The experimental rate law for the decomposition of nitrous oxide (N20) to N2 and Oz is Rate = k[N2OF. Which of the following could be a correct mechanism? Briefly explain why. %3D N20-N2 +O N20 +0-N2 + O2 (slow) (fast) N20 N2 +O N20 +0-N2 + O2 ii (fast) (slow) 2N20 N,O2 N4O2-2N2 + O2 (fast) (slow) II.

Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter11: Chemical Kinetics
Section: Chapter Questions
Problem 11.48PAE: 11.48 The following data were collected for the decomposition of NT),-: Time, f (min) [N2Os] (mol...
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The experimental rate law for the decomposition of nitrous
oxide (N20) to N2 and Oz is Rate = k[N2O]F. Which of the following could be
a correct mechanism? Briefly explain why.
N20-N2 + O
N20 + 0-N2 + O2
(slow)
(fast)
ii
N20 N2 + O
N20 +0-N2 + O2
(fast)
(slow)
2N20 N,02
N4O2-2N2 + O2
iI.
(fast)
(slow)
Transcribed Image Text:The experimental rate law for the decomposition of nitrous oxide (N20) to N2 and Oz is Rate = k[N2O]F. Which of the following could be a correct mechanism? Briefly explain why. N20-N2 + O N20 + 0-N2 + O2 (slow) (fast) ii N20 N2 + O N20 +0-N2 + O2 (fast) (slow) 2N20 N,02 N4O2-2N2 + O2 iI. (fast) (slow)
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