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The Experimental Rate Law for the Decomposition of Nitrous Oxide \to

Question 20

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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 \to N2 + O
N2O + O \to 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  \to  N<sub>2</sub> + O N<sub>2</sub>O + O  \to  N<sub>2</sub> + O<sub>2</sub> II.2N<sub>2</sub>O   N<sub>4</sub>O<sub>2</sub> N<sub>4</sub>O<sub>2</sub>  \to  2N<sub>2</sub> + O<sub>2</sub> Which of the following could be a correct mechanism? A) Mechanism I,with the first step as the rate-determining step. B) Mechanism I,with the second step as the rate-determining step as long as the first step is a fast equilibrium step. C) Mechanism II,with the second step as the rate-determining step if the first step is a fast equilibrium step. D) None of the choices (A-C) could be correct. E) At least two of the above choices (A-C) could be correct. N4O2
N4O2 \to 2N2 + O2
Which of the following could be a correct mechanism?


A) Mechanism I,with the first step as the rate-determining step.
B) Mechanism I,with the second step as the rate-determining step as long as the first step is a fast equilibrium step.
C) Mechanism II,with the second step as the rate-determining step if the first step is a fast equilibrium step.
D) None of the choices (A-C) could be correct.
E) At least two of the above choices (A-C) could be correct.

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