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 11.Calculate the concentration of S 2– ions in a saturated solution of FeS that contains 0.010 M Fe 2+ ions. Answer: 1.6  10 –17 M Standardized Test Preparation Interpreting Graphics Chapter 14

121. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 12.What will happen if a solution containing 1  10 –10 M Na 2 CO 3 is mixed with an equal volume of a solution containing 1  10 –10 M MgCl 2 and 1  10 –10 M ZnCl 2 ? F.No precipitate will form. G.MgCO 3 will precipitate out of the solution. H.ZnCO 3 will precipitate out of the solution. I.MgCO 3 and ZnCO 3 will precipitate out of the solution. Standardized Test Preparation Interpreting Graphics Chapter 14

122. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 12.What will happen if a solution containing 1  10 –10 M Na 2 CO 3 is mixed with an equal volume of a solution containing 1  10 –10 M MgCl 2 and 1  10 –10 M ZnCl 2 ? F.No precipitate will form. G.MgCO 3 will precipitate out of the solution. H.ZnCO 3 will precipitate out of the solution. I.MgCO 3 and ZnCO 3 will precipitate out of the solution. Standardized Test Preparation Interpreting Graphics Chapter 14