Guide for Reading. Vocabulary reversible reaction chemical equilibrium equilibrium position Le Châtelier s principle equilibrium constant

Transcription

1 18.2 Reversible Reactions and Equilibrium 18.2 Connecting to Your World For years, scientists tried to produce nitrogen compounds for fertilizers to increase the amount of available foodstuffs. Unfortunately, none of these efforts proved to be commercially successful. Finally, in the early 1900s, two German chemists, Fritz Haber and Karl Bosch, refined the process of making ammonia from elemental nitrogen and hydrogen. Their success came from controlling the temperature and pressure under which the two gases are reacted. In this section, you will learn how changing the reaction conditions can influence the yield of a chemical reaction. Reversible Reactions From what you have already learned, you may have inferred that chemical reactions go completely from reactants to products, just as the chemical equation indicates. This is not usually the case. Some reactions are reversible. A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously. One example of a reversible reaction is the following: Forward reaction: 2SO 2 (g) O 2 (g) 2SO 3 (g) Reverse reaction: 2SO 2 (g) O 2 (g) 2SO 3 (g) In the first reaction, which is read from left to right, sulfur dioxide and oxygen produce sulfur trioxide. In the second reaction, which is read from right to left, sulfur trioxide decomposes into oxygen and sulfur dioxide. The first reaction is called the forward reaction. The second is called the reverse reaction. The two equations can be combined into one using a double arrow. The double arrow tells you that this reaction is reversible. Figure 18.9 shows that this equation represents two opposite reactions. SO 2 and O 2 (not at equilibrium) 2SO 2 1 g 2 + O 2 1 g 2 JE PJ 2SO 3 1 g 2 Sulfur Oxygen Sulfur dioxide trioxide 2SO 2 O 2 (at equilibrium) DJ JE 2SO 3 SO 3 (not at equilibrium) Guide for Reading Key Concepts How do the amounts of reactants and products change in a chemical system at equilibrium? What three stresses can cause a change in the equilibrium position of a chemical system? What does the value of K eq indicate about the position of equilibrium? Vocabulary reversible reaction chemical equilibrium equilibrium position Le Châtelier s principle equilibrium constant Reading Strategy Previewing Before you read this section, rewrite the headings as how, why, and what questions about reversible reactions and equilibrium. As you read, write answers to the questions. Figure 18.9 Molecules of SO 2 and O 2 react to give SO 3. Molecules of SO 3 decompose to give SO 2 and O 2. At equilibrium, all three types of molecules are present in the mixture. 1 FOCUS Objectives Describe how the amounts of reactants and products change in a chemical system at equilibrium Identify three stresses that can change the equilibrium position of a chemical system Explain what the value of K eq indicates about the position of equilibrium. Guide for Reading Build Vocabulary LINCS Have students use the LINCS strategy to learn what Le Châtelier s principle means. In LINCS exercises, students list what they know about each term, imagine a picture that describes the term, note a reminding sound-alike word, connect the terms to the sound-alike word by making up a short story, and then perform a brief self-test. Reading Strategy Monitor Your Understanding Ask students if the flow chart relating cause and effect helped them read Section Ask them to consider where they would use that technique in this section. 2 INSTRUCT Section Resources Print Guided Reading and Study Workbook, Section 18.2 Core Teaching Resources, Section 18.2 Review Transparencies, T199 T202 Laboratory Manual, Lab 38 Small-Scale Chemistry Laboratory Manual, Lab 29 Section 18.2 Reversible Reactions and Equilibrium 549 Technology Interactive Textbook with ChemASAP, Animation 23; Simulation 24; Problem- Solving 18.6, 18.7, 18.9; Assessment 18.2 Point out that the two elements involved in production of ammonia are gases. Ask, Do you think that controlling temperature and pressure of liquids or solids in a similar reaction would have had a similar affect? Why? (Temperature might, but liquids and solids aren t sensitive to pressure difference the way gases are.) Reaction Rates and Equilibrium 549

