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NAME____________________________________ PER____________ DATE DUE____________ ACTIVE LEARNING I N C HEMISTRY E DUCATION "ALICE" CHAPTER 20 CHEMICAL EQUILIBRIUM Keq 20-1 ©1997, A.J. Girondi NOTICE OF RIGHTS All rights reserved. No part of this document may be reproduced or transmitted in any form by any means, electronic, mechanical, photocopying, or otherwise, without the prior written permission of the author. Copies of this document may be made free of charge for use in public or nonprofit private educational institutions provided that permission is obtained from the author . Please indicate the name and address of the institution where use is anticipated. © 1997 A.J. Girondi, Ph.D. 505 Latshmere Drive Harrisburg, PA 17109 [email protected] Website: www.geocities.com/Athens/Oracle/2041 20-2 ©1997, A.J. Girondi SECTION 20.1 Introduction to Equilibrium Systems In most of the chemical reactions we have studied so far, it appears as though all of the reactants are converted to products before a reaction stops. In truth, however, experiments show that the conversion of reactants into products is often incomplete in chemical reactions. This is the case no matter how long the reaction is allowed to continue. As a reaction progresses, the concentrations of the reactants decrease, while the concentrations of the products increase. Eventually, a state is established in which the concentrations of products and reactants no longer change. This is known as the state of equilibrium. You have already studied many examples of equilibrium. The most obvious example was probably encountered during your study of solutions. At that time, you performed some activities that involved the use of saturated solutions. Equilibrium is a balance between two equal and opposing processes or forces. When a system reaches equilibrium, it does not undergo additional change unless the equilibrium is somehow disturbed. A saturated solution is in equilibrium because there is a balance between the two opposing forces of dissolving and crystallizing. (See Figure 20.1 below.) This is an example of a physical equilibrium since dissolving and crystallizing are physical changes. An equation can be written to represent what is occurring in this saturated solution after equilibrium has been reached: CaCl2(s) Ca2+(aq) + 2 Cl1-(aq) REACTANTS PRODUCTS The (aq) subscript refers to "aqueous" which means dissolved in water. The double arrow indicates that the forward reaction is occurring at the SAME RATE as the reverse reaction. Therefore, no overall change occurs in the system. To an observer, it appears as though nothing whatsoever is happening in the saturated CaCl2 solution. Actually, the change shown at right is occurring constantly in both directions. Note that the equation above shows chlorine as 2 Cl1-, not as Cl2. Chloride ions (Cl1-) do not exist in diatomic form. This is because these ions already have a stable octet of electrons. Only chlorine atoms form diatomic molecules: (Cl2). This is also true for the other diatomic elements. WRONG! ---> CaCl2(s) <===> Ca2+(aq) + Cl2(aq) <--- WRONG! Saturated CaCl2 Solution Ca2+ Cl1- Cl1- Figure 20.1 Solution Equilibrium CaCl2(s) <===> Ca2+(aq) + 2 Cl1-(aq) The type of change shown in Figures 20.1 and 20.2 is referred to as dynamic equilibrium because, although no changes can be observed in the system, the two opposing changes are happening constantly. Figure 20.2 illustrates two more examples of physical equilibria. Examine what is in each beaker, and note how a double arrow is used in the equilibrium equation for each situation. Note, too, how the state (solid, liquid, gas, or aqueous) of each substance is represented using subsscripts. Finally, notice that coefficients are used to balance the equations properly. 20-3 ©1997, A.J. Girondi Saturated PbI 2 Solution Saturated AlCl3 Solution Pb 2+ Al3+ I 1- I 1- Cl1- Cl1Cl1- PbI 2 PbI 2 (s) <===> AlCl3 Pb2+(aq) Figure 20.2 + 2 I 1-(aq) AlCl3(s) <===> Al3+(aq) + 3 Cl1-(aq) Equilibrium Systems in Saturated Solutions Problem 1. The three equations below represent equilibrium systems created by dissolving solids in water. Complete the equilibrium equations below. Include any charges on ions and the state of the ions which in this case is "aqueous" or (aq): a. NaCl(s) <====> _____________________________ b. Fe2(SO4)3(s) <====> c. BaBr2(s) <====> _____________________________ ______________________________ The formula for water is not included in the examples above because dissolving is not a chemical change and, therefore, water is not consider a reactant. When substances dissolve in water they do not chemically react with it. They merely come part in the water. Later in this chapter you will study reactions that involve chemical equilibrium, since the changes are chemical, rather than physical. There are several criteria that must be satisfied before an equilibrium can be achieved. Let's take a closer look at some of these. Compare the two systems shown below (beakers A and B). Which beaker contains a system at equilibrium?{1}_________ What do you think prevents the system in the other beaker from forming an equilibrium system? {2}____________________________________________ H2O(g) H2O(g) H2O(l) H2O(l) BEAKER "A" BEAKER "B" Figure 20.3 Closed and Open Systems H2O(l) H2O(g) 20-4 ©1997, A.J. Girondi Could equilibrium ever be achieved in an open system such as that shown in beaker B? {3}________ Why or why not? {4}___________________________________________________________________ Hopefully, you have arrived at the conclusion that an equilibrium can be established in a system such as this only if the system is closed. In addition, this system must be held under conditions of constant temperature, pressure, and volume. If one or more of these factors is altered, the equilibrium changes as well. (Not all equilibrium systems are affected by pressure or volume, but all equilibrium systems are affected by temperature. We will discuss this later in the chapter.) Equilibrium equations give an overall picture of the types of chemical changes that are occurring. The equations do not, however, provide any information about the actual amounts of reactants and products present. The equilibrium equation for a saturated sodium