lecnotes33 - MIT OpenCourseWare http/ocw.mit.edu 5.111...

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MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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33.1 5.111 Lecture 33 Kinetics Topics: Reaction Mechanisms Chapter 13 p 549-552 (p 502-505 in 3rd ed) Investigating Reaction Mechanisms (Ch 13.8) 2NO (g) + O 2 (g) 2NO 2 (g) It is experimentally determined that the rate of formation of NO 2 is k obs [NO] 2 [O 2 ] Overall order = ? Is a one step mechanism likely? Proposed mechanism k 1 Step 1 NO + NO N 2 O 2 forward rate = k -1 order= molecularity= reverse rate = order= molecularity= k 2 Step 2 O 2 + N 2 O 2 NO 2 + NO 2 rate = order= molecularity= What is the rate of NO 2 formation? NO 2 is formed in step 2 and the rate equals: rate of formation of NO 2 = 2k 2 [O 2 ][N 2 O 2 ] (The factor of 2 appears because two molecules of NO 2 are formed; so the concentration of NO 2 increases twice as fast as the concentration of N 2 O 2 decreases). but this expression includes an intermediate, [N 2 O 2 ], and is therefore not acceptable. Solve for [N 2 O 2 ] in terms of reactants, products, and rate constants: net rate of formation of N 2 O 2 = k 1 [NO] 2 - k -1 [N 2 O 2 ] - k 2 [N 2 O 2 ][O 2 ]
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33.2 At this point, we use the steady-state approximation. Steady-state approximation
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lecnotes33 - MIT OpenCourseWare http/ocw.mit.edu 5.111...

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