Bimolecular rate law, 14.47b and 14.49


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Isha Bagga 3J
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Joined: Fri Sep 26, 2014 2:02 pm

Bimolecular rate law, 14.47b and 14.49

Postby Isha Bagga 3J » Tue Feb 17, 2015 11:59 pm

For 14.47b, I thought that the answer was supposed to be rate =k*[NO2]^2*[O] for step 2, but the answer showed rate =k*[NO2]*[O]. Also, for 14.49, I thought the answer would be rate =k*[NO]^2*[Br], but again it is just rate =k*[NO]*[Br]. Can anybody explain why this is? Thank you!

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: Bimolecular rate law, 14.47b and 14.49

Postby Neil DSilva 1L » Wed Feb 18, 2015 1:04 am

Problem 14.47 refers you to Table 14.3 (on page 581) and that might help you understand how to write the rate law for an elementary step.

For elementary reactions, we use the coefficients to write the rate law for the step. So in the case of 14.47b, the solutions manual is correct. Since each of the reactants have a coefficient of 1, .

Another thing that might help you is that since the step is bimolecular, the overall order has to be 2. With , the overall order is 3, so it doesn't match.

Brandon Jarrold4H
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Joined: Fri Sep 26, 2014 2:02 pm

Re: Bimolecular rate law, 14.47b and 14.49

Postby Brandon Jarrold4H » Thu Feb 19, 2015 1:27 pm

For 14.49, I think the rate law is k[NO][Br2] instead of k[NO]^2[Br2], because they identify the slow step, which is the only step that affects the rate law. Correct me if I am wrong though.

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

Re: Bimolecular rate law, 14.47b and 14.49

Postby Justin Le 2I » Fri Feb 20, 2015 1:36 am

For 14.47b, why do you think that rate =k*[NO2]^2*[O]? Neil's explanation is correct.

For 14.49, the answer is k[NO][Br2]. I think you might have read the solutions guide wrong.


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