chloride solution: NaCl(s) <====> Na1+(aq) + Cl1-(aq) merely indicates the identities of the ions and molecules involved in the equilibrium, and the double arrows show that equilibrium prevails. There is absolutely no way that you can tell what the concentrations of any of the products or reactants are just by looking at this equation. The reaction container may have a few crystals of solid NaCl present or it may have a large pile of solid salt. SECTION 20.2 Solving Problems Involving Equilibrium Systems There is a simple relationship between the concentrations of the products and the reactants. Through much experimental study, scientists were able to come up with a law of chemical equilibrium. Consider the hypothetical equilibrium system: A + B <===> C + D Brackets such as, [ ], are used to denote molar (M) concentration. So when you see something like [Ag 1+], it means "molar concentration of silver ions." This law of chemical equilibrium states that if you find the product of the concentrations of the products, [C] [D], and divide that result by the product of the concentrations of the reactants, [A] [B], the result will be a constant value: constant value = [C] [D] [A] [B] Now, let's look at a real example: 4 NO(g) + 6 H2O(g) <====> 4 NH3(g) + 5 O2(g) Since this system is at equilibrium, this means that the RATE of the forward reaction (--->) is equal to the RATE of the reverse (<---) reaction. From the equation for the equilibrium system, we can write what is called the equilibrium expression for this reaction: [NH3 ] 4 [O 2 ]5 equilibrium expression = [NO]4 [H 2 O]6 You should notice two things about the expression above. First, note that the numerator contains the product of the concentrations of the products, and the denominator is the product of the concentrations of the reactants. Second, you should see that the coefficients in the equation become exponents in the equilibrium expression. The reasons why we turn coefficients into exponents can be explained using kinetic–molecular theory and collision theory. However, we will simply accept this, and save the explanation for a more advanced course. If you know the concentrations of all the reactants and products in this equilibrium system, you can plug those values into this expression and solve it. The value you will 20-5 ©1997, A.J. Girondi get is a constant. Why? Well, if you somehow manage to change the concentration of NH3, for example, the concentrations of the other substances in the system will change, too. The net effect will be that the expression will maintain the same value. Since it is a constant value, it is given the symbol K. Since it involves equilibrium, it is often called Keq. (Other authors may designate it as K, Kc, etc.) So, K eq = [NH 3 ]4 [O 2 ]5 [NO]4 [H 2 O]6 Problem 2. Write the equilibrium expressions for each of the chemical situations given below. a. 2 NO2(g) <===> N2O4(g) K eq = b. N2(g) + 3 H2(g) <===> 2 NH3(g) K eq = c. Ag1+(aq) + 2 NH3(aq) <===> Ag(NH3)21+(aq) K eq = d. 2 NO(g) + 2 H2(g) <===> N2(g) + 2 H2O(g) K eq = Setting up an equilibrium expression is an extremely useful skill after you become aware of what information it can provide. In other words, what good is Keq? One way to determine the value of Keq for a particular reaction is to allow the reaction to proceed at a given temperature in such a way as to allow the products to accumulate in the reaction container. After a period of time, the reaction will reach equilibrium. At this point, it may be possible to experimentally determine the concentrations of the reactants and products in the container. The concentration values are then substituted into the equilibrium expression which is then solved for Keq. Let's consider an equilibrium system involving only gases and assume that some method is available to determine the concentrations of reactants and products at equilibrium. Equilibrium equations are frequently accompanied by the temperature at which the reaction was allowed to achieve equilibrium. Since temperature always affects equilibrium, it also affects the value of Keq. For example: H2(g) + I 2(g) <===> 2 HI (g) (at 250oC) Let's look at how the value of Keq for this system can be determined (at this temperature). 1. Start by writing the correct formula for the Keq of this reaction: 20-6 K eq = [HI] 2 [H 2 ] [I 2 ] ©1997, A.J. Girondi 2. Below are the experimentally determined equilibrium concentrations of each reactant and product for the system above. The brackets mean molarity (moles / liter of sol'n). Temp. [H2] [I 2] [HI] Experiment 1: 250 oC 0.00560 0.000590 0.01270 Experiment 2: 250 oC 0.00460 0.000970 0.01476 Notice that two experiments were probably performed under slightly different conditions, producing different equilibrium concentrations. Temperature was constant. 3. Substituting this data into the equilibrium expression: Experiment 1: K eq = (0.0127) 2 (0.0056) (0.00059) = 48.8 Experiment 2: K eq = (0.01476)2 (0.0046) (0.00097) = 48.8 Note that while the concentrations of reactants and products changed from experiment 1 to experiment 2, the value of Keq did not change. Complete the following problems. First, write the expression for Keq for each reaction, and then use the given data to calculate the value of Keq. Show all work. Problem 3. Reaction: N2O4(g) <===> 2 NO2(g) (at 520oC) Equilibrium Concentrations: [N2O4] = 0.014; [NO2] = 0.072 Keq = ____________ Problem 4. Reaction: N2(g) + 3 H2(g) <===> 2 NH3(g) (at 583oC) Equilibrium Concentrations: [N2] = 0.20; [H2] = 0.20; [NH3] = 0.016 K eq = ____________ 20-7 ©1997, A.J. Girondi Problem 5. Reaction: 2 NO(g) + O2(g) <===> 2 NO2(g) (at 500oC) Equilibrium Concentrations: [NO] = 3.49 X 10-4; [O2] = 0.80; [NO2] = 0.25 K eq = ____________ Problem 6. Reaction: NCl3(g) + Cl2(g) <===> NCl5(g) (at 520oC) Equilibrium is established when there is 0.0350 mole of NCl3, 0.0200 mole Cl2, and 0.0110 mole of NCl5 in a volume of 2.50 liters. (Hint: you must first calculate the concentrations in moles per one liter of sol'n.) K eq = _____________ SECTION 20.3 LeChatelier's Principle Earlier in this chapter it was mentioned that certain factors can disrupt an equilibrium. If one or more of these factors is altered, the equilibrium is momentarily upset. These factors are changes in: 1. 2. 3. 