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Name:______________________________ Date:_____________ CH105 Pre-Lab 7: Chemical Equilibrium Directions: Read the Goals, Background, Safety, and Procedure sections for this experiment, then answer the following questions in the space provided. For short answer questions, write complete sentences and provide a reason for the answer. For calculation questions, show all work and report answers in a box with the appropriate significant figures and units. Pencil is acceptable for this assignment. 1. Write the forward reaction for the following equilibrium equation. Fe3+ + NCS- → ← FeNCS2+ 2. Write the reverse reaction for the following equilibrium equation. Fe3+ + NCS- → ← FeNCS2+ 3. Write the Keq expression for the following equilibrium equation. 3 N2H4 (g) + 4 ClF3 (g) → ← 12 HF (g) + 3 N2 (g) + 2 Cl2 (g) 4. The equilibrium constant for the following reaction is 0.212 at 100°C. N2O4 (g) → ← 2 NO2 (g) Is the forward or reverse reaction favored at 100°C?____________________________ Explain. 5. If you cool the equilibrium system from #4 and the K value decreases, is the forward reaction endothermic or exothermic?_________________ Explain. 1 PCC Experiment 7: Chemical Equilibrium Goals: • Represent chemical equilibrium with models. • Determine an equilibrium constant using models. • Observe Le Chatelier’s Principle for the iron-thiocyanate and acetic acid-sodium acetate systems. • Based on observations, determine whether the forward reaction is endothermic or exothermic for the iron-thiocyanate equilibrium system. Purpose: Explore chemical equilibrium through models and chemical reactions. Visualize Le Chatelier’s Principle with chemical systems. Background: Dynamic Chemical Equilibrium Defined Until this point in your chemistry career, almost all of the reactions you have studied involved one strong arrow pointing to the right (→). This arrow showed you that reactions began with reactants and ended with products. While this way of representing chemical equations is useful to understanding the concepts of reactants and products, it is slightly deceptive. In reality, not all chemical reactions proceed only in a forward direction. Some chemical reactions can reverse and turn products back into reactants. For instance, imagine you are making a sandwich in your kitchen. You take 2 pieces of bread for every slice of cheese and make a cheese sandwich. 2 slices of bread + 1 slice of cheese → 1 cheese sandwich Logically, your reactants (bread and cheese) can become a sandwich, if you provide the reaction energy. This reaction represents the forward reaction. This is not the only direction the reaction can proceed, though. Imagine you have a cheese sandwich, but you want to eat only the bread. You could break down your cheese sandwich into 2 slices of bread and 1 slice of cheese. 1 cheese sandwich → 2 slices of bread + 1 slice of cheese This second reaction would be considered the reverse reaction because you took a product (cheese sandwich) and turned it back into reactants. 2 PCC Chemical Equilibrium • 3 Standing in your kitchen, you could create an equilibrium system by making cheese sandwiches and breaking them back down into bread and cheese. 2 slices of bread + 1 slice of cheese → ← 1 cheese sandwich Many chemical systems establish chemical equilibrium in a similar manner: reactants create products which in turn create reactants. On a macroscopic scale, it appears that no reaction is occurring; however, on the molecular level, the process is occurring with reactants creating products as fast as products are creating reactants. In your kitchen, this equilibrium would make it appear as if the total number of sandwiches, slices of bread, and slices of cheese were not changing (macroscopic); however, you would still be breaking apart sandwiches as fast as you were making sandwiches (molecular). This constant action on the molecular level is the reason chemical equilibrium is frequently referred to as dynamic chemical equilibrium. Chemical Equilibrium and Concentration In most chemical systems, the concentration of reactants and products are not equal at equilibrium. What is equal at equilibrium is the rate (or speed) of the forward and reverse reactions. In your kitchen, you might have 25 cheese sandwiches made and only 2 slices of bread and 1 slice of cheese. As long as the rate of sandwich making and sandwich separation are equal, you will have 25 cheese sandwiches, 2 slices of bread, and 1 slice of cheese. The concentrations of reactants and products do not have to be equal at equilibrium. Consequently, chemists try to account for these differences in