2 Section 18.2 (continued) Reversible Reactions Use Visuals Figure 18.9 Explain that chemical reactions do not always go completely to products because they are often reversible. Once the product(s) form, they may decompose to form the reactants. Ask, Does the reaction continue when the products and reactants are at equilibrium? (Yes; reactants change into products, and products reform into reactants, but their net amounts stay the same.) Interpreting Graphs a. Initial concentrations are at the far left, and equilibrium concentrations are at the far right. b. SO 3 c. SO 3 is much higher than SO 2 and O 2 ; O 2 is lowest. Enrichment Question L3 Would these graphs change if you changed volume, pressure, or temperature? Explain. (Yes, you can change the concentration of the gas by changing the volume of its container, pressure, and temperature.) Word Origins Disequilibrium for a chemical reaction means the opposite of being in balance. Therefore, the amount of reactant and product is still changing. L1 Figure These graphs show how the concentrations of O 2, SO 2, and SO 3 vary with time. Left: Initially, SO 2 and O 2 are present. Right: Initially, only SO 3 is present. INTERPRETING GRAPHS a. Navigate Where on the graphs can you find the initial concentrations of the reactants and products? The equilibrium concentrations? b. Read Which gas is most abundant at equilibrium? c. Interpret How do the equilibrium concentrations of O 2, SO 2, and SO 3 compare? Word Origins Equilibrium comes from the Latin word aequilibrium, meaning in balance. In a chemical system at equilibrium, the rate of the forward reaction is balanced by the rate of the reverse reaction. If the prefix dis-means opposite or absence of, what does a state of disequilibrium indicate? Concentration uy Changes in Concentrations of Reactants and Products Time uy Equilibrium concentrations SO 3 SO 3 SO 2 SO 2 O 2 O 2 Time uy What actually happens when sulfur dioxide and oxygen gases are mixed in a sealed chamber? The two reactants begin to react to form sulfur trioxide at a particular rate. Because no sulfur trioxide is present at the beginning of the reaction, the initial rate of the reverse reaction is zero. As sulfur trioxide begins to form, however, the decomposition of sulfur trioxide begins. This reverse reaction proceeds slowly at first, but its rate increases as the concentration of sulfur trioxide increases. Simultaneously, the rate of the forward reaction decreases because sulfur dioxide and oxygen are being used up. Eventually sulfur trioxide is decomposing to sulfur dioxide and oxygen as fast as sulfur dioxide and oxygen are forming sulfur trioxide. When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium. Changes in concentrations of the three components during the course of the reaction are shown in the graphs in Figure The graph on the left shows the progress of a reaction that starts with specific concentrations of SO 2 and O 2, but with zero concentration of SO 3. The graph on the right shows concentrations for a reaction that begins with an initial concentration of SO 3 and zero concentrations for SO 2 and O 2. Notice that after a certain time, all concentrations remain constant. At chemical equilibrium, no net change occurs in the actual amounts of the components of the system. The amount of SO 3 in the equilibrium mixture is the maximum amount that can be produced by this reaction under the conditions of the reaction. The unchanging amounts of SO 2, O 2, and SO 3 in the reaction mixture at equilibrium might cause you to think that both reactions have stopped. This is not the case. Chemical equilibrium is a dynamic state. Both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in their concentrations. The escalators in Figure are like the double arrows in a dynamic equilibrium equation. The number of people using the up escalator must equal the number of people using the down escalator for the number of people on both floors to remain constant. Concentration uy Equilibrium concentrations Checkpoint Why is chemical equilibrium called a dynamic state? 550 Chapter Chapter 18

3 Although the rates of the forward and reverse reactions are equal at chemical equilibrium, the concentrations of the components on both sides of the chemical equation are not necessarily the same. In fact, they can be dramatically different. Figure shows the equilibrium concentrations of SO 2, O 2, and SO 3. The concentration of SO 3 is significantly greater than the concentrations of SO 2 and O 2. The relative concentrations of the reactants and products at equilibrium constitute the equilibrium position of a reaction. The equilibrium position indicates whether the reactants or products are favored in a reversible reaction. If A reacts to give B and the equilibrium mixture contains significantly more of B say 1% A and 99% B then the formation of B is said to be favored. A B JEP 1% 99% On the other hand, if the mixture contains 99% A and 1% B at equilibrium, then the formation of A is favored. A B PJ E 99% 1% Notice that the equilibrium arrows are not of equal length; the longer of the two arrows indicates the favored direction of a reaction. In principle, almost all reactions are reversible to some extent under the right conditions. In practice, one set of components is often so favored at equilibrium that the other set cannot be detected. If one set of components (reactants) is completely converted to new substances (products), you can say that the reaction has gone to completion, or is irreversible. When you mix chemicals expecting to get a reaction but no