EXPERIMENT 17 CHEMICAL EQUILIBRIUM

EXPERIMENT 17 CHEMICAL EQUILIBRIUM INTRODUCTION Complete conversion of reactants into products is not characteristic of all chemical changes. It is more common for a reaction to reach a state in which both reactants and products are present and their concentrations no longer change. This situation is called chemical equilibrium. The equilibrium is a dynamic one; it is established when the rate at which products are being reconverted to reactants equals the rate at which products are forming, and so there is no net change in the amounts of reactants and products. Equations for such reactions are often written with two arrows to indicate that both the forward and reverse directions of reaction occur. For example: N2(g) + O2(g) 2NO(g) (17-1) An equilibrium constant K quantitatively describes the relative amounts of reactants and products present at equilibrium. This constant states the ratio of product concentrations to reactant concentrations with each concentration raised to the power that is the substance s coefficient from the balanced equation. The value of K is characteristic of a particular reaction at a particular temperature. For EQUATION 17-1 2 [NO] K = = 0.05 at 2100 C (17-2) [N ][O ] 2 2 For EQUATION 17-1, K < 1, meaning that at equilibrium the concentration of product is small compared to the concentrations of reactants remaining unreacted. We say that the position of equilibrium lies to the left, meaning that when chemical equilibrium is established, the substances on the left side of the equation (reactants) predominate in the reaction mixture. For many reactions, K > 1, meaning that product concentrations are greater than reactant concentrations when equilibrium is established. For these reactions we say that the position of equilibrium lies to the right. A change in temperature changes the rates of forward and reverse directions of a reaction by different amounts. Consequently, a change in temperature changes the relative amounts of products and reactants in an equilibrium mixture and, therefore, the value of the equilibrium constant. For EQUATION 17-1, K = 0.05 at 2100 C. This temperature can be reached in the cylinders of an auto- mobile engine; nitrogen and oxygen in the air that enters the cylinders can react extensively enough to form enough NO, nitrogen(ii) oxide, to be a serious cause of air pollution. However, at normal room or outdoor temperatures, the equilibrium constant for this reaction is much smaller. At 25 C, K = 10 31 ; the position of equilibrium is extremely far to the left. This effect of temperature can be predicted using Le Chatelier s principle: If a stress is applied to a system at equilibrium, the system will change in a way that relieves the stress and establishes a new equilibrium state. If the stress is a temperature increase, the direction of reaction that requires heat (is endothermic) is favored and occurs more extensively. If the stress is a temperature decrease, the direction of reaction that evolves heat (is exothermic) is favored. We say that the position of equilibrium shifts with temperature, meaning that the concentrations of

reactants and products change with temperature. The forward direction of the reaction of EQUATION 17-1 is endothermic; consequently, it occurs more extensively as the temperature increases, and K increases as the temperature increases. Other changes in conditions can also shift the position of equilibrium. If, to an equilibrium mixture, more of a reactant or product is added, reaction occurs more extensively in the direction using that substance. When equilibrium is re-established, the concentrations of substances in the mixture will be different from those in the previous equilibrium mixture, but the concentration ratio described by K will be the same. Only a change in temperature changes the value of K for a particular reaction. Removing some of a substance from an equilibrium mixture also shifts the position of equilibrium; the direction of reaction producing more of that substance occurs more extensively. For reactions involving gases, changes in the volume of a reaction container shift the position of equilibrium. A decreased volume favors the direction of reaction for which there is a decrease in the number of moles of gas. An increased volume favors the direction of reaction for which there is an increase in the number of moles of gas. (But neither effect causes a change in the value of K.) In this experiment you will work with reversible reactions that occur in aqueous solution. You will investigate the effects that addition of a substance, removal of a substance, and temperature changes have on position of equilibrium. EQUIPMENT NEEDED balance beakers hotplate 10-mL graduated cylinder disposable pipets ring stand test-tube holder test-tube rack medium test tubes CHEMICALS NEEDED 2 M AgNO 3 0.5 M Co(NO 3) 2; cobalt(ii) nitrate 0.5 M Cr(NO 3) 3; chromium(iii) nitrate 0.5 M Cu(NO 3) 2; copper(ii) nitrate 6 M HCl; hydrochloric acid 12M HCl; concentrated hydrochloric acid 6 M HNO 3; nitric acid ice K 2CrO 4; potassium chromate K 2Cr 2O 7; potassium dichromate KNCS; potassium thiocyanate 6 M KNCS 6 M NaOH; sodium hydroxide CAUTION! Concentrated hydrochloric acid (12 M) and 6 M NaOH are extremely corrosive. Nitric acid reacts very quickly with proteins in skin leaving a yellow stain. Also, 6 M HCl is very corrosive. If you spill any of these on your skin, wash it off immediately under lots of cold, running water. If you spill any of these on the outside of the bottle or on the table top, sponge it off and wash out the sponge.

