03 PreLab_Kinetics2-Erio_KEY (2)

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Document ID: 104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision: 2/9/24 Page 1 of 5 Kinetics of Erioglaucine with Sodium Hypochlorite Part 2: Pre-Lab Activity ANSWER KEY SAMPLE DATA To familiarize you with the data processing you will complete for this lab, sample data has been created for a hypothetical reaction involving two reactants, A and B. Each line of the table below represents a single run of the experiment, and each run is comprised of many absorbance readings over a period of time. For the sake of space, the absorbance vs. time data sets are not provided here. Instead, processed data (similar to what you would read off the LabQuest output) is provided as the basis for these problems. Run [A] at t = 0 sec (M) [B] (M) Temp (°C) 1 7.24 x 10 -5 1.250 23.4 2 7.22 x 10 -5 1.250 23.3 3 3.61 x 10 -5 1.250 23.4 4 3.63 x 10 -5 1.250 23.4 5 7.25 x 10 -5 0.625 23.2 6 7.24 x 10 -5 0.625 23.5 7 7.23 x 10 -5 1.250 13.5 8 7.22 x 10 -5 1.250 13.4 METHOD OF INITIAL RATES (Part B of Lab) 1. In order to determine the kinetics of the reaction via the Method of Initial Rates, what data will you plot for each of your eight runs? Concentration of A vs. Time 2. You will take the first 10 seconds worth of data on each these plots and perform linear fits to these 10 data points. What valuable information can be obtained from this line? How does this relate to the kinetics of the reaction? Slope = ∆𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 ∆𝑻𝑻𝑪𝑪𝑻𝑻𝑪𝑪 This slope is used as the initial rate. 3. The table below provides examples of linear curve fit data for each of the eight sample runs. Use this information to determine the initial rate for each run. Record these rates in the table below. Run [A] at t = 0 sec (M) [B] (M) Temp (°C) Equation of Line (t = 0s to t = 10s) Rate (M/s) 1 7.24 x 10 -5 1.250 23.4 y = (4.32 x 10 -7 ) x + 7.24 x 10 -5 4.32 x 10 -7 2 7.22 x 10 -5 1.250 23.3 y = (4.31 x 10 -7 ) x + 7.22 x 10 -5 4.31 x 10 -7 3 3.61 x 10 -5 1.250 23.4 y = (1.09 x 10 -7 ) x + 3.61 x 10 -5 1.09 x 10 -7 4 3.63 x 10 -5 1.250 23.4 y = (1.08 x 10 -7 ) x + 3.63 x 10 -5 1.08 x 10 -7 5 7.25 x 10 -5 0.625 23.2 y = (2.14 x 10 -7 ) x + 7.25 x 10 -5 2.14 x 10 -7 6 7.24 x 10 -5 0.625 23.5 y = (2.17 x 10 -7 ) x + 7.24 x 10 -5 2.17 x 10 -7 7 7.23 x 10 -5 1.250 13.5 y = (2.33 x 10 -7 ) x + 7.23 x 10 -5 2.33 x 10 -7 8 7.22 x 10 -5 1.250 13.4 y = (2.35 x 10 -7 ) x + 7.22 x 10 -5 2.35 x 10 -7

Document ID: 104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision: 2/9/24 Page 2 of 5 4. From the initial rates you listed in the table, determine: a. Order of reaction in A Second b. Order of reaction in B First c. Rate law Rate = k[A] 2 [B] d. Rate constant at room temperature Using data from Run 1, k = 65.9 M -2 s -1 k = Rate [A] 2 [B] = 4.32 × 10 −7 M/s (7.24 × 10 −5 M) 2 (1.250 M) = 𝟔𝟔𝟔𝟔 . 𝟗𝟗 𝐌𝐌 −𝟐𝟐 𝐬𝐬 −𝟏𝟏 Determination of Activation Energy (Parts A and B of Lab) Kinetics studies provide the opportunity to determine activation energies for reactions through the use of the Arrhenius equation. By determining the value of k at two different temperatures, the Arrhenius equation can be arranged into this combined equation: 𝐥𝐥𝐥𝐥 � 𝒌𝒌 𝟐𝟐 𝒌𝒌 𝟏𝟏 � = 𝑬𝑬 𝑪𝑪 𝐑𝐑 � 𝟏𝟏 𝑻𝑻 𝟏𝟏 − 𝟏𝟏 𝑻𝑻 𝟐𝟐 � where: k 1 is the rate constant at temperature T 1 k 2 is the rate constant at temperature T 2 E a is the activation energy in J/mol R is the universal gas constant (8.314 J/mol K) For each of the methods of analysis you use in this lab, you will be asked to determine the activation energy. Refer to the equation above when you are asked to complete this calculation. 5. Using the data from the table provided in Question 3 above, find the activation energy for this reaction. k 2 = Rate [A] 2 [B] = 2.33 × 10 −7 M/s (7.23 × 10 −5 M) 2 (1.250 M) = 35.7 M −2 s −1 𝑬𝑬 𝑪𝑪 = R × � ln � 𝑘𝑘 2 𝑘𝑘 1 � � 1 𝑇𝑇 1 − 1 𝑇𝑇 2 � � = (8.314 J mol −1 K −1 ) � ln � 35.7 M −2 s −1 65.9 M −2 s −1 � � 1 296.5 K − 1 286.6 K � � = 43,745 J mol = 𝟒𝟒𝟒𝟒 . 𝟕𝟕 𝐤𝐤𝐤𝐤 / 𝐦𝐦𝐦𝐦𝐥𝐥

