Camille West Ea and Catalysis

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West 1 E a and Catalysis Camille West Date of Experiment: February 10 Lab Partners: James Foley My document identifies all sources used and assistance received in completing this assignment. I promise to abide by the principles of academic integrity outlines in the Wake Tech syllabus for this class and in the student handbook. I understand that students found guilty of plagiarism or cheating will receive a grade of "0" (zero) on the work submitted, and the Dean of Students will be notified immediately. I did not use any sources or assistance requiring documentation in completing this assignment. Name: ________________Camille West_________________________________ Signature: ______________Camille West________________________________

West 2 Purpose: The purpose of this experiment was to determine the activation energy (Ea) for the reaction between the potassium iodate, KIO 3 -1 (aq), and sodium hydrogen bisulfite with starch, NaHSO 3 -1 (aq), solution. The other purpose of this experiment is to understand the change in the rate of this reaction by introducing catalysts. A series of trials run at various conditioned temperatures was used to determine the activation energy of the reaction. To understand the effects of catalysts on a reaction, at room temperature another series of trials are conducted and compared against run B of part I to find which metal cations increased the rate of the reaction. Methods: The lab procedure is taken from the updated Wake Technical Community College General Chemistry II Laboratory Manual handout. In the first part of this experiment, four trials were administrated at various conditioned temperatures for the uncatalyzed reaction underwent. Each trial acquired two beakers; one 150 mL beaker contains 70.0 mL of deionized water and 20.0 mL of the 0.024 M potassium iodate solution. The other 50 mL beaker holds only 10.0 mL of the 0.016 M sodium hydrogen sulfite with a starch solution. In part IA, the reaction is conducted at low temperatures; so fill a 1000 mL beaker halfway with ice and the rest with tap water until it is ¾ full. Leave the beakers in the cold bath for ten minutes. For runs C and D place the beakers in their appropriate hot water baths; C is at 35°C and D is at 60°C respectively for ten minutes as well. After the ten minutes, keep the beakers in their bath and pour the contents of the second beaker into the first beaker. While the reaction is running, stir the solution and time the reaction. Record the time and temperature it took for the solution to complete, which is indicated by the change in color. Since run B is at room temperature, run the reaction without waiting ten minutes and then record the time and temperature values again. This reaction is happening during each trail as the timer is running is shown below:

West 3 2IO 3 -1 (aq) + 5HSO 3 -1 (aq) → I 2 (aq) + 5SO 4 (aq) -2 + H 2 O (l) +3H +1 (aq) Once the reaction mixture turns blue for runs A, B, and C and a faint brown or black color for run D the reaction is complete. The second part of this experiment is for the catalyzed reaction. In this section repeat the procedure part IB as before and replace the old 70.0 mL of deionized water and 20.0 mL of 0.024 M potassium iodate solution. With the new 65.0 mL of deionized water, 5.0 mL of one of the cation solutions and 20.0 mL of the potassium iodate solution to a 150 mL beaker. Test each cation individually and run the reaction for as long it took for Run B to finish the reaction. Data: Part I: Uncatalyzed reaction For the first part of this experiment, the temperature and time values from the reaction are recorded for all four runs, as seen in Table 1. For runs A, B, and C the solution turns a blue color, while run D turned a black color because starch is unable to bind to iodine at higher temperatures. The temperature will be used to find the x-value points of Figure 1. The reaction time will find the rate by using the ratio one over reaction time. Table 1: recorded temperature and time values Run Temperature (°C) Time (min:s) Reaction Time (s) A 4.3°C 17 min: 22 secs 1042 secs B 21.8°C 7 min: 16 secs 436 secs C 39.2°C 4 min: 17 secs 257 secs D 61.1°C 2 min: 10 secs 130 secs Part II: Catalyzed reaction

