Pre-Equilibrium

Pre-Equilibrium

• when the rates of formation of the intermediate and its decay back into reactants are much faster than the rates of formation of products, you can assume that reactants and intermediates are in equilibrium throughout the course of the reaction=pre-equilibrium

$R \xrightleftharpoons[k_{-1}]{k_1}I \hspace{20pt} k_1 \gg k_2$

$I \xrightleftharpoons{k_2}P \hspace{20pt} k_{-1} \gg k_2$

The overall rate of reaction here is:

v2=k2[I]

Write the rates of each elementary step:

• elementary step #1: v1=k1[R]
• elementary step #1: v1'=k'[I]
• elementary step #2: v2=k2[I]

We also have pre-equilibrium here since R and I are in equilibrium throughout the reaction and k1>> k2

Because of this we can write an expression K=$$$$$$$$$\frac{k1}{k'}=\frac{\left[I\right]}{\left[R\right]}$

Now we can solve for our intermediate (I):

$\frac{k1\left[R\right]}{k'}=\left[I\right]$

Now take the overall rate of reaction we have and plug in the expression for the intermediate:

v=v2=k2[I]

v=v2=k1k2[R]/k'

Could let K=k1k2/k' to simplify it:

v(overall)=K[R]

This was much faster than standard state approximation! We got the same answer by using pre-equilibrium and we were able to do this because there was an equilibrium between the reactants and intermediate!