Elementary rate law

  • Thread starter Maylis
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  • #1
Maylis
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Hello, I am having some confusion over elementary rate laws. This is a hydrodealkylation reaction.

testb.13.gif


The specific reaction rates k1 and k4 are defined w.r.t. H2.

If I want to write the rate law for the hydrogen radical for the termination step, would the elementary rate law be ##r_{H \bullet} = -2k_{4}C_{H \bullet}^2## or ##r_{H \bullet} = -k_{4}C_{H \bullet}^2##. I think it is the former because of the part of k4 being with respect to hydrogen. If it was with respect to the hydrogen radical, then it would be the latter? Thanks
 
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  • #2
Bystander
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Is one hydrogen radical distinguishable from another?
 
  • #3
Maylis
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No, but I don't see how it relates to the rate law
 
  • #4
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Not only are you being confused, so am I --- I took off the wrong direction with how you came up with "-2k4CH⋅2, and thought you were making distinctions among hydrogen radicals.
I think it is the former
And rereading the problem statement for the twentieth time, I'll agree, because ...

The specific reaction rates k1 and k4 are defined w.r.t. H2.
, and this is the source of the confusion, I don't recall ever seeing a rate constant referred to the product. I'm a bit of a dinosaur, and conventions in kinetics may have evolved since I last had any use or interest for the field.
 
  • #5
epenguin
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I wouldn't worry about it. I would just write
##r_{H \bullet} = -k_{4}C_{H \bullet}^2##.
myself, so as not to have to remember about the 2, and just get on with the problem.
It would be quite reasonable to have an equilibrium constant defined as K = [H.]2/[H2].
If it turns out later the other definition would have been more convenient, I can always change definition.
 

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