Document ID: 104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision: 2/9/24 Page 3 of 5 ANALYSIS BY INTEGRATED RATES (Part B of Lab) In order to determine the kinetics of the reaction via the Method of Integrated Rates, you will plot three different types of plots and observe which one produces a linear relationship. 6. Use the information organizer below to summarize what these three different types of plots and the information each provides. Plotted Data If Plot is Linear: Order of Reactant Slope [A] t vs. t Zero - k obs ln[A] t vs. t First - k obs 1/[A] t vs. t Second k obs 7. For the hypothetical reaction between A and B, the plot of 1/[A] t vs. t for Run 1 is linear with a curve- fit equation of y = 82.375 x + 13,812. The same plot for Run 5 is y = 41.095 x + 13,835. Based on this information, determine: a. Order of reaction in A Second Plot of 1/[A] t vs. t is LINEAR b. The value of k obs for Run 1 82.4 M -1 s -1 Slope = k obs c. The value of k obs for Run 5 41.1 M -1 s -1 Slope = k obs 8. Determine the order of the reaction in B. Clearly show your reasoning/work. (Simple investigation or using the equation listed as Equation (8) in the Background section of this lab manual entry). 𝑏𝑏 = ln( 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜2 ) − ln ( 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜1 ) ln[ 𝐵𝐵 ] 2 − ln[ 𝐵𝐵 ] 1 𝑏𝑏 = ln( 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜2 ) − ln ( 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜1 ) ln[ 𝐵𝐵 ] 2 − ln[ 𝐵𝐵 ] 1 = 𝐥𝐥𝐥𝐥 ( 𝟒𝟒𝟏𝟏 . 𝟏𝟏 ) − 𝐥𝐥𝐥𝐥 ( 𝟖𝟖𝟐𝟐 . 𝟒𝟒 ) 𝐥𝐥𝐥𝐥 ( 𝟎𝟎 . 𝟔𝟔𝟐𝟐𝟔𝟔 𝐌𝐌 ) − 𝐥𝐥𝐥𝐥 ( 𝟏𝟏 . 𝟐𝟐𝟔𝟔𝟎𝟎 𝐌𝐌 ) = 𝟏𝟏 𝑭𝑭𝑪𝑪𝑪𝑪𝑭𝑭𝑪𝑪 𝑪𝑪𝑪𝑪𝒐𝒐𝑪𝑪𝑪𝑪 𝑪𝑪𝑪𝑪 𝑩𝑩

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Document ID: 104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision: 2/9/24 Page 4 of 5 9. What is the rate law for this reaction? What is the rate constant? a. Rate law Rate = k[A] 2 [B] b. Rate constant at room temperature Using data from Run 1, k = 65.9 M -2 s -1 k obs = k [B] b = k [B] 1 𝑘𝑘 = 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜 [ B ] = 82 . 375 M −1 s −1 ( 1 . 250 M ) = 𝟔𝟔𝟔𝟔 . 𝟗𝟗 𝐌𝐌 −𝟐𝟐 𝐬𝐬 −𝟏𝟏 10. The linear plot of 1/[A] t vs. t for Run 7 has a curve-fit equation of: y = 44.627 x + 13,831. What is the rate constant for Run 7 using the Method of Integrated Rates? Using data from Run 7, k = 35.7 M -2 s -1 k obs = k [B] b = k [B] 1 𝑘𝑘 = 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜 [ B ] = 44 . 627 M −1 s −1 ( 1 . 250 M ) = 𝟒𝟒𝟔𝟔 . 𝟕𝟕 𝐌𝐌 −𝟐𝟐 𝐬𝐬 −𝟏𝟏 11. Find the activation energy for this reaction. k 1 and T 1 are from Run 1. k 2 and T 2 are from Run 7. 𝑬𝑬 𝑪𝑪 = R × � ln � 𝑘𝑘 2 𝑘𝑘 1 � � 1 𝑇𝑇 1 − 1 𝑇𝑇 2 � � = (8.314 J mol −1 K −1 ) � ln � 35.7 M −2 s −1 65.9 M −2 s −1 � � 1 296.5 K − 1 286.6 K � � = 43,745 J mol = 𝟒𝟒𝟒𝟒 . 𝟕𝟕 𝐤𝐤𝐤𝐤 / 𝐦𝐦𝐦𝐦𝐥𝐥 ANALYSIS BY HALF-LIFE (Part B of Lab) One more way to process kinetics data is through analysis of half-life. In the Kinetics, Part 1 pre-lab activity, you created tables of collected data for the reaction between crystal violet and hydroxide, then determined the half-life by inspecting your collected data. The results from that part are presented below. (You did so for the reaction between erioglaucine and hypochlorite in the actual lab.) For the following questions, refer to “TABLE for RUN 1” 12. Based on the information in TABLE for RUN 1, which statement below is true? a. The half-life decreases by a factor of two for every time interval. b. The half-life remains constant for every time interval. c. The half-life increases by a factor of two for every time interval. 13. What, therefore, is the reaction order with respect to crystal violet? first order _ (Refer to the information the background section of this experiment for information on how to answer this question.)

