Lab 19: The Iodine Clock PDF

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Lab 19: Iodine Clock. A Chemical Kinetics Lab...

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Elizabeth Teper CHEM 106 12/19 Lab 19: The Iodine Clock-A Chemical Kinetics Lab Introduction: The iodine clock reaction is an experiment that found out that a reaction does not always have to react right away but can occur after a certain amount of time. In this lab, we will be determining the rate constant and the activation energy of persulfate. When mixing persulfate with a clocking reagent thiosulfate, both colorless liquids, the solution will eventually turn dark blue1. By adding different amounts of thiosulfate, we can change the time of the reaction. We will also use different temperatures that effects the activation energy and as a result, changes the time of the reaction as well. This reaction will not be an elementary step reaction where the rate law can be determined from one step, but will be a multi-step reaction. After observing and recording our data, we will find the rate constant and the activation energy, as well as plot our data to prepare an Arrhenius plot. Materials: - 1M KI - 0.1M Na2S2O3 - water - 1% starch - 3.0mL 0.15M Na2S2O8 - 5 large test tubes - 10mL graduated cylinder - Thermometer - Timer - 600mL beaker - Hot-plate - Ice-bath - Stirring rod Observations/Experimental: 2I-(aq)+ S2O82-(aq)→I2(aq)+2SO4-(aq) I2(aq)+ S2O32-(aq)→ S4O32-(aq)+2I-(aq) starch+I2(aq)→ blue complex concentration of iodine before color change doesn’t change. rate=k[I-][ S 2O82-] Part1: Determination of the Rate Constant Room Temperature Determ- mL Moles Moles ination Na2S2O3 Na2S2O3 S2O82consumed consumed

[S2O82-] Moles S2O82remaining remaining

ln[S2O82-] Elapsed Elapsed Temp, ℃ remaining time, s time, min

1

0.0003

-3.69

3.0

0.0003

0.00015

0.025

250

4.17

23.1

2 3 4 5

2.5 2.0 1.5 1.0

0.00025 0.0002 0.00015 0.0001

0.000125 0.0001 0.00075 0.0005

0.000325 0.000035 0.000375 0.0004

0.027 0.029 0.031 0.033

-3.61 -3.54 -3.47 -3.41

168 136 84 55

2.80 2.27 1.40 0.92

23.1 23.1 23.1 23.1

Sample Calculations: mol Na2S2O3 consumed: (0.1M)(0.003L)=0.0003 mol Na2S2O3 mol S2O82- consumed: mol Na2S2O3 consumed* (1mol S2O82- /2mol S2O3)→0.00015 mol S2O82mol S2O82- remaining: 0.00045mol total- 0.00015mol S2O82- consumed= 0.0003 mol S2O82[S2O82-] remaining: mol S2O82-/Total Volume: 0.0003mol/0.012L= 0.025 ln[S2O82- ] remaining: ln(0.025M)=-3.688

Rom temp: ln[S2O8^2-] vs. Time Elapsed time in min 0.00

1.00

2.00

3.00

4.00

5.00

-3.35 y = -0.0854x - 3.3481 R² = 0.9771

ln[S2O8^2-] remaining

-3.40 -3.45

ln[S2O8^2-] remaining

-3.50 -3.55

Linear (ln[S2O8^2] remaining)

-3.60 -3.65 -3.70 -3.75

y = -0.0854x - 3.3481 K’=k*initial conc. of I K=k’/initial conc.

k’=-0.0854 k’=k[I0-]→[I0-]=(1M KI)(45mL/180mL)= 0.25M k=0.0854min-1/0.25M→ 0.354min-1

Part 2: Determination of the Activation Energy Low temperature Determ- mL Moles Moles ination Na2S2O3 Na2S2O3 S2O82consumed consumed

[S2O82-] Moles S2O82remaining remaining

ln[S2O82-] Elapsed Elapsed Temp, ℃ remaining time, s time, min

1 2 3 4 5

0.0003 0.000325 0.000035 0.000375 0.0004

-3.69 -3.61 -3.54 -3.47 -3.41

3.0 2.5 2.0 1.5 1.0

0.0003 0.00025 0.0002 0.00015 0.0001

0.00015 0.000125 0.0001 0.00075 0.0005

0.025 0.027 0.029 0.031 0.033

705 525 265 185 125

11.75 8.75 4.42 3.08 2.08

13.6 13.6 13.6 13.6 13.6

Low temp: ln[S2O8^2-] vs.Time Elapsed time in min 0.00

2.00

4.00

6.00

8.00

10.00

12.00

14.00

ln[S2O8^2-] remaining

-3.35

y = -0.0261x - 3.388 R² = 0.9495

-3.4 -3.45

ln[S2O8^2-] remaining

-3.5

Linear (ln[S2O8^2-] remaining)

-3.55 -3.6

-3.65 -3.7 -3.75

y = -0.0261x - 3.388 k’=0.0261 k=0.0261/0.25→0.104min-1 High temperature Determ- mL Moles Moles ination Na2S2O3 Na2S2O3 S2O82consumed consumed

