Clarification on Chem/Orgo Chapter 6 passage question

Post Reply
nhuebschmann
Posts: 6
Joined: Sun Aug 04, 2019 5:19 pm

Clarification on Chem/Orgo Chapter 6 passage question

Post by nhuebschmann » Thu Aug 29, 2019 7:59 pm

Hi there,

I was hoping to get some clarification for question number 4 in the practice passage for chapter 6 of the Chemistry and Organic Chemistry book. On page 135, it explains why the rate law for mechanism II is not correct, but how do you know that the rate is first order in both nitrogen monoxide and oxygen without doing an initial rate analysis like in question 1? Thanks!

- Nathan
NS_Tutor_Mathias
Posts: 257
Joined: Sat Mar 30, 2019 8:39 pm

Re: Clarification on Chem/Orgo Chapter 6 passage question

Post by NS_Tutor_Mathias » Fri Aug 30, 2019 3:45 pm

These are proposed mechanisms, so we can't test their rates empirically the way we did in the experiment anyway. We do however know that the order for each reactant will depend on the stoichiometric coefficient in the rate-limiting step.

So for each option, we only really need to look at the stoichiometric coefficients in the rate-limiting step (the slow one!) and make a decision from there. Note that for RN I, there is a fast step producing N2O2 feeding directly into our rate-limiting step, so the reaction really is third order but ever so slightly disguised. The answer key gives you the somewhat rigorous mathematical expression for this, but that is unnecessary, as long as you realize that "fast" means you can essentially substitute 2 NO in place of N2O2 and determine the reaction order from there.

That is of course a bit of a shortcut, but in general, the MCAT will test your understanding of kinetics and rate limiting steps more than have you do a whole lot of algebra with them.
nhuebschmann
Posts: 6
Joined: Sun Aug 04, 2019 5:19 pm

Re: Clarification on Chem/Orgo Chapter 6 passage question

Post by nhuebschmann » Sun Sep 01, 2019 12:03 pm

Great, thank you.

So to clarify, we can use the stoichiometric coefficients to determine the order with respect to a reactant in the slow step of a multi-step reaction mechanism or if the reaction is stated as being elementary?
NS_Tutor_Mathias
Posts: 257
Joined: Sat Mar 30, 2019 8:39 pm

Re: Clarification on Chem/Orgo Chapter 6 passage question

Post by NS_Tutor_Mathias » Sun Sep 01, 2019 10:54 pm

The stoichiometric ratio will always give you the reaction order of a single-step reaction (which is the same as an elementary reaction). With the caveat of course that like in RN I, we account for how those reagents came to be.

The slow-step is also always the rate-determining step.
Post Reply