products can be detected, you can say that there is no reaction. Reversible reactions occupy a middle ground between the theoretical extremes of irreversibility and no reaction. A catalyst speeds up both the forward and the reverse reactions equally because the reverse reaction is exactly the opposite of the forward reaction. The catalyst lowers the activation energy of the reaction by the same amount in both the forward and reverse directions. Catalysts do not affect the amounts of reactants and products present at equilibrium; they simply decrease the time it takes to establish equilibrium. Animation 23 Take a close look at a generalized reversible reaction. Discuss Point out that, in many cases, the concept of reactants turning into products, as described by a chemical equation, is an oversimplification. Many reactions reach an equilibrium state in which the reaction mixture contains both reactant and product particles. The percent of reactants converted to products varies considerably. When hydrogen and oxygen combine to form water, for example, nearly all of the molecules react. In the reaction of hydrogen and nitrogen to form ammonia, the percent of reactants converted to product is much smaller. Stress that equilibrium refers to the rates of the forward or reverse reactions, not to the quantities of reactants and products. Checkpoint What is chemical equilibrium? Figure If the rate at which shoppers move from the first floor to the second is equal to the rate at which shoppers move from the second floor to the first, then the number of shoppers on each floor remains constant. Applying Concepts Is it necessary that an equal number of shoppers be on each floor? Explain. 551 Answers to... Figure No, the numbers can be different but the number of shoppers moving between floors must be the same. Checkpoint because both the forward and reverse reactions continue Checkpoint a situation in which the forward and reverse reactions occur at the same rate Reaction Rates and Equilibrium 551

4 Section 18.2 (continued) Factors Affecting Equilibrium: Le Châtelier s Principle Discuss L1 Refer to the formation of banana oil, which was described on page 546. Write on the board the general word equation for the formation of an ester (organic acid + alcohol ester + water). Add a reverse arrow to show the possibility of the opposite reaction. Remind students that the reaction was effective in the test tube to which anhydrous calcium sulfate had been added. Calcium sulfate reacted with water to form a hydrate, which limited the reverse reaction by removing a product. TEACHER Demo Revisiting Banana Oil Demonstration You may want to repeat the banana oil demonstration on page 546. This time, add concentrated sulfuric acid to both test tubes before heating. However, add the anhydrous calcium sulfate to only one tube. Ask students to predict which reaction will yield the most product. Then compare the results qualitatively by strength of odor or quantitatively if you have access to gas phase chromatography (GC) equipment. Expected Outcome The tube containing anhydrous calcium sulfate yields the larger amount of product. Factors Affecting Equilibrium: Le Châtelier s Principle A delicate balance exists in a system at equilibrium. Changes of almost any kind can disrupt this balance. When the equilibrium of a system is disturbed, the system makes adjustments to restore equilibrium. However, the equilibrium position of the restored equilibrium is different from the original equilibrium position; that is, the amounts of products and reactants may have increased or decreased. Such a change is called a shift in the equilibrium position. The French chemist Henri Le Châtelier ( ) studied how the equilibrium position shifts as a result of changing conditions. He proposed what has come to be called Le Châtelier s principle: If a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress. Stresses that upset the equilibrium of a chemical system include changes in the concentration of reactants or products, changes in temperature, and changes in pressure. The following examples of applications of Le Châtelier s principle all involve reversible reactions. For simplicity and clarity, the components to the left of the reaction arrow will be considered the reactants and the components to the right of the reaction arrow will be considered the products. Blue arrows indicate the shifts resulting from additions to or removals from the system. The arrows always point in the direction of the resulting shift in the equilibrium position that is, toward the favored side. Concentration Changing the amount, or concentration, of any reactant or product in a system at equilibrium disturbs the equilibrium. The system adjusts to minimize the effects of the change. Consider the decomposition of carbonic acid (H 2 CO 3 ) in aqueous solution to form the products carbon dioxide and water. The system has reached equilibrium. At equilibrium the amount of carbonic acid is less than 1%. Add CO 2 d Direction of shift H 2 CO 3 1 aq 2 CO 2 1aq 2 + H 2 O1 l 2 JJJJJJE PJJJJ Remove CO 2 1% Direction of shift S 99% 552 Chapter Chapter 18