[S2O82-] Moles S2O82remaining remaining

ln[S2O82-] Elapsed Elapsed Temp, ℃ remaining time, s time, min

1 2 3 4 5

0.0003 0.000325 0.000035 0.000375 0.0004

-3.69 -3.61 -3.54 -3.47 -3.41

3.0 2.5 2.0 1.5 1.0

0.0003 0.00025 0.0002 0.00015 0.0001

0.00015 0.000125 0.0001 0.00075 0.0005

0.025 0.027 0.029 0.031 0.033

158 79 59 50 46

2.63 1.32 0.98 0.83 0.77

31.2 31.2 30.7 32.6 31.8

High Temp: ln[S2O8^2-] vs. Time Elapsed time in min 0.00

0.50

1.00

1.50

2.00

2.50

3.00

ln[S2O8^2-] remaining

-3.35

y = -0.1256x - 3.3811 R² = 0.7796

-3.4 -3.45

ln[S2O8^2-] remaining

-3.5 -3.55

Linear (ln[S2O8^2-] remaining)

-3.6

-3.65 -3.7 -3.75

y = -0.1256x - 3.3811k’=6.2072 k=0.1256/0.25→0.5024min-1 Arrhenius Graph: Low Temp. Room Temp. High Temp.

T(K) 286.6 296.1 304.5

1/T 0.003489 0.003377 0.00328

k'(min-1) 0.104 0.354 0.5024

ln(k') -2.263 -1.038 -0.688

Arrhenius Plot 1/T 0.00325 0

0.0033

ln(k')

-0.5 -1

0.00335

0.0034

0.00345

0.0035 y = -7623x + 24.451 R² = 0.9294 ln(k') Linear (ln(k'))

-1.5

-2 -2.5

Ln(k)=(-)+ln(A) Ea=x-R Ea= -7623*-0.008314kJ/mol-K Ea= 63.378kJ/mol

Conclusion: After completing all three experiments, it can be determined that the rate law, the time it takes for the iodine to turn into a dark blue color, can be affected by factors such as temperature. In the lab, it was observed that as the temperature decreases, the speed of the reaction slowed down and the rate constant, k, was larger than the room temperature rate constant. When put into a higher temperature, the reaction sped up and the rate constant was much smaller. As the temperature increase, the rate constant did as well which can be seen in the Arrhenius plot. The graphs of all four parts were linear with a negative slope. The iodine clock reaction had a rate of the second order. Rate=k[I-][S2O82-] Although our data was not too accurate and we did not have enough class time to get all the data for each experiment, we used the data of our classmates and achieved accurate and precise results. Some errors that may have occurs where not having the solutions the same temperature as the ice/hot water baths. References: 1. Smeureanu, Gabriela, General Chemistry 106 Laboratory Manual, 2021 Focus Questions: Part 1: 1. If we don't measure the concentration of persulfate at the clock point, how can we know its concentration? - If concentration isn’t measured in the beginning, we can still find it out from the reaction equation. 2. What is the value of the rate constant at room temperature for the iodine clock reaction? - The value of k at room temperature is 0.354min-1 Part 2: 1. If the temperature is increased or decreased by 10 ℃, by what factor does the iodine-clock reaction speed up or slow down? - When temperature is increased or decreased, the reaction speeds up or slows down by roughly a factor of 2.5. 2. What is the value of the activation energy for the iodine clock reaction? - The activation energy is 63.378kJ/mol Post-Lab Questions: 1. A group of students forgot to add thiosulfate to one test tube. What will they observe? - Without the presence of thiosulfate, the reaction will not delay at all and will turn blue immediately. 2. When performing Part 2 of this lab a group of students forgot to place the test tubes into the hot water bath for 10 min before starting the reaction. Only after adding persulfate they placed the test tubes into the hot water bath. How will this mistake affect their data (elapsed time) and their results(rate constant)? - The mistake of not heating up the test tubes initially will result in the reaction taking a longer time to change color. Since heating up the solution will speed up the rxn, the fact that the test tubes are not at a higher temperature, the elapsed time will take longer until the solution heats up. The rate constant will be higher than intended. 3. Is it possible to determine k’ with a single measurement? A) Explain how.

B) Why did we bother to perform multiple measurements to determine k’ if we can do it with a single measurement? Be specific. - A) It would be possible to measure k’ with a single measurement by using the instantaneous rate of reaction as well as the change in reactants. B) By performing multiple measurements, it allows us to get more accurate data and being able to identify the source of error. 4. Using your Arrhenius plot, determine the rate constant at 75℃. How many seconds would it take for determination 5 to clock at 75℃? - T=75℃→348K ln(k)=-(Ea/R)(1/T)+lnA 1/T=0.00287K-1 -2.5458=(7.623/8.314)(1/348)+lnA Y= -7623x + 24.451=2.5458 lnA=-0.2548 =7623(0.00287)- 24.451=-2.5458 A=0.078 sec lnK=-2.5458 k=e-2.5458=0.0763...