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Chem Chp 6 - End of Chp Test

Posted: Mon Jan 27, 2020 4:20 pm
by bbiebelberg
Hello!

Chemister Chp 6 - End of Chp Test

First, the passage states that Reaction 2 is studied, but the caption for Table 1 states the data describe Reaction 1.  This caused a great deal of confusion initially — I assume it is an error, as the table seems to describe Reaction 2.

Question 4.   I’m confused by the whole question.
First, answer (III) is one equation, not a set of two.  Why is it included when the question stem says “The following two-step mechanisms…”?  Doesn’t a two-step mechanism by definition include two equations, one for each step?

Second, I don’t understand the basis on which certain of the proposed mechanisms are eliminated?  Can you please help me understand?  I find the explanation confusing…it suggests choice B is consistent with the data in Table 1.  So why would choice B be the one that is eliminated?

Independent questions 5 and 7 (Q12 and Q14 overall).
These questions asks about rates of an elementary reaction.  I was able to remember from my Gen Chem class that for elementary reactions, the exponents in the rate law correspond to the stoichiometric coefficients.
But rates of elementary steps are not discussed in the Chapter 6 text.  In fact, I don’t think elementary steps are mentioned at all.  Did I miss this somewhere? Are we responsible for knowing it?
I only recall seeing rate laws for overall reactions, in which the exponents depend on the order of the reaction.

Thank you!

Re: Chem Chp 6 - End of Chp Test

Posted: Tue Jan 28, 2020 5:39 am
by NS_Tutor_Mathias
Question 4:
Reaction III is indeed a single step. You are just expected to recognize that this cannot be a viable two-step mechanism by virtue of it lacking a second step. Not everything is complicated!

The rest of the question wants you to fundamentally understand that the slow step of a multi-step reaction will be rate determining. That means that we can predict the rate law of any of these combined reactions solely on hand of the slow step. Additionally, fast steps that form reactants can essentially be treated as replacing the reactant in the slow step as far as reaction order is concerned - since they create the input for the slow step at a rate dependent on the input of reactants into the fast step. In other words, and without resorting to math, it is creating N2O2 so fast that really the concentration of NO is the driving factor.

So for reaction I:
2 NO + O2 -> 2 NO2

(rate determining step re-written to account for the rapidly formed N2O2 input)
Making this a third order net reaction that is second order for NO and first order for O2. Nota bene that this approach of simply plugging in the reactants of the fast step is a gigantic informality on my part, but this is what the manipulations of rate constants and equilibrium in the book boil down to. The MCAT will encourage you heavily to think qualitatively about problems like this, so I try to encourage this approach.

For reaction II:
Only the top step needs to be considered at all here, making this a second order reaction. First order for both NO and O2.

Now we can compare this with the data provided and see that for a doubling in O2, the reaction rate doubles. This is consistent with both possibilities. For a doubling in NO, we find the reaction rate increases significantly more than linearly (you won't have to do much more than estimate these things on the MCAT). We can safely assume that the reaction is second order for NO. This in turn is only consistent with reaction I.

Question 5 & 7:
Elementary reaction just means single-step reaction. It means you can rest safe in the assumption that there are no invisible intermediate steps hidden in some multi-step reaction of which you are only seeing the end product. Any reaction that is single-step (elementary) in turn obeys exactly the simple coefficient -> exponent rule of determining order that you've learned.

This is a case where our materials are actually good, in that you will frequently encounter slightly arcane language and notation for concepts you are very familiar with on the MCAT. Part of test-taking skill will most certainly be recognizing these familiar concepts in slightly unfamiliar terms.