High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Chemistry Textbook > Kinetics

Pre-Equilibrium

Steady State Approximation Previous Section
Oxidation States Next Section
Popular Courses
Find My Course
0:00 / 0:00

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⇌k−1k1Ik1≫k2R \xrightleftharpoons[k_{-1}]{k_1}I \hspace{20pt} k_1 \gg k_2
I⇌k2Pk−1≫k2I \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

Wize Concept
Since the reactants and intermediate are in equilibrium, recall that for equilibrium we know:
the rate of the forward reaction=rate of the backward reaction
In this case: k1[R]=k'[I]

Because of this we can write an expression K=k1k′=[I][R]\frac{k1}{k'}=\frac{\left[I\right]}{\left[R\right]}
Now we can solve for our intermediate (I):

k1[R]k′=[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!

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
Write a rate law for the formation of NH3BH3 which involves only the concentration of reactants and rate constants.
B2H6+2NH3→2H3N−BH3‾\underline{B_2H_6+2NH_3\to2H_3N-BH_3}
B2H6⇋k−1k12BH3k1≫k2B_2H_6 \xleftrightharpoons[k_{-1}]{k_1}2BH_3 \hspace{20pt} k_1 \gg k_2
2NH3+2BH3→k22H3N−BH3k−1≫k22NH_3+2BH_3 \xrightarrow{k_2}2H_3N-BH_3 \hspace{20pt} k_{-1} \gg k_2