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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: capybara on March 06, 2015, 01:36:25 AM
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Hi, I was wondering how you would derive the observed rate expression kobs for an acid catalyzed aquation of a tris(bipy) complex. The kinetic scheme and kobs expression is attached. The intermediate (second molecule in the scheme) is assumed to follow SS kinetics
Thank you
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Okay, so far I've got k1 = k2 + k3 + k4 [H+] from the steady state assumption
The problem is linking kobs in with the above equation, substituting back from kobs would give something like k1(k1-k2)/k1?
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Okay, so far I've got k1 = k2 + k3 + k4 [H+] from the steady state assumption
That is not true. Write the differential equation for the concentration of the intermediate.
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Ah yes, thanks for the tip
So d{B}/dt = k1{A} - k2{B} - k3{B} - k4{B}{H+} = 0
Rearranging gives k1{A} = {B}(k2 + k3 + k4{H+})
Therefore {B} = k1{A}/(k2 + k3 + k4{H+})
Since kobs = d{C}/dt = k3{B} + K4{B}{H+} = {B}(k3 + k4{H+})
Subsittuting for {B} gives kobs = k1{A}(k3 + k4{H+})/(k2 + k3 + k4{H+})
Which looks sort of correct? Sorry for the messy formatting
However there is still the [A] term, I was wondering if it was relevant as the concentration used remained constant throughout the experiment.
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Subsittuting for {B} gives kobs = k1{A}(k3 + k4{H+})/(k2 + k3 + k4{H+})
No, d[C]/dt = kobs[A] = ...
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It makes sense now! Thank you for taking the time to explain