Document ID: 104 Pre-Lab, Kinetics of Erioglaucine #2, KEY Date of last revision: 2/9/24 Page 5 of 5 14. Using the half-life of the first interval in the TABLE for RUN 1 and using the information in Table 2 of the Background section of this experiment, determine the value of k obs for this set of experimental conditions: 0.030 s -1 (We will refer to this value as k obs 1 below.) 𝑪𝑪 𝟏𝟏 / 𝟐𝟐 = 𝐥𝐥𝐥𝐥 𝟐𝟐 𝒌𝒌 For the following question, refer to “TABLE for RUN 3” . 15. Using the half-life of the first interval in the TABLE for RUN 3 and using the information in Table 2 of the Background section of this experiment, determine the value of k obs for this set of experimental conditions: 0.060 s -1 (We will refer to this value as k obs 2 .) 16. Determine the reaction order with respect to [OH - ] by using the values of k obs1 and k obs2 you found in question 14 and 15. Clearly show your reasoning/work. (Simple investigation or using the equation listed as Equation (8) in the Background section of this lab manual entry). First order 𝑏𝑏 = ln( 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜2 ) − ln ( 𝑘𝑘 𝑜𝑜𝑜𝑜𝑜𝑜1 ) ln[ 𝑂𝑂𝑂𝑂 − ] 2 − ln[ 𝑂𝑂𝑂𝑂 − ] 1 = 𝐥𝐥𝐥𝐥 ( 𝟎𝟎 . 𝟎𝟎𝟔𝟔𝟎𝟎 𝐬𝐬 −𝟏𝟏 ) − 𝐥𝐥𝐥𝐥 ( 𝟎𝟎 . 𝟎𝟎𝟒𝟒𝟎𝟎 𝐬𝐬 −𝟏𝟏 ) 𝐥𝐥𝐥𝐥 ( 𝟎𝟎 . 𝟐𝟐𝟎𝟎𝟎𝟎 𝐌𝐌 ) − 𝐥𝐥𝐥𝐥 ( 𝟎𝟎 . 𝟏𝟏𝟎𝟎𝟎𝟎 𝐌𝐌 ) = 𝟏𝟏 17. Because of the flooding technique used, k obs = k [OH - ] b . What is the value of k ? 0.30 M -1 s -1 18. What is the rate law expression for this reaction? rate = k [CV + ] a [OH - ] b & a = 1, b = 1 gives rate = k [CV + ][OH - ] 19. The lab partners run the crystal violet and hydroxide experiment once more, this time after cooling their reaction mixture down to 11.9 °C. They determine the rate constant at this temperature to be 0.095 M -1 s -1 . Using the value of k determined above (which was measured at a reaction temperature of 22.2 °C), determine the activation energy of this reaction in kJ/mol. 𝑬𝑬 𝑪𝑪 = 𝐑𝐑 × � 𝐥𝐥𝐥𝐥 � 𝒌𝒌 𝟐𝟐 𝒌𝒌 𝟏𝟏 � � 𝟏𝟏 𝑻𝑻 𝟏𝟏 − 𝟏𝟏 𝑻𝑻 𝟐𝟐 � � = ( 𝟖𝟖 . 𝟒𝟒𝟏𝟏𝟒𝟒 𝐤𝐤 𝐦𝐦𝐦𝐦𝐥𝐥 −𝟏𝟏 𝐊𝐊 −𝟏𝟏 ) � 𝐥𝐥𝐥𝐥 � 𝟎𝟎 . 𝟎𝟎𝟗𝟗𝟔𝟔 𝐌𝐌 −𝟏𝟏 𝐬𝐬 −𝟏𝟏 𝟎𝟎 . 𝟒𝟒𝟎𝟎 𝐌𝐌 −𝟏𝟏 𝐬𝐬 −𝟏𝟏 � � 𝟏𝟏 𝟐𝟐𝟗𝟗𝟔𝟔 . 𝟒𝟒 𝐊𝐊 − 𝟏𝟏 𝟐𝟐𝟖𝟖𝟔𝟔 . 𝟎𝟎 𝐊𝐊 � � = 𝟕𝟕𝟖𝟖 , 𝟏𝟏𝟏𝟏𝟔𝟔 𝐤𝐤 𝐦𝐦𝐦𝐦𝐥𝐥 = 𝟕𝟕𝟖𝟖 𝐤𝐤𝐤𝐤 / 𝐦𝐦𝐦𝐦𝐥𝐥