Le Chatelier s Principle and Equilibrium

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Le Chatelier s Principle and Equilibrium Discussion A chemical equilibrium is a condition in which the rate of formation of products from reactants equals the rate of formation of reactants from products. In other words, the rate of the forward reaction equals the rate of the reverse reaction, and the relative proportions of the reactants to products remains unchanged. Reactants Products Le Chatelier s Principle states that: A system at equilibrium when placed under stress will react to counteract the stress and in so doing restore the equilibrium to a new position. What would be considered a stress on a chemical reaction at equilibrium? Anything that changes the position of equilibrium or balance is a stress. Changes in concentration(s)of reactants and/or products, and variations in temperature, pressure, etc. cause stress on a reaction. Consider a saturated solution of KNO 3, where solid KNO 3 is in contact with the solution. This dissolution is an endothermic process as shown below. KNO 3 (s) + Heat K + (aq) + NO 3 - (aq) At equilibrium the rate of dissolution of KNO 3 (s) equals the rate of its crystallization. Though no net change occurs, the system is still dynamic, that is, ions are dissolving and recrystallizing at the same rate. If NaNO 3 (s) is added to this system, it will dissolve and increase the concentration of - NO 3 (aq) causing a stress on the right side of the reaction. Na + (aq) will act as a "spectator" and have no direct involvement with the equilibrium). The equilibrium shifts to the left. In order words the increase in the concentration of nitrate ions speeds up the reverse reaction when compared to the forward reaction rate. The net ionic equation that describes the change is NO 3 - (aq) + K + (aq) KNO 3 (s) The observed effect is that KNO 3 will precipitate out of solution until the rate of the forward reaction catches up with the reverse reaction, and a new equilibrium is established. What would happen if you increased the temperature of the saturated solution of KNO 3? Answer according to Le Chatelier s Principle: Experimental Procedure Caution: Care must be taken to avoid cross-contamination of solutions. Do not use a stirring rod to mix unless it has been cleaned after each use. Pour 20 ml of ethanol into a 25-mL or 50-mL Erlenmeyer flask. Place several pieces of solid cobalt(ii) chloride, CoCl 2 ` 6H 2 O in one of the wells of a 24-well culture plate. Note its color. Color of cobalt chloride crystals Add approximately 10 crystals of cobalt(ii) chloride in the ethanol until a dark blue color persists.

2 Add more crystals if needed. Stir until all the crystals have dissolved. Transfer 2-mL of the blue solution to each of four wells in the culture plate(1 ml is approximately 20 drops). Be sure to leave a small amount of solution in the beaker. Place a white paper underneath the culture plate and label each filled well by writing #1,2,3 and 4 on the paper above each well. Once dissolved the following equilibrium is established. It involves two complex ions, hexaaquo cobalt(ii) and tetrachloro cobaltate(ii): 2+ Co(H 2 O) 6 + 4Cl - 2- CoCl 4 + 6H 2 O pink blue hexaaquo cobalt(ii) tetrachloro cobaltate(ii) Which side, reactants or products, is favored at equilibrium? How do you know? To well #1 add 5 drops of deionized water, one drop at a time. Stir. Record you observations. Well #1 Observations: Write the net equation that describes the observed change: Repeat this step for well #2 so that they both exhibit the same color. Use well #4 as a standard for comparison purposes. Carefully add one drop of 12 M HCl at a time to well #1 until five drops have been added and stir. Caution: 12 M HCl is caustic and corrosive. Avoid contact and immediately rinse all spills with lots of water. Well #1 + HCl(aq) Observations: Comment on the difference in the intensity or hue of the color and its relationship to the equilibrium. Write the net equation that describes the observed change: To well #2 add a 2-3 spatula tips (approximately 0.06-0.07 g) of granular anhydrous calcium chloride and stir. You should observe a change. Well #2 + CaCl 2 Observations: Write the net equation that describes the observed change:

3 To well #3 add 10 drops of 0.1 M silver nitrate, AgNO 3. Caution: Silver Nitrate will stain clothing and skin. Well #3 + AgNO 3 (aq) Observations: What is the precipitate formed? (If you do not know, find out by checking a solubility table in your textbook.) Write the net equation(s) that describes the observed change: To the remaining solution in the flask add enough deionized water dropwise until a purple color is achieved that is half-way between blue and pink. Place the beaker on a hot plate and warm until a color change appears. Record your observations. Observations: Write the net equation that describes the observed change (be sure to include heat term): Chill the beaker in an ice bath for several minutes and again record observations. Observations: Write the net equation that describes the observed change change(be sure to include heat term): Explain the color changes that resulted from heating or cooling. Is the original equilibrium reaction exothermic or endothermic? Where would heat go in 2+ Co(H 2 O) 6 + 4Cl - 2- CoCl 4 + 6H 2 O Quantitative Analysis of Equilibria You will calculate the K eq for the reaction between aqueous iron(iii) nitrate, Fe(NO 3 ) 3 and potassium thiocyanate, KSCN. The reaction forms a red-orange complex, FeSCN 2+. The net-ionic equation is shown below. Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq)