TECHNIQUE AND PROCEDURE A. A Reversible Reaction A solution that is prepared by dissolving a salt containing CrO 4 or by dissolving a salt containing Cr 2O 7 can actually have both of these ions present. If the solution is basic, CrO 4 predominates; if the solution is acidic, Cr 2O 7 predominates. Because the colors of these two ions are different, it is possible to tell which ion is present in a solution, or at least to tell which ion is present in a greater amount when both ions might be present. Equations for the reversible conversion of one ion into the other are Cr2O7 2- + OH HCrO4 - + CrO4 2- (17-3) HCrO4 - + OH CrO4 2- + H2O (17-4) These reactions can be summarized- in one equation that overlooks the presence of HCrO 4 : This simplification will be used in this experiment. Cr2O7 + 2OH 2 CrO4 2- + H2O (17-5) EQUATION 17-5 makes it clear that Cr 2O 7 in solution can be converted to CrO 4 by the addition of OH, that is, by adding a base. You will use NaOH and watch for a color change indicating that the forward reaction given in EQUATION 17-5 is occurring. It is not as obvious from EQUATION 17-5 how CrO 4 in solution can be converted to Cr 2O 7 Keep in mind that a solution prepared from one of the ions will also contain at least a trace of the other ion because of the interconversion. If a solution is prepared from CrO 4, the extent to which the reverse reaction of EQUATION 17-5 occurs will be increased if OH is removed from solution; this is the same as saying that the position of equilibrium will be farther to the left. The easiest way to remove OH is to add an acid (a source of hydronium ions, H 3O + ) to the solution, so that the following reaction will occur: H3O + + OH 2 H2O (17-6) As OH - is removed, the extent to which the reverse reaction in EQUATION 17-5 occurs will increase, and the accompanying color change will be your visible evidence of the formation of Cr 2O 7. Dissolve 0.3 g of K 2Cr 2O 7 in 20 ml of distilled water. Dissolve 0.4 g of K 2CrO 4 in 20 ml of distilled water. Record the color of each solid and of each solution. The two solutions contain different masses of solute and different concentrations of chromium-containing ions, but both solutions have the same concentration of chromium atoms. 1. Pour about one-fourth of the K 2Cr 2O 7 solution into a 20-mL test tube. Save the rest of the solution for color comparisons. To the solution in the test tube, add 6 M NaOH dropwise; after each drop shake the tube to mix the contents thoroughly. Add NaOH until a definite color change occurs. Record the amount of base required. Account for the color change in terms of the preceding equations. Then add 6 M HNO 3 dropwise to this basic solution; as before, mix the solution thoroughly after you add each drop. Add acid until a definite color change occurs; explain this color change in terms of the appropriate equations. Again, add base to the solution, and

