Kinetics at Work

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University of Nebraska, Lincoln *

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Chemistry

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Apr 3, 2024

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Experiment 4 Kinetics at Work Total Points: 43 Crystal Violet Experimental Data: (17 pts) Print out three graphs for each of the concentrations of NaOH (0.2 M, 0.1 M, and 0.05 M). Print the Data Table for constructing your 0.2 M Molarity vs. Time graph. Label the graphs and the axes on the graphs: Molarity vs. Time ln Molarity vs. Time 1/Molarity vs. Time Use linear regression of the linear graphs to determine the best fit. Then show the rate constant, k, of the reaction. NOTE: You must show your work for all calculations; no work, no credit. (2 pts) Calculate the molarity of crystal violet at time t = 0 s for the 0.2 M NaOH. A=E*b*c 0.443= (5x10^4 cm-1 M-1) (1cm) (c) c=0.443 / 5x10^4 cm-1 M-1 c= 8.86x10^-6 M (2 pts) Crystal violet reacting with hydroxide observations: The crystal violet and hydroxide mic started out as a dark purple color, as time passed and the reaction proceeded the solution turned transparent. (3 pts) What is the order of the reaction with respect to the crystal violet and what is the experimental rate constant? The reaction is first order with respect to crystal violet. The experimental rate constant, according to the graph is, 0.0112. Post-lab Questions NOTE: You must show your work for all calculations; no work, no credit.
Experiment 4 Kinetics at Work Total Points: 43 1. (9 pts) Table sugar, sucrose (C 12 H 22 O 11 ), breaks down in the presence of an aqueous acid to give the two simple sugars fructose and glucose. Although they differ in structure, fructose and glucose have the same written formulas: C 6 H 12 O 6 . Suppose the following data table represented the breakdown of sucrose Calculate the values for columns three and four in the table, calculate R 2 , and label the order of the reaction for each column. Time in Minutes M of Sucrose 1/[Sucrose] ln [Sucrose] 0 1.018 0.9823 0.0178 36 0.970 1.0309 -0.03 78 0.939 1.065 -0.063 135 0.888 1.1261 -0.119 300 0.808 1.2376 -0.0213 R 2 0.96284 0.984439 0.974892898 Order of the Reaction zero second first 1/(1.018) = 0.9823 Ln(1.018) = 0.0178 According to the table, what is the overall order of the breakdown reaction? (A calculator will give the most accurate results.) Second Order 2. (4 pts) Ammonium cyanate (NH 4 NCO) reacts in water to form urea, (NH 2 ) 2 CO. Using the rate law for the reaction, predict the amount of ammonium cyanate remaining after 70 minutes from a starting amount of 0.940 mol/L. Rate law for the reaction: NH 4 NCO (NH 2 ) 2 CO Rate = k[NH 4 NCO] 2 The value of k = 0.00511 L/mol min. 1/(NH4NCO)t=kt+1/(NH4NCO) 1/(NH4NCO)t= (0.00511 L/mol min) (70 min) +1/(0.940 mol/L) (NH4NCO)= 0.77059 M 3. (6 pts) In the reaction in which NO 2 and CO produce NO and CO 2 as shown below: NO 2 (g) + CO (g) NO (g) + CO 2 (g) Two steps have been proposed: Step one: (slow) NO 2 (g) + NO 2 (g) NO 3 (g) + NO (g)
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