5 Adding more carbon dioxide disturbs the equilibrium. It increases the concentration of CO 2 in the mixture and causes the rate of the reverse reaction to increase. As more reactant (H 2 CO 3 ) is formed, the rate of the forward reaction also begins to increase. In time, the rates of the forward and reverse reactions again become equal, and a new equilibrium is established with a higher concentration of reactant (H 2 CO 3 ). Adding a product to a reaction at equilibrium pushes a reversible reaction in the direction of reactants. If, on the other hand, carbon dioxide is removed, the concentration of CO 2 decreases. This causes the rate of the reverse reaction to decrease. As less reactant (H 2 CO 3 ) is being formed, the rate of the forward reaction also begins to decrease. When the rates of the forward and reverse reactions again become equal, equilibrium is restored but at a different equilibrium position. Removing a product always pushes a reversible reaction in the direction of products. Farmers use this technique to increase the yield of eggs laid by hens. Hens lay eggs and then proceed to hatch them. If the eggs are removed after they are laid (removing the product), the hen will lay more eggs (increasing the yield). Similarly as products are removed from a reaction mixture, the system continually changes to restore equilibrium by producing more products. But because the products are being removed, the reaction can never reestablish equilibrium. The reaction continues to produce products until the reactants are used up. Another example of this concept is found in the body. Blood contains dissolved carbonic acid in equilibrium with carbon dioxide and water. The body uses the removal of products to keep the concentration of carbonic acid within a safe range. When the athletes in Figure exhale carbon dioxide, the equilibrium shifts toward carbon dioxide and water, thus reducing the amount of carbonic acid. The same principle applies to adding or removing reactants. When a reactant is added to a system at equilibrium, the reaction shifts in the direction of the formation of products. When a reactant is removed, the reaction shifts in the direction of formation of reactants. Simulation 24 Simulate Le Châtelier s principle for the synthesis of ammonia. Relate Write the equation for the dissociation of H 2 CO 3 on the board. Explain that our bodies expel CO 2 to maintain the correct equilibrium between CO 2 and H 2 CO 3. During and after vigorous exercise, more CO 2 than usual needs to be removed to maintain equilibrium. Ask, What are some signs that the body is trying to reestablish equilibrium after exercising? (a faster heartbeat and breathing rate) Checkpoint According to Le Châtelier s principle, how does a system at equilibrium respond to a stress? Figure The rapid exhalation of CO 2 during and after vigorous exercise helps reestablish the body s correct CO 2 :H 2 CO 3 equilibrium. This keeps the acid concentration in the blood within a safe range. 553 Facts and Figures Carbon Monoxide Molecules such as carbon monoxide (CO) that are structurally similar to the oxygen molecule can bind to the hemoglobin molecule. Therefore, they compete with oxygen for the same binding locations. This competition is the cause of carbon monoxide poisoning. Because their affinity for the heme group is greater, carbon monoxide molecules can displace oxygen molecules from hemoglobin. HbO 2 + CO ΗbCO + O 2 The equilibrium in this reaction lies far to the right because the affinity of carbon monoxide for heme is approximately 150 times stronger than that of oxygen. Answers to... Checkpoint It moves in the direction that relieves the stress. Reaction Rates and Equilibrium 553

6 Section 18.2 (continued) Use Visuals L1 Figure Ask students to study the diagrams and explain how equilibrium in a relates to equilibrium in c. (Increasing the pressure on the system in a caused the reaction to shift to the right producing more molecules of product in c.) Ask, What kind of substances does a change in pressure affect? (gases) Relate The industrial production of ammonia provides a practical application of Le Châtelier s principle. Write the equilibrium equation on the board. Ask, Based on the equilibrium equation, select ideal temperature and pressure conditions for the highest possible yield of ammonia. (high pressure and low temperature) Tell students that the industrial process devised by Carl Bosch actually uses both high temperature and pressure plus an iron-based catalyst. Ask, Why doesn t Bosch s process use the conditions predicted by Le Châtelier s principle? (A higher-than-ideal temperature (plus the catalyst) is needed to increase the reaction rate.) Point out that an efficient industrial process must often strike a balance