Lowell High School AP Chemistry Spring 2009 REACTION KINETICS EXPERIMENT

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1 Lowell High School AP Chemistry Spring 2009 REACTION KINETICS EXPERIMENT Complete the following for Pre-Lab on a clean sheet of paper: (1) In your own words, explain the following: a. why the I 2 concentration from the 1 st elementary step doesn t build up initially. b. what triggers the dark blue color to appear. c. why we vary the concentrations of KI and Na 2 S 2 O 8. d. what role Na 2 S 2 O 3 plays. e. what role the starch plays. f. what we can get from varying the temperature of the reaction (2) Look at Part 1 of the Procedure section of this protocol. For your assigned Student Pair (the number is the same as your lab table number in class), you will be performing a dilution starting with either 0.20M (NH 4 ) 2 S 2 O 8 or 0.20M KI to make a total of 60.0 ml of the desired diluted concentration. Looking at the Dilution Chart, the volumes of solutions needed to dilute are given. Show your work to explain how those solutions will mathematically arrive at the desired diluted concentration for your group s part. Theory: In this experiment, we shall study the rate of oxidation of iodide ion (I - ) by peroxydisulfate ion (S 2 O 2 8 ), which is a rather slow reaction, and consider the specific effects on the rate of: Part I Concentration of Reactions Part II Temperature Part III A Catalyst (aqueous cupric ion) The Iodine clock reaction is a classic chemical clock demonstration experiment to display chemical kinetics in action; it was discovered by Hans Heinrich Landolt in The elementary reactions are: 2I - (aq) + S 2 O 8 (aq) I 2 (aq) + 2SO 4 (aq) very slow I 2 (aq) + 2S 2 O 3 (aq) 2I - (aq) + S 4 O 6 (aq) very fast where S 2 O 8 = peroxydisulfate, S 2 O 3 = thiosulfate, SO 4 = sulfate, S 4 O 6 = dithionate Starch is added as an indicator to detect the presence of a build-up of Iodine (I 2 ). In the 1 st, rate limiting, step, two clear solutions are mixed and at first there is no visible reaction, but after a short time delay, the liquid suddenly turns to a shade of dark blue. In the 2 nd step, Sodium thiosulfate (Na 2 S 2 O 3 ) rapidly reduces Iodine (I 2 ) back to Iodide (I - ) before the Iodine can complex with the starch to form the characteristic blue-black color. Once all the thiosulfate (S 2 O 3 ) is consumed, the Iodine (I 2 ) concentration will build up and then will form the dark blue complex with the starch. In Part I, you will be changing the concentration of the 2 reactants in the 1 st step (I - and S 2 O 8 ) to change the rate of that 1 st step. For every reaction, you will be using a fixed 10 ml of M Na 2 S 2 O 3 solution for a total of L * mol/l = mol of S 2 O 3 ions. Timing to the color change tells you how long it takes the 1 st elementary step to produce a set number of moles of I 2, enough I 2 made in the first step to consume all moles of S 2 O 3 ions in the second, fast step. Once that set amount of I 2 has been produced and all S 2 O 3 ions have been consumed, then the concentration of I 2 can increase and form the dark blue complex with the starch. Under these conditions, S 2 O 8 and S 2 O 3 do not react with each other. The rate of the first reaction can be measured by the amount of time (t) it takes for the S 2 O 3 to be consumed and the starch to turn blue. 1

