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.

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu

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

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

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

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

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.

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

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)

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

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

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

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Reaching an equilibrium state. In most reversible reactions balance points exist between the forward and backward reactions. Reactants and products appear together The reaction appears to have stopped Neither forward or backward reaction is complete This is chemical equilibrium

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Dynamic Equilibrium In dynamic equilibrium the forward and backwards reactions continue at equal rates so the overall effect does not change. On a molecular scale there is continuous change. On the macroscopic scale nothing appears to be happening. The system needs to be closed – isolated from the outside world.

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.

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.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Homework 1.Read pgs Solve Section Review pg.501, Qs. 1-4 Homework 1.Read pgs Solve Section Review pg.501, Qs. 1-4 Chapter 14 Section 1 Reversible Reactions and Equilibrium

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section Review, pg Which of the following equations best represents a reaction that goes to completion? a. Reactants products b. reactants products c. Reactants products 2. At equilibrium, what is the relationship between the rates of the forward and reverse reactions? The rate of the reactions are equal.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3. Explain what each reaction below shows. Describe the relationship between the amounts of products and reactants for each case. a. reactants products b. Reactants products a.The reaction is reversible and at equilibrium. The relative amounts of products and reactants cannot be determined. b. This reversible reaction is also at equilibrium; however, there are more products than reactants. 4. What is the difference between dynamic and static equilibria? During static equilibrium, nothing changes. During dynamic equilibrium, there are no net changes.

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

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

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.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu

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. 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.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Equilibrium Constant Visual Concepts Chapter 14 The equilibrium constant always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with.

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.

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.

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

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.

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1. For the system involving N 2 O 4 and NO 2 at equilibrium at a temperature of 100  C, the product concentration of N 2 O 4 is 4.0  10  mol/L and the reactant concentration of NO 2 is 1.4  10  1 mol/L. What is the K eq value for this reaction? Sample Problem A, practice pg. 506 H.W. #2

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu

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

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

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu

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

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.

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Sample Problem B, practice pg. 506 HW. #2 Chapter 14

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, mol of CaF 2 will dissolve in 1 L of water to give a saturated solution. If you try to dissolve mol of CaF 2 in 1 L of water, mol of CaF 2 will remain undissolved. Section 2 Systems at Equilibrium Chapter 14

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

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.

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

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.

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

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

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

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

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.

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1)Copper(I) bromide is dissolved in water to saturation at 25  C. The concentration of Cu  ions in solution is 7.9  10  5 mol/L. Calculate the K sp for copper(I) bromide at this temperature. ( ans. 6.2E-9 ) 2) What is the K sp value for Ca 3 (PO 4 ) 2 at 298 K if the concentrations in a solution at equilibrium with excess solid are 3.42  10  7 M for Ca 2  ions and 2.28  10  7 M for PO 4 -3  ions? (ans.2.08E-33) Sample Problem C, practice pg. 509 HW. # 3

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 − ] = Chapter 14 Section 2 Systems at Equilibrium

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.

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 3) Calculate the concentration of Pb 2  ions in solution when PbCl 2 is dissolved in water.The concentration of Cl  ions in this solution is found to be 2.86  10  2 mol/L. At 25  C, the K sp of PbCl 2 is 1.17  10  5. ans ) What is the concentration of Cu + ions in a saturated solution of copper(I) chloride given that the K sp of CuCl is 1.72  10  7 at 25  C? ans. 4.15E-4 Sample Problem D, practice pg. 510 HW. # 1,2

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1)The K sp for silver carbonate is 8.4  10  12 at 298 K. The concentration of carbonate ions in a saturated solution is 1.28  10  4 M. What is the concentration of silver ions? (2.6E-4) 2) Lead-acid batteries employ lead(II) sulfate plates in a solution of sulfuric acid. Use data from Table 3 to calculate the solubility of PbSO 4 in a battery acid that has an SO 4   concentration of 1.0 M. (1.8E-8) Sample Problem D, practice pg. 510

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu HW. Section Review, pg. 511, # 1-7 HW. Section Review, pg. 511, # 1-7

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 1.Giving an example, explain how to write an expression for K eq from a chemical equation. 2. Which species are left out from the K eq expression, and why? 3. To which chemical systems can a K sp be assigned? 4. When does K sp not apply? Section 14.2 Review, pg. 511

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 5) For the reaction in which hydrogen iodide is made at 425  C in the gas phase from its elements, calculate [HI], given that [H 2 ]  [I 2 ]  4.79  10  4 and K eq  54.3.(ans. 3.53E-3) Section 14.2 Review, pg. 511

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 6) Given that the Ksp value of CuS is 1.3  10  36, what is [Cu 2  ] in a saturated solution ? (ans. 1.14E-18) Section 14.2 Review, pg. 511

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 7) Write the equation for the reaction in which solid carbon reacts with gaseous carbon dioxide to form gaseous carbon monoxide. At equilibrium, a 2.0 L reaction vessel is found to contain 0.40 mol of C, 0.20 mol of CO 2, and 0.10 mol of CO. Find the K eq. (ans. 2.5E-2) Section 14.2 Review, pg. 511

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 8.The K sp for calcium carbonate is 2.8  and the K sp value for calcium sulfate is 9.1  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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu 8.The K sp for calcium carbonate is 2.8  and the K sp value for calcium sulfate is 9.1  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

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

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 M G M H.0.84 M I.1.31 M Standardized Test Preparation Interpreting Graphics Chapter 14

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 M G M H.0.84 M I.1.31 M Standardized Test Preparation Interpreting Graphics Chapter 14

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

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

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 M Fe 2+ ions. Standardized Test Preparation Interpreting Graphics Chapter 14

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 M Fe 2+ ions. Answer: 1.6  10 –17 M Standardized Test Preparation Interpreting Graphics Chapter 14

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

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