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)