[capybara]

Hi, I was wondering how you would derive the observed rate expression k_obs for an acid catalyzed aquation of a tris(bipy) complex. The kinetic scheme and k_obs expression is attached. The intermediate (second molecule in the scheme) is assumed to follow SS kinetics

Thank you

[billnotgatez]

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[capybara]

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?

[mjc123]

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.

[capybara]

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.

[mjc123]

Subsittuting for {B} gives kobs = k1{A}(k3 + k4{H+})/(k2 + k3 + k4{H+})

No, d[C]/dt = k_obs[A] = ...

[capybara]

It makes sense now! Thank you for taking the time to explain