Kinetics Rate

CHEM 1112
Kinetics of the Persulfate – iodide Clock Reaction The purpose of this experiment is to determine the rate law and the activation energy for the reaction between persulfate ion, S2O82-, and iodide ion, I-: S2O82-(aq) + 2 I-(aq)  2 SO42-(aq) + I2(aq)

The rate law can be written as
Reaction rate = (1)

Where m and n are the orders with respect to S2O82- and I-, respectively, and k is the rate constant. Determining the rate law involves determining the values of m and n. The temperature dependence of the rate constant is given by (2)

Equation (2) is called the Arrhenius Equation, where A is the pre-exponential factor, E is the activation energy with units of J/mol, T is the absolute temperature, and R is the gas constant (R= 8.3145 J K-1 mol-1). Equation (2) can also be written in the form:

(3) According to equation (3) the activation energy can be obtained by measuring the rate constant at several temperatures, and then plotting ln k versus 1/T. Method Reaction (1) will be carried out in the presence of thiosulfate ion, S2O32- and starch. The concentration of thiosulfate ion will be maintained at a much lower value than that of either the persulfate ion or the iodide ion. The reactions that occur are: S2O82-(aq) + 2 I-(aq)  2 SO42-(aq) + I2(aq) Slow (4)
I2¬(aq) + 2 S2O32-(aq)  2 I- + S4O62-(aq) Fast (5)
I2¬(aq) + starch  blue complex Slow (6)

Reaction (4) is much slower than reaction (5), and, as a result, the I2 in formed in reaction (4) is immediately consumed by reaction (5), and the concentration of I2 remains at a very low value as long as thiosulfate ion is present. Once all of the thiosulfate ion is used up, the concentration of I2 from reaction (4) increases. The presence of I2 is detected by the formation of a blue complex resulting from the reaction of I2 with starch, reaction (6). The rate of reaction (6) does not become significant until the concentration of I2 becomes appreciable. A characteristic of this reaction is that the reaction mixture remains colorless for several minutes after the reactants are mixed. During this time both reactions (4) and (5) are occurring. The solution remains colorless because the I2 from reaction (4) is being consumed by reaction (5), and can’t react with the starch. As soon as the thiosulfate ion is used up, the I2 reacts with the starch, and an abrupt and dramatic color change, from colorless to blue, occurs. The rate of reaction (4) is the rate of consumption of S2O82- ion. Notice that for every S2O82- ion used up two S2O32- ions are consumed. Thus, we have: Rate of reaction (4) = (7) If (S2O32-)o is the initial concentration of S2O32 ion, and t is the time interval from the start of the reaction until the solution changes color, then in equation (7),
(S2O32-) = 0 - (S2O32-)o , and if the initial concentration of S2O32- is constant then equation (7) becomes Rate of reaction (4) = (8)

Combining equations (1) and (8) and taking the logarithms of both sides gives: (9) Or (10) From equation (9) we see that if we carry out two or more experiments with a constant concentration of I- we obtain (10) And a graph of versus will give a straight line with slope = m. Similarly, for two or more experiments with constant concentrations of S2O82-, a graph of versus ln(I-) will give a straight line with slope = n. To obtain the activation energy, we combine equations (3) and (9): (11) If the concentrations of reactants are constant then equation (11) becomes: (12) A graph of versus will give a straight line with slope = -

Procedure The following solutions will be provided by the Stockroom: A 0.100 M (NH4)2S2O8
B 0.100 M (NH4)2SO4
C 0.010 M Na2S2O3
D 0.100 M KI in 0.066 M (NH4)2SO4 1 % starch solution

Into each of four clean, labeled beakers pour about 85 mL of each of the solutions A, B, C and D. These solutions will be used to perform the following experiments. Do not pour any solutions back into the stock bottles. All waste can go down the drain. Part 1. Observation
To a test tube add 2 mL of solution A, 1 drop of starch solution and 2 mL of solution D. What do you observe? Now add 4 drops of solution C. What do you observe? Add 4 more drops of solution C, and record your observations.

