2 NO + Br2 --> 2 NOBr
Two step Mechanism
NO + Br2 <--> NOBr2 (fast)
NOBr2 + NO --> 2 NO2 (slow)
1. What is the Rate Law for this reaction as predicted by this mechanism assuming the Equilibrium Approx. can be applied to the first step in this mechanism?
is it R=K[NO]^2[Br2] ?
2. As a single step (the first equation only), What is the Rate Law?
is it R=K[NOBr][NO] ?
3. Which of the rate laws is more satisfying?
I have no clue.
Two step Mechanism
NO + Br2 <--> NOBr2 (fast)
NOBr2 + NO --> 2 NO2 (slow)
1. What is the Rate Law for this reaction as predicted by this mechanism assuming the Equilibrium Approx. can be applied to the first step in this mechanism?
is it R=K[NO]^2[Br2] ?
2. As a single step (the first equation only), What is the Rate Law?
is it R=K[NOBr][NO] ?
3. Which of the rate laws is more satisfying?
I have no clue.
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The rate law is based on the rate determining step, the slow step. We treat it as an elementary step, and can therefore, use the law of mass action to determine the rate law. In this case the rate determining step includes an intermediate.
The rate law will therefore, be
rate = k[NO]^2[Br2]
... which is the same as if the first equation (2NO + Br2 --> 2NOBr) occurred as a single step.
rate = k[NO]^2[Br2]
Since both rate laws are the same, neither is more "satisfying" than the other, except that the first approach shows a more likely two particle collision (NOBr2 and NO) collision rather than a less likely three particle collision (2 NO and Br2).
The rate law will therefore, be
rate = k[NO]^2[Br2]
... which is the same as if the first equation (2NO + Br2 --> 2NOBr) occurred as a single step.
rate = k[NO]^2[Br2]
Since both rate laws are the same, neither is more "satisfying" than the other, except that the first approach shows a more likely two particle collision (NOBr2 and NO) collision rather than a less likely three particle collision (2 NO and Br2).