between multiple goals, in this case, yield and reaction time. CONCEPTUAL PROBLEM 18.1 Answers 6. a. favors reactants b. favors reactants c. favors products d. favors products Practice Problems Plus The gases dinitrogen tetroxide and nitrogen dioxide exist in equilibrium according to the equation N 2 O 4 (g) + 58 kj 2NO 2 (g) What effect do the following changes have on the equilibrium position? a. addition of heat (favors products) b. decrease in pressure (favors products) c. addition of NO 2 (favors reactants) d. removal of N 2 O 4 (favors reactants) Figure Pressure affects a mixture of nitrogen, hydrogen, and ammonia at equilibrium. a The system is at equilibrium. b Equilibrium is disturbed by an increase in pressure. c A new equilibrium position is established with fewer gas molecules. Interpreting Diagrams What effect does a decrease in volume have on the number of gas molecules? 554 Chapter 18 Ammonia molecule (NH 3 ) Hydrogen molecule (H 2 ) Nitrogen molecule (N 2 ) Temperature Increasing the temperature causes the equilibrium position of a reaction to shift in the direction that absorbs heat. The heat absorption reduces the applied temperature stress. For example, consider the following exothermic reaction that occurs when SO 3 is produced from the reaction of SO 2 and O 2. Add heat d Direction of shift 2SO 2 1 g 2 + O 2 1 g 2 2SO 3 1 g 2 + heat PJJJJJJ JJJJJJJE Remove heat (cool) Direction of shift S Heat can be considered to be a product, just like SO 3. Heating the reaction mixture at equilibrium pushes the equilibrium position to the left, which favors the reactants. As a result, the product yield decreases. Cooling, or removing heat, pulls the equilibrium to the right, and the product yield increases. Pressure A change in the pressure on a system affects only gaseous equilibria that have an unequal number of moles of reactants and products. An example is the reaction you read about in Connecting to Your World: hydrogen and nitrogen react to form ammonia. Imagine that the three gases are at equilibrium in a cylinder that has a piston attached to a plunger similar to a bicycle pump but with the hose sealed. A catalyst has been included to speed up the reaction. What happens to the pressure when you push the plunger down? The pressure on the gases momentarily increases because the same number of molecules is contained in a smaller volume. The system immediately relieves some of the pressure increase by reducing the number of gas molecules. For every two molecules of ammonia made, four molecules of the reactants are used up (three molecules of hydrogen and one of nitrogen). Therefore, the equilibrium position shifts to make more ammonia. There are then fewer molecules in the system. The pressure decreases, although it will not decrease all the way to the original pressure. As you can see in Figure 18.13, increasing the pressure on the system results in a shift in the equilibrium position that favors the formation of product. a b c Initial equilibrium condition (11 gas molecules) Increase pressure Direction of shift S N 2 1 g 2 + 3H 2 1 g 2 2NH 3 1 g 2 PJJJJJJ JJJJJJJE Reduce pressure d Direction of shift Pressure increased, equilibrium disturbed New equilibrium condition at increased pressure (9 gas molecules) 554 Chapter 18

7 The equilibrium position for this reaction can be made to favor the reactants instead of the product. Imagine pulling the plunger of the piston device back up so the volume containing the gases increases. This increase in volume decreases the pressure on the system. To restore the higher starting pressure, the system can produce more gas molecules by the decomposition of some ammonia molecules. Decomposition of two molecules of gaseous NH 3 produces four molecules of reactants (three H 2 and one N 2 ). Pressure at the new equilibrium is higher than when the pressure was first decreased, but not as high as it was at the starting equilibrium. Lowering the pressure on the system thus results in a shift of the equilibrium to favor the reactants. CONCEPTUAL PROBLEM 18.1 Applying Le Châtelier s Principle What effect do each of the following changes have on the equilibrium position for this reversible reaction? PCl 5 (g) heat m PCl 3 (g) Cl 2 (g) a. addition of Cl 2 b. increase in pressure c. removal of heat d. removal of PCl 3 as it is formed Analyze Identify the relevant concepts. a. d. The stress placed on each system is known. The effect of each stress is unknown. By Le Châtelier s principle, the equilibrium system will shift in a direction that minimizes the imposed stress. Analyze the effect of each change on the reaction. Solve Apply concepts to this situation. a. The addition of Cl 2, a product, shifts the equilibrium to the left, forming more PCl 5. Practice Problem 6. How is the equilibrium position of this reaction affected by the following changes? C(s) H 2 O(g) heat m CO(g) H 2 (g) a. lowering the temperature b. increasing the pressure c. removing hydrogen