Experiment 7: SIMULTANEOUS EQUILIBRIA

Experiment 7: SIMULTANEOUS EQUILIBRIA Purpose: A qualitative view of chemical equilibrium is explored based on the reaction of iron(iii) ion and thiocyanate ion to form the iron(iii) thiocyanate complex ion. Le Châtelier s Principle is used to explain the shifts in the equilibrium as various types of stress is placed on the system. Introduction: Chemical reactions occur to different extents depending on the conditions of the reaction and on the thermodynamics of the chemical reaction itself. Balanced chemical reactions provide information about the molar ratios of reactants consumed and products formed if a reaction goes to completion. They do not show the extent to which the reaction occurs or if it will occur at all. Some reactions do proceed to completion, but many other reactions do not and the amount of product formed is less than the theoretical yield. These incomplete reactions proceed until the forward and reverse reactions occur at the same rate, resulting in product and reactant concentrations that no longer change. This state is referred to as an equilibrium. Equilibrium is indicated by using a set of double arrows to show that the reaction is occurring in both directions. It is this type of reactions that we will focus on in this lab. Consider the reaction, H 2 (g) + I 2 (g) 2HI(g) Equation 1 At 450ºC and normal atmospheric pressure, the reaction between H 2 and I 2 does not go to completion. Some H 2 and I 2 remain unreacted, resulting in a mixture of HI, H 2 and I 2. Under the same conditions, the reverse reaction (Equation 2) also will not go to completion. At some point before all the HI is consumed it ceases to show activity, likewise giving a mixture of reactants and products. 2HI(g) H 2 (g) + I 2 (g) Equation 2 Equation 1 is called the forward reaction and Equation 2, the reverse reaction. equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. At H 2 (g) + I 2 (g) 2HI(g) Equation 3 Three types of general observations are made about reversible reactions. First, reversible reactions reach equilibrium with both products and reactants present. Second, when the reaction reaches equilibrium under specified conditions of temperature and pressure, the ratio of the concentrations of the products to reactants is always constant when expressed in the following way: aa + bb cc +dd Equation 4 [C] c [D] d K eq = [A] a [B] b K eq is called the equilibrium constant. The K eq expression for Equation 3 is 71

72 EXPERIMENT 7: SIMULTANEOUS EQUILIBRIA eq 2 [HI] K = Equation 5 [H ][I ] 2 2 The form of the equilibrium constant expression is always the product of the molar concentrations of the products divided by the product of the molar concentration of the reactants. The molar concentration of each species is placed in brackets [ ], and the concentration is raised to an exponent equal to the stoichiometric coefficient of that species in the balanced equation. Note that when the rate of the forward reaction equals to the rate of the reverse reaction, K eq is NOT necessarily equal to 1. A common misconception amongst general chemistry students is to think the concentration of reactants must equal to the concentration of products at equilibrium. Some equilibria favor the products and reach an equilibrium with K eq much larger than 1, and conversely, some favor the reactants and reach an equilibrium with K eq much smaller than 1. The third observation concerns Le Châtelier s Principle, which states that when a system at equilibrium is disturbed, the system will respond by favoring the direction of reaction that restores equilibrium. In simpler terms, if the equilibrium is disturbed, the reaction will shift to compensate for the disturbance. Changes in temperature, pressure, and concentration can all disrupt equilibrium. The products are always written on the right side of a chemical equation. If more products are formed the equilibrium is said to shift to the right. If more reactants are formed the reaction is said to shift to the left. Consider the reaction described in Equation 3. When more H 2 or I 2 is added to an established equilibrium of H 2, I 2 and HI, the equilibrium will shift to the right in order to deplete the amount of the H 2 and I 2 and build up the amount of HI until a new equilibrium is reached. If more HI is added to an established equilibrium, it will shift to the left to remove the extra HI and form more H 2 and I 2. Removing some H 2 or I 2 favors the reverse reaction (H 2 or I 2 formation) while removing some HI favors the forward reaction (HI formation). It is also possible to have multiple equilibria occurring simultaneously. If any of the substances are found in more than one equilibrium reaction, then changes in the concentration of that substance will affect multiple reactions. This is referred to as the common ion effect. Over the next two weeks, we will study the reaction of the iron(iii) cation complexing with a thiocyanate anion (SCN ) to form the iron(iii) thiocyanate complex, Fe(SCN) 2+ (Equation 6). The first week you will investigate Le Châtelier s Principle by observing the effect of changing various reaction conditions on the equilibrium. The second week you will determine, quantitatively, the equilibrium constant for the reaction. Note that Equation 6 is the net ionic equation. If you do not remember the differences between molecular, ionic and net ionic equations you should review this in your textbook.