4. temperature pressure volume of the reaction container concentration of a reactant and or a product The exact effect of any change in reaction conditions on equilibrium was studied extensively by Henri LeChatelier, a French chemist. LeChatelier was mainly interested in what occurred when changes in the four factors listed above were made in a system at equilibrium. LeChatelier found that changes in these factors could put a "stress" on the system at equilibrium, causing it to move away from the state of equilibrium. This stress causes a change in the rate of either the forward or reverse reaction. After the system moves away from equilibrium, he observed that the amounts of reactants and products adjusted in order to restore the system to equilibrium. The results of his observations allowed him to formulate the following law, or principle, known as LeChatelier's Principle: When a stress is placed on a system at equilibrium, the system will adjust to relieve the stress and to restore equilibrium in the system. The stress changes the RATE of either the forward or the reverse reaction. LeChatelier's principle states that when a stress is applied, the RATES of the reactions will adjust so that the RATES once again become equal. 20-8 ©1997, A.J. Girondi As an example, let's look at the equilibrium system: 2 NO(g) + O2(g) <===> 2 NO2(g) Assume we mix some O2 molecules with an excess (more than needed) of NO molecules: original equilibrium concentrations NO O2 NO2 + (Excess NO present) length of arrows indicates relative rates of forward and reverse reactions LeChatelier's principle enables us to predict the direction in which the equilibrium would shift when the concentration of one or more of the products or reactants is changed. Suppose more oxygen were added to the reaction container. We could ask, "What effect would this change have on the rates of the forward and reverse reactions, and would this concentration increase affect the concentrations of the other components in the system?" You are putting a "stress" on a system in equilibrium whenever you change the concentration of any of the products or reactants. This means that, as a result of the change, the system is no longer in equilibrium. LeChatelier's principle can be explained using the collision theory. Part of this theory states that a collision must occur before a chemical reaction can take place between reactants. Furthermore, the collisions must produce enough energy and the particles must often collide in just the right way. Not all collisions result in a reaction. When more oxygen is added, there will be more collisions between the NO and O2 molecules. This will increase the rate of the forward (---->) reaction: NO O2 NO2 + (Excess NO present) rate of forward reaction is now greater some oxygen is added The system will try to remove the stress by getting rid of the excess oxygen. In this way, an equilibrium condition can be restored. The only way for the system to get rid of the extra oxygen is to have it react with NO to produce more product, NO2. So, momentarily the system "shifts to the right" and produces more product: NO O2 NO2 + (Excess NO present) more product is now produced Looking at it another way, the added oxygen molecules temporarily increase the rate of the forward (--->) reaction. However, since there are now more product molecules (NO2), they collide more often which increases the rate of the reverse reaction (<-----), and eventually a new equilibrium is established. However, this equilibrium situation is different from the original equilibrium situation because the concentrations of the reactants and the products are not the same as they were originally. The rates of the 20-9 ©1997, A.J. Girondi reactions are again equal, but they are not the same as the original rates. Because of the "shift to the right" the system now has more product than it did before more oxygen was added. Notice, too, that the shift caused the concentration of a reactant (NO) to decrease. The stress was relieved in the sense that at least some of the added oxygen has been removed. new equilibrium concentrations NO O2 NO2 + (Excess NO present) new equilibrium rates Since oxygen was added, how do you think the rates of the forward and reverse reactions in the new equilibrium compare to those in the old equilibrium? {5}______________________________________ Using these principles, predict the direction of the shift (forward ---->) or (reverse <----) in equilibrium when the concentration of Cl2 is increased in the system below, and how this will affect the concentrations of each of the other products and reactants. 4 HCl(g) + O2(g) <===> 2 H2O(g) + 2 Cl2(g) When the concentration of Cl2 is increased, the equilibrium will shift to the {6}_______________. The concentration of HCl will {7}____________________ . As the system begins to shift after the addition of Cl2, the concentration {9}_________________. of Cl 2 will {8}_________________. The concentration of O2 will The concentration of H2O will {10}_________________. Problem 7. Complete Table 20.1 below. The first blank has been completed as an example. Table 20.1 Direction of Equilibrium Shift Equation Added Substance a. CO(g) + 2 H2(g) <===> CH3OH(g) CO b. PCl5(g) <===> PCl3(g) + Cl2(g) PCl3 ________________ c. N2(g) + 3 H2(g) <===> 2 NH3(g) H2 ________________ d. CO(g) + H2O(g) <===> CO2(g) + H2(g) H2 ________________ e. 4 NO(g) + 6 H2O(g) <===> 4 NH3(g) + 5 O2(g) H2O ________________ f. 2 SO2(g) + O2(g) <===> 2 SO3(g) SO 3 ________________ g. 2 NCl3(g) <===> N2(g) + 3 Cl2(g) Cl2 ________________ h. CH3COOH(aq) <===> H1+(aq) + CH3COO1-(aq) i. C2H6(g) <===> H2(g) + C2H4(g) CH3COOH H2 20-10 Equilibrium Shift to Right or Left ------> ________________ ________________ ©1997, A.J. Girondi ACTIVITY 20.4 Lechatelier's Principle and Changes in Concentration Now that you are able to predict the direction of an equilibrium shift, you will perform an experiment that will allow you to apply this skill. The shifting of an equilibrium will be examined experimentally by using the reaction for the formation of a "complex ion" with the formula Fe(SCN)2+ by the combination of Fe 3+ ions with an SCN1- ion: Fe3+(aq) + (pale yellow) SCN1-(aq) (colorless) <===> Fe(SCN)2+(aq) (reddish) This is a good example to use because a solution