Part 2. Determining the Rate Law
Fill a plastic tub with room temperature water to create a constant temperature bath. Add about 250 mL of room temperature water to each of two 400 mL beakers, and place them in bath. Use a graduated cylinder to accurately measure 8 mL of solution A into a 100 mL beaker, and add two drops of starch solution. You can use a dropper to adjust the amount of liquid in the cylinder. Place this beaker in the 400 mL beaker in the bath. The smaller beaker will be prevented from tipping by the walls of the larger beaker. Again using a graduated cylinder place into a second 100 mL beaker, 2 mL of solution B, 5 mL of solution C and 10 mL of solution D. It is not necessary to wash the graduated cylinder between measurements of B, C and D since these solutions do not react. Place this beaker in the water in the other 400 mL beaker in the bath. Check the temperature of the solutions in the two 100 mL beakers with a thermometer. Use a different thermometer for each solution. When the two solutions are at the same temperature, record the temperature, and pour the contents of the beaker containing solutions B, C and D into the beaker containing solution A and the starch. Measure and record in your notebook, the time interval from the time of mixing to the appearance of the blue color. The volumes just described correspond to Experiment 1 in Table 1, below. Repeat the procedure for the remaining experiments in the Table. Table 1: Solution volumes in mL
Experiment Solution A
S2O82- Solution B (NH4)2SO4 Solution C S2O32- Solution D I-
1 8 2 5 10
2 10 0 5 10
3 5 5 5 10
4 10 5 5 5
5 10 2 5 8 Table 2. Sample Data Table for Part 2 Temperature of reactants:
Experiment Time to color change (S2O82-), mol/L (I-), mol/L 1
2
3
4
5 Part 3. Determining the Activation Energy
To determine the activation energy, the rate of the reaction will be measured at two additional temperatures. Repeat the procedure in Part 2, using the solution volumes corresponding to Experiment 2 in Table 1 at temperatures of about 10 oC and 40 oC. The lower temperature can be obtained by adding ice to tap water in the plastic tub. Fill the tub with warm tap water to obtain the higher temperature. The two solutions must be at the temperature of the bath before they are mixed. Again record the temperature, and the time to the appearance of the blue color.
Perform two trials at each temperature if time permits. Table 3. Sample Data table for Part 3
Temperature of Reactants Time to color change, t Trial 1 Trial 2
Temp. 1 (use your data from Part 2, Exp. 2 here)
Temp. 2
Temp. 3 Calculations Table 4. SUMMARY OF EQUATIONS AND CALCULATIONS: (10a) Plot versus ln(S2O82-) for Experiments 1 – 3. The slope = m (10b) Plot versus ln(I-) for Experiments 2, 4 and 5. The slope = n (12) Plot versus 1/T for Experiment 2, of Part 2, and for Experiments in Part 3.
Slope = -E/R (R = 8.3145 J/K) Calculations for Part 2: For each of the experiments, 1 through 5, calculate the initial concentrations in the reaction mixture (before reaction occurs) of S2O82- and I-, and enter them in a table. Hint: M1V1 = M2V2. Enter your values of t and these concentrations into an Excel spreadsheet, and prepare graphs of ln(1/t) versus ln (S2O82-), for experiments with constant initial values of (I-) (Experiments 1, 2 and 3), and of ln(1/t) versus ln (I-) for experiments with constant initial values of (S2O82-) (Experiments 2, 4 and 5). The equations to be graphed in this part are summarized in Table 4 (Equations 10a and 10b) Use the add trendline feature to determine the slope of each graph. Right click on a data point in the graph. Left click on add trendline. Pick linear. Click on “Options” tab. Choose “Display equation on chart”. What are the values of m and n in equation (1)? Calculations for Part 3: For Experiment 2 of Part 2, and for each of the experiments in Part 3, enter t and the absolute temperature into an Excel spreadsheet, and prepare a graph of ln(1/t) versus 1/T. The equation to be graphed in this part is Equation (12) in Table 4. Use the add trendline feature, as before, to determine the slope of the line and calculate the activation energy. Conclusion Answer the following questions in complete sentences. 1. Explain your observation in Part 1 of the procedure in terms of reactions (4), (5) and (6). 2. For reaction (4), what is the order with respect to each of the reactants? What is the rate law. 3. What is the activation energy for reaction (4)?