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

Question 64

Multiple Choice

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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