of Fe3+ ions has a faint yellow color, while a solution of SCN1- ions has no color. When these two solutions are mixed, a deep-red color results due to the formation of the Fe(SCN) 2+ ion. In this activity you will examine what happens when certain stresses are placed on the equilibrium that exists between these three ions. Wear safety glasses. Before you begin note that: There are two concentrations of Fe(NO3)3 solutions of the materials shelf for this activity. Be sure to use 0.1M Fe(NO3)3 in step 1 and 0.2M Fe(NO3)3 in step 5, part b. Procedure: 1. Look at the bottle of 0.1M Fe(NO 3)3 solution. This solution contains Fe 3+ ions and NO31- ions. The color of this solution is due to the Fe3+ ions. What's the color of the aqueous Fe3+ ions? {11}__________ (The NO31- ions are only spectator ions here, so we will ignore them.) 2. Look at the bottle of 0.1M KSCN solution. The KSCN solution contains K1+ and SCN1- ions. Based on the appearance of this solution, what can you conclude about the color of the K 1+ and SCN1- ions? ___________________________ (The K1+ ions are only spectator ions here, so we will ignore them.) 3. In a test tube, mix 1 or 2 mL of the Fe(NO 3)3 solution with 1 or 2 mL of the KSCN solution. The resulting color is due to the product, FeSCN2+. What's the color of the FeSCN2+ ion?{12}________________ Look again at the equilibrium equation for this reaction. Note that FeSCN2+ is the product of the reaction between Fe3+ and SCN1-. 4. To a 100 or 150 mL beaker, add 1 mL of 0.1M Fe(NO3)3 solution. (Be sure it is the 0.1 M solution.) Next, add 2 mL of 0.1M KSCN solution. Finally, add 75 mL of distilled water. Stir well. This solution contains some reactants (Fe3+ and SCN1-) and some product (FeSCN2+). The reddish color of the product (FeSCN2+) is not very evident at this point, since it is diluted. Fe3+(aq) (pale yellow) + SCN1-(aq) (colorless) <===> Fe(SCN)2+(aq) (reddish) 5. Now place four 4-mL portions of the solution formed in step 4 into separate clean standard-sized (150 mm) test tubes. (Make sure the test tubes all have the same internal diameter.) Label the tubes A through D and perform the following tests: a.) Do nothing to tube A. Use it as a control in order to compare its color to the other tubes. b.) To tube B, add 15 drops of 0.2M Fe(NO3)3 solution and compare the color of test tube B to test tube A. Has the color gotten lighter or darker?{13}____________________ Does this color change indicate an increase or a decrease in the concentration of Fe(SCN)2+.{14}_________________________ 20-11 ©1997, A.J. Girondi When you added Fe(NO3)3, you were adding Fe 3+ ions to the equilibrium system. This resulted in an increase in the concentration of Fe3+ ions, [Fe3+], in the system. Explain the effect (on the equilibrium) of increasing the Fe 3+ concentration in the system. {15}_______________________________________ ______________________________________________________________________________ c.) To tube C, add several drops of 6M NaOH (dangerous). Handle this solution with great care. If you get any on you, wash with lots of water. Be sure to wear safety glasses! Mix well. Does the solution's color get lighter or darker?{16}______________ concentration of Does this change indicate an increase or a decrease in the Fe(SCN) 2+?{17}___________________________ Thus, adding NaOH has the overall effect of DECREASING the concentration of Fe3+ ions in the equilibrium system: Fe3+(aq) + SCN1-(aq) <===> Fe(SCN)2+(aq) (pale yellow) (colorless) (reddish) Explain the effect on the equilibrium system which resulted from the addition of NaOH: {18}___________ ______________________________________________________________________________ d.) To tube D, add about 1 mL (20 drops) of 0.1M AgNO3 solution. (Avoid getting this solution on your hands. After several hours, a dark stain can result.) Is the color of this mixture lighter or darker than that of tube A? {19}________________ Does this change indicate an increase or a decrease in the concentration of Fe(SCN)2+? {20}___________________ When you add AgNO3 to the equilibrium 11system, some of the SCN ions react with it. The SCN that reacted this way is removed from the equilibrium system. So, adding AgNO3 has the effect of DECREASING the concentration of SCN1- ions in the equilibrium system. What is the effect of this change on the equilibrium system? {21}________________________________ Problem 8. Using your observations from the activities above, complete Table 20.2 for the Fe(SCN)2+ equilibrium system. Table 20.2 Shifts in the Fe(SCN) 2+ Equilibrium System Test Substance Added Effect a. nothing none b. Fe(NO3)3 increases Fe3+ ___________________ c. NaOH decreases Fe 3+ ___________________ d. AgNO 3 SECTION 20.5 decreases Shift Forward (--->) Toward Products, or Shift Reverse (<---) Toward Reactants none SCN1- ___________________ LeChatelier's Principle and Changes in Pressure Now let's "shift" our discussion to the effects of pressure changes on equilibrium systems. For reactions which occur in liquid solutions, a change in pressure will not affect the equilibrium to any great 20-12 ©1997, A.J. Girondi extent because the volume of the liquid solution will not change very much even if extreme pressure is placed on the system. However, when a gas is involved, changing pressure will have an effect on the equilibrium. Consider the equilibrium system below: N2O4(g) <===> 2 NO2(g) If a container full of these two gases is at equilibrium and the pressure is changed, a stress is placed on the system. If we increase the pressure on the system, it will adjust itself to reduce the pressure. This is in accord with LeChatelier's principle. Since the pressure of the system is directly proportional to the number of gas molecules present, the only way to reduce the pressure (at constant temperature) is to reduce the total number of molecules in the system. This can occur if two NO2 molecules combine to form one N 2O4 molecule. Therefore, to relieve the strain caused by increasing the pressure, the equilibrium will shift to the left toward the reactant which is N2O4. Increasing the pressure on a gaseous system at equilibrium causes the equilibrium to shift to the side with the fewest number of molecules. On the other hand, if the pressure is decreased, more gas molecules