1aa3-2010-tut11-answerkey

1aa3-2010-tut11-answerkey - CHEMISTRY 1AA3 March 29-April...

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CHEMISTRY 1AA3 March 29-April 2, 2010 TUTORIAL PROBLEM SET 11 SOLUTIONS ________________________________________________________________________ Chemistry 1AA3 1. The decomposition of N 2 O 5 in the gas phase at room temperature was studied. Using the following data verify that the rate law is first order in [N 2 O 5 ] and calculate the rate constant. Time (s) [N 2 O 5 ] (M) 0 0.100 50 0.0707 100 0.0500 200 0.0250 300 0.0125 400 0.00625 SOLUTION: Plot ln [N 2 O 5 ] versus time this gives a straight line confirming that the reaction is first order in N 2 O 5 . Plot: ln [N 2 O 5 ] t = - k t + ln [N 2 O 5 ] 0 slope = - k = -6.93 x10 -3 s -1 2. Using the data in the previous question, calculate [N 2 O 5 ] at 150 s after the start of the reaction. Since 150 s is half-way between 100 s and 200 s (two time points at which [N 2 O 5 ] was measured) is it possible to use the average of the two values to obtain the [N 2 O 5 ] at 150s? If your answer is no, explain why. SOLUTION:
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CHEMISTRY 1AA3 March 29-April 2, 2010 TUTORIAL PROBLEM SET 11 SOLUTIONS ________________________________________________________________________ Chemistry 1AA3 It is not possible to use the average to determine the [N 2 O 5 ] because it is the ln [N 2 O 5 ] that is proportional to time not simply [N 2 O 5 ]. You must use the integrated rate equation for first order reactions: ln([A] t ) = - k t + ln([A] 0 ). Thus, ln [N 2 O 5 ] t = - k t + ln[N 2 O 5 ] 0 t= 150 s k = 6.93 x10 -3 s -1 [N 2 O 5 ] 0 = 0.100 M [N 2 O 5 ] t = exp(-3.343) = 0.0353 M Notice that this value is not halfway between 0.0500 and 0.0250 M.
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1aa3-2010-tut11-answerkey - CHEMISTRY 1AA3 March 29-April...

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