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Topic: Correcting rate equations for uncatalyzed back ground reactions.  (Read 3479 times)

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Offline Tinus

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Correcting rate equations for uncatalyzed back ground reactions.
« on: September 07, 2012, 10:25:47 AM »
In catalysis, how does one correct a measured rate equation for the uncatalyzed competitive background reaction?

Lets say we have compound A being converted to compound B using a catalyst. At the same time compound A is also uncatalytically converted to compound B so that there are two competitive reaction pathways with different rate constants.

How can you mathematically eliminate the uncatalyzed rate equation so that the observed kinetic behaviour can be attributed to the catalyst only? Simple addition/substraction doesnt work because both are correlated to eachother.

For simple first order models it can be solved like this:
 
Let say we have two independent first order reactions

R1:   A -> B, with reaction rate constant k1
R2:   A -> B, with reaction rate constant k2

Both can be described by the following simple first order reactions

R1:   d[A]/dt = -k1[A]
R2:   d[A]/dt = -k2[A]

When both differential equations are integrated:

R1   [A]t = [A]0 exp(-k1 t)
R2   [A]t = [A]0 exp(-k2 t)

Now assume we perform an experiment where both reactions are present:

The reaction rate equation and the integrated rate equations then become:

R3   d[A]/dt = -k1[A]-k2[A]
R3   [A]t = [A]0 exp(-(k1 + k2) t)

The total integrated equation R3 can be corrected for R2 by simple multiplication of the reverse of the integrated rate equation of R2, because mathematically:

[A]t = [A]0 exp(-(k1 + k2) t) * exp(k2 t) = [A]0 exp(-k1 t)

Eliminating the presence of R2 and only leaving you the kinetic behaviour of R1.

For more complex integrated rate equations however this does not suffice.

Now in my more complex model I can describe the total integrated rate equation as:

R3   [A]t = C1 exp(-t / c2) + C3 t + C4

Where C1, C2, C3 and C4 are constants obtained by curve fitting measured data points.

I know that the uncatalyzed back ground reaction can be described by a perfect 1st order rate equation:

R2   [A]t = C1 exp(-C2 t)

The big question being: "How do I obtain the data of R1 using these expressions of R2 and R3?

Or is there a completely other basic method of correcting measured data for uncatalyzed background reactions? I thought it would be more trivial but it has been keeping me busy way too long now...

Hope I’m being clear!

Tinus

Offline curiouscat

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Re: Correcting rate equations for uncatalyzed back ground reactions.
« Reply #1 on: September 07, 2012, 11:13:01 AM »
I don't exactly understand what you mean.

Catalyzed expt. gives you C1 C2 C3 C4
Uncatalyzed expt. gives you C1 C2

What more do you want?

Offline Tinus

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Re: Correcting rate equations for uncatalyzed back ground reactions.
« Reply #2 on: September 10, 2012, 04:40:06 AM »
Yes, but the catalyzed experiment is still a combination of the uncatalyzed experiment and catalyzed experiment. I've added some graphs in the attachment to clarify.

So in the first graph you see the uncatalyzed conversion of CeIV to CeIII. After ~4500 seconds the conversion is complete. Its a perfect 1st order reaction and its decrease in time can be described by 0,0015*exp(-0,0007*t) as you can see in the graph.

Now the second graph shows the results when the catalyst is added. You again see the conversion of CeIV to CeIII but in the presence of three different concentrations of the same catalyst.

Now this has to be corrected for the uncatalyzed conversion which is competitively reducing Cerium in paralel during the reaction.

The constants C1, C2, etc. I've mentioned are not related by the way, that was just an example.

Offline Tinus

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Re: Correcting rate equations for uncatalyzed back ground reactions.
« Reply #3 on: September 10, 2012, 05:33:00 AM »
Oops... I think the answer is way more trivial then I thought...

For who is interested:

the total reaction rate can be described by:

R: = -d[Ce]/dt = k1[Ce] + k2[Ce][Cat]

which gives:

d[Ce]/dt = (k1 + k2[Ir])[Ce]

kobs = k1 + k2[Ir]

In other words: The reaction rate constant I measure is a linear expression of both reaction reaction rates present where k1 can be found in the intercept and k2 in the slope.

Idea of my colleague ;)

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