must be formed to bring the system's pressure back to equilibrium. In this case, some N2O4 molecules will decompose into two NO2 molecules. This will increase the number of gas molecules present, thereby increasing the pressure in the system. Decreasing the pressure on a gaseous system at equilibrium causes the equilibrium to shift to the side with the larger number of molecules. Consider a container in which the reaction shown at right has come to equilibrium: In which direction will the equilibrium decreased? {22}_____________ Explain: N2(g) + 3 H2(g) <===> 2 NH3(g) shift if the pressure on the system above is {23}____________________________________________ ______________________________________________________________________________ How should the pressure be changed on the system above in order to produce a larger amount of ammonia, NH3? {24}____________________ Explain: {25}_________________________________ ______________________________________________________________________________ Shown at right is another reaction which you studied previously involving the formation of HI gas by the reaction: H2(g) + I2(g) <===> 2 HI(g) How will the equilibrium in the reaction above be affected by an increase in pressure? Explain: {26}_____________ {27}_____________________________________________________________________ In part a of Problem 9 below, the equation shows 1 molecule of N2O4 on the left side and 2 molecules of NO2 on the right side. Therefore, a decrease in pressure would cause a shift of the equilibrium to the right (toward more molecules). In part b of Problem 9, the equation reveals 2 molecules on the left side of the arrow and only 1 molecule on the right side. Indicate the direction of shift when the pressure decreases in the space provided. Then, complete Problem 9. 20-13 ©1997, A.J. Girondi Problem 9. Complete Table 20.3 below. Indicate the direction in which the equilibrium will shift if the pressure is changed in the manner indicated. Table 20.3 Pressure Changes and Equilibrium Shifts Equilibrium Equations Pressure Shifts a. N2O4(g) <===> 2 NO2(g) b. PCl3(g) + Cl2(g) <===> PCl5(g) c. 2 SO3(g) <===> 2 SO2(g) + O2(g) d. 2 CO(g) + O2(g) <===> 2 CO2(g) e. N2(g) + O2(g) <===> 2 NO(g) f. 2 H2(g) + O2(g) <===> 2 H2O(g) g. C3H8(g) + 5 O2(g) <===> 3 CO2(g) + 4 H2O(g) h. 2 N2O(g) <===> 2 N2(g) + O2(g) i. 2 HBr(g) <===> H2(g) + Br2(g) j. CH4(g) + 2 O2(g) <===> CO2(g) + 2 H2O(g) decreased decreased decreased decreased increased decreased decreased increased increased increased ----> _____ _____ _____ _____ _____ _____ _____ _____ _____ SECTION 20.6 LeChatelier's Principle and Changes in Volume LeChatelier's principle can also be used to predict the effect of changes in the volume of the reaction container on equilibrium. Equilibrium shifts caused by volume changes are similar to those caused by pressure changes. When the volume of a particular reaction container is reduced, the molecules get crowded together and collide more frequently. This stress can be relieved by decreasing the number of molecules present. Look again at the nitrogen dioxide equilibrium: 2 NO(g) + O2(g) <===> 2 NO2(g) A decrease in the volume of the reaction container can be compensated for by forming fewer molecules. The result is that the equilibrium above shifts to the right to form more molecules of NO2. In the process, three molecules of reactants will become two molecules of product. In general terms, this relationship can be stated as follows: For reactions in which there is a change in the number of gas molecules, a decrease in the volume favors the reaction that produces fewer molecules. An increase in volume favors the reaction that produces the larger number of molecules. If the forward and reverse reactions of an equilibrium system produce the same number of molecules, then changes in volume H 2(g) + Cl 2(g) <===> 2 HCl (g) have no effect on the system. Consider the system shown at right: Note that the forward reaction (--->) produces two molecules of HCl, while the reverse reaction (<---) also produces two molecules – one H 2 and one Cl2. (This is also true for pressure changes.) 20-14 ©1997, A.J. Girondi Problem 10. Based on this generalization, predict whether equilibrium shifts toward the products (--->) or the reactants (<---) in each example below when the volume of the reaction container is decreased: a. PCl5(g) <===> PCl3(g) + Cl2(g) _____________________________ b. N2(g) + 3 H2(g) <===> 2 NH3(g) _____________________________ c. 2 CO(g) + O2(g) <===> 2 CO2(g) _____________________________ Problem 11. In which direction will the systems below shift if the volume of the reaction container is increased: a. H2(g) + I 2(g) <===> 2 HI (g) _____________________________ b. CO(g) + 2 H2(g) <===> CH3OH(g) _____________________________ c. C3H8(g) + 5 O2(g) <===> 3 CO2(g) + 4 H2O(g) _____________________________ Problem 12. Study the equations in Table 20.4. Determine the direction of equilibrium shift. Answer by writing either "forward" or "reverse," or by using arrows: ---> or <---. These equations are a little more complicated, because they involve liquids and solids in addition to gases. Pressure changes do not have much, if any, effect on liquids and solids, so you should only consider molecules of gases in an equilibrium system when effects of pressure changes are being evaluated. Therefore, substances which are liquids or solids with subscripts (l) or (s) in the equations should be ignored in Table 20.4. Remember: As you complete Table 20.4, ignore any substances below which are solids or pure liquids. Table 20.4 Equilibrium Shifts and Pressure Changes Equilibrium Equation Pressure Change Shift Direction a. CO(g) + H2O(l) <===> CO2(g) + H2(g) increase ___________ b. 4 FeS(s) + 7 O2(g) <===> 2 Fe2O3(s) + 4 SO2(g) increase ___________ c. 4 NH3(g) + 5 O2(g) <===> 4 NO(g) + 6 H2O(l) decrease ___________ d. C(s) + O2(g) <===> CO2(g) decrease ___________ When an equilibrium system involving gases is present in a closed container, changes in volume cause changes in pressure. If the size of the container is decreased (volume is decreased) the pressure goes up. For example, imagine that you are squeezing a balloon which contains gases into a smaller volume. To relieve the increased pressure, the system will shift in the direction of fewer gas atoms or molecules. If the size of the container is increased, the system will shift in the direction which will provide more gas atoms or molecules which can occupy the added volume of space. Changes in volume – like changes in pressure – do not affect solids or pure liquids. 