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?

Question

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

The experimental rate law for the decomposition of nitrous oxide (N<sub>2</sub>O) to N<sub>2</sub> and O<sub>2</sub> is Rate = k[N<sub>2</sub>O]<sup>2</sup>. Two mechanisms are proposed: I. N<sub>2</sub>O → N<sub>2</sub> + O N<sub>2</sub>O + O → N<sub>2</sub> + O<sub>2</sub><sub></sub><sub></sub> II. 2N<sub>2</sub>O N<sub>4</sub>O<sub>2</sub><sub></sub> N<sub>4</sub>O<sub>2</sub> → 2N<sub>2</sub> + O<sub>2</sub><sub></sub> Which of the following could be a correct mechanism? A) Mechanism I with the second step as the rate-determining step. B) Mechanism II with the second step as the rate-determining step. C) Mechanism I with the first step as the rate-determining step. D) Two of these could be correct. E) None of these could be correct.

N₄O₂
N₄O₂ → 2N₂ + O₂
Which of the following could be a correct mechanism?

Quizplus · Accepted Answer

In Mechanism II, two molecules of nitrous oxide are involved in the slow, rate-determining step, which matches the experimentally observed rate law. In Mechanism I, only one molecule of nitrous oxide is involved in the slow step, which does not match the observed rate law. Therefore, Mechanism II is the correct choice.

The Experimental Rate Law for the Decomposition of Nitrous Oxide