d. adding water vapor A stress on this physical equilibrium could result in a loss of balance. b. The equation shows 2 mol of gaseous product and 1 mol of gaseous reactant. The increase in pressure is relieved if the equilibrium shifts to the left, because a decrease in the number of moles of gaseous substances produces a decrease in pressure. c. The removal of heat causes the equilibrium to shift to the left, because the reverse reaction is heat-producing. d. The removal of PCl 3 causes the equilibrium to shift to the right to produce more PCl 3. Problem-Solving 18.6 Solve Problem 6 with the help of an interactive guided tutorial. Section 18.2 Reversible Reactions and Equilibrium 555 TEACHER Demo Temperature and Equilibrium Purpose Students observe the effect of temperature changes on the equilibrium position of an exothermic reaction. Materials 3 g copper turnings, 3 Pyrex test tubes, 10 ml 6M nitric acid, onehole stopper, 2 solid stoppers, glass delivery tube, rubber tubing, warm water bath, ice water bath Preparation Place a small amount (3 g or less) of copper turnings in a test tube, and add 10 ml of 6M nitric acid (HNO 3 ). Insert a one-hole stopper, fitted with a glass delivery tube and rubber tubing, into the test tube. Fill two Pyrex test tubes with the brown gas that forms. When the color is the same as in the reacting test tube, stopper the tubes. Once sealed, the tubes can be saved indefinitely for repeated use, or discharged in a fume hood. Safety Use a fume hood. Nitric acid is corrosive, and NO 2 is a toxic gas. Procedure Set up an ice bath and a warm water bath (70 80 C). Hold up the tubes for students to see and tell them that each tube contains NO 2 gas. Place one tube in warm water and the other in cold water. After a few minutes, remove the tubes and have students compare their colors. Now reverse the positions of the tubes in the water baths and after a few minutes, have students again observe the colors. Write the reaction equation on the board: 2NO 2 (g) N 2 O 4 (g) + heat Explain that nitrogen dioxide is brown and dinitrogen tetroxide is colorless. Ask students to explain the effect of changing temperature. (Adding heat shifts the equilibrium in the direction that absorbs heat, to the left. Removing heat shifts the equilibrium in the direction that produces heat, to the right.) Answers to... Figure Decreasing the volume shifts the equilibrium in the direction of fewer gas molecules. Reaction Rates and Equilibrium 555

8 Section 18.2 (continued) Equilibrium Constants Discuss Point out that whenever a chemical system reaches equilibrium, the reactants and products have a fixed numerical relationship. This relationship is stated in the equilibrium constant expression. The value of the equilibrium constant can be used to predict the position of equilibrium. Because the equilibrium constant refers to an expression in which products are divided by reactants, a K eq value greater than 1 favors the products. Conversely, a K eq value less than 1 favors the reactants. Figure Dinitrogen tetroxide is a colorless gas; nitrogen dioxide is a brown gas. The flask on the left is in a dish of hot water; the flask on the right is in ice. Interpreting Illustrations How does an increase in temperature affect the equilibrium of a mixture of these gases? Equilibrium Constants Chemists express the position of equilibrium in terms of numerical values. These values relate the amounts of reactants to products at equilibrium. In a general reaction, a mol of reactant A and b mol of reactant B react to give c mol of product C and d mol of product D at equilibrium. aa bb m cc dd The equilibrium constant (K eq ) is the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation. The expression for the equilibrium constant can be written this way. K eq 3C4c 3D4 d 3A4 a 3B4 b The exponents in the equilibrium-constant expression are the coefficients in the balanced chemical equation. The square brackets indicate the concentrations of substances in moles per liter (mol/l). The value of K eq depends on the temperature of the reaction. If the temperature changes, the value of K eq also changes. The size of the equilibrium constant shows whether products or reactants are favored at equilibrium. A value of K eq greater than 1 means that products are favored over reactants; a value of K eq less than 1 means that reactants are favored over products. K eq > 1, products favored at equilibrium K eq < 1, reactants favored at equilibrium When the numerical value of an equilibrium constant is calculated, the cancellation of units may or may not lead to a unit for the constant. Chemists have agreed to report all equilibrium constants without stated units. Sample Problem 18.1 shows how to calculate the equilibrium constant for the reaction illustrated in Figure Nitrogen dioxide (NO 2 ) Dinitrogen tetroxide (N 2 O 4 ) Warm Cool 556 Chapter Chapter 18

