## Chem/Ochem book, p. 131 #4

cperz448
Posts: 12
Joined: Fri Jan 04, 2019 5:51 pm

### Chem/Ochem book, p. 131 #4

For comparing mechanisms based off of data from the provided table, why are you able to use the coefficients in the fast step as the exponents for determining the rate fwd and rate rev? For example, the solution provides the Rate fwd as k1[NO]^2, but the book clearly says that the exponent is not the coefficient, rather it must be experimentally determined. It seems like for these mechanisms, the rate was determined for each of them using the reactants' coefficients that were present in the reaction. Could you explain why this is allowed under these circumstances?
NS_Tutor_Will
Posts: 766
Joined: Fri May 25, 2018 9:15 am

### Re: Chem/Ochem book, p. 131 #4

We actually use the data, which was experimentally derived, to determine the rate law for the reaction. Table 1 shows that as O2 increases and NO is held constant, the rate increases linearly with O2. When O2 is held constant and NO is varied, the rate increases with the square. As NO triples the rate increases 9x. That implies that the rate law should be:

K[O2][NO]^2

When you're asked to find a rate law in the context of the MCAT, it'll either be given directly in the passage or, more likely, you'll have a table just like Table 1 that gives you experimental results that you have to interpret.

Thanks for the question!
TessaO
Posts: 1
Joined: Thu Dec 20, 2018 1:49 pm

### Re: Chem/Ochem book, p. 131 #4

I was also really confused about this and I think the answer given here is misunderstanding the question. For the explanation in question four it is explaining why roman numeral one cannot be eliminated based on the information in data table one. So it says you have to set the forward reaction and reverse reaction equal to each other for the reaction 2NO->N2O2 and gives the rate law for this as Rate fwd=k1[NO]^2 and Rate rev=k-1[N2O2] even though for this reaction there is no table and we can't know that the forward reaction is second order or that the reverse reaction is first order. So how was the order for these reactions found?
cperz448
Posts: 12
Joined: Fri Jan 04, 2019 5:51 pm

### Re: Chem/Ochem book, p. 131 #4

Tessa, I was really confused by this as well, especially by the solution provided by the book. I watched a Khan academy video (https://www.khanacademy.org/science/che ... ining-step) and this helped me make sense of it I think. The slow step is your rate determining step (RDS), however both the slow and fast steps are considered elementary reactions. With elementary reactions, you are able to use the reactant coefficients to determine the rate law, which is what I believe the book solution was doing to some extent. For a general reaction, the exponents need to be determined experimentally, but that does not seem to be the case with elementary reactions.
NS_Tutor_Will
Posts: 766
Joined: Fri May 25, 2018 9:15 am

### Re: Chem/Ochem book, p. 131 #4

Thanks for sharing the video—that is definitely helpful! And thanks to both of you for raising the question. I think I have a better sense of what the original question was now!

I've written to our content team to see if we might make the explanation in the book clearer for the next edition.

Keep up the good work!