EXPERIMENT 7: SIMULTANEOUS EQUILIBRIA 73 Fe 3+ (aq) + SCN (aq) Fe(SCN) 2+ (aq) Equation 6 pale yellow dark red This is an interesting reaction to study. You are probably familiar with precipitation reactions, which have equations that look somewhat like the one shown above. However, there is one big difference; the product here is not a precipitate (not a solid), but what is called a complex ion, which stays in solution. The signal that a reaction has taken place is not the appearance of cloudiness or fogginess in the reaction mixture, but a change in color. Note that the reaction mixture may be colored, but will remain clear as no solids are forming. Copy the Data Table into your lab notebook prior to the Pre-Lab. Original Reagents 0.01 M Fe(NO 3 ) 3 0.01M KSCN Well #1 (1mL Stock Solution) Well #2 1 ml Stock Soln + 5 drops 0.1 M Fe(NO 3 ) 3 (5 mins) Well #3 1 ml Stock Soln + 5 drops 0.1 M KSCN (5 mins) Well #4 1 ml Stock Soln + 10 drops 0.1 M NaOH (5 mins) then + 5 drops 0.1 M KSCN (5 mins) Well #5 1 ml Stock Soln + 10 drops 0.1 M NaOH (5 mins) then + 10 drops 0.1 M HCl (5 mins) then + 5 drops 0.1 M KSCN (5 mins) COLOR. Procedure: PART I: Working with the Equilibrium of NH 3 This section of the experiment will be done as a demonstration. Take careful notes as your instructor explains how your observations are consistent with Le Châtelier s Principle. Use the same terminology in your own explanations when you get to Part II. 1. Write the equilibrium involved when gaseous ammonia reacts with water. 2. Phenolphthalein is an indicator which is colorless when acidic and pink when basic. Place about 2 ml of 0.1 M aqueous ammonia in a large test tube and add 1 drop of phenolphthalein. Record the color. 3. Using a test tube holder, hold the test tube over a Bunsen burner and heat it gently until there is a color change. Record your observations. 4. Explain what you observed in terms of Le Châtelier s Principle.

74 EXPERIMENT 7: SIMULTANEOUS EQUILIBRIA PART II: Working with the Equilibrium of Fe(SCN) 2+ Work with one partner. Follow the instructions carefully or you will not get the correct results. When the entire class is finished with the procedure, the class will gather and your instructor will go over your observations and hold a discussion on how to explain the observations with Le Châtelier s Principle. You will then complete the worksheet at home based on the discussion. First, note that in this experiment there are two different concentrations (0.01 M and 0.1 M) of the same reagent. Be very careful you are using the correct concentration for the step you are at. You will be stirring reaction mixtures with toothpicks. Be sure you do not get your toothpicks mixed up and cross-contaminate your samples. 1. This experiment will be done on a microscale level to eliminate excessive waste of material. Obtain a multi-well plate. Clean and dry the wells with the help of a cotton swab. If the wells are wet, the residual water could change your concentration drastically. 2. Examine the 0.01 M iron (III) nitrate and the 0.01 M potassium thiocyanate, and record the color of each solution in your notebook. 3. Prepare a stock solution that you will use for the entire experiment. To a clean and DRY graduated cylinder, using the dropper bottles, place 2.0 ml of 0.01 M iron(iii) nitrate, 2.0 ml of 0.01 M potassium thiocyanate. Please be very careful you do not contaminate the droppers. Always avoid touching the tips of the droppers to anything. Next, add enough deionized water to bring the total volume to 10.0 ml. Pour the entire mixture into a clean and DRY 50-mL beaker and with a graduated disposable pipet, stir the mixture to mix thoroughly. Leave the pipet in the beaker as you will be using it in the next step. Note: the 0.01 M concentrations are used ONLY in this step. The next steps will require the 0.1 M concentrations. 4. Using the graduated disposable pipet in step 3, place 1 ml of the stock solution into each of five wells. Well #1 will serve as a control to compare with Wells #2 #5. Nothing else will be added to Well #1. Record the color of the solution. It should be a dark red. See your instructor if it is not. You can save a lot of time if you plan carefully and do Steps 5 and 6 together. Gather the reagents needed for these 2 steps before you proceed. 5. To Well #2 add 5 drops of 0.1 M iron(iii) nitrate stirring well with a toothpick. Quickly proceed to the next step. 6. To Well #3 add 5 drops of 0.1 M potassium thiocyanate stirring well with a toothpick. Wait 5 minutes and record any change in color in Wells #2 and #3. In recording your observations, you should be noting not only just a change in color (red to yellow) but also whether colors are getting darker or lighter.