concentration by writing equilibrium constant expressions. These expressions help chemists to identify whether the forward or reverse reactions are favored from a concentration standpoint. Consider the general equation below: aA +bB → ← cC + dD A chemist would look at that equation and write an equilibrium constant [C]c [D]d expression (Keq) that looks like: K eq [A]a [B]b If the value of Keq is greater than 1, then the products (produced by the forward reaction) are favored. If the value of Keq is less than 1, then the reactants (produced by the reverse reaction) are favored. PCC 4 • Experiment 7 Let’s return to our cheese sandwich example. To write the Keq expression, we must look at our equation. 2 slices of bread + 1 slice of cheese → ← 1 cheese sandwich Here, the Keq expression would be: K eq [cheese sandwich]1 [bread]2 [cheese]1 When the coefficient in an equilibrium expression is 1, it is not usually written as an exponent. Any number to the first power is equal to itself. The Keq expression could be rewritten as: Keq [cheese sandwich] [bread]2 [cheese] The value of the Keq expression could be calculated from the equilibrium concentration values for bread, cheese, and cheese sandwiches. Earlier, the equilibrium concentration of cheese sandwiches was said to be 25, while there were 2 slices of bread and 1 slice of cheese. Plugging these numbers into the Keq expression, we find the value of Keq is 6.25. K eq [25] 6.25 [2]2 [1] Because the Keq value is greater than 1, the products (or forward reaction) is favored. The equilibrium is said to be “towards the right.” We already suspected that because we knew the number of sandwiches was greater than the amount of bread and cheese; however, the Keq value confirmed it. This is extremely important at times that we do not know the concentrations of reactants or products, but we do know the Keq value. When the equilibrium constant expression is written for the ionization reaction between a weak acid and water, it is called a Ka. Acetic acid, HC2H3O2, is a weak acid. When placed in water, the following equilibrium reaction occurs: HC2H3O2 (aq) + H2O (l) → ← C2H3O2 (aq) + H3O+ (aq) The concentration of water will not change significantly because it is the solvent for this reaction. Liquid and solid concentrations are not included in equilibrium constant expressions; only gas and aqueous phases, which can change concentration, are included. As long as some liquid or solid is present, the reaction will occur. Chemical Equilibrium • 5 Consequently, the Ka expression for acetic acid is: Ka [C 2 H 3O -2 ] [H 3O ] [HC 2 H 3O 2 ] Several strong acids exist. Among these is hydrochloric acid. When placed in water, hydrochloric acid completely ionizes as shown in the following reaction: HCl (aq) + H2O (l) → Cl- (aq) + H3O+ (aq) Notice the arrow in this reaction is pointing only to the right. This reaction is not an equilibrium system. The reaction is said to go to completion. Consequently, the number of chloride ions and hydronium ions formed by this reaction are equal to the number HCl molecules dissolved in water. That is not the case for weak acids. For weak acids, the number of anions and hydronium ions produced from each molecule of weak acid has to be determined from the pH of the solution, known Ka value, or percent ionization data. Le Chatelier’s Principle Once a chemical system establishes equilibrium, it will resist changes introduced as stresses to the chemical system. These stresses can include changes in concentration, pressure, volume, heat, temperature, or presence of a catalyst. Changes in Concentration Whatever stress is introduced, the system will work to counteract the stress and reestablish equilibrium. Consequently, if more reactant or product is added to a chemical system at equilibrium, the chemical system will shift to reestablish equilibrium by using part of the added substance. The shift in the favored reaction happens because the value of Keq at a given temperature is a constant. In order to maintain the Keq for the temperature, the chemical system has to make products (when extra reactants are present or some products are removed) and use products (when extra products are present or some reactants are removed). Table 1 summarizes the various concentration stresses a chemical system at equilibrium can experience as well as the responses of the system in order to maintain Keq at a given temperature. PCC 6 • Experiment 7 Table 1. Summary of Concentration Stresses, Responses, and Keq Effects. Stress Response Keq increase in a reactant use reactants to produce products does not change as long as decrease in a reactant use products to make