2.9: Kinetic and Thermodynamic Control
- Page ID
- 101153
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- Understand kinetic and thermodynamic control
Reactants can sometimes give rise to a variety of products.
Consider the nitration of nitrobenzene:
The relative portion of the products before reaching equilibrium is given by the ratio of the rates of production.
\[A+B→P_{1}\]
\[A+B→P_{2}\]
Here, before equilibrium:
| \(\frac{[P_{2}]}{[P_{1}]}=\frac{k_{r,2}}{k_{r,1\) |
This is called kinetic control, and it is dictated by reaction rates.
As opposed to thermodynamic control, which is dictated by reaction equilibrium (after a long time):
Say we have the system:
If \(k_{e1},k_{e2}\text{<<}k_{r1},k_{r2}\)
Then at any time before the equilibrium reaction start severely affecting product concentration, the reaction simplifies to:
| \(\frac{[P_{1}]}{[P_{2}]}=\frac{k_{r1}}{k_{r2}}\) |
The reaction is kinetically controlled: the amount of products depends on the rates of reaction.
Proof to show: \(\frac{[P_{1}]}{[P_{2}]}=\frac{k_{r1}}{k_{r2}}\)
\begin{align*}
r_{P1}=k_{r1}[A][B]&=\frac{d[P1]}{dt}\\
r_{P2}=k_{r2}[A][B]&=\frac{d[P2]}{dt}
\end{align*}
Say that both \(P_{1}\), \(P_{2}\) start at a concentration of 0. We can express the change in concentration for \(P_{1}\) and \(P_{2}\) at any time before [A][B] reaches 0. Note that once [A][B] reaches 0, the equilibrium reaction starts to dominate as we no longer have forward reactions that consume A and B to produce P1 and P2.
\[\frac{[P1]}{[P2]}=\frac{\frac{d[P_{1}]}{dt}}{\frac{d[P_{2}]}{dt}}=\frac{k_{r1}[A][B]}{k_{r2}[A][B]}=\frac{k_{r1}}{k_{r2}}\]
If \(k_{e1},k_{e2}>>k_{r1},k_{r2}\)
Then at any given time, this reaction simplifies to:
| \(\frac{[P_{1}]}{[P_{2}]}=\frac{k_{e1}}{k_{e2}}\) |
The reaction is thermodynamically controlled : the amount of products depends on the equilibrium state.
Proof to show: \(\frac{[P_{1}]}{[P_{2}]}=\frac{k_{e1}}{k_{e2}}\):
At equilibrium:
Forward reaction rate: \(\frac{d[P2]}{dt}=k_{e2}[P1]\)
Reverse reaction rate: \(\frac{d[P1]}{dt}=k_{e1}[P2]\)
At equilibrium, the forward and reverse reaction rates are equal:
\begin{align*}
k_{e2}[P1] & = k_{e1}[P2]\\
\frac{[P_{1}]}{[P_{2}]} & = \frac{k_{e1}}{k_{e2}}
\end{align*}