4 yellow-orange red-orange The equilibrium constant for this reaction is: In order to find the equilibrium concentration a spectrometer will be used. Since a colored complex is formed, the concentration of FeSCN 2+ can be determined by measuring the absorbance. The amount of light absorbed by the sample is proportional to the concentration of FeSCN 2+. For our experiment the constant is 5.00 x 10 3. [FeSCN 2+ ] = Absorbance/constant To reduce the probability of large errors you will determine K eq three times for different initial. Experimental Label three 25 ml flasks 1,2 and 3. Obtain approximately 15 ml of the 2.00 x 10-3 M Fe(NO 3 ) 3 solution from the stock bottle and place in a properly labeled flask. Also obtain approximately 10 ml of 2.00 x 10-3 M KSCN samples and place in a properly labeled flask. Using a 10-mL graduated pipet, transfer the indicated volumes of 0.00200 M Fe(NO 3 ) 3 into each of three small flasks. Be sure to rinse the pipet with deionized water and a sample of the next solution before each transfer. Add the required amounts of 1.00M HNO 3 and deionized water in each flask. Do not add the KSCN solution until you are ready to measure the absorbance for each flask. Flask ml of 0.00200M ml of 1.00M HNO 3 ml of d.i. water ml of KSCN Fe(NO 3 ) 3 in 1.00M HNO 3 0.00200 M 1 4.50 0.50 3.50 1.50 2 3.00 2.00 3.50 1.50 3 1.50 3.50 2.00 3.00 Volumes will be additive for a total volume of 10.00 ml of solution. Stir each solution well before poring into a cuvette. Place a portion of each solution in a spectrophotometer cell and measure the absorbance to three significant figures if possible at a wavelength of 447nm (preset). Record the values. Absorbance [FeSCN 2+ ] (calculated) M Test tube 1

5 Test tube 2 Test tube 3 Set up your equilibrium calculations just as if you were doing an equilibrium problem in class. You can attach an extra page with your calculations. Remember molarity = moles/volume(l). Note that the initial volume of reactants is different from the final volume at equilibrium. Be sure to correct for this as you calculate the initial. Note that in each of the test tubes : the moles of Fe 3+ reacted = moles of FeSCN 2+ at equilibrium. Test Tube 1 Calculations: Fe 3+ (aq)+scn - (aq) FeSCN 2+ (aq) [Fe 3+ ] [SCN - ] [ FeSCN 2+ ] initial (corrected to total volume) change in equilibrium Test Tube 2 Calculations: initial (corrected to total volume) change in equilibrium [Fe 3+ ] [SCN - ] [ FeSCN 2+ ] Test Tube 3 Calculations: [Fe 3+ ] [SCN - ] [ FeSCN 2+ ]

6 initial (corrected to total volume) change in equilibrium K eq (test tube 1) K eq (test tube 2) K eq (test tube 3) K eq = ± (Average ± average deviation) Show setup below: Post-Lab Questions: 1. Suppose you have P 2 (g) in a sealed flask. At a constant temperature the following

7 equilibrium takes place P 4 (g) Ø 2 P 2 (g) a. What is the value of Q for the equilibrium as written? b. In which direction will the reaction proceed? c. Is K p greater, lesser or equal to K eq? Once equilibrium is attained you decide to do the following changes. In which direction will the equilibrium shift if you: d. add P 4 (g) at constant temperature e. increase the volume of the container at constant temperature f. increase the pressure of the system at constant temperature g. increase the temperature of the system at constant pressure 2. Imagine yourself the size of atoms inside a beaker containing the equilibrium which you studied in this lab which has a K eq greater than one. Co(H 2 O) 6 2+ + 4Cl - CoCl 4 2- + 6H 2 O Write a brief, nanoscale description of what you observe around you before and after the addition of more water to the mixture. 3. The reaction of oxygen to produce ozone proceeds according to the reaction below and has an equilibrium constant of 6.3 x 10-58. The concentration of oxygen in air is about 9.4 x 10-3 M. Determine the concentration of ozone in an equilibrium mixture. 3O 2 (g) 2O 3 (g)