Exam I Review Problems

Exam I Review Problems - Sample Review Problems Chapters 12...

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Sample Review Problems Chapters 12 and 13
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Determining Reaction Order from Initial Rate Data continued 0.080 0.0050 rate 2 rate 1 [NO 2 ] 2 [NO 2 ] 1 m = k [NO 2 ] m 2 [CO] n 2 k [NO 2 ] m 1 [CO] n 1 = 0.40 0.10 = m ; 16 = 4 m and m = 2 k [NO 2 ] m 3 [CO] n 3 k [NO 2 ] m 1 [CO] n 1 [CO] 3 [CO] 1 n = rate 3 rate 1 = 0.0050 0.0050 = 0.20 0.10 n ; 1 = 2 n and n = 0 The reaction is 2nd order in NO 2 . The reaction is zero order in CO. rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2
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PLAN: SOLUTION: Determining Reaction Concentration at a Given Time PROBLEM: At 1000 0 C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87s -1 , to two molecules of ethylene (C 2 H 4 ). (a) If the initial C 4 H 8 concentration is 2.00M, what is the concentration after 0.010 s? (b) What fraction of C 4 H 8 has decomposed in this time? Find the [C 4 H 8 ] at time, t, using the integrated rate law for a 1st order reaction. Once that value is found, divide the amount decomposed by the initial concentration. Use ln[A] t = - k t + ln[A] 0 ; ln 2.00 [C 4 H 8 ] = (87s -1 ) (0.010s) [C 4 H 8 ] = 0.83mol/L ln [C 4 H 8 ] 0 [C 4 H 8 ] t = k t (a) (b) [C 4 H 8 ] 0 - [C 4 H 8 ] t [C 4 H 8 ] 0 = 2.00M - 0.87M 2.00M = 0.58
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PLAN: SOLUTION: Determining the Energy of Activation PROBLEM: The decomposition of hydrogen iodide, 2H I ( g ) H 2 ( g ) + I 2 ( g ) has rate constants of 9.51x10 -9 L/mol*s at 500. K and 1.10x10 -5 L/mol*s at 600. K. Find E a .
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Exam I Review Problems - Sample Review Problems Chapters 12...

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