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I received a question from someone about how how you get the rate<br>
law in a multi-step mechanism and how you know which step is the<br>
slowest step.<br>
<br>
You always start by writing the the rate law in terms of the
REACTANTS<br>
in the slowest step. Then if there are any intermediates you try to
get<br>
rid of them by going back through the preceding fast steps, assuming<br>
each reaches an equilibrium (which means the rates of the forward
and<br>
reverse steps are equal so you can set the rate law for the forward
and<br>
reverse steps equal to each other and solve for what you need).
This is<br>
often referred to as the "steady-state equilibrium approximation"
method,<br>
which I used in lecture (and it's done in the book).<br>
<br>
How do you determine which step is the slowest step? It depends on<br>
how the question is presented.<br>
<br>
1) You might be given the mechanism and explicitly told which step
is the<br>
slowest step and asked to determine the rate law based on the
information<br>
given. Start with the step that's labeled as the slow step.<br>
<br>
2) You might be given a problem like the examples I did in class.
In that problem<br>
you weren't told ahead of time which step was the slowest step.
However, I<br>
gave the experimental rate law. In this case you start by
assuming the first<br>
step is the slowest step and write the rate law based on the
REACTANTS<br>
in that step. See if the rate law agrees with the given exp.
rate law. If not that<br>
means that step is not the rate-determining step (slowest
step). Then you<br>
write the rate law in terms of the reactants in the 2nd step.
If there are<br>
intermediates in the rate law based on this step you need to go
back to the<br>
preceding step(s) and try get rid of those intermediates by
expressing them in<br>
terms of the reactants and products from these preceding steps.
If the 2nd step<br>
doesn't give the exp. rate law you go on to the next step and so
on.<br>
<br>
In either case your <b>rate law will depend only on the slowest
step and any of the</b><b><br>
</b><b>preceding fast steps</b>. Any steps after the slowest step
don't affect the rate law.<br>
They are there to get rid of intermediates which may still be
present, get rid of<br>
reactants, produce products or reproduce catalysts.<br>
<br>
When you are done you <b>may have reactants from the slowest step
and reactants</b><b><br>
</b><b>and maybe products from preceding steps</b>. You can have <b>reactants
(including</b><b><br>
</b><b>catalysts) and products in a rate law</b>.<br>
<br>
Thinking about this can you answer the following (which I discussed
in class):<br>
<br>
1) Can a reactant which appears for the first time after the slow
step wind up in the<br>
rate law?<br>
<br>
2) If a product is in the experimental rate law could the first
step be the slowest step<br>
based on the steady-state equilibrium approximation (equil. is
reached in the fast<br>
steps which proceed the slowest step)?<br>
<br>
After doing the assigned problems from the textbook concerning
mechanisms try the<br>
one (#5) on the Extra Ch 14 Homework problems found on my web page.
This problem<br>
gives the exp. rate law but doesn't tell you what the slowest step
is (there are others in<br>
the book which do the same but this one is a little more challenging
- try a couple of<br>
textbook problems first).<br>
<br>
<a
href="https://www.asc.ohio-state.edu/zellmer.1/chem1220/homewk/ch14hmwk_add.pdf"
moz-do-not-send="true"><b>CH 14 - Kinetics, Extra Problems and
Solutions</b></a><br>
<br>
Dr. Zellmer
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