explain any color change. Remember that dilution will make the colors slightly lighter than they were in the original solutions. 2. Pour about one-fourth of the K 2CrO 4 solution into a 20-mL test tube. To this solution, add 6 M HNO 3 dropwise until a definite color change occurs. Then add NaOH until a color change occurs. Then add nitric acid, as before. Account for your observations in terms of the appropriate equations. When you complete this part of the experiment, dispose of any unused NaOH and HNO 3 by combining them and washing the solution down the drain. Dispose of the K 2Cr 2O 7 and K 2CrO 4 solutions as directed by your instructor; do not wash these Cr-containing solutions down the drain. B. Relative Positions of Equilibrium Transition metal ions exist in aqueous solution as complex ions in which water molecules are bound by coordinate covalent bonds to the metal. For cobalt(ii), chromium(iii), and copper(ii), these complex ions have the formulas Co(H 2O) 2+ 6, Cr(H 2O) 3+ 6, and Cu(H 2O) 2+ 6. The water molecules can be replaced by other ions or molecules forming different complex ions. If two complex ions of a particular metal differ in color, it is easy to tell which ion is present in a solution. You will attempt to convert the aqua complex ions listed above to chloro complex ions, using concentrated HCl as a source of Cl. For Co 2+, the reaction is Co(H2O)6 2+ + 4 Cl CoCl4 + 6 H2O (17-7) A general equation for various metal ions would be M(H2O)x n+ + y Cl MCly n-y + x H2O (17-8) In working with the three metal ions chosen for this experiment, you must decide whether or not there is any visible evidence of formation of a complex ion within a short time period. You will then list the ions on the basis of how extensive the reaction of EQUATION 17-8 is for a particular amount of Cl, that is, on the basis of the size of the equilibrium constant for the reaction of EQUATION 17-8. After comparing chloro complex ions of the three different metal ions, you will compare two different complex ions of one metal ion. You will investigate the reactions in EQUATIONS 17-7 and 17-9 to determine which reaction is more extensive for a particular amount of a complexing agent, Cl or NCS ; that is, which reaction has a larger equilibrium constant. Co(H2O)6 2+ + 4 NCS Co(NCS)4 + 6 H2O (17-9) 1. Pour 5.0 ml of 0.5 M Co(NO 3) 2 into each of two 20-mL test tubes. Pour 5.0 ml of 0.5 M Cr(NO 3) 3 into each of a second set of two 20-mL test tubes. Pour 5.0 ml of 0.5 M Cu(NO 3) 2 into each of a third set of 20-mL test tubes. Record the color of each solution. These solutions contain the hexaaqua ions for which the formulas are given above. Set aside one test tube of each solution to use as a control for color comparisons. Obtain ~20 ml of 12M HCl in a 50 ml beaker. Caution: 12M HCl is highly corrosive. Use a disposable pipet to add the HCl solution dropwise to each aqua metal ion solution until a definite color change (compared to the control) is observed. Record (to the nearest 0.5 ml) the volume of HCl that is required to produce a definite color change in each solution. You can either use the volume markings on the barrel of the pipet, or use the conversion 20

drops = 1 ml to determine the volume used. Do not add more than 10.0 ml of acid to each solution. If you are not sure whether the solution has changed color, add to the color comparison tube a volume of water equal to the volume of acid that you used; a slight color change will occur just from the dilution. 2. Mix 3.0 ml of 0.5 M Co(NO 3) 2 and 3.0 ml of 0.5 M Cu(NO 3) 2 in a 20-mL test tube. Add 12 M HCl dropwise until a definite color change occurs; record the volume of acid that you use. Does the chloro complex of cobalt or the chloro complex of copper form first? Does this agree with your previous conclusions about the positions of equilibrium for the reactions of these two metal ions with Cl? The color that you see in this mixture will not be exactly like the colors you observed for the solutions containing only one metal ion. However, the colors of CoC1 4 and CuC1 4 are sufficiently intense and different enough that you should be able to recognize either one despite the presence of another colored material. 3. Obtain ~10 ml each of 6M KNCS and 6M HCl. Pour 5.0 ml of 0.5 M Co(NO 3) 2 into each of two 20-mL test tubes. To one test tube, add 6 M KNCS dropwise until a definite color change occurs. To the second test tube, add 6 M HCl until a definite color change occurs. Do not add more than 10 ml of either solution to the Co(NO 3) 2; record the volume of solution that is required in each case. The colors of the two complex ions CoCl 4 and Co(NCS) 4 are very nearly the same. When you complete this part of the experiment, save any unused Co(NO 3) 2 and KNCS solutions for the next part of the experiment. All other solutions may be washed down the drain. C. Shifting the Position of Equilibrium The colors of Co(H 2O) 2+ 6 and Co(NCS) 4 are quite distinct, so you can easily tell which of these complex ions is present in the greater amount in a solution. If similar amounts of these two ions are present, the solution will have an intermediate color that can be described as reddish-purple. If anything is added to this solution that causes the forward reaction of EQUATION 17-9 to occur to a greater extent (that is, shifts the position of equilibrium to the right), the color becomes bluer. If a change in conditions causes the reverse reaction of EQUATION 17-9 to occur to a greater extent (that is, shifts the position of equilibrium to the left), the color becomes pinker. You will investigate how the position of equilibrium for this reaction is affected (1) by adding each of two materials that are involved in the reaction, (2) by removing one of the materials, and (3) by changing the temperature. In a small beaker, add 7 ml of 6 M KNCS to 3 ml of 0.5 M Co(NO 3) 2. The solution should have the intense blue color of Co(NCS) 4. Dilute this solution with 15-18 ml of distilled water until it has a purplish color that is intermediate between the colors of Co(H 2O) 2+ 6 and Co(NCS) 4. This intermediate color indicates that a moderate amount of each complex is present. Divide this solution as evenly as possible among seven 20-mL test tubes. Set tube 1 aside to use for color comparisons. 1. Effect of concentration of a reactant. To tube 2, add 1 ml of distilled water. Record any color change; that is, does tube 2 appear more pink or more blue compared to tube 1? To tube 3, add 5 ml of distilled water. Compare the colors of the solutions in tubes 1 and 3. Again, your observation should be whether the contents of tube appear more pink or more blue than those in test tube 1. To tube 4, add about 0.5 g of KNCS. Compare the colors of the solutions in tubes 1 and 4.

