340_probset3_s02_answers

340_probset3_s02_answers - CHEM 340.001–Debye–Spring...

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Unformatted text preview: CHEM 340.001–Debye–Spring 2002 NAME: ANSWERS PROBLEM SET 3 This Problem Set is given out Tuesday, April 23 at 8:00am and is due back on Tuesday, April 30 at 9:00am . You must include all work used to obtain your answers including graphs, data points generated and/or plotted, equations of lines used to fit data, etc. All work must be your own although you may collaborate with your classmates in deciding on strategies for answering the problems. ...................................................................................................................................................................... Consider the simple irreversible reaction A --> C to proceed by the following mechanism: k 1 i. A ---> B k 2 ii. B ---> C Assume that the reaction is run with [ A ] o =a o and [ C ] o =0 . The exact concentrations of the three chemical species as a function of time (t) are then given by: [ A ] = a o exp(-k 1 t ) k 1 a o exp(-k 2 t ) {exp[(k 2-k 1 ) t ] – 1} [ B ] = ------------------------------------------ k 2 – k 1 [ C ] = a o – [A] – [B] The notation is that exp[f( t )] = e f (t ) . For Problems #1 through #4 below, assume that k 1 =0.00500 s-1 , k 2 =0.00250 s-1 , and [ A ] o =0.100 moles/L 1. Using the above exact expressions, make a graph of [ B ] vs. time from the beginning of the reaction to the time when the reaction is 80% completed. Use enough points to define your graph well. Problem 1: [B] Time (in seconds) [B] 100 200 300 400 500 600 700 800 900 .0 .0 5 .0 1 .0 1 5 .0 2 .0 2 5 .0 3 .0 3 5 .0 4 .0 4 5 .0 5 .0 5 5 With an intermediate, the reaction is 80% complete based on the product formation (900 seconds, here) , and not based on the reactant disappearance (320 seconds, here). CHEM 340.001–Debye–Spring 2002 2. Again using the above exact expressions, make a graph of [ C...
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340_probset3_s02_answers - CHEM 340.001–Debye–Spring...

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