## Question 15.51

$K = \frac{k_{forward}}{k_{reverse}}$

Victoria Draper 1G
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### Question 15.51

In this homework problem why is the answer rate=k[NO][Br2] instead of rate=k[NO]^2[Br2]? Because in the end you are left with two NO molecules on the reactant side of the equation.

Maeve Gallagher 1J
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### Re: Question 15.51

In a reaction mechanism, it is only the slowest elementary step that determines the rate law of the formation of product. In this case, the slow step is NO + Br2 ---> NOBR2. Since this is an elementary reaction we can write the rate law using stoichiometric coefficients of only step 1, since that is the slow step. This gives Rate = k[NO][BR2].

Gurkriti Ahluwalia 1K
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### Re: Question 15.51

can someone elaborate on what the solutions manual is talking about when we say that the second elementary reaction is fast and it does not affect the reaction order??

Seth_Evasco1L
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### Re: Question 15.51

If the second elementary reaction is significantly faster than the first, then it has no effect in the overall reaction order. We've discussed that the slow step (in this case, the first elementary reaction) is the one that determines the overall reaction order because the overall reaction cannot proceed any faster than its slowest elementary step.

204918982
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### Re: Question 15.51

Gurkriti Ahluwalia 1K wrote:can someone elaborate on what the solutions manual is talking about when we say that the second elementary reaction is fast and it does not affect the reaction order??

the rate law is determined by the slowest elementary step, which is step 1 in this case, so step 2 doesn't have any effect on it because its faster than step 1

allyz1F
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### Re: Question 15.51

Another main point to add to this is that the slow step must occur before other elementary steps in order for the overall reaction rate to only depend upon this slowest step; in other words, if there is another elementary step occurring before the slow step, this must be included in the overall rxn rate as well. When the slow step occurs first, it takes so much longer than the faster steps following it that when it is finally over, the faster steps occur instantaneously and therefore do not contribute to the overall rate.

Gurkriti Ahluwalia 1K
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### Re: Question 15.51

allyz1F wrote:Another main point to add to this is that the slow step must occur before other elementary steps in order for the overall reaction rate to only depend upon this slowest step; in other words, if there is another elementary step occurring before the slow step, this must be included in the overall rxn rate as well. When the slow step occurs first, it takes so much longer than the faster steps following it that when it is finally over, the faster steps occur instantaneously and therefore do not contribute to the overall rate.

then what would you say for a reaction where the fast step occurs first? is this the bottleneck concept??

Seth_Evasco1L
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### Re: Question 15.51

Yes, that would be the bottleneck concept where the intermediate is built up due to the fact that a fast step occurs before the slow step. In this case, you would use the pre-equilibrium approximation approach.

Timothy Kim 1B
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### Re: Question 15.51

The slow step determines the rate of the reaction so that is where you get your rate law from.