Question about simple kinetics and stoichiometry

  • Thread starter dRic2
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  • #1
dRic2
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Hi, I'm having trouble with this simple reaction (1):

$$A + A → I$$

with ##R = kC^2_A##

I'm assuming it takes place in a Batch reactor so the mass balance should yield:

$$\frac {dC_A} {dt} = -2R = -2kC^2_A$$

But, as I recall (I should be wrong though), stoichiometry in thermodynamics and kinetics is just "math": it does not have any "physical" meaning (unlike in chemistry), but it only serves to assure mass is conserved during the reaction.

So I can re-write reaction (1) like:

$$ A → \frac 1 2 I$$

But now:

$$\frac {dC_A} {dt} = -R = -kC^2_A$$

What am I missing ?
 

Answers and Replies

  • #2
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Your k will depend on how you write the reaction.
 
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What am I missing ?
The rate equation is given for a specific chemical equation. That means A+A→I and A → ½ I have different rate equations.
 
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  • #4
dRic2
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That makes a lot of sense! Thank you very much :)
 
  • #5
DrDu
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Or to put it differently, the kinetics of a reaction cannot be read off from the overall reaction equation. Even reactions of the from A->B seldomly follow an unimolecular reaction kinetics. You will have to find the elementary steps and the kinetic can be calculated from these elementary steps. In unimolecular reactions,
the steps may be: A+M -> A* +M, where M is a collision partner, and A*->B. The total kinetics will depend on which of the two steps is faster.
 
  • #6
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You will have to find the elementary steps and the kinetic can be calculated from these elementary steps.
That's another topic. If I understand the OP correctly than he already knows the kinetics of the reaction A + A → I and wonders what that means for the same reaction written as A → ½ I. In that case the kinetics R1 = k·[A]² for the first chemical equation just turns into R2 = 2·k·[A]² for the second one.
 
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  • #7
dRic2
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Yes, that was what I was asking about. Sorry for may late reply and thanks a lot again!
 
  • #8
It’s dA/dt = k(A)^2

And the integrated form or solution is 1/A = 1A0 + 2kt
 

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