20-15 ©1997, A.J. Girondi Problem 13. Complete Table 20.5 below. Reminder: Ignore any substances which are solids (s) or pure liquids (l). Table 20.5 Equilibrium Shifts and Volume Changes Equilibrium Equation Volume Change a. CO(g) + H2O(l) <===> CO2(g) + H2(g) increase ___________ b. 4 FeS(s) + 7 O2(g) <===> 2 Fe2O3(s) + 4 SO2(g) increase ___________ c. 4 NH3(g) + 5 O2(g) <===> 4 NO(g) + 6 H2O(l) decrease ___________ d. C(s) + O2(g) <===> CO2(g) decrease ___________ SECTION 20.7 Shift Direction Equilibrium Systems Involving Solids and / or Liquids Many of the equilibrium equations that you have seen so far in this chapter have reactants and products which are either gases (g) or water (aqueous) solutions (aq). Changes in volume or pressure have an effect on gases. Changes in concentration have an effect on both gases or aqueous solutions. However, changes in volume, pressure, or concentration in equilibrium systems do not affect solids or pure liquids. They are not variables that play a role in determining the value of Keq. We will save an indepth discussion of the reasons for this for a future chemistry course! Solids and pure liquids are never included in K eq expressions. Examine the following examples. CaO (s) + CO 2(g) H 2 O(l) + HF(g) <===> CaCO 3(s) <===> H 3O 1+(aq) + F 1- (aq) K eq = 1 [CO 2 ] K eq = [H 3 O1+ ] [F 1- ] [HF] Problem 14. Write the equilibrium (Keq) expressions for the following systems. substances which are solids (s) or pure liquids (l). a. BaCO3(s) <===> BaO(s) + CO2(g) K eq = b. HCN(aq) + H2O(l) <===> H3O1+(aq) + CN1-(aq) K eq = c. CuSO 4•3H 2O(s) + 2 H2O(g) <===> CuSO4•5H 2O(s) K eq = 20-16 Do not include ©1997, A.J. Girondi SECTION 20.8 LeChatelier's Principle and Changes in Temperature Changes in temperature will also create a stress on a system at equilibrium. In this case, the effect is more complicated than that for stresses caused by concentration and pressure changes. This is because K eq is temperature dependent, meaning that its value will be numerically different at different temperatures for a given reaction. Thus, heating or cooling an equilibrium system will result in a shifting of the equilibrium forward (to the right) or reverse (to the left), depending on which of the two reactions is exothermic and which is endothermic. Let's first consider a reaction that is exothermic (as most reactions tend to be). Such a reaction can be written as: reactants <===> products + heat energy Suppose the temperature of the system is increased. This involves adding heat energy to the system. This is like increasing the amount of a product of the reaction. (Since heat appears on the right side of the equation, it can be considered a product.) You might think of this as placing a stress on the right side of the equation. To relieve this stress, the equilibrium will shift to the left. The concentrations of products will then {28}_____________ , and the concentrations of reactants will {29}_______________. . Lowering the temperature of an exothermic reaction that has come to equilibrium should have the opposite effect. Lowering the temperature amounts to removing heat energy from the system. This stress will result in the equilibrium shifting to the right. This will products and {31}________________ {30}______________the concentration of the concentration of reactants. Next, let's consider an endothermic reaction that can be written as: reactants + heat energy ----> products The heat energy can be regarded as part of the reaction, in this case as one of the reactants. Adding heat energy by increasing the temperature of this system at equilibrium amounts to placing a stress on the left side of the equation. To relieve the stress, the concentration of reactants will decrease by forming more {32}___________. (The system shifts to the right.) reaction will cause the equilibrium to shift to the Decreasing the temperature of this endothermic {33}_____________. In summary, you know that in an equilibrium system one of the reactions is exothermic and the other is {34}______________. Raising the temperature (which amounts to adding heat) will cause an increase in the rate of both reactions. In terms of collision theory, why would this be true? {35}_________ ______________________________________________________________________________ However, when heat is added, the rate of the endothermic reaction (the reaction which uses heat) will generally increase more than the rate of the exothermic reaction (which gives off heat). In other words, adding heat causes a shift in favor of the endothermic reaction. Removing heat, causes a shift in favor of the exothermic reaction. 20-17 ©1997, A.J. Girondi Problem 15. Complete Table 20.6 by indicating the direction of the equilibrium shift when the temperature is changed as indicated. Table 20.6 The Effect of Temperature Changes on Equilibrium Systems Equilibrium Equation a. H2(g) + Cl2(g) <===> 2 HCl(g) + 44184 kJ b. 50.2 kJ + H2(g) + I2(g) <===> 2 HI (g) c. CH4(g) + 2 O2(g) <===> CO(g) + 2 H2O(l) + 887 kJ d. C(s) + O2(g) <===> CO2(g) + 393 kJ e. N2(g) + 3 H2(g) <===> 2 NH3(g) + 46 kJ f. 2376 kJ + 8 SO2(g) <===> S8(s) + 16 O2(g) g. 75.3 kJ + CH4(g) <===> C(s) + 2 H2(g) ACTIVITY 20.9 Temperature Change Shift Direction decrease decrease increase decrease increase decrease increase __________ __________ __________ __________ __________ __________ __________ Testing LeChatelier's Principle With Cobalt Ions This next activity will allow you to study the effects of concentration and temperature changes on an equilibrium system that exists between two different cobalt complexes. In water solutions, the Co2+ ion is pink. The pink color is due to the existence of a complex ion with the formula: Co(H 2O)62+. This is the form in which cobalt normally exists in water. When Cl1- ions are also present in high concentrations, the Co(H2O)62+ is converted to Co(H2O)4Cl2, which is blue: Co(H2O)62+(aq) + 2 Cl1-(aq) <===> Co(H2O)4Cl2(aq) + 2 H2O(l) pink blue The two colored Co 2+ species can be converted to one another by appropriate changes in the concentration of Cl 1- ion or of water and by changes in temperature. Follow the procedure below, and be sure to wear your glasses. 