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West 4 For the second part of the experiment time and temperature are recorded for all five cations to find which act as catalysts, as seen in Table 2. The time then compared to the rate of reaction IB. The time will be used to find the rate of each cation. The rates of the new catalyzed reaction are found by the ratio one over the reaction time. Table 2: recorded temperature and time values Cation Temperature (°C) Time (min:s) Reaction (s) Cu+2 22.3 2 min:11 secs 131 Fe+2 22.1 0 min:11 secs 11 Fe+3 22.0 1min: 04 secs 64 Ca+2 22.1 7min: 11secs 431 Li+1 22.1 4min: 56secs 294 Results/Discussion Part I: Uncatalyzed Reaction The recorded time (min: s) from Table 1 is converted to seconds to find the rate of the reaction. Then the concentrations of iodate and bisulfite solutions are determined by the dilution formula, M 1 V 1 =M 2 V 2 . The rate constant (k) was resolved to equal one over the time multiplied by the concentrations of the diluted solutions. Listed in Table 3 are the values needed to determine the y –values (ln (k)) of Figure 1, which will be used to find the Arrhenius equation to figure out the activation energy of this reaction. Table 3: calculated concentration, rate constant, and rate time values Run Rate (1/time s -1 ) [ IO 3 -1 ] M [HSO 3 -1 ] M Rate Constant

West 5 A 9.59x10 -4 0.0048M 0.0016M 125 B 2.29x10 -3 0.0048M 0.0016M 299 C 3.89x10 -3 0.0048M 0.0016M 507 D 7.69x10 -3 0.0048M 0.0016M 1002 A ratio was used to convert reaction time (secs) from Table 1 into the rate of reaction. For example, using the data collected from run AI: rate = 1 1042 sec = 9.59 x 10 − 4 s − 1 The dilution formula yields the concentrations of the iodate and bisulfite solutions. For example, based on this formula the concentration of the diluted iodate solution is 0.0048 M. M 1 means the concentration of the first solution (0.024 M). V 1 is the volume of the first solution in milliliters (20.0 mL). M 2 is the concentration of the new solution and the unknown variable. V 2 is the total volume of the new solution in milliliters (100.0 mL). Now rearrange the formula to find the new concentration. M 1 V 1 = M 2 V 2 M 2 = 0.024 M x 20.0 ml 100.0 ml = 0.048 M IO 3 − 1 Since the rate of this reaction is monitored by how long it takes iodine to appear. In the presence of starch, iodine forms a blue solution at low temperatures and a black solution at high temperatures to help procure the inverse relationship between time and rate. The rate constant (k) can be derived using the following formula: k = 1 time x [ IO 3 − 1 ] x [ HSO 3 − 1 ]

West 6 For example, for run B of part I the rate constant was found to be 299 s − 1 M − 2 . k = 1 ( 436 x 0.048 x 0,016 ) = 299 s − 1 M − 2 The Ea or activation energy is found by the Arrhenius equation in its linearized form using the temperature and rant constant values from Table 1. k = Ae − Ea RT ln ( k ) = − Ea R x 1 T + ln ( A ) In this equation, Ea is the activation energy in units of kJ/mole. R is the universal gas constant (0.008314 kJ/mole-k). T represents the temperature, which is always in Kelvin. Finally, A is the Arrhenius factor of the reaction. Ea can be found by plotting the natural log of the rate constant (y) versus one over the temperature in kelvin (x) and fitting the plots in linear regression. Table 4 contains the calculate points of x and y. Table 4: 1/T and ln(k) values of Figure 1. Run 1/T ln(k) A 3.60 x 10 -3 4.83 B 3.39 x 10 -3 5.70 C 3.20 x 10 -3 6.23 D 2.99 x 10 -3 6.90 As said previously the x is found by taking one over the temperature in Kelvin. However, y is found by the natural log of the rate constant (k). Shown below is an example of how to determine the data points of Figure 1 by using data Run IC:

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West 7 x = 1 ( 39.2 + 273 ) = 3.20 x 10 − 3 y = ln ( 507 ) = 6.23 Graph the data points from Table 4 to get the y = mx + b equation. The equation is given in Figure 1 once a trendline is added to the graph. The slope of Figure 1 equals the negative activation energy (Ea) divided by the ideal gas constant of the Arrhenius relationship. Figure 1 demonstrates the relationship between the rate constant and the temperature of a first-order reaction. Figure 1: Rate constant (ln (k)) versus Temperature (1/T) 0 0 0 1 2 3 4 5 6 7 8 f(x) = − 3343.64 x + 16.95 R² = 0.99 ln(k) versus 1/T 1/T (1/K) ln(k) Rearrange the equation found from Figure 1 to determine the Ea of this reaction. Ea equals the slope of the equation multiplied by the ideal gas constant, which was found to be 27.8 kJ/mole. Ea = 3343.6 x 0.008314 = 27.8 kJ / mole

