2.9: Kinetic and Thermodynamic Control

Learning Objectives

By the end of this section, you should be able to:

• 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*}