Chem 2 - Chemical Equilibrium II: The Reltionship Between Kinetics and the Equilibrium Constant K

1. Chemical Equilibrium (Pt. 2)
The Relationship Between
Kinetics and the Equilibrium
Constant K
By Shawn P. Shields, Ph.D.
This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.

2. Recall: Fast Equilibrium Step in a Kinetic
Mechanism
What is a fast equilibrium step?
For the reversible reaction
A B
The forward reaction is A B
The backward reaction is A B
(or alternatively B A)
k1
k1
k1
k1
k1

3. Our Example: Gas Phase Equilibrium
For the reversible reaction
2 NO2 N2O4
The forward reaction is
2 NO2 N2O4
The backward reaction is
N2O4 2 NO2

4. Write the Reaction as a Fast Equilibrium
2 NO2 N2O4
The rate law for the forward reaction is
Rate = k1[NO2]2
The rate law for the backward reaction is
Rate = k1[N2O4]
k1
k1

5. Reaction Rates and Equilibrium
2 NO2 (g) N2O4 (g)
At the beginning of the reaction, the
forward reaction rate is much faster
than the back reaction rate!
Reactant is converted to product faster
than product is converted to reactant.

6. Time (s)
P (atm)
0.75
0.50
0.25
1.0
NO2 is converted to N2O4 at a much
faster rate than the reverse at the
beginning of the reaction.
𝐏 𝐍𝐎 𝟐
𝐏 𝐍 𝟐 𝐎 𝟒
Recall Experiment 1: Start with 1.0 atm NO2 (g)

7. 𝐏 𝐍𝐎 𝟐
𝐏 𝐍 𝟐 𝐎 𝟒
The “kinetics ”
part of the
experiment
Kinetics versus Equilibrium
The “equilibrium” part
of the experiment
Time (s)
P (atm)
0.75
0.50
0.25
1.0

8. What is going on at Equilibrium?
2 NO2 N2O4 (g)
Once the reaction is at equilibrium,
the forward reaction rate is equal to
the back reaction rate!

9. What is going on at Equilibrium?
2 NO2 N2O4 (g)
At equilibrium, both of these processes
are occurring microscopically at the
same rate!
2 NO2 N2O4
N2O4 2 NO2

10. Deriving the Equilibrium Constant K
2 NO2 N2O4
The forward reaction rate is equal to the
backward rate at equilibrium, so set them
equal to each other.
k1[NO2]2 = k1[N2O4]
rate of fwd rxn = rate of back rxn
k1
k1

11. Deriving the Equilibrium Constant K
2 NO2 N2O4
Rearrange to solve for
𝑘1
𝑘−1
𝒌 𝟏 𝐍𝐎 𝟐
𝟐
= 𝒌−𝟏 𝐍 𝟐 𝐎 𝟒
Divide both sides by NO2
2
𝒌 𝟏 𝐍𝐎 𝟐
𝟐
𝐍𝐎 𝟐
𝟐
=
𝒌−𝟏 𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐
k1
k1

12. Deriving the Equilibrium Constant K
2 NO2 N2O4
𝒌 𝟏 𝐍𝐎 𝟐
𝟐
𝐍𝐎 𝟐
𝟐
=
𝒌−𝟏 𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐
Divide both sides by k1
𝒌 𝟏
𝒌−𝟏
=
𝒌−𝟏 𝐍 𝟐 𝐎 𝟒
𝒌−𝟏 𝐍𝐎 𝟐
𝟐
k1
k1
Next ...

13. Deriving the Equilibrium Constant K
2 NO2 N2O4
𝐊 =
𝐤 𝟏
𝐤−𝟏
=
𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐
The equilibrium constant K gives the
fundamental relationship between kinetics
and equilibrium.
k1
k1

14. Deriving the Equilibrium Constant K
2 NO2 N2O4
𝐊 =
𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐
The equilibrium constant K is the ratio of
the rate constants for a reversible
reaction.
k1
k1

15. Summary of the Characteristics
of Equilibrium
I. At equilibrium, macroscopic
observables have stopped
changing.
II. There is a balance between the
forward reaction rate and the
backward (reverse) reaction rate.

16. Summary of the Characteristics
of Equilibrium
III. The equilibrium state is
completely independent of the
initial concentrations or partial
pressures of reactants and
products.

17. Summary of the Characteristics
of Equilibrium
IV. The equilibrium state does not
depend on reaction rates! The
reaction can go to equilibrium either
quickly or slowly.

18. Next up,
The Law of Mass Action
(LOMA) (Pt 3)