9 SAMPLE PROBLEM 18.1 Expressing and Calculating K eq The colorless gas dinitrogen tetroxide (N 2 O 4 ) and the dark brown gas nitrogen dioxide (NO 2 ) exist in equilibrium with each other. N 2 O 4 (g) m 2NO 2 (g) A liter of a gas mixture at equilibrium at 10 C contains mol of N 2 O 4 and mol of NO 2. Write the expression for the equilibrium constant and calculate the equilibrium constant (K eq ) for the reaction. Analyze List the knowns and the unknowns. Knowns [N 2 O 4 ] mol/l [NO 2 ] mol/l K eq 3C4c 3D4 d 3A4 a 3B4 b Unknowns K eq (algebraic expression)? K eq (numerical value)? Calculate Solve for the unknowns. To write the equilibrium constant expressions, place the concentration of the product in the numerator and the concentration of the reactant in the denominator. Raise each to the power equal to its coefficient in the chemical equation. Substitute the given concentrations and calculate K eq. K eq 3NO mol/l)2 3N 2 O mol/l mol/l mol/ mol/l Practice Problems Evaluate Does the result make sense? Each concentration is raised to the correct power. The numerical value of the constant is correctly expressed to two significant figures. Math Handbook For help with using a calculator, go to page R62. Sample Problem 18.1 Answers 7. K eq = K eq = , One is the inverse of the other. Practice Problems Plus Analysis of an equilibrium mixture of hydrogen, iodine, and hydrogen iodide contained in a 10.0-L flask at a certain temperature gives the following results: hydrogen, 0.15 mol; iodine, 0.15 mol; hydrogen iodide, 0.87 mol. Calculate K eq for the reaction: H 2 (g) + I 2 (g) 2HI(g) (K eq = 34) Math Handbook For a math refresher and practice, direct students to using a calculator, page R62. Practice Problems 7. The reaction N 2 (g) 3H 2 (g) m 2NH 3 (g) produces ammonia, a fertilizer. At equilibrium, a 1-L flask contains 0.15 mol H 2, 0.25 mol N 2, and 0.10 mol NH 3. Calculate K eq for the reaction. 8. For the same mixture, under the same conditions described in Problem 8, calculate K eq for this reaction 2NH 3 (g) m N 2 (g) 3H 2 (g). How is the K eq for a forward reaction related to the K eq for a reverse reaction? Problem Solving 18.7 Solve Problem 7 with the help of an interactive guided tutorial. Ammonia (NH 3 ), produced by a reversible reaction, is used extensively as a fertilizer. 557 Answers to... Figure An increase in temperature causes the equilibrium to shift toward the formation of more NO 2. Reaction Rates and Equilibrium 557

10 Section 18.2 (continued) Discuss Show students how to set up and solve problems involving equilibrium constants. Stress the need to clearly establish what is given and what is being asked. Break down the process into easily followed steps. Point out that the only quantities that can go into the brackets used in the equilibrium constant expression are concentrations, or quantities like partial pressures, which are related to concentrations. Warn students not to forget to subtract the number of moles that react from the number of moles originally present in order to determine the number remaining at equilibrium. Point out that it is this value for the concentration of a reactant that is substituted in the denominator of the equation for K eq. Sample Problem 18.2 Answers 9. K eq = K eq = 0.79 Math Handbook For help with using a calculator, go to page R62. SAMPLE PROBLEM 18.2 Finding the Equilibrium Constant One mol of colorless hydrogen gas and 1.00 mol of violet iodine vapor are sealed in a 1-L flask and allowed to react at 450 C. At equilibrium, 1.56 mol of colorless hydrogen iodide is present, together with some of the reactant gases. Calculate K eq for the reaction. H 2 (g) I 2 (g) m 2HI(g) Analyze List the knowns and the unknowns. Knowns [H 2 ] (initial) 1.00 mol/l [I 2 ] (initial) 1.00 mol/l [HI] (equilibrium) 1.56 mol/l K eq 3C4c 3D4 d 3A4 a 3B4 b Practice Problems mol/l) 2 K eq 0.22 mol/l 0.22 mol/l Unknowns K eq (algebraic expression)? K eq (numerical value)? Calculate Solve for the unknowns. The balanced equation indicates that 1.00 mol H 2 and 1.00 mol I 2 are needed to form 2.00 mol HI. Making 1.56 mol of HI, therefore, consumes 1.56 mol of each or 0.78 mol of H 2 and 0.78 mol of I Calculate how much H 2 and I 2 remain in the flask at equilibrium. mol H 2 mol I 2 (1.00 mol 0.78 mol) 0.22 mol 3HI4 2 Write the expression for K eq : K eq 3H 2 4 3I 2 4 Substitute the equilibrium concentrations of the reactants and products into the equation and calculate mol/l 1.56 mol/l mol/l 0.22 mol/l Evaluate Does the result make sense? Each concentration is raised to the correct power. The numerical value of the constant is correctly rounded to two significant figures. Math Handbook For a math refresher and practice, direct students to using a calculator, page R62. Problem Solving 18.9 Solve Problem 9 with the help of an interactive guided tutorial. Practice Problems 9. Suppose the following system reaches equilibrium. N 2 (g) O 2 (g) m 2NO(g) Analysis of the equilibrium mixture in a 1-L flask gives the following results: N mol, O mol, and NO mol. Calculate K eq for the reaction. 10. At 750 C the following reaction reaches equilibrium in a 1-L flask. H 2 (g) CO 2 (g) m H 2 O(g) CO(g) Analysis of the equilibrium mixture gives the following results: H mol, CO mol, H 2 O mol, and CO mol. Calculate K eq for the reaction. 558 Chapter Chapter 18

