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Author Question: The experimental rate law for the decomposition of nitrous oxide (N2O) to N2 and O2 is Rate = ... (Read 15 times)

madam-professor

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on: Mar 21, 2021
The experimental rate law for the decomposition of nitrous oxide (N2O) to N2 and O2 is Rate = k[N2O]2. Two mechanisms are proposed:
I.
N2O → N2 + O
 
N2O + O → N2 + O2
II.
2N2O N4O2
 
N4O2→ 2N2 + O2
Which of the following could be a correct mechanism?

​


◦ Mechanism I, with the first step as the rate-determining step.
◦ Mechanism I, with the second step as the rate-determining step as long as the first step is a fast equilibrium step.
◦ Mechanism II, with the second step as the rate-determining step if the first step is a fast equilibrium step.
◦ None of the listed choices could be correct.
◦ At least two of the listed choices could be correct.
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Marked as best answer by madam-professor on Mar 21, 2021

ghepp

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Reply #1 on: Mar 21, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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madam-professor

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Reply 2 on: Mar 21, 2021
YES! Correct, THANKS for helping me on my review

at

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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