1. Mark two 50 mL Erlenmeyer flasks or 50 mL beakers "1" and "2." 2. Prepare the following two solutions: Solution 1: Dissolve 0.50 g of CoCl2•6H 2O in 10 mL of 6M HCl (hydrochloric acid). Handle the HCl with care. The high concentration of Cl1- in the HCl pushes the equilibrium to the right and most of the cobalt in this mixture is in the form of Co(H2O)4Cl2. Stir the solution until the solid is completely dissolved. What is the color of Co(H2O)4Cl2 in solution 1? {36}______________ Solution 2: Dissolve 0.50 g of CoCl2•6H 2O in 15 mL of water. The high concentration of H 2O in this mixture pushes the equilibrium to the left and most of the cobalt in this mixture is in the form of Co(H2O)62+. Stir the solution until the solid is completely dissolved. What is the color of Co(H2O)62+ in solution 2? {37}_______________ Adding HCl (and therefore Cl 1-) to this system causes it to shift to the turns more {39}___________. {38}____________ and the color Adding H2O to this system causes it to shift to the {40}____________and the color turns more {41}________________. 20-18 ©1997, A.J. Girondi 2. Add 5 mL of water (or more if necessary) to solution 1 until a color change occurs. Now what is the color of the solution? {42}_______________ Heat the flask of solution 1 on a hotplate or with a burner until a color change occurs. What is the color of the heated solution 1?{43}______________ 3. What do you think will happen to the color of the solution if it is cooled? {44}__________________ Now place the flask of solution 1 into an ice water bath. Allow the flask to remain in the ice water bath until a change occurs. Was your prediction correct?{45}_________________ 4. Keeping in mind that solution 2 contains CoCl2•6H 2O, what two things could you do to solution 2 to get it to form more Co(H2O)4Cl2? {46}______________________________________________________ Now do both of these two things – simultaneously – to a 15 mL portion of solution 2 in a 50 mL flask or beaker. Describe the result: {47}______________________________________________________ Did you manage to make more Co(H2O)4Cl2 in solution 2? {48}_________________________________ How do you know? {49}_____________________________________________________________ 5. Based on your results, is the forward (--->) reaction for this cobalt system endothermic or exothermic? {50}________________. Explain how you know: {51}_______________________________________ ______________________________________________________________________________ If you have a little time left try this. Add some solid CoCl2 to a small volume of water in a test tube and shake to dissolve. Pour the solution onto a piece of filter paper. Note that it is red in color. Now hold the wet filter paper with your crucible tongs and warm it gently over the flame of a lab burner. Be careful not to ignite the paper as it drys. Note the color change as you evaporate the water out of the system. Wet the paper withplain water to restore the red color. SECTION 20.10 A Review of LeChatelier's Principle 1. State LeChatelier's principle: {52}___________________________________________________ ______________________________________________________________________________ 2. In which direction will an equilibrium system shift if the concentration of one of the products is decreased (at constant T and P)? {53}__________________________________________________ 3. In which direction will an equilibrium shift if a reaction has more gas molecules on the left (reactant side) than on the right (product side) and if the pressure of the system is increased? {54}__________________ 4. Suppose for a hypothetical equilibrium system such as A(g) <===> B (g), the forward reaction (from left to right) is exothermic. In which direction (forward --->) or (<--- reverse) will the equilibrium shift if the temperature is increased? {55}___________________ 5. Will an equilibrium reaction shift forward (--->) or reverse (<---) if the concentration of one of the reactants is decreased (at constant T and P)? {56}__________________ 6. For a reaction involving equal numbers of gas molecules on both sides of the equation, will the equilibrium shift forward or reverse if the pressure is decreased? {57}____________________________ 7. For an endothermic reaction, will the equilibrium shift toward products or reactants if the temperature is increased? {58}_______________________________ 20-19 ©1997, A.J. Girondi SECTION 20.11 Using K e q Values to Make Predictions Equilibrium constants (Keq) are quite useful to chemists because they provide a clue about the amount of product that can be produced in a given chemical reaction. Normally, a chemical reaction is carried out because an experimenter wants to obtain and use the product. Ideally, one would like to get a 100% yield, which means that all of the reactants would be converted into products. However, 100% yields are not always possible. Instead, a system may go to equilibrium resulting in less than a 100% yield. We are able to predict the extent to which reactants will be converted into products based on the size of K eq. Remember that Keq is related to a ratio involving products over reactants. Look at the K eq values calculated below: K eq = 100 2 = 50; K eq = 10 = 1 X 10 4 ; 0.001 K eq = 50 0.02 = 2.5 X 10 3 1. Suppose the three Keq values above represent very similar reactions. Which K eq value represents the reaction which produced the most product?{59}_________________ 2. Which Keq value represents the reaction which produced the least product?{60}__________________ 3. Explain how Keq values can be used to determine which of a series of similar reactions will produce the most product? {61}________________________________________________________________ 4. For the gaseous reaction A + B <===> C, a chemist is interested in getting as large a yield of the product C as possible. He varies the reaction temperature which is the one variable that can change the K eq value of a system. At 300 oC he experimentally calculates that K eq for the system is 26.2. At 10oC he finds that Keq = 0.012. To maximize the amount of C produced, should the chemist heat the reaction container or cool it?