Question 18.6: Consider the following sequential reaction scheme:A→I→P Assu...
Consider the following sequential reaction scheme:
A\overset{k_{A}}{\rightarrow } I\overset{k_{I}}{\rightarrow } PAssuming that only reactant A is present at t=0, what is the expected time dependence of \left[ P \right] using the steady-state approximation?
The differential rate expressions for this reaction were provided in Equations (18.47), (18.48), and (18.49):
\frac{d\left[ A \right]}{dt}=-k_{A}\left[ A \right] (18.47)
\frac{d\left[ I \right]}{dt}=k_{A}\left[ A \right]-k_{I}\left[ I \right] (18.48)
\frac{d\left[ P \right]}{dt}=k_{I}\left[ I \right] (18.49)
Applying the steady-state approximation to the differential rate expression for I and substituting in the integrated expression for \left[ A \right] of Equation (18.51) yield
\left[ A \right]=\left[ A \right]_{0}e^{-k_{A}t} (18.51)
\frac{d\left[ I \right]}{dt}=0=k_{A}\left[ A \right] -k_{I}\left[ I \right]
\frac{k_{A}}{k_{I}}\left[ A \right]=\frac{k_{A}}{k_{I}}\left[ A \right]_{0}e^{-k_{A}t}=\left[ I \right]
Substituting the preceding expression for \left[ I \right] into the differential rate expression for the product and integrating yield
\frac{d\left[ P \right]}{dt}=k_{I}\left[ I \right]=\frac{K_{A}}{k_{I}}\left( k_{I}\left[ A \right]_{0} e^{-k_{A}t}\right)\int_{0}^{\left[ P \right]}d\left[ P \right]=k_{A}\left[ A \right]_{0}\int_{0}^{t}e^{-k_{A}t}dt
\left[ P \right]=k_{A}\left[ A \right]_{0}\left[ \frac{1}{k_{A}}\left( 1-e^{-k_{A}t} \right) \right]
\left[ P \right]=\left[ A \right]_{0}\left( 1-e^{-k_{A}t} \right)
This expression for \left[ P \right] is identical to that derived when the decay of A is treated as the rate-limiting step in the sequential reaction [Equation (18.59)].
\underset{k_{I}\gg k_{A}}{\lim} \left[ P \right]=\underset{k_{I}\gg k_{A}}{\lim}\left( \left( \frac{k_{A}e^{-k_{I}t}-k_{I}e^{k_{A}t}}{k_{I}-k_{A}} +1\right)\left[ A \right]_{0} \right)=\left( 1-e^{-k_{A}t} \right)\left[ A \right]_{0} (18.59)