EXPERIMENT 7: SIMULTANEOUS EQUILIBRIA 75 Wells #4 and #5 both have multiple steps and it is important that you perform all of them in the correct sequence, recording color changes after each step. Again gather all the reagents needed before proceeding. 7. To both Wells #4 and #5 add 10 drops of 0.1 M sodium hydroxide stirring thoroughly. Wait 5 minutes and record the color change in each well. 8. To Well #5 (only) add 10 drops of 0.1 M hydrochloric acid. Stir well, wait 5 minutes, and record any color changes. 9. Next, to both Wells #4 and #5, add 5 drops of 0.1 M potassium thiocyanate stirring well with a toothpick. Wait 5 minutes and record any change in color in each well. Preparation for Pre-Lab Quiz: 1. Ammonia is a gas. You should have learned in the first semester of General Chemistry why ammonia is considered a base even though it does not contain OH. Look up what happens to ammonia when it is placed in water. Either use your chemistry textbook or do an Internet search. Write the equilibrium involved when ammonia reacts with water. 2. State Le Châtelier s Principle. Look it up in your textbook if you do not know how. 3. Consider the equilibrium A B. Describe two different ways to make this equilibrium shift to the left and explain your answers using Le Châtelier s Principle. 4. The molecular equation for the reaction is as follows: Fe(NO 3 ) 3 (aq) + KSCN (aq) Fe(SCN)(NO 3 ) 2 (aq) + KNO 3 (aq) a) Give the total ionic equation for the reaction. Don t forget these equations have no meaning if you don t include physical states for all species. b) Give the net ionic equation for the reaction. c) What are the spectator ions in this reaction? 5. In Step 3 you will be preparing the stock solution. Give the formulas of all the species that you are expecting to be in the stock solution. Species refer to cations, anions, and molecules. 6. Consider your answer to Question #5. In Well #4 you will be adding a solution of NaOH to the stock solution. How might the ions of the stock solution react with the ions in NaOH? Write down ALL the combinations of the cations and anions that may be possible in this mixture and then look up the solubility of the various combinations and predict what product (if any) may result in this mixture. For example, could K + from the potassium thiocyanate form a product with the OH from the sodium hydroxide? Hint: How many types of cations and how many types of anions are in this mixture? How many different combinations can you come up with? Remember that cations cannot combine with cations!

76 EXPERIMENT 7: SIMULTANEOUS EQUILIBRIA Assignment: There are no calculations in this experiment. Your grade for this experiment will be mostly based on your answers on the Assignment Page which involves filling in the blanks. It serves as an example of the terminology chemists use to explain Le Châtelier s principle. your grade will also be based on answering the post-lab questions below. Learn from the structure of the answers in the Assignment Page as you answer the post-lab questions. Post-Lab Questions: Answers must be typed. Preliminary Observations: K 2 CrO 4 dissolved in water forms a yellow solution. K 2 Cr 2 O 7 dissolved in water forms an orange solution. When aqueous HCl is added to a solution of K 2 CrO 4, the color changes from yellow to orange. Rxn 1 2 CrO 4 2 + 2 H + Cr 2 O 7 2 + H 2 O Experiment: When aqueous solutions of Ba(NO 3 ) 2 and K 2 CrO 4 are mixed, a yellow precipitate is observed. This is due to the formation of insoluble BaCrO 4. Rxn 2 Ba 2+ (aq) + CrO 4 2 (aq) BaCrO 4 (s) When aqueous HCl is added the yellow precipitate dissolved and an orange solution is formed. Explain what happened (why the precipitate dissolved and why we ended up with an orange solution) based on Le Châtelier s Principle by answering the questions below: Answers must be typed and numbered. 1) Identify the equilibrium involved. Write out the net ionic equation.(hint: Rxn 1) 2) Identify exactly what the stress is on the equilibrium when HCl is added. 3) Predict what the stress would do to the equilibrium. Would it cause the equilibrium to shift to the left or to the right? 4) What would this shift do to the concentration of each of the reactants and products? Would they increase or decrease? List the effect on each and refer to them by formula (CrO 4 2, H +, Cr 2 O 7 2, H 2 O). 5) If more than one equilibrium are involved, how would this shift affect the other equilibrium (Rxn 2)? The concentration of which species is/are expected to be affected and how would it/they be affected (increase or decrease)? 6) Would the change in concentration be observable? Explain fully. 7) Is the above consistent with what was actually observed? Optional: To see this in action type the following URL in your browser: http://tinyurl.com/chromatedemo

