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Feb_12_2010 - Important Newofficehours:MonandFri1112;Wed.45...

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Important…. New office hours: Mon and Fri 11 – 12; Wed. 4 – 5   Chemistry Tutor Room – Fleming 236 Exam 1 will cover Ch. 12, 13 and 16 through integrated  rate laws.
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ln k 2 k 1 = E a R - 1 T 2 1 T 1 - The Arrhenius  Equation ln k 2  = ln A – (E a /R)(1/T 2 ) ln k 1  = ln A – (E a /R)(1/T 1 ) Two different rate constants at two different temperatures, T 1   and T 2 . ln k 1  – ln k 2  
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Figure 16.11 Graphical determination of the activation energy. ln  k  = (- E a / R  )(1/ T ) + lnA
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Sample Problem 16.8 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 . Use the modification of the Arrhenius equation         to find  E a ln k 2 k 1 = E a - R 1 T 2 1 T 1 - E = -  R ln k 2 k 1 1 T 2 1 T 1 - -1 1 600 K 1 500 K - ln 1.10x10 -5  L/mol*s 9..51x10 -9 
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