2 In Part I, the order of the reaction (n and m in this equation) Rate = - [S 2 O ] / t = k[s 2 O 8 ] n [ I - ] m is determined by applying the method of initial rates. This involves keeping the concentration of one reactant in the 1 st, rate limiting, elementary step constant and in excess, and varying the concentration of the other reactant and examining the effect on the reaction rate. Then one does the same for the other reactant. Since k varies with ionic strength of solution, dilutions of reactants will not be made with distilled water but will be made with equal concentrations of other spectator ions - NaNO 3 and (NH 4 ) 2 SO 4. In Part II, The reaction rate constant, k, is determined for the one case where a large excess of both reactants is present (mixture 1), so that one can assume that there is no change in concentration of reactant during the timing. The activation energy, E a, is determined as follows in the Arrhenius equation: k = Ae -Ea/RT Or ln k = (E a /RT) + ln A Without getting into the mathematical proof, the reaction rate, k, is inversely proportional to time, t, when concentration of reactants are kept constant. Thus k=c/t where c is some constant of proportionality. Then ln k = ln c ln t and substituting into the above logarithmic equation, we will get ln t = ln c ln A (E a /RT) or ln t = (E a /RT) + ln c/a On comparing this equation with y = mx + b, we see that on graphing ln t versus (1/T), we should get a straight line whose slope is E a /RT. Remember to represent T in degrees Kelvin and that R = J/K mol. Procedure: There will be eight different student pairs (number 1 through 8). Part I: Effect of Changing Concentration of Reactants Teacher ( Mixture 1 ) 25.0 ml of 0.20M KI 25.0 ml of 0.20M (NH 4 ) 2 S 2 O ml of 0.010M Na 2 S 2 O ml of 0.2% starch solution Student Pair #1 same as Mixture 1 except use 25.0 ml of 0.15M (NH 4 ) 2 S 2 O 8 Student Pair #2 same as Mixture 1 except use 25.0 ml of 0.10M (NH 4 ) 2 S 2 O 8 Student Pair #3 same as Mixture 1 except use 25.0 ml of 0.067M (NH 4 ) 2 S 2 O 8 Student Pair #4 same as Mixture 1 except use 25.0 ml of 0.050M (NH 4 ) 2 S 2 O 8 Student Pair #5 same as Mixture 1 except use 25.0 ml of 0.15M KI Student Pair #6 same as Mixture 1 except use 25.0 ml of 0.10M KI Student Pair #7 same as Mixture 1 except use 25.0 ml of 0.067M KI Student Pair #8 same as Mixture 1 except use 25.0 ml of 0.050M KI Dilutions Needed See Chart at Bottom of Page Pipettes will be provided for you to measure out the 5.00 and ml portions for the mixtures. Follow your teacher s instructions explicitly when using them. To prepare the diluted reactants, large quantities of 0.200M (NH 4 ) 2 S 2 O 8 and 0.200M KI will be provided along with burets on different desks, so that you can measure out the appropriate amounts of each. Dilutions must be made with 0.200M (NH 4 ) 2 SO 4 and 0.200M NaNO 3, respectively (also on other desks). 2

3 Student Pair # (NH 4 ) 2 S 2 O 8 Dilution Chart (NH 4 ) 2 SO 4 KI NaNO 3 Teacher Note: (NH 4 ) 2 S 2 O 8 is diluted with (NH 4 ) 2 SO 4 KI is diluted with NaNO 3 Place 25.0 ml of the (NH 4 ) 2 S 2 O 8 in a beaker using the buret provided. (In the case where a dilution was made, pipet out 25.0 ml, using a volumetric pipet, from the 60 ml mixture). Place the other three reactants (starch, Na 2 S 2 O 3, and 25.0 ml KI (again, if dilution was necessary, use a 25-mL pipet as described above for the peroxydisulfate) in a 250 ml Erlenmeyer flask (can be wet with distilled water). Use a stopwatch or timing device that shows seconds elapsed, and start timing when your partner pours the contents of the beaker into the Erlenmeyer flask. Swirl the flask a few times and then wait for the liquid to turn blue. DO THE ABOVE TWICE FOR EACH MIXTURE, and record the average time on the whiteboard for all to see and record. Part II: Effect of Changing Temperature Since the concentration of all reactants must be kept constant, use 25.0 ml of 0.20M KI (provided with burets) 25.0 ml of 0.20M (NH 4 ) 2 S 2 O ml of 0.010M Na 2 S 2 O ml of 0.2% starch solution Use a pneumatic trough for cooling down the two solutions (ice provided) or heating up the solutions (hot water also provided). Mix as before, but record the exact temperature of solutions while you hold the Erlenmeyer flask in the trough. DO THE ABOVE TWICE FOR EACH MIXTURE, and record the time and temperature for each on the whiteboard (report the average only if the temperature is the same in each trial). 3