To tube 5, add 1 ml of 2 M AgNO 3. The white precipitate that forms is insoluble AgNCS; the addition of Ag + causes the removal of NCS from the solution. Mix thoroughly, allow the precipitate to settle, and compare the color of the solution in tube 5 to the solution in tube 2, to which 1 ml of distilled water was added. 2. Effect of temperature. Place tube 6 in a beaker of boiling water. After several minutes, compare the color of the solution in tube 6 to the original solution in tube 1. Place tube 7 in a beaker of ice and water. After several minutes, compare the color of the solution in tube 7 to the color of the original solution in tube 1, which remained at room temperature, and to the color of the solution in tube 6, which was heated. RESULTS A. A Reversible Reaction On the REPORT SHEET, provide the requested information about the colors of the original solids and solutions and about the color changes that provide visible evidence of the reaction that is occurring. B. Relative Positions of Equilibrium 1. Provide the requested information about colors that is the basis for comparing the extent of the reactions in EQUATION 17-8 for the three metal ions. In addition, calculate the concentration of Cl that has been added to each solution; this is the total concentration of Cl that would be present if none reacted with a metal ion. If no significant color change occurred, meaning that no significant amount of chloro complex formed, calculate the chloride ion concentration that is present after the addition of the entire 10 ml of HCl. As an example of such a calculation, assume that a color change occurs when 2.5 ml of 12 M HCl is added to 5.0 ml of the original solution; the total volume is 7.5 ml. In the total solution - - (12 moles Cl /1 L HCl) (0.0025 L HCl) [Cl ] = = 4.0 M 0.0075 L of solution (17-10) 2. Because it is difficult to assign specific terms to the descriptions of colors, especially when they are mixtures, not all students will have the same answers here. However, your conclusions about positions of equilibrium should be the same, regardless of how you describe the colors of the solutions. 3. In addition to describing your observations for the reactions in EQUATIONS 17-7 and 17-9, calculate the concentrations of Cl and NCS that you put into the solutions. C. Shifting the Position of Equilibrium Describe the appearance of the solution in each tube.

EXPERIMENT 17 REPORT SHEET Name: Date: A. A REVERSIBLE REACTION Cr2O7 2- + 2OH 2 CrO4 2- + H2O (17-5) 1. Color of Cr2O7 2- solid and in solution ml of 6M NaOH 6M HNO3 to above solution 6M NaOH to above solution 2. Color of CrO4 2- solid and in solution ml of 6M HNO3 6M NaOH to above solution 6M HNO3 to above solution

B. RELATIVE POSITIONS OF EQUILIBRIUM M(H2O)x n+ + y Cl MCly n-y + x H2O (17-8) 1. Co 2+ Cr 2+ Cu 2+ Color of aquo complex Does chloro complex form? If so, its color ml of 12M HCl to form complex [Cl ] to form complex 2. Mixture containing Co(H2O)6 2+ and Cu(H2O)6 2+ Original color of solution ml of 12M HCl 3. CoCl4 and Co(NCS)4 Color of Co(H2O)6 2+ solution ml of 6M HCl Total [Cl ] of solution ml of 6M KNCS Total [NCS ] of solution

EXPERIMENT 17 REPORT SHEET (CONT.) C. SHIFTING THE POSITION OF EQUILIBRIUM Co(H2O)6 2+ + 4 NCS Co(NCS)4 2- + 6 H2O (17-9) 1. Effect of reactant concentration Color in tube 1 Color in tube 2 Color in tube 3 (relative to test tube 1) Color in tube 4 (relative to test tube 1) Color in tube 5 (relative to test tube 2) 2. Effect of temperature Color in tube 6 (relative to test tube 1) Color in tube 7 (relative to test tube 1)