{62}_______________ Explain: {63}_____________________________________ ______________________________________________________________________________ Problem 16. For the equilibrium system: 2 SO 2(g) + O2(g) <===> 2 SO 3(g) the value of K eq at room temperature is 30.0. Predict the concentration of SO3 gas in the system when it is at equilibrium if the other concentrations are: [SO 2] = 0.20M and [O2] = 0.30M. __________M 20-20 ©1997, A.J. Girondi SECTION 20.12 Review Problems Problem 17. A five-liter flask contains the system: CO(g) + Cl2(g) <===> COCl2(g). The flask contains 1.50 moles of CO, 1.00 mole of Cl2, and 4.00 moles of COCl2. (These are all gases.) Calculate the value of the equilibrium constant for this system. (Remember values used must be in moles per 1.00 liter.) ________________ Problem 18. In a 1-liter flask the following system is at equilibrium: C (s) + H2O(g) <===> CO(g) + H2(g). The amounts of substances present in the 1-liter flask are 0.16 mole of C, 0.58 mole of H2O, 0.15 mole of CO, and 0.15 mole of H 2. Calculate the value of Keq for this system. (Note that C is a solid while the other substances are in the gas phase.) ________________ There is a supplementary discussion of another type of equilibrium constant known as the solubility product constant, Ksp, in Appendix E of your ALICE materials. Ask your teacher if you should study that Appendix, or if you should end Chapter 20 here. Go To Appendix E??? (Ask the Instructor) 20-21 ©1997, A.J. Girondi SECTION 20.13 Learning Outcomes Equilibrium is an extremely important topic in chemistry and in all of the sciences. It will be very useful to you in the upcoming chapter on acids and bases. Look over the learning outcomes and make sure that you have mastered each of them. Check them off once you are satisfied. Arrange to take any quizzes or exams on Chapter 20, and then move on to Chapter 21. _____1. Define equilibrium and state the general characteristics of a system in equilibrium. _____2. Write equilibrium expressions for chemical systems involving solids, liquids, and gases. _____3. Calculate Keq values given the equilibrium concentrations of the products and reactants. _____4. State LeChatelier's principle in general terms. _____5. Use LeChatelier's principle to predict the direction an equilibrium will shift if there is a change in pressure, concentrations, temperature, or volume of the reaction container. _____6. Given the Keq values for two or more similar systems in equilibrium, predict which system contains the greater concentration of products. The following learning outcomes are to be included only if you studied the material concerning Ksp in Appendix E. _____7. Determine the identity of unknown chemicals by testing and comparing them with a set of known chemicals. _____8. Given the solubility of a substance, calculate its Ksp value. _____9. Given the Ksp value and the concentration of one ion, calculate the concentration of the other ion. 20-22 ©1997, A.J. Girondi SECTION 20.14 Answers to Questions and Problems Questions: {1} beaker A; {2} molecules are escaping; {3} no; {4} molecules that escape cannot return to liquid phase; {5} they will be greater; {6} left; {7} increase; {8} decrease; {9} increase; {10} decrease; {11} amber (depends on your color vision); {12} deep red (depends on your color vision); {13} darker; {14} increase; {15} causes more collisions between Fe3+ ions and SCN1- ions and shifts system toward right (products); {16} lighter; {17} decrease; {18} system shifted toward the left (toward reactants); {19} lighter; {20} decrease; {21} system shifts toward the left (toward reactants}; {22} shift to left toward reactants; {23} decreased pressure is a stress, so system shifts to the left to form more molecules to help increase the pressure; {24} increase the pressure; {25} increasing the pressure will create a stress which the system will try to relieve by forming fewer molecules (shift to the right) which will help lower the pressure; {26} no effect; {27} since both sides of equation have same number of molecules, shifting would not change pressure; {28} decrease; {29} increase; {30} increase; {31} decrease; {32} products; {33} left toward reactants; {34} endothermic; {35} more collisions between particles occur at higher temperatures; {36} blue (depends on your color vision); {37} pink (depends on your color vision); {38} right; {39} blue; {40} left; {41} pink; {42} pink; {43} blue; {44} will turn pink; {45} I hope so! {46} heat it and add more HCl; {47} solution should shift toward blue; {48} yes; {49} because of the change in color; {50} endothermic; {51} because adding heat speeds up an endothermic reaction more than it speeds up an exothermic one; {52} when a stress is placed on a system at equilibrium, the system will adjust to relieve the stress and to restore equilibrium in the system; {53} shifts to the right toward products; {54} shifts to the right toward products; {55} shift to the left toward reactants; {56} shifts to the left (reverse) toward reactants; {57} neither, it will not shift either way; {58} shift to the right toward products; {59} 1X 10 4; {60} 50; {61} bigger Keq value means more products; {62} heat it; {63} since K eq is greater at the higher temperature, the forward reaction which forms C is endothermic Problems: 3. a. NaCl(s) <===> Na1+(aq) + Cl1-(aq); b. Fe2(SO4)3(s) <===> 2 Fe3+(aq) + 3 SO42-(aq) c. BaBr2(s) <====> Ba2+(aq) + 2 Br1-(aq) a. Keq = [N2O4] / [NO2]2; b. Keq = [NH3]2 / [N2] [H2]3; c. Keq = [Ag(NH3)21+] / [Ag1+] [NH3]2 d. Keq = [N2] [H2O] 2 / [NO]2[H2]2 0.37 4. 0.16 5. 6.4 X 105 6. 39.3 7. a. --->; b. <---; c. --->; d. <---; e. --->; f. <---; g. <---; h. --->; i. <--- 8. a. none; b. --->; c. <---; d. <--- 9. a. --->; b. <---; c. --->; d. <---; e. no effect; f. <---; g. --->; h. <---; i. no effect; j. no effect 1. 2. 10. a. <---; b. --->; c. ---> 11. a. no effect; b. <---; c. ---> 12. a. <---; b. --->; c. <---; d. no effect 13. a. --->; b. <---; c. --->; d. no effect 14. a. Keq = [CO2]; b. Keq = [H3O1+] [CN1-] / [HCN]; c. Keq = 1 / [H2O] 2 15. a. --->; b. <---; c. <---; d. --->; e. <---; f. <---; g. ---> 20-23 ©1997, A.J. Girondi 16. 0.60 17. 13.3 18. 3.9 X 10-2 20-24 ©1997, A.J. Girondi