key for Exam II - SP10 - Version A - michel

key for Exam II - SP10 - Version A - michel - 1 CHEMISTRY...

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1 CHEMISTRY 1032 – Michel EXAM II Name __________________________________ signature ____________________ (print clearly) Show all of your work. Partial credit cannot be given for numerical questions if you write your final answer without showing how you arrived at your answer. If you need more room for your work, then use page 7, but indicate clearly that you have done so. Read the problems carefully. Give numerical answers to the correct number of significant figures. The numbers in the [ ] at the end of each question are the point values. Check your work! No graphing or memory accessible calculators are allowed at your examination desk. No cell phones or PDAs are allowed on your persons. Violation of these department rules will be brought to the attention of the Chemistry Department Chairman, the Dean and the Provost, and may result in significant disciplinary action for violation of university standards of academic integrity. Useful information and a periodic table are provided on the last 2 pages.
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2 1. The decomposition of dinitrogen pentoxide: 2 N 2 O 5 ( g ) 4 NO 2 ( g ) + O 2 ( g ), was monitored at 50 o C. The results are shown below. Use this data to answer the questions that follow on the next page. [N 2 O 5 ] vs time 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.0 10.0 20.0 30.0 40.0 50.0 60.0 time, hr [N 2 O 5 ], M ln[N 2 O 5 ] vs time -3.500 -3.000 -2.500 -2.000 -1.500 -1.000 -0.500 0.000 0.0 10.0 20.0 30.0 40.0 50.0 60.0 time, hr ln[N ] target
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3 (a) The rate law for the reaction is: (circle your answer): [2] Rate = – t ] O N [ 5 2 Rate = k[N 2 O 5 ] Rate = k[N 2 O 5 ] 2 k = 2 5 2 2 4 2 ] O N [ ] O [ ] NO [ (b) What is the instantaneous rate of decomposition of N 2 O 5 at 24.0 hours? Show your work clearly. [4] 1) can solve from slope of target from graph 1, or 2) rate = k[N 2 O 5 ] 24 hr = .43hr -1 *0.18M = .0078Mhr -1 (see part e for calculation of k) rate = (-1/2)*( Δ [N 2 O 5 ]/ Δ t), so Δ [N 2 O 5 ]/ Δ t = -2 * rate = -0.016Mhr -1 (c) What is the instantaneous rate of formation of NO 2 at 24.0 hours? [3] rate=(1/4)* ( Δ [NO 2 ]/ Δ t);  Δ [NO 2 ]/ Δ t = 4*0.078Mhr -1  = 0.031Mhr -1 (d) What is the approximate half life of the reaction (to the nearest hour)? [3] t [ ] t 1/2  = 16hr                                              from graph  
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key for Exam II - SP10 - Version A - michel - 1 CHEMISTRY...

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