EXPERIMENT 7: SIMULTANEOUS EQUILIBRIA 77 Assignment Page: Name: CHEM 124 Sec: Lab Partner s Name: PLEASE KEEP THIS NEAT BY USING PENCIL PART I: Equilibrium of NH 3 Rxn 1 NH 3 (aq) NH 3 (g) Rxn 2 NH 3 (aq) + H 2 O (l) NH + 4 (aq) + OH (aq) 1. Which directly causes the color of phenolphthalein to become pink? CIRCLE ONE: NH 3 (g) H 2 O (l) NH + 4 (aq) OH (aq) 2. Consider the effect of temperature on the solubility of gases. (Is a gas more or less soluble when the temperature is increased?) Is Rxn 1 endothermic or exothermic? ANS. 3. Write + heat to either the left side or the right side of Rxn 1 to correspond to your answer to Question #1. 4. When heat (the stress) is applied, do you expect the equilibrium in Rxn 1 to shift to the left or the right? ANS. 5. The shift in equilibrium in Rxn 1 created a stress in Rxn 2. What is the stress? The concentration of what species was affected? Be sure to include physical states! ANS. The concentration of was decreased. This caused the equilibrium in Rxn 2 to shift to the (left or right), thus causing the concentration of OH to (increase or decrease), and the pink color to fade. PART II Equilibrium of Fe(SCN) 2+ Rxn 1 Fe 3+ (aq) + SCN (aq) Fe(SCN) 2+ (aq) Rxn 2 Fe 3+ (aq) + 3OH (aq) Fe(OH) 3 (s) 6. The stock solution is a mixture of aqueous solutions of Fe(NO 3 ) 3 and KSCN. Since the reagents did not have any color, why does the stock solution have color? Give the formula of the species that gave it color. ANS. 7. In Well #2, what is the effect of adding Fe(NO 3 ) 3? ANS. It placed a stress on the equilibrium of Rxn 1 by increasing the concentration of. This caused the equilibrium of Rxn 1 to shift to the (left or right), thus (increasing or decreasing) the concentration of, which causes the color to (fade or darken). 8. In Well #3, what is the effect of adding KSCN? ANS. It placed a stress on the equilibrium of Rxn 1 by increasing the concentration of. This caused the equilibrium of Rxn 1 to shift to the (left or right), thus (increasing or decreasing) the concentration of, which causes the color to (fade or darken).

78 EXPERIMENT 7: SIMULTANEOUS EQUILIBRIA Name: Rxn 1 Fe 3+ (aq) + SCN (aq) Fe(SCN) 2+ (aq) Rxn 2 Fe 3+ (aq) + 3OH (aq) Fe(OH) 3 (s) Rxn 3 H + (aq) + OH (aq) H 2 O (l) 9. In Well #4, addition of NaOH did what to the equilibrium in Rxn 2? ANS. It placed a stress on the equilibrium in Rxn 2 by increasing the concentration of OH. This caused the equilibrium to shift to the (left or right). In turn, this shift caused a stress in the equilibrium in Rxn 1 by (increasing or decreasing) the concentration of in Rxn 1. This stress in Rxn 1 is relieved by a shift of the equilibrium to the (left or right), thus causing the concentration of Fe(SCN) 2+ to (increase or decrease). This is supported by our observation in that the color (faded or darkened). 10. In Step 9, aqueous KSCN is added to Well #4, we might expect the color to darken because the increase in SCN should caused the equilibrium in Rxn 1 to shift (left or right). However, there was hardly any noticeable change in color because... 11. In Step 8, aqueous HCl was then added to Well #5. Which reaction did it initially affect? ANS. The HCl reacted with OH and caused the equilibrium in Rxn 3 to shift to the (left or right). In so doing this caused a stress in Rxn 2 by (increasing or decreasing) the concentration of. The result is a shift in the equilibrium in Rxn 2 to the (left or right). In so doing, it caused the concentration of Fe 3+ to (increase or decrease). In turn, this created a stress in Rxn 1 which shifted to the (left or right). This shift is supported by our observation in that the color (faded or darkened). 12. In Step 9, aqueous KSCN is then added to Well #5. Unlike Well #4, the color darkened. This is due to a shift of the equilibrium in Rxn 1 to the (left or right). The reason for the difference is in Well #5, is now free to react, thus allowing the equilibrium to shift. PLEASE TYPE answers to post-lab questions on your own paper.