Question 15.51


Moderators: Chem_Mod, Chem_Admin

Victoria Draper 1G
Posts: 52
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time

Question 15.51

Postby Victoria Draper 1G » Mon Mar 12, 2018 9:39 pm

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
Posts: 56
Joined: Fri Sep 29, 2017 7:07 am

Re: Question 15.51

Postby Maeve Gallagher 1J » Mon Mar 12, 2018 10:28 pm

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
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

Re: Question 15.51

Postby Gurkriti Ahluwalia 1K » Mon Mar 12, 2018 11:34 pm

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
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

Re: Question 15.51

Postby Seth_Evasco1L » Tue Mar 13, 2018 12:25 am

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
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

Re: Question 15.51

Postby 204918982 » Tue Mar 13, 2018 5:43 pm

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
Posts: 58
Joined: Sat Jul 22, 2017 3:00 am

Re: Question 15.51

Postby allyz1F » Wed Mar 14, 2018 12:04 am

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
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

Re: Question 15.51

Postby Gurkriti Ahluwalia 1K » Thu Mar 15, 2018 12:04 am

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
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

Re: Question 15.51

Postby Seth_Evasco1L » Thu Mar 15, 2018 3:44 pm

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
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: Question 15.51

Postby Timothy Kim 1B » Fri Mar 16, 2018 3:42 pm

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


Return to “Reaction Mechanisms, Reaction Profiles”

Who is online

Users browsing this forum: No registered users and 1 guest