Fast or Slow

$K = \frac{k_{forward}}{k_{reverse}}$

ClaireHW
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Fast or Slow

Is there a way to tell if a step is slow or fast in a proposed mechanism? What sort of data would need to be given? And will this sort of question be asked?

Claire Woolson Dis 1K

Vincent Kim 2I
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Re: Fast or Slow

I'm pretty sure this fact will be given in a problem

Clarisse Wikstrom 1H
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Re: Fast or Slow

The problem may ask you to determine which is the rate-limiting step, which we know is the slow step. If so, you have to analyze each step to see if it's individual rate law matches the overall rate law. Remember, you may have to use the pre-equilibrium approach in order to do so.
Whichver step's rate law matches the overall rate law, then that is the slow(est) step of the reaction.

diangelosoriano
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Re: Fast or Slow

Also, if you look at a reaction profile, you can see which is the slow step / fast step based on how high the energy barrier is. A smaller barrier would indicate fast step, and a larger one, slow step.

Jessica Benitez 1K
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Re: Fast or Slow

The step that has a rate law that matches the overall rate law is the slow step.
When finding the rate law for a step remember to look out for intermediates as they should not be in the rate law for the step.
If we happen to have an intermediate present in the rate law for the step we have to replace it with an expression in terms of the equilibrium constant for a step prior to the one we are analyzing as it is faster and other concentrations.

Rachel Wang
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Re: Fast or Slow

So far the instances I've seen are:

1. If given a free energy vs time graph, the step with higher activation energy (higher camelback) is the rate-limiting step.
2. If one of the steps involves breaking bonds and the other doesn't, breaking bonds requires more energy.
3. If given the rate law and asked to identify the rate limiting step from student a, b, c's prediction.

Clarissa Molina 1D
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Re: Fast or Slow

I'm still sort of confused about the pre-equilibrium approach, can somebody please explain it?

ClaireHW
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Re: Fast or Slow

For the pre-equilibrium approach, there are 2 ways to do it. I'm going to do it Lavelle's way.
For: A + B -> C (fast)
C -> D (slow)
Write out the rate law for the slow step: Rate = k[C]
Since C is an intermediate and it cancels out, you don't want it in your final rate. You can replace it using the equilibrium constant (K).
K = [C]/[A][B]
Solve for [C] = K[A][B] and substitute that for [C]
The rate law then is Rate = kK[A][B]

Its important to keep in mind the difference between k and K.

Tim Nguyen 2J
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Re: Fast or Slow

The pre-equilibrium method is essentially writing the rate law for the slow step and substituting any intermediate terms with equivalents derived from previous or consequent steps' equilibrium constants.