Exploring Chemical Kinetics In Our Lab

Lab 19 and 20 Introduction Lab 19 In lab 19 we will be exploring chemical kinetics in our lab, where we will be looking at an iodine clock reaction and then determining the rate constant and activation energy using the necessary calculations. We can find the rate constant using our given rate constant formula in order to plot the lnk vs 1/T data and find the slope of our graph. With slope information in this kind of graph, we will then be able to find the activation energy because of the relationship between slope and activation energy in this type of graph. In part 1 of the lab, where we are focusing on the rate constant, something we want to look for is the concentration of persulfate at the checkpoint. In part 2 we will be looking at how temperature will speed our or slow down our iodine clock reaction. Lab 20 In lab 20, our focus will be on kinetics, which is defined as the study of rates of reaction. In part 1 of our experiment, we will be focusing on determining the rate law when ferroin dissociates. We will be discussing the mechanisms that occur during this process as well in order to gain an understanding on the reaction in which it dissociates. In part 2 of our experiment we will be determining the activation energy for the reaction, where we will be copying data into a spreadsheet, determining the half-lives, initial rates, rate constants. Using the rate constants which we found through our graphs of different runs in our experiment, will then allow us to plot our arrhenius plot, which will tell us the activation energy through its slope. Materials Lab 19 1. Graduated Cylinders 2. Test Tube 3. Water 4. Beaker 5. Hot bath 6. Ice bath 7. Reagents 8. Lab manual Lab 20 1. Ferroin 2. Sulfuric Acid 3. Lab manual Observation and Experimental Lab 19 pt 1 mL Na2S2O 3 Moles Na2S2O 3 Consum ed Moles S2O82Consum ed Moles S2O82Remaini ng [S2O82-] Remaini ng ln[S2O8 2] Remaini ng Elapsed Time in s Elapsed Time in min 1 3 .0002 .00015 .0003 .0250 -3.69 312.6 5.21 2 2.5 .000125 .000125 .000325 .0271 -3.61 246 4.1 3 2 .0002 .0001 .00035 .0292 -3.53 198 3.3 4 1.5 .00015 .000075 .000375 .03125 -3.47 138.6 2.31 5 1 .0001 .00005 .0004 .0333 -3.40 97.8 1.63 Determi nation Sample Calc Moles Na2S2O3 Consumed (M Na2S2O3 x V Na2S2O3) (.1M)(3mL/1 L) = .0003 moles Na2S2O3 Moles S2O82- Consumed (M Na2S2O3/2) (.003mol)/(2) = .00015 mol S2O82Moles S2O82- Remaining (S2O82- fin - S2O82- initial) (.15M)(.003L) - (.00015mol) = .003 moles [S2O82-] Remaining (S2O82- Remaining / Total V) (.003)/(.012ml) = .025M [S2O82-] ln[S2O82-] Remaining ln(.025) = -3.69 Elapsed Time in s (m x 60) (5.21m)(60s) = 312.6sec k’=k[I-]0 k’ = .0801 [I-]0 = 1M x .25ml = .25M k = .0801/.25 = .320 Lab 19 pt 2 Determi nation mL Na2S2O 3 1 1 Moles Na2S2O 3 Consum ed Moles S2O82Consum ed Moles S2O82Remaini ng [S2O82-] Remaini ng ln[S2O8 2] Remaini ng Elapsed Time in s Elapsed Time in min .0002 .00015 .0003 .0250 -3.69 85.8 1.43 2 2 .000125 .000125 .000325 .0271 -3.61 64.8 1.08 3 3 .0002 .0001 .00035 .0292 -3.53 48.6 .81 4 4 .00015 .000075 .000375 .03125 -3.47 30 .5 5 5 .0001 .00005 .0004 .0333 -3.40 21 .35 K = .2598 / .25 = 1.04 Moles Na2S2O 3 Consum ed Moles S2O82Consum ed Moles S2O82Remaini ng [S2O82-] Remaini ng ln[S2O8 2] Remaini ng Elapsed Time in s Elapsed Time in min 1 .0002 .00015 .0003 .0250 -3.69 432 7.2 2 2 .000125 .000125 .000325 .0271 -3.61 330 5.5 3 3 .0002 .0001 .00035 .0292 -3.53 252 4.2 4 4 .00015 .000075 .000375 .03125 -3.47 187.8 3.13 5 5 .0001 .00005 .0004 .0333 -3.40 126 2.10 Determi nation mL Na2S2O 3 1 K = .0569 / .25 = .2276 Hot Room: 35.5 C + 273K = 308.5 K K = 1.04 K’ = .2598 Lnk’ = -1.35 1/t = .00324 Cold Room: 15.5 + 273K = 288.5 K K = .2276 K’ = .0569 Lnk’ = -2.87 1/t = .00347 Room temp: 25.5 + 273K = 298.5K K = .320 K’ = .08 Lnk’ = -2.53 1/t = .00335 Slope = - Ea / R Slope x R = = -EA -6551.6 x 8.314 = 54470 J → 54.7 kJ/mol Lab 20 part 1 Run 1 Half Life: 10 minutes Rate Constant: ln A vs. T → .0672 Initial Rate: y = .6034e-.067x → y’ = .0404 x 1 = .04043 min-1 Run 2 Half Life: 10 minutes Rate Constant: ln A vs. T → .0683 Initial Rate: y = .6005e-.068x→ y’ = .04083 min -1 Rate = k[ferroin]1 The initial rates are almost the same, which tells us that our reactions have begun consuming reactants at similar paces. The rate law does not represent an elementary step. Slow/rate limiting: Fe(phen)32+ ⇌ Fe(phen)2+ 2+ + phen Fast: Fe(phen)32+ ⇌ Fe2+ + 2phen Lab 20 Part 2 Run 3 Half Life: 53 minutes Rate Constant: ln A vs. T → .0124 Initial Rate: y = .6002e-.012x→ y’ = .007 min -1 Run 4 Half Life: 2.5 minutes Rate Constant: ln A vs. T → .2847 Initial Rate: y = .6239e-.285x→ y’ = .178 min -1 Run 5 Half Life: .65 minutes Rate Constant: ln A vs. T → 1.15 Initial Rate: y = .6345e-.1.15x→ y’ = .73 min -1 Arrhenius Plot 1 40C→ 313K, 1/T→ .00319 K→ .0672, lnk = 4.39 3 30C→ 303K, 1/T→ .00330 K→ .0124, lnk = 2.7 4 50C→ 323K, 1/T→ .00310 K→ .2847, lnk = 1.26 5 60C→ 333K, 1/T→ .00300 K→ 1.15, lnk = .139 Ea = -m x R -15178 x 8.314 = -126, 190 J/mol → 126.190 kJ/mol Discussion and Conclusion Lab 19 In this experiment, our main concern was on getting introduced to the concept of the iodine clock reaction. In the reaction we were studying, we were able to see how a reagent was able to affect the results of our reaction, where changes in thiosulfate concentration affected the amount of our end product. Another goal of this experiment was to learn more about activation energy, and how to obtain it using a graph. Sources of error could have resulted from calculation errors. Lab 20 In our experiment for lab 20, we measured various variables in 5 runs of the dissociation of ferroin which occurred at different temperatures. In order to obtain these variables, such as rate constant and half-life, we needed to graph various versions of the runs, for example, in the form of 1/A vs T, A vs T. and ln A vs T. We were then able to graph lnK vs 1/A in order to obtain the activation energy, which I found to be 126.190 kJ/mol, which was found through multiplying the slope and R constant. Sources of error could have occurred as a result of calculation errors or could have occured when graphing our points. References Smeureanu, G. Geggier, S. (2020). General Chemistry. New York, NY Focus Questions Lab 19 1. If we don’t know the concentration at the clockpoint by measuring it using the equation ln[S2O82-] = ln[S2O82-]0 - k’t 2. The rate constant would be .320 Lab 19 part 2 1. If the temperature is increased by 10 degrees celsius then the reaction would speed up by a factor of 2, and if it is decreased by 10 degrees celsius, then the reaction would decrease by a factor of 2. 2. 54.7 kJ is the value of the activation energy for the clock reaction Lab 20 1. Rate = K [ferroin]1 2. Reaction Mechanism a. Fe(phen)32+ + 3H3O+ → Fe2+ + 3(phen) H+ + 3H2O 3. 126.190 kJ/mol would be the activation energy Post Lab Questions Lab 19 1. Because the group of students forgot to add the thiosulfate to the test tube, no iodide will be converted and we won’t see a color change. 2. A higher temperature will cause an increased rate of reaction and will cause an increase of rate constant as the elapsed time decreases. 3. K’ a. We can determine the k’ by using the equation that uses the slope value plotted on the graph. K’ = e slope/t b. We take multiple measurements to reduce error and obtain a more precise value. 4. y= -6551.6x + 19.72 a. 75 + 273 = 348, 1/t = .00287, 6551.6(.00287) - 19.72 = -.917, lnk = -.917, k=.400 b. -.917 = (6.5516 / 8.314)(1/348) ln A → lnA = -.915→ e -.915 = .4 seconds Lab 20 1. Since the elementary steps don’t require rate-limiting reactions within them, it’s coefficients will always equal the reaction order of its rate law. 2. This type of reaction is one that will proceed spontaneously, but slowly. A reaction that is kinetically favorable is once that will proceed quickly, and a reaction that is thermodynamically favorable is one that will have negative value in free energy change. 3. Mechanism a. A + D → F b. The catalyst is C, and the intermediates are B and E c. Molecularity and Rate law i. 1: Bimolecular, Rate = k1 [A][C] ii. 2: Bimolecular, Rate = k2 [B][D] iii. 3: Unimolecular, Rate = k3 [E] d. The mechanism is consistent with the actual rate law.