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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.
At equilibrium…
Macroscopic observables have stopped changing.The forward and reverse reaction rates are equal.
Recall: Equilibrium Constant K and Rate Constants
2 NO2 N2O4
k1[NO2]2 = k1[N2O4]
The equilibrium constant K is the ratio of the forward and reverse rate
constants.
k1
k1
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.)
The Law of Mass Action
Consider the generalized reaction
Reactants A and B
Products C and D
coefficients
The Law of Mass Action
For the reaction
The relationship between the value of the equilibrium constant K and the concentrations of reactants and products is
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
The Law of Mass Action
What happens if we reverse the reaction?
The relationship between the value of the equilibrium constant K and the concentrations of reactants and products is still products over reactants!
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
The Law of Mass Action
This relationship is true no matter the initial distribution
(relative amounts) of reactants and products!
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
𝐏𝐍𝐎𝟐
𝐏𝐍𝟐𝐎𝟒
“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
𝐏𝐍𝐎𝟐
𝐏𝐍𝟐𝐎𝟒
Equilibrium Constant K is Unitless
Why is the value for K unitless?
Each concentration in the equilibrium constant expression is divided by a “standard concentration” of 1.0 M.
All molarity units
cancel out!
Equilibrium Constant K is Unitless
When we divide by a standard concentration, we are left with “effective concentrations” or “activities” (a).
“activity” can be though of as the “presence” of the
reactant or product
Gases and the Equilibrium Constant Expression
The Law of Mass Action holds for gases in equilibrium, as well. represents the partial pressure of gas C, etc.
Gases and the Equilibrium Constant Expression
Each partial pressure is divided by a standard pressure (1 atm), so the K for gaseous systems is also unitless!
Partial pressures for gases are given in “activities”, just
like solutions.
Example 1: The Equilibrium Constant Expression
2 NO2 N2O4
Write the equilibrium constant expression for the following reaction:
Example 1 Solution: The Equilibrium Constant Expression
2 NO2 N2O4
Write the equilibrium constant expression for the following reaction:
𝐊=[𝐍𝟐𝐎𝟒 ][𝐍𝐎𝟐 ]𝟐
The coefficient for N2O4 is 1
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.
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?
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:
Remember, K is unitless!
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?
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…
This value is just K-1 for the previous reaction!
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.
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:
NOTE: These are
not the same experimental conditions from the previous problem.
Next up, Properties of the
Equilibrium Constant K (Pt 4)
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