West 8 Part II: Catalyzed Reaction For the second part of this experiment, again determine the rate of the reaction by converting the recorded time (min: s) into reaction time (secs) for each cation solution. In Table 5 the reaction time for each run is compared to part IB, to see which metal cation increased the rate of the reaction at room temperature. Table 5: Rate of reaction with cation Cation Rate (1/time s -1 ) Cu +2 7.63 x 10 -3 Fe +2 9.09 x 10 -2 Fe +3 1.56 x 10 -2 Ca +2 2.32 x 10 -3 Li +1 3.40 x 10 -3 A catalyst lowers the activation energy and increases the rate of a reaction because it offers a new reaction pathway. Based on Tables 2 and 5 the cations that were determined to be a catalyst for this reaction are Cu +2 , Fe +2 , and Fe +3 . These cations took only one or two minutes to complete the reaction compared to the reaction IB. This means the reactant molecules were able to break and reform more due to the new reaction pathway provided by the metal cations. Two cations were determined to not be a catalyst; they were Ca +2 and Li +1 . Ca +2 took seven minutes to complete the reaction just like run B, meaning the metal cation did not provide a new reaction pathway. On the other hand, Li +1 did provide a new reaction pathway for the reactant molecules to break and reform, but the reaction pathway was not low enough for the reactant molecules to break and reform with the energy provided from the reaction. Conclusion

West 9 A first-order reaction between 70.0 mL of deionized water and the two solutions (20.0 mL of 0.024 M potassium iodate and 10.0 mL of 0.016 M bisulfate) help determine the activation energy of an uncatalyzed reaction. For the first part of this experiment, four trials are conducted at various temperatures to find the Arrhenius relationship of the reaction. Once the timed reaction is finished; the reaction changes from a clear color to an intense blue or black. The temperatures and rate constant values are plotted on a linear regression graph to find Ea. Where y equals the natural log of the rate constant (ln (k)) and x equals one over the temperature in Kelvin (1/T). After extracting the Ea value ( -3343.6 kJ/mole), from the slope of the graph and multiplying it by the ideal gas constant (0.008314 kJ/mole-K); the activation was determined to be 27.8 kJ/mole. The reported activation energy was 30.0 kJ/mole giving a percent error of 7.33%. The error most likely from run B of part I, as Figure 1 depicts the data point above the trend line. This happened because the reaction finished earlier than the intended time, based on the trajectory of the trend line. After splitting the difference, the rate constant should have been 260 s -1 M -2 suggesting the reaction was to finish at eight-minute and twenty seconds. This mistake occurred while the two solutions were mixing. The reactant molecules are more energized from using the same beakers throughout each reaction without waiting for them to heat up for the next trial after removing them from the cold water bath. Cold temperatures tend to slow down the reaction. So when the beakers were placed on the lab bench, they were slowly reaching room temperature, creating a new reaction pathway. This new pathway provided the reactant molecules with enough energy to overcome the activation barrier at a faster rate. As for the second part of this experiment, five trials were run between 10.0 mL bisulfate solution and 65.0 mL deionized water with 5.0 mL of one of the five metal cations, and 20.0 mL of the iodate solution to recognize the effects of a catalyst on a reaction. These reactions are timed and

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West 10 compared to IB to find which metal cations act as a catalyst. The rate of the reaction helped determine this since catalysts are known to speed up the reaction by lowering the activation barrier. The new reaction pathway allows the reactant molecules to break and reform more since the energy is not influenced by other stresses such as, temperature. The catalysts are found to be Cu +2 , Fe +2 , and Fe +3 , for they completed the reaction within one to two minutes compared to Ca +2 and Li +1 , who took roughly seven minutes to complete. References Wake Technical Community College: General Chemistry II (CHM 152) Laboratory Experiments ; Wake Tech: North Carolina, 2019; pp 36-43.