reactants temperature is increase in a product use products to make reactants constant decrease in a product use reactants to produce products This process is impossible to observe if the reactants and products all look alike; however, when the reactants and products have different colors, the shift of the chemical system can be seen with the naked eye. Changes in Volume and Pressure Volume and pressure changes can be treated in much the same way as changes in concentration when the equilibrium system involves gases. As long as the system maintains constant temperature, the Keq value will not change. Consequently, as the volume of a gaseous mixture that was at equilibrium is increased, the chemical system will shift to minimize this increase in volume by increasing the pressure. Remember, volume and pressure are inversely related. When the volume of a system increases, the pressure is lowered. The chemical system increases pressure by shifting to the side of the reaction with more moles of gas. As the number of gas molecules increase, the pressure increases because these newly formed molecules will run into the walls of the container more frequently than the lower number of molecules did. Similarly, when the volume of a chemical system at equilibrium is decreased, the equilibrium will shift to the side of the reaction with fewer moles of gas in order to lower the pressure. Table 2 summarizes the results of volume and pressure stresses on a chemical system at equilibrium. Table 2. Summary of Volume and Pressure Stresses, Responses, and Keq Effects. Stress Response Keq volume increase does not change shift to side with more moles of gas as long as pressure decrease temperature is volume decrease shift to side with fewer moles of gas constant pressure increase Changes in Heat Heat can be treated as a reactant if the chemical system is endothermic or a product if the chemical system is exothermic. An addition of heat will shift the equilibrium in the direction that the heat is used. Heat is not accounted for in the Keq expression. Consequently, the Keq value will actually change with as the amount of heat changes. As the reaction Chemical Equilibrium • 7 compensates for the change in heat, the concentrations of reactants and products will change. This change will minimize the effect of changing the heat of the system; however, because heat is not part of the Keq equation, the Keq value will change to reflect the new concentrations of the reactants and products. (Before reading the next section on temperature changes, remember, heat and temperature are not the same thing. A substance at its boiling point or freezing point (constant temperatures) can experience a change in heat without experiencing a change in temperature. Most equilibrium systems are not studied at their boiling point or freezing point, though. Consequently, a change in heat generally corresponds to a change in temperature and vice versa. Also note, heat can be written into an equation as a reactant or a product, as appropriate. Occasionally, the heat for the reaction is written quanitatively to the right of the equation as a H value. If the H value is positive, the forward reaction is endothermic. If the H value is negative, the forward reaction is exothermic.) Changes in Temperature The temperature of a reaction will affect the Keq value. As the temperature is increased, the reaction that counteracts that effect (endothermic) will become dominant. As the temperature is decreased, the reaction that counteracts that effect (exothermic) will become dominant. Consequently, by observing a chemical system at equilibrium at two different temperatures, the forward reaction can be identified as endothermic or exothermic depending on the favored conditions. For instance, the following equilibrium is established when cobalt (II) chloride, CoCl2, is dissolved in hydrochloric acid, HCl (aq): → Co(H2O)2+ 6 (aq) + 4 Cl (aq) ← CoCl4 (aq) + 6 H2O (l) pale pink deep blue The Co(H2O)2+ 6 and CoCl4 (aq) are called complex ions. They have unique colors that allow the forward and reverse directions of a reaction to be distinguished from one another. At room temperature, the equilibrium condition for this reaction looks violet because both ions exist in solution. When the reaction is heated and the temperature increases, more CoCl-4 (aq) ions form and the reaction looks more blue. When the reaction is cooled and the temperature is lowered, more Co(H2O)2+ 6 (aq) ions form and the reaction looks pink. PCC 8 • Experiment 7 If this chemical system at equilibrium is working to counteract the stress of increased heat by producing more CoCl-4 (aq) ions and counteract