CHAPTER 13: CHEMICAL EQUILIBRIA
Chem
1212
Dr. Aimée Tomlinson
Section 13.1
Equilibrium State
Graphical Representation
N 2O4( g )  2 NO2( g )
colorless
brown
After some time:
[N2O4] stops decreasing
[NO2] stops increasing
Solution from clear to brown
and back again
Rate_forward = rate_reverse
2 NO2( g )  N 2O4( g )
brown
colorless
After some time:
[N2O4] stops increasing
[NO2] stops decreasing
Solution from brown to clear
and back again
Rate_forward = rate_reverse
Chemical Equilibrium
Defn: a phenomenon in which the concentrations of reactants and
products remain constant over time
N 2 O4( g )
2 NO2( g )
NOTE: this does not mean the concentrations go to zero!
Section 13.2
The Equilibrium Constant, Kc
Equilibrium Constant Expression
AKA Law of Mass Action: the chemical equilibrium
expression will give rise to a characteristic value for a
given temperature
For the General case:
aA  bB
cC  dD
[C ]c [ D]d
Kc 
[ A]a [ B]b
Kc is the equilibrium constant in terms of concentrations
Example
Recall: aA  bB
cC  dD
[C ]c [ D]d
Kc 
[ A]a [ B]b
Write the Kc for both the forward and reverse directions for the
following:
1.) 4A  3B2
𝐶 2 [𝐷
𝐾𝑐 =
𝐴4𝐵3
2.) 3H 2  N 2
𝑁𝐻3 2
𝐾𝑐 =
𝐻2 3 [𝑁2
2C  D
4 𝐵 3
𝐴
𝐾𝑐′ =
𝐶 2 [𝐷
2NH 3
3
𝐻
[𝑁2
2
𝐾𝑐′ =
𝑁𝐻3 2
Mass Action Rule #1
Kc in the forward direction is the inverse for the
reverse direction
Kc  1
K
'
c
Section 13.3
Equilibrium Constant, Kp
Generalized Kp
aA g   bB g 
cC g   dD g 
( PC )c ( PD ) d
Kp 
( PA ) a ( PB )b
Example: What is Kc and Kp for the reaction below.
2 NO( g )  Cl2( g )
2 NOCl( g )
𝑁𝑂𝐶𝑙 2
𝐾𝑐 =
𝑁𝑂 2 [𝐶𝑙2
𝐾𝑝 =
𝑃𝑁𝑂𝐶𝑙
2
𝑃𝑁𝑂 2 (𝑃𝐶𝑙2
Numeric Example
A reaction vessel contains an eq mix of the following: PSO2 = 0.0018
atm, PO2 = 0.0032 atm, and PSO3 = 0.0166 atm. What is the eq constant for
the following reaction?
2 SO2( g )  O2( g )
2SO3( g )
How are Kp & Kc Related?
Using Previous Example:
2 NO2( g )  2 NO( g )  O2( g )
We start with the ideal gas Law:
n
1
P  RT  M RT  M  P
V
RT
2
[ NO] [O2 ]
Kc 
[ NO2 ]2
Kp 
( PNO ) 2 ( PO2 )
( PNO2 ) 2
Next, we plug into the Kc expression
2
P
 NO   PO2 
 RT   RT 
2
[ NO] [O2 ] 
 

Kc 

2
[ NO2 ]2
 PNO2 


RT


( PNO ) 2 ( PO2 ) 1
1
Kc 
 Kp
2
RT
RT
( PNO2 )
or , K p  K c RT
Kp & Kc Generalized Relationship
K p  Kc ( RT )
n
Where n = moles(g)products - moles(g)reactants
For previous example we had:
2 NO2( g )  2 NO( g )  O2( g )
So n = 3-2 = 1 hence Kp = KcRT
Another Example
For which of the following will Kp = Kc?
1.) 2 SO2( g )  O( g )
2 SO3( g )
n  2  3  1
2.) Fe( s )  CO2( g )
FeO( s )  CO( g )
n  1  1  0
3.) H 2 O( g )  CO( g )
H 2( g )  CO2( g )
So only 2.) & 3.) will work.
n  2  2  0
One More Example
What is the Kp of the reaction below at 325C given Kc = 5.0?
Cl2( g )  CO( g )
COCl2( g )
Section 13.4
Heterogeneous
Equilibrium
The Two Types of Equilibria
Heterogeneous
when the reactants/products are in more than one phase
Homogeneous
when the reactants/products are all in the same phase
For Eq we never include:
Solids
H 2O( g )  C( s )
CO( g )  H 2( g )
Kc 
[CO][ H 2 ]
[ H 2O ]
Pure Liquids
CO2( g )  H 2O( l )
H 2CO3( aq )
[ H 2CO3 ]
Kc 
[CO2 ]
Handling Combined Equations
(1) N 2( g )  O2( g )
2 NO( g )
(2) 2 NO( g )  O2( g )
2 NO2( g )
N 2( g )  O2( g )  2 NO( g )
2 NO( g )  2 NO2( g )
Overall : N 2( g )  2O2( g )
[ NO]2
K1 
[ N 2 ][O2 ]
2 NO2( g )
[ NO2 ]2
K2 
[ NO]2 [O2 ]
[ NO]2
[ NO2 ]2
K overall
[ NO2 ]2
K1  K 2 