11 You can also calculate the equilibrium concentrations of the reactants and products if you know the equilibrium constant and if you have some information about the initial or final concentration of one of the constituents. For example, bromine chloride (BrCl) decomposes to form bromine and chlorine according to this equation. 2BrCl(g) m Br 2 (g) Cl 2 (g) At a certain temperature, the equilibrium constant for this reaction is Suppose that pure BrCl is placed in a 1-L container and allowed to come to equilibrium. Analysis then shows that the reaction mixture contains 4.00 moles of Cl 2. How many moles of Br 2 and BrCl are also present in the equilibrium mixture? According to the equation, when BrCl decomposes, equal numbers of moles of Br 2 and Cl 2 are formed. At equilibrium, 4.00 mol of Cl 2 is present, so 4.00 mol of Br 2 must also be present. Because the volume of the container is 1 L, the concentrations of Br 2 and Cl 2 are each 4.00 mol/l. These concentrations can be used in the equilibrium constant expression to calculate the concentration of BrCl Br 4 3Cl BrCl4 2 Rearrange the equation to solve for [BrCl] 2 and substitute the known values. 3BrCl4 2 3Br 2 4 3Cl BrCl mol 2 /L mol/l The concentration of the reactant (BrCl) is less than the concentrations of the products, so the products (Br 2 and Cl 2 ) are slightly favored at equilibrium. This is to be expected because K eq > 1. Yellow chlorine gas and brown liquid bromine with its vapor are shown in Figure Section Assessment (4.00 mol/ (4.00 mol/ Key Concept How do the amounts of reactants and products change after a reaction has reached chemical equilibrium? 12. Key Concept What are three stresses that can upset the equilibrium of a chemical system? 13. Key Concept What does the value of the equilibrium constant tell you about the amounts of reactants and products present at equilbrium? 14. How can a balanced chemical equation be used to write an equilibrium-constant expression? 15. Can a pressure change shift the equilibrium position in every reversible reaction? Explain. 16. Using the following equilibrium constants for several reactions, determine in which reactions the products are favored. Why? a. K eq b. K eq c. K eq mol 2 /L 2 Section 18.2 Reversible Reactions and Equilibrium 559 Section 18.2 Assessment Figure Chlorine and bromine are two of the halogens. Comparing and Contrasting How are they similar? In what ways are they different? Percent Yield Recall the concept of percent yield of a chemical reaction in Section Could the fact that a reaction is reversible and establishes equilibrium affect the percent yield of a reaction? Explain. Assessment 18.2 Test yourself on the concepts in Section They do not change. 12. changes in concentrations of reactants and products, changes in temperature, and changes in pressure 13. K eq > 1 favors products; K eq < 1 favors reactants. 14. An equilibrium constant expression is a ratio. In the numerator, product concentrations are multiplied. In the denominator, reactant concentrations are multiplied. Each concentration is raised to a power equal to the coefficient for that species in the balanced equation. 15. no; only in reversible reactions in which the mole ratios of gaseous reactants and products are unequal 16. Products are favored in reactions a and c because K eq > 1. 3 ASSESS Evaluate Understanding Write several reversible chemical equations on the board. Have students use Le Châtelier's principle to predict how the yield of a reaction product could be increased. Ask them to write the equation for calculating an equilibrium constant using the generalized expression aa + bb cc + dd K eq = [C]c [D] d [A] a [B] b Reteach L1 If students have trouble writing the correct equilibrium constant expression from a balanced equation, list the steps on the board. Use as an example the equilibrium process for the production and decomposition of ozone in the stratosphere. Explain that stratospheric ozone screens out about 95% of the UV radiation from the sun. Within the ozone layer, ozone is continually formed and consumed through an equilibrium process: 3O 2 (g) 2O 3 (g) Connecting Concepts Yes; 100% yield is obtained only when all reactants are converted to products. If a reversible reaction stops before all reactants are used up, it can never reach 100% yield. If your class subscribes to the Interactive Textbook, use it to review key concepts in Section with ChemASAP Answers to... Figure Both are in Group 7A, form ions with 1 charge, and exist as diatomic molecules. Chlorine is a yellow gas; bromine is a brown liquid at ordinary temperatures. Reaction Rates and Equilibrium 559