the stress of decreased heat by producing more Co(H2O)2+ 6 (aq) ions, heat must be a reactant. → Co(H2O)2+ 6 (aq) + 4 Cl (aq) + heat ← CoCl4 (aq) + 6 H2O (l) As this reaction at equilibrium is heated by an outside source, the chemical system works to reduce the heat by using it and produces more of the product ions as a result. As the reaction at equilibrium is cooled by an outside source, the chemical system works to increase the heat and makes more reactant ions as a result. Consequently, we are able to deduce that the forward reaction is endothermic (uses heat), while the reverse reaction is exothermic (produces heat). Table 3 summarizes the relationship between heat and temperature stresses on an equilibrium system and the possible responses of the system depending on whether the forward reaction is endothermic or exothermic. Table 3. Summary of Heat/Temperature Stresses, Possible Responses of the Equilibrium System Based on Nature of the Forward Reaction, and Resulting Change in Keq. Stress Possible Responses Forward Reaction is… Keq make more products endothermic ↑ heat/raise temperature make more reactants exothermic ↓ make more products exothermic ↑ cool/lower temperature make more reactants endothermic ↓ Presence of a Catalyst When a catalyst is added to a reaction, it does not actually react. It merely lowers the activation energy of the reaction. That allows both the forward and reverse reactions to occur more quickly. The concentration of reactants and products do not change when a catalyst is added. Consequently, the equilibrium does not shift when a catalyst is added. On the macroscopic scale, no visible changes occur to a system at equilibrium to which a catalyst has been added. On the molecular scale, the rate of changing between reactants and products is faster. To go back to the cheese sandwich analogy, a catalyst would enable you to take apart and reassemble the cheese sandwiches faster. The number of cheese sandwiches, slices of bread, and slices of cheese would not change. Chemical Equilibrium • 9 Reactions in today’s experiment While examining Le Chatelier’s Principle, you will make the ironthiocyanate equilibrium system by reacting 1 M Fe(NO3)3 (aq) and 1 M NH4NCS (aq). The net ionic equation for this reaction is shown below. Fe3+ (aq) + NCS- (aq) → ← FeNCS2+ (aq) yellow colorless red Depending on the Keq value for this reaction at room temperature, this equilibrium system may be yellow (Keq less than 1; reactants favored), red (Keq greater than 1; products favored), or orange (Keq approximately equal to 1; reactants and products both present). You will keep some of this solution as a color reference as you perform several mini-experiments on this equilibrium system. As you perform the experiments, watch to see if the color changes to more yellow (reverse reaction favored and more reactants produced) or more red (forward reaction favored and more products produced). In preparation for the lab, you might want to predict which type of color change you would expect from the following reactions: If you add Fe(NO3)3 (aq), you will essentially add more Fe3+ to the system at equilibrium. What type of stress is that? If you said, “It’s a concentration stress, I am adding more reactant ions.” You would be correct. What effect will that have on the chemical system? Would you expect the solution to become more yellow or more red? Will the equilibrium shift to the right (make more products) or to the left (make more reactants)? If you add 1 M NH4NCS, you will essentially add more NCS- (aq) to the system at equilibrium. What type of stress is that? What effect will that have on the chemical system? Would you expect the solution to become more yellow or more red? Will the equilibrium shift to the right or to the left? If you add 0.1 M SnCl2 (aq), the tin (II) ions will reduce iron (III) ions to iron (II) ions as seen in the equation below: Sn2+ (aq) + 2 Fe3+ (aq) → ← 2 Fe2+ (aq) + Sn4+ (aq) You will essentially remove some Fe3+ from the system at equilibrium. What type of stress is that? What effect will that have on the chemical system? Would you expect the solution to become more yellow or more red? Will the equilibrium shift to the right or to the left? PCC 10 • Experiment 7 If you add 0.1 M AgNO3 (aq), the silver ions will react with the thiocyanate ions to give a white precipitate of silver thiocyanate as seen in the equation below: Ag+ (aq) + NCS- (aq) → ← AgNCS (s) You will essentially remove some NCS- from the system at equilibrium. What type of stress is that? What effect will that have on the chemical system? Would you expect the solution to become more yellow or more red? Will