4 Part II: Temperature Effect Student Assignments Student Pair #1: Under 10ºC Student Pair #2: Between 10 and 15ºC Student Pair #3: Between 15 and 20ºC Student Pair #4: Between 21 and 25ºC Student Pair #5: Between 26 and 30ºC Part III: Effect of a Catalyst The effect of a catalyst can be measured as follows use the same concentration of reactants as in Part II and use all solutions at room temperature but add to the solution of (NH 4 ) 2 S 2 O 8 : Student Pair #6: one drop of 0.02M Cu(NO 3 ) 2 Student Pair #7: two drops of 0.02M Cu(NO 3 ) 2 Student Pair #8: three drops of 0.02M Cu(NO 3 ) 2 Data: Part 1 Data Table Effect of Changing Concentrations of Reactants Student # [KI] [(NH 4 ) 2 S 2 O 8 ] Time (s) Rate (1/Time) Teacher 0.20M 0.20M M 0.15M M 0.10M M 0.067M M 0.050M M 0.20M M 0.20M M 0.20M M 0.20M Part 2 Data Table Effect of Changing Temperature Student # Temperature Time (s) Rate (1/Time) 1 < 10ºC ºC ºC ºC ºC Part 3 Data Table Effect of a Catalyst Student # Drops of 0.02M Cu(NO 3 ) 2 Time (s) Rate (1/Time)

5 Data Analysis / Graphs / Calculations / Discussion Questions: Put the answers to the following questions in the appropriate lab report sections. 1. Using the method of initial rates and the data from Part I, determine the exponents, n and m, in the rate law expression on page 1. Describe mathematically your reasoning. Determine and write the correct rate law. 2. Using the answer in Question #1 above calculate [S 2 O 8 ] / t. This is done by calculating how many moles of the S 2 O 8 are consumed when all the S 2 O 3 is gone. Then convert moles to molarity and then divide by elapsed time. 3. From the data in Part II, graph ln t versus 1/T. Draw the best straight line you can (preferably through two of the points). Use the coordinates of those two points to calculate the slope, calculate the activation energy of the reaction (E a ) in kj/mole (see page 1). 4. Make a labeled potential energy vs. reaction coordinate diagram, approximately to scale, for this reaction: 2 I - (aq) + S 2 O 8 (aq) I 2 (aq) + 2SO 4 (aq) kj a similar graph can be found on Chemistry, Zumdahl, 7 th Ed., p. 553, Figure In general, what was the effect of the catalyst? What was the effect of increasing the concentration of the catalyst (was any rate increase proportional to the increase in concentration)? 6. What is the approximate concentration of Cu +2 (aq) in reaction mixture used by Student Pair #7, assuming there are 20 drops per ml. Conclusion: - Use Collision Theory to describe how temperature, concentration, surface area, degree of randomness and catalysts affect the rate of reaction. For temperature, concentration, and catalysts, use lab data to support your answers. - Follow normal lab report guidelines for the rest of the conclusion. 5

6 Reference Tables Part 1: Concentration Changes Student Pair # In Beaker In Erlenmeyer Flask Total ml 1 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.2M KI 2 25 ml of 0.15M (NH 4 ) 2 S 2 O 8 25 ml of 0.2M KI 3 25 ml of 0.10M (NH 4 ) 2 S 2 O 8 25 ml of 0.2M KI 4 25 ml of 0.067M (NH 4 ) 2 S 2 O 8 25 ml of 0.2M KI 5 25 ml of 0.050M (NH 4 ) 2 S 2 O 8 25 ml of 0.2M KI 6 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.15M KI 7 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.10M KI 8 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of M KI 10 ml of 0.01 M Na 2 S 2 O 3 Part 2: Temperature Changes / Part 3: Effect of a Catalyst Student Pair # In beaker In Erlenmeyer flask Total ml 1 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.10M KI 2 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.10M KI 3 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.10M KI 4 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.10M KI 5 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.10M KI 6 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 1 drop of Cu(NO 3 ) 2 25 ml of 0.10M KI 7 25 ml of 0.2M (NH 4 ) 2 S 2 O 8 25 ml of 0.10M KI 2 drops of Cu(NO 3 ) ml of 0.2M (NH 4 ) 2 S 2 O 8 3 drops of Cu(NO 3 ) 2 25 ml of 0.10M KI 6

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