mult-step mechanism questions

[robert zellmer]

I received a question from someone about how how you get the rate
law in a multi-step mechanism and how you know which step in is the
slowest step.

You always start by writing the the rate law in terms of the REACTANTS
in the slowest step.  Then if there are any intermediates you try to get
rid of them by going back through the preceding fast steps, assuming
each reaches an equilibrium (which means the rates of the forward and
reverse steps are equal so you can set the rate law for the forward and
reverse steps equal to each other and solve for what you need). This is
often referred to as the "steady-state approximation" method, which I
used in lecture (and it's done in the book).

How do you determine which step is the slowest step?  It depends on
how the question is presented.

1)  You might be given the mechanism and explicitly told which step is the
     slowest step and asked to determine the rate law based on the 
information
     given.

2) You might be given a problem like the examples I did in class. In 
that problem
     you weren't told ahead of time which step was the slowest step. 
However, I
     gave the experimental rate law.  In this case you start by assuming 
the first
     step is the slowest step and write the rate law based on the REACTANTS
     in that step.  See if the rate law agrees with the given exp. rate 
law.  If not that
     means that step is not the rate-determining step (slowest step).  
Then you
     write the rate law in terms of the reactants in the 2nd step. If 
there are
     intermediates in the rate law you need to go back to the preceding 
step(s)
     and try get rid of those intermediates by expressing them in terms 
of the
     reactants and products from these preceding steps.  If the 2nd step 
doesn't
     give the exp. rate law you go on to the next step and so on.

In either case your *rate law will depend only on the slowest step and 
any of the**
**preceding fast steps*.  Any steps after the slowest step don't affect 
the rate law.
They are there to get rid of intermediates which may still be present, 
get rid of
reactants, produce products or we'll see later reproduce catalysts.

When you are done you *may have reactants from the slowest step and 
reactants**
**and maybe products from preceding steps*.  You can have *reactants 
(including**
**catalysts) and products in a rate law*.

Thinking about this can you answer the following (which I discussed in 
class):

1)  Can a reactant which appears for the first time after the slow step 
wind up in the
     rate law?

2)  If a product is in the experimental rate law could the first step be 
the slowest step
     based on the steady-state approximation (equilibrium is reached in 
the fast steps
     which proceed the slowest step)?

After doing the assigned problems from the textbook concerning 
mechanisms try the
one on the Extra Ch 14 Homework problems passed out in class on Friday and
also found on my web page.  This problem gives the exp. rate law but 
doesn't tell
you what the slowest step is (there are others in the book which do the 
same but
this one is a little more challenging).

Dr. Zellmer
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <https://lists.osu.edu/mailman/private/cbc-chem1220/attachments/20140629/5f1f99a9/attachment.html>