2
[ N 2 ][O2 ] [ NO] [O2 ] [ N 2 ][O2 ]2
[ NO2 ]2

[ N 2 ][O2 ]2
Mass Action Rule #2
When adding up equilibrium equations to get
an overall the product of their equilibrium
constants will give rise to the Kc of the overall
equation
K c,overall  K1  K 2 
Example Problem
Calculate the Kc for 2D
Using the data below:
A  2B
C
2D
C
Kc  3.3
Kc  0.041
A  2B
How does factor multiplication
impact Kc?
2D
3  2D
A  2B
[ A][ B]2
K1 
[ D ]2
A  2B   6D
 K 2  ( K1 )3
3A  6B
[ A]3[ B]6
K2 
[ D ]6
Mass Action Rule #3
Multiplying a chemical by a factor leads to
exponential factor for the Kc
K   Kc 
'
c
n
Summary of Mass Action Rules
Example Problem
If Kc = 2.4 x 10-3 for 2 SO2(g)  O2(g)
what is it for the following?
1.) SO2(g)  21 O2(g)
2.) 2 SO3(g)
3.) SO3(g)
SO3(g)
2 SO2(g)  O2(g)
SO2(g)  21 O2(g)
2 SO3(g)
Section 13.5
Using the Equilibrium
Constant
Judging the Extent of Reaction
[products]
Kc 
[reactants]

Kc > 103 products predominate over reactants:



Kc < 10‾3 reactants predominate over products:



reaction proceeds nearly to completion
very little reactant is left
reaction hardly proceeds at all
very little product is produced
10‾3 < Kc < 103 neither dominates both are
present at eq
Reaction Quotient Q
Q looks just like Kc but it not guaranteed to be at eq
aA  bB
cC  dD
[C ]c [ D]d
Qc 
[ A]a [ B]b
Using Q for Direction Prediction
Need to consume reactants
• Go forward
• To the right
• Toward products
At Eq we go in neither
direction
Need to consume products
• Go reverse
• To the left
• Toward reactants
Example – Practice with Q
Given the data below is the reaction in equilibrium
and if not in which direction will need to go in order
to reach eq?
A
B
K  22,  A  0.10 M ,  B   2.0 M
Finding Equilibrium Concentrations
We use the "ICE" table:
"I"nitial
"C"hange
"E"q
ICE Example I
The value of Kc = 0.0900 at 298K for the reaction below, determine the
eq concentrations if initially [H2O] = 0.00432 M and [Cl2O] = 0.00442 M.
H 2O( g )  Cl2O( g )
2 HOCl( g )
ICE Example II
The value of Kc for the thermal decomposition of hydrogen sulfide
2 H 2 S( g )
2 H 2( g )  S 2( g )
is 2.2 x 10-6 at 1400K. A sample of gas in which [H2S] = 0.600M is heated
to 1400K in a sealed vessel. After chemical eq has been achieved,
what is the value of [H2S]? Assume no H2 and S2 was present in the
original sample.
ICE Example III
What are the eq concentrations of each of the species in the following
reaction, given the Kc = 5.1 at 700K and the initial concentration of all
species is 0.050 M?
CO( g )  H 2 O( g )
CO2( g )  H 2( g )
Section 13.6-13.9
Le Châtelier’s Principle
Le Châtelier’s Principle
Defn: when a stressor is applied to a system at
equilibrium, the system will adjust to counteract
the stressor in order to reestablish equilibrium
Stressors include:



Adding or removing reactants or products
Changing the pressure or the volume
Changing the temperature
Stressor I: Concentration Changes
A
B
Increase in A or decrease in B:
[ B]
K
[ A]
We will need to consume A (or produce B) to get back to eq
We go:
Q





Forward
Toward product
To the right
Increase in B or decrease in A:
[ B]
K
[ A]
We will need to consume A (or produce B) to get back to eq
We go:
Q





Reverse
Toward reactant
To the left
Stressor II: P or V Changes
2 A g 
B g 
Decrease V or increase P:
Q

We increase the number of collisions between molecules



P will increase and V will decrease
Number of moles appears to increase
We go:



Forward
Toward product
To the right
Increase in V or decrease in P:
Q

[ B]
K
2
[ A]
We decrease the number of collisions between molecules



[ B]
K
2
[ A]
P will decrease and V will increase
Number of moles appears to decrease
We go:



Reverse
Toward reactant
To the left
LCP Example I
For each scenario predict the direction the reaction goes to attain eq:
CO(g)  2H 2(g)
a.) CO is added
b.) CH3OH is added
c.) Pressure is reduced
d.) Volume is increased
CH 3 OH (g)
Stressor III: Temperature Changes
Unique since this stressor will also change the value for Kc

For exothermic reaction, heat is a product
Reactants → Products + ΔH



Removal of heat will force the reaction to go:

Forward

Toward product

To the right
Addition of heat will force the reaction to go:

Reverse

Toward reactant

To the left
For endothermic reaction, heat is a reactant
ΔH + Reactants → Products


Addition of heat will force the reaction to go:

Forward

Toward product

To the right
Removal of heat will force the reaction to go:

Reverse

Toward reactant

To the left
LCP Example II
In what direction will the eq shift when each of the following changes
are made to the system at eq?
N 2 O4 g 
2 NO2 g 
a.) N2O4 is added
b.) NO2 is removed
c.) Pressure is increased by adding N2
d.) Volume is decreased
e.) Temperature is decreased
H  58.0 kJ
Section 13.10
Catalytic Effect on Eq
There really isn’t one
Section 13.11
Chemical Eq & Kinetics
Another Gen Chem Lie
aA  bB
cC  dD
rate forward  ratereverse
assuming this is an single step reaction:
rate forward  k f  A  B   ratereverse  kr C   D 
a
b
c
d
kr C   D 


 Kc
a
b
k f  A  B 
c
d