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ClkRxn

Course: CHEM 6, Spring 2005
School: Dartmouth
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6 LECTURE Chem DEMONSTRATION #1 (CLOCK REACTION) The aqueous reaction between two anions, peroxydisulfate (S2O2) and thiosulfate 8 (S2O2), to form sulfate (SO2) and dithionate (S4O2) anions is accelerated by iodide (I). In the 3 4 6 course of the reaction, iodide is converted to triiodide (I3) temporarily, then I3 is returned to I. Consequently, I3 is an intermediate and iodide is a catalyst. The net reaction...

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6 LECTURE Chem DEMONSTRATION #1 (CLOCK REACTION) The aqueous reaction between two anions, peroxydisulfate (S2O2) and thiosulfate 8 (S2O2), to form sulfate (SO2) and dithionate (S4O2) anions is accelerated by iodide (I). In the 3 4 6 course of the reaction, iodide is converted to triiodide (I3) temporarily, then I3 is returned to I. Consequently, I3 is an intermediate and iodide is a catalyst. The net reaction is, therefore, 2 S2O8 + peroxydisulfate 2 S2O2 3 thiosulfate 2 SO2 + S4O2 4 6 sulfate dithionate while the two reactions involving the consumption and re-generation of the catalyst are 2 2 S2O8 + 3 I 2 SO4 + I3 2 S2O2 + I3 S4O2 + 3 I 3 6 which, youll note, add to the net reaction. Our goal is to find the kinetic rate law, which well assume has the general form d [S2O2] 8 Rate = = k [S2O2]m [I]n 8 dt where k is the rate constant, and m and n are reaction order exponents. These are the quantities we seek, and heres how well do it. Triiodide (itself colorless) forms a colored complex with starch, but, since the step that destroys triiodide is faster than its reaction with starch, no color can form as long as some of the thiosulfate reactant is present. As soon as the thiosulfate is gone, the color rapidly appears (as long as some peroxydisulfate and iodide are present, which well ensure is the case). Thus, our primary data are initial concentrations and the time required for the color to appear after the solution is initially mixed. Clock reaction 1 If we make the initial amount of peroxydisulfate much larger than the initial of thiosulfate, amount we will be able to measure a very good approximation to the initial rate: d [S2O2] [S2O2]t = [S2O2]t = 0 8 8 8 Rate = dt Well measure for three different reaction mixtures: Reagent Mixture A Mixture B Mixture C 0.01 M Na2S2O3 100 mL 100 mL 100 mL 0.2 M KI 250 mL 250 mL 125 mL Starch Indicator 25 mL 25 mL 25 mL 0.2 M Na2S2O8 250 mL 125 mL 250 mL 0.2 M Na2SO4 0 mL 125 mL 0 mL 0.2 M KNO3 0 mL 0L 125 mL Note that the total volume of each solution is the same, 625 mL. The KNO3 added in mixture C is chemically inert. It is there simply to keep the total ion concentration constant in all three mixtures. The table on the next page should be completed in advance so that experimental data can be added in class as they are collected. Clock reaction 2 Clock Reaction Data and Calculations The table below should be completed in advance, except for the boxes with the thicker outline which rely on data we will measure in lecture. MIXTURE A MIXTURE B MIXTURE C [S2O2]t = 0 3 [S2O2]t = 0 8 [I]t = 0 [S2O2]t = 8 /s Rate/ (units?) k/ (units?) Once has been measured for each mixture, the Rate can be calculated from the expression on the previous page. Comparison between experiments with different initial concentrations allow us to find the reaction orders m and n. Once we know these orders, the rates, and the initial concentrations, we can find k and deduce its units. We end with the full rate expression. (You should fill in the boxes.) d [S2O2] 8 Rate = = dt Clock reaction [S2O2] 8 [I] 3
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