Chem 2 - Chemical Equilibrium III: The Equilibrium Constant Expression and the Law of Mass Action (LOMA)

1. Chemical Equilibrium (Pt. 3)
The Equilibrium Constant
Expression and the Law of
Mass Action (LOMA)
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. At equilibrium…
Macroscopic observables have
stopped changing.
The forward and reverse reaction
rates are equal.

3. Recall: Equilibrium Constant K
and Rate Constants
2 NO2 N2O4
k1[NO2]2 = k1[N2O4]
𝐊 =
𝐤 𝟏
𝐤−𝟏
=
𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐
The equilibrium constant K is the ratio of
the forward and reverse rate constants.
k1
k1

4. The Equilibrium Constant Expression
Experiments done by Guldburg and
Waage (1864-1879) demonstrated
the ratio of products to reactants is
always constant
under a certain set of experimental
conditions.
(This is a simplified statement.)

5. The Law of Mass Action
Consider the generalized reaction
aA + bB ⇌ cC + dD
Reactants
A and B
Products
C and D
coefficients

6. The Law of Mass Action
For the reaction
𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃
The relationship between the value of
the equilibrium constant K and the
concentrations of reactants and
products is
𝐊 =
𝐂 𝐜
𝐃 𝐝
𝐀 𝐚 𝐁 𝐛
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬
𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

7. The Law of Mass Action
What happens if we reverse the reaction?
𝐜𝐂 + 𝐝𝐃 ⇌ 𝐚𝐀 + 𝐛𝐁
The relationship between the value of the
equilibrium constant K and the concentrations
of reactants and products is still products over
reactants!
𝐊 =
𝐀 𝐚
𝐁 𝐛
𝐂 𝐂 𝐃 𝐝
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬
𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

8. The Law of Mass Action
𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃
This relationship is true no matter the
initial distribution (relative amounts)
of reactants and products!
𝐊 =
𝐂 𝐜
𝐃 𝐝
𝐀 𝐚 𝐁 𝐛
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬
𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

9. 𝐏 𝐍𝐎 𝟐
𝐏 𝐍 𝟐 𝐎 𝟒
“kinetics ”
Remember… Once we reached the “equilibrium”
part of the experiment, we ended
up with 0.50 atm NO2 and 0.25
atm N2O4 …
And it didn’t matter if we started
with all products or all reactants!
Time (s)
P (atm)
0.75
0.50
0.25
1.0
𝐏 𝐍𝐎 𝟐
𝐏 𝐍 𝟐 𝐎 𝟒

10. Equilibrium Constant K is Unitless
Why is the value for K unitless?
𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃
Each concentration in the equilibrium constant
expression is divided by a “standard
concentration” of 1.0 M.
𝐊 =
𝐂 𝐜
𝟏. 𝟎
𝐃 𝐝
𝟏. 𝟎
𝐀 𝐚
𝟏. 𝟎
𝐁 𝐛
𝟏. 𝟎
All molarity
units cancel
out!

11. Equilibrium Constant K is Unitless
𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃
When we divide by a standard concentration,
we are left with “effective concentrations”
or “activities” (a).
𝐊 =
𝒂 𝑪
𝐜
𝒂 𝑫
𝐝
𝒂 𝑨
𝐚 𝒂 𝑩
𝐛
“activity” can be though of as
the “presence” of the reactant
or product

12. Gases and the Equilibrium Constant
Expression
𝐚𝐀 𝒈 + 𝐛𝐁 𝒈 ⇌ 𝐜𝐂 𝒈 + 𝐝𝐃(𝒈)
The Law of Mass Action holds for gases in
equilibrium, as well.
𝑷 𝑪
𝐜
represents the partial pressure of gas C, etc.
𝐊 =
𝑷 𝑪
𝐜
𝑷 𝑫
𝐝
𝑷 𝑨
𝐚
𝑷 𝑩
𝐛

13. Gases and the Equilibrium Constant
Expression
𝐚𝐀 𝒈 + 𝐛𝐁 𝒈 ⇌ 𝐜𝐂 𝒈 + 𝐝𝐃(𝒈)
Each partial pressure is divided by a
standard pressure (1 atm), so the K for
gaseous systems is also unitless!
𝐊 =
𝑷 𝑪
𝐜
𝑷 𝑫
𝐝
𝑷 𝑨
𝐚
𝑷 𝑩
𝐛
Partial pressures for gases are
given in “activities”, just like
solutions.

14. Example 1: The Equilibrium Constant
Expression
2 NO2 N2O4
Write the equilibrium constant
expression for the following reaction:

15. Example 1 Solution: The Equilibrium
Constant Expression
2 NO2 N2O4
Write the equilibrium constant
expression for the following reaction:
𝐊 =
𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐
The coefficient
for N2O4 is 1

16. Calculating the Value of the
Equilibrium Constant
2 NO2 N2O4
The value for the equilibrium constant
K can be calculated by inserting the
equilibrium concentrations (or partial
pressures) into the equilibrium constant
expression.

17. Example 2: Calculating the Value of the
Equilibrium Constant
2 NO2 N2O4
What is the value of K when the
concentration of NO2 (at equilibrium)
is 0.0165 M and the concentration of
N2O4 is 0.0417 M?

18. Example 2 Solution: Calculating the Value
of the Equilibrium Constant
2 NO2 N2O4
The equilibrium NO2 conc is 0.0165 M and
N2O4 is 0.0417 M:
𝐊 =
𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐 =
𝟎.𝟎𝟒𝟏𝟕
𝟎.𝟎𝟏𝟔𝟓 𝟐 = 𝟏𝟓𝟑 Remember, K
is unitless!

19. Example 3: Reversing the Reaction and
the Equilibrium Constant Expression
N2O4 2 NO2
The equilibrium NO2 conc is still 0.0165
M and N2O4 is 0.0417 M. What is the
value for K?

20. Example 3 Solution: Reversing the Reaction
and the Equilibrium Constant Expression
N2O4 2 NO2
The equilibrium NO2 conc is still 0.0165
M and N2O4 is 0.0417 M…
𝐊 =
𝐍𝐎 𝟐
𝟐
𝐍 𝟐 𝐎 𝟒
=
𝟎.𝟎𝟏𝟔𝟓 𝟐
𝟎.𝟎𝟒𝟏𝟕
= 𝟔. 𝟓𝟑 × 𝟏𝟎−𝟑
This value is just K-1 for the previous reaction!

21. Example 4: The Equilibrium Constant Value
and Expression with Partial Pressures
2 NO2 N2O4
The equilibrium NO2 partial pressure is
1.26 atm and N2O4 is 0.199 atm:
NOTE: These are not
the same experimental
conditions from the
previous problem.

22. Example 4 Solution: The Equilibrium
Constant Expression with Partial Pressures
2 NO2 N2O4
The equilibrium NO2 partial pressure is
1.26 atm and N2O4 is 0.199 atm:
𝐊 =
𝑷 𝑵 𝟐 𝑶 𝟒
𝑷 𝑵𝑶 𝟐
𝟐 =
𝟎.𝟏𝟗𝟗
𝟏.𝟐𝟔 𝟐 = 𝟎. 𝟏𝟐𝟓 NOTE: These are not
the same experimental
conditions from the
previous problem.

23. Next up,
Properties of the Equilibrium
Constant K (Pt 4)