the equilibrium shift to the right or to the left? If you add 0.1 M Na2HPO4 (aq), the hydrogen phosphate ions form a complex ion with iron (III) ions as shown in the equation below: + Fe3+ (aq) + HPO24 (aq) → ← FeHPO4 (aq) How will that affect the system at equilibrium? What type of stress is that? Would you expect the solution to become more yellow or more red? Will the equilibrium shift to the right or to the left? If you add 1 M NH3 (aq), it will react as a base to form a precipitate or a colloidal suspension of iron (III) hydroxide when mixed with iron (III) ions as shown in the equations below: + NH3 (aq) + H2O (l) → ← NH4 (aq) + OH (aq) Fe3+ (aq) + 3 OH (aq) → ← Fe(OH)3 (s) How will that affect the system at equilibrium? What type of stress is that? What effect will that have on the chemical system? Would you expect the solution to become more yellow or more red? Will the equilibrium shift to the right or to the left? Procedure: Modeling Equilibrium Constants Equilibrium #1: Consider the equation A → ← B. For this Keq = [B]/[A] and the value of Keq is 8.0 (or 8:1). 1. Obtain 2 petri dishes and 27 BB’s. One Petri dish will represent A, the reactant side of the reaction, and the other will represent B, the product side of the reaction. 2. Place all 27 BB’s on the reactant side (in A). 3. Move the BB’s from the reactant side to the product side until the Keq expression as a value of 8.0. (Hint: Move one BB to petri dish B. What’s the ratio of B:A? If it is not 8:1, then move another BB and recheck. Keep going until the ratio is 8:1.) Chemical Equilibrium • 11 4. At this point the system has reached equilibrium. How many BB’s are on the reactant side at equilibrium? the product side? Record these values on your data sheet. 5. Add 18 new BBs to the product side. Find the new equilibrium values for A and B and record them on your data sheet. Remember, Keq = 8.0, regardless of how many total BB’s you have, the system will shift the BB’s (with your help) to maintain the Keq value. Equilibrium #2: Consider the equation C → ← D. For this Keq = [D]/[C] and the value of Keq is 1/15 (or 0.067). 1. Begin with 32 BB’s on the product side. Move the BB’s to establish equilibrium, then record the number of reactant and product BB’s on your data sheet. 2. Start again with all 32 BB’s on the reactant side. Move the BB’s to establish equilibrium, then record the number of reactant and product BB’s on your data sheet. 3. Keq for A → ← B was 120 times larger than Keq for C → ← D. What does the size of the Keq value have to do with the amount of reactants present at equilibrium? Equilibrium #3: Sometimes two equilibrium systems share reactants or products. These equilibria will affect each other. They cannot be written as one equation because they are separate processes that happen to overlap. Consider the equation E → ← F. For this Keq = [F]/[E] and the value of Keq is 1/4 (or 0.25). Also, consider the equation F → ← G. For this equation, Keq = [G]/[F] and the value of Keq is 5.0. 1. Obtain 3 petri dishes to represent E, F, and G positions. Begin with 30 BB’s all representing E molecules. Move the BB’s from E to F and from F to G to establish both equilibria. (Hint: Move one BB to petri dish F. What’s the ratio of F:E? If it is not 1:4, then move another BB and recheck. Keep going until the ratio is 1:4. Keep in mind, you will also need to have enough BB’s in petri dish G that the ratio of G:F is 5:1.) 2. Record the values of E, F, and G at equilibrium on your data sheet. 3. Add 10 new BB’s to the F dish. Find the new equilibrium values of E, F, and G. Record these values on your data sheet. PCC 12 • Experiment 7 4. Do the two Keq values allow you to predict which compounds are most abundant at equilibrium? Explain. Le Chatelier’s Principle Iron-thiocyanate system 1. Add 1 drop each of 1 M Fe(NO3)3 and 1 M NH4NCS to 25 mL of distilled water. Mix well. Fe3+ + NCS- → ← FeNCS2+ yellow colorless red 2. Add a few drops of this solution to each of seven wells in a spot plate. One well will serve as a color standard against which to judge color changes in the other wells. The other six wells will be used to introduce an external stress to this equilibrium system. 3. Add one drop of 1 M Fe(NO3)3 to one of the wells, mix, and observe. Record your observations on the data sheet. 4. Add one drop 1 M NH4NCS to a second well, mix, and observe. Record your observations on the data sheet. 