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

[NO]0.05 at 2100C[N][O]K== o

222

(17-2) 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 AgNO3 0.5 M Cr(NO3)3; chromium(III) nitrate 6 M HCl; hydrochloric acid 6 M HNO3; nitric acid K2CrO4; potassium chromate KNCS; potassium thiocyanate 6 M NaOH; sodium hydroxide CAUTION!

0.5 M Co(NO3)2; cobalt(II) nitrate 0.5 M Cu(NO3)2; copper(II) nitrate 12M HCl; concentrated hydrochloric acid ice K2Cr2O7; potassium dichromate 6 M KNCS

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 CrO42 or by dissolving a salt containing Cr2O72 can actually have both of these ions present. If the solution is basic, CrO42 predominates; if the solution is acidic, Cr2O72 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

Cr2O72- + OH HCrO4- + OH

HCrO4- + CrO42-

(17-3)

CrO42- + H2O

(17-4)

These reactions can be summarized- in one equation that overlooks the presence of HCrO4:

Cr2O72 + 2OH

2 CrO42- + H2O

(17-5)

This simplification will be used in this experiment. EQUATION 17-5 makes it clear that Cr2O72 in solution can be converted to CrO42 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 CrO42 in solution can be converted to Cr2O72 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 CrO42, 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, H3O+) 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 Cr2O72. Dissolve 0.3 g of K2Cr2O7 in 20 mL of distilled water. Dissolve 0.4 g of K2CrO4 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 K2Cr2O7 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 K2CrO4 solution into a 20-mL test tube. To this solution, add 6 M HNO3 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 HNO3 by combining them and washing the solution down the drain. Dispose of the K 2Cr2O7 and K2CrO4 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(H2O)62+, Cr(H2O)63+, and Cu(H2O)62+. 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 Co2+, the reaction is

Co(H2O)62+ + 4 Cl

CoCl42 + 6 H2O

(17-7)

MClyn-y + x H2O

(17-8)

A general equation for various metal ions would be

M(H2O)xn+ + y Cl

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)62+ + 4 NCS

Co(NCS)42 + 6 H2O

(17-9)

1. Pour 5.0 mL of 0.5 M Co(NO3)2 into each of two 20-mL test tubes. Pour 5.0 mL of 0.5 M Cr(NO3)3 into each of a second set of two 20-mL test tubes. Pour 5.0 mL of 0.5 M Cu(NO3)2 into each of a third set of 20-mL test tubes. Record the color of each solution. These solutions contain the aqua ions for which the formulas are given above. Set aside one test tube of each solution to use for color comparisons. To each of the three remaining test tubes, add 12 M HCl. Use a medicine dropper to add the acid dropwise from a 10-mL graduated cylinder containing 10.0 mL of the acid. Record (to the nearest 0.5 mL) the volume of HCl that is required to produce a definite color change in each solution. After you think a color change has occurred, add another drop or two of acid to decide whether you are seeing a distinct color rather than a mixture of two colors. For example, if a

solution changes color from red to orange, the orange might be the color of the product or it might be from a mixture of red-colored reactant and yellow-colored product. 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(NO3)2 and 3.0 mL of 0.5 M Cu(NO3)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 CoC142 and CuC142 are sufficiently intense and different enough that you should be able to recognize either one despite the presence of another colored material. 3. Pour 5.0 mL of 0.5 M Co(NO3)2 into each of two 20-mL test tubes. To one test tube, add 6 M KNCS dropwise until a definite color change occurs; continue to add KNCS solution until the color no longer is changing. 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(NO3)2; record the volume of solution that is required in each case. The colors of the two complex ions CoCl42 and Co(NCS)4 are very nearly the same. When you complete this part of the experiment, save any unused Co(NO3)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(H2O)62+ and Co(NCS)42 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(NO3)2. The solution should have the intense blue color of Co(NCS)42. Dilute this solution with 15-18 mL of distilled water until it has a purplish color that is intermediate between the colors of Co(H2O)62+ and Co(NCS)42. 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 AgNO3. 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

[Cl- ] =

(12 moles Cl- /1 L HCl) ´ (0.0025 L HCl) = 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 Cr2O72- + 2OH

2 CrO42- + H2O

1. Color of Cr2O72- solid and in solution

Color after addition of

mL of 6M NaOH

Color after addition of 6M HNO3 to above solution

Color after addition of 6M NaOH to above solution

2. Color of CrO42- solid and in solution

Color after addition of

mL of 6M HNO3

Color after addition of 6M NaOH to above solution

Color after addition of 6M HNO3 to above solution

(17-5)

B. RELATIVE “POSITIONS” OF EQUILIBRIUM M(H2O)xn+ + y Cl

MClyn-y + x H2O Co2+

1. 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)62+ and Cu(H2O)62+ Original color of solution

Color after addition of

3.

mL of 12M HCl

CoCl42 and Co(NCS)42 Color of Co(H2O)62+ solution Color after addition of

mL of 6M HCl

Total [Cl] of solution

Color after addition of Total [NCS] of solution

mL of 6M KNCS

Cr2+

(17-8) Cu2+

EXPERIMENT 17 REPORT SHEET (CONT.)

C. “SHIFTING THE POSITION” OF EQUILIBRIUM Co(H2O)62+ + 4 NCS

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)

Co(NCS)42- + 6 H2O

(17-9)