5. Add one drop 0.1 M SnCl2 to a third well, mix, and observe. Record your observations on the data sheet. Tin (II) ions reduce iron (III) ions to iron (II) ions as seen in the equation below: Sn2+ + 2 Fe3+ → ← 2 Fe2+ + Sn4+ 6. Add one drop 0.1 M AgNO3 to a fourth well, mix, and observe. Record your observations on the data sheet. Silver ions react with the thiocyanate ions to give a white precipitate of silver thiocyanate as seen in the equation below: Ag+ + NCS- → ← AgNCS (s) 7. Add one drop 0.1 M Na2HPO4 to a fifth well, mix, and observe. Record your observations on the data sheet. Hydrogen phosphate ions form a complex ion with iron (III) ions as shown in the equation below: + Fe3+ + HPO24 → ← FeHPO4 Chemical Equilibrium • 13 8. Add one drop of 1 M NH3 (probably labeled NH4OH) to a sixth well, mix, and observe. Record your observations on the data sheet. Any base will form a precipitate or a colloidal suspension of iron (III) hydroxide when mixed with iron (III) ions as shown in the equations below: + NH3 + H2O → ← NH4 + OH Fe3+ + 3 OH → ← Fe(OH)3 (s) 9. Pour 4-5 mL of the iron-thiocyanate solution made in step 1 into two test tubes. Set one tube aside as a color standard against which to judge color changes in the other tube. 10. Gently warm the other tube in a hot water bath on a hot plate. Do not boil the solution. Observe and record your observations on your data sheet. 11. After you have determined which way the reaction shifted, cool the tube in a beaker of ice water. Record your observations on your data sheet. Do these results indicate that the forward reaction is exothermic or endothermic? 12. Make sure the reaction is reversible. Place the test tube in the warm water bath again. Record your observations on your data sheet. 13. Place the test tube in the ice water bath again. Record your observations on your data sheet. 14. Pour all of the solutions from this part of the experiment into the waste container labeled “used iron-thiocyanate solutions.” PCC Name:______________________________ Date:_____________ CH105/Sp06 Data Sheet 7: Chemical Equilibrium Directions: Record the data as it is collected onto this sheet in BLUE or BLACK ink. Do not use white out. Correct mistakes by making a single line through the error and writing the new information above or beside the mistake. Modeling A B Total 27 BB on A side initially 18 new BB added to B side 27 45 C D Total 32 BB on D side initially 32 BB on C side initially 32 32 Keq for A → ← B was 120 times larger than Keq for C → ← D. What does the size of the Keq value have to do with the amount of reactants present at equilibrium? E F G 30 BB on E side initially 10 new BB added to F Total 30 40 Do the two Keq values allow you to predict which compounds are most abundant at equilibrium? Explain. 14 PCC Experiment 7 Data Sheet Iron-thiocyanate system Fe3+ + NCS- → ← FeNCS2+ yellow colorless red If you add a “stress” and the solution turns more yellow than the standard, the reaction shifted to the left. If you add a “stress” and the solution turns more red than the standard, then the reaction shifted to the right. If you add a “stress” and the solution looks the same, then no shift occurred. Stress Color/Observations Equilibrium Shift (Left or Right?) Color/Observations Equilibrium Shift (Left or Right?) none + 1 drop Fe(NO3)3 + 1 drop NH4NCS + 1 drop SnCl2 + 1 drop AgNO3 + 1 drop Na2HPO4 + 1 drop NH3 Stress heat cool heat cool Is the forward reaction endothermic or exothermic?____________________________ Explain. 15 PCC Name:______________________________ Date:_____________ CH105/Sp06 Post Lab 7: Chemical Equilibrium Directions: Answer the following questions in the space provided. For short answer questions, write complete sentences and provide a reason for the answer. For calculation questions, show all work and report answers in a box with the appropriate significant figures and units. Pencil is acceptable for this assignment. 1. The following reaction was examined at 250°C: PCl5 (g) → ← PCl3 (g) + Cl2 (g) At equilibrium, [PCl5] = 4.2 x 10-5 M, [PCl3] = 1.3 x 10-2 M, and [Cl2] = 3.9 x 10-3 M. Calculate the equilibrium constant for the reaction. 2. Is the forward or reverse reaction favored in question 1?____________________ Explain. 3. If the concentration of Cl2 (g) is increased at 250°C, will the concentration of PCl3 (g) increase, decrease, or stay the same?______________________________________ Explain. 4. If the concentration of Cl2 (g) is increased at 250°C, will the concentration of PCl5 (g) increase, decrease, or stay the same?______________________________________ Explain. 5. The equilibrium constant for the following reaction is 0.212 at 100°C. N2O4 (g) → ← 2 NO2 (g) H = 57.2 kJ* *Did you see the H value information on page 7 of the lab? As the temperature is lowered, what will the equilibrium constant increase, decrease, or stay the same?______________ Explain. 16 PCC