Chapter 13: Chemical Equilibrium
Numeric Example : A reaction vessel contains an eq mix of the following: P
SO2
= 0.0018 atm, P
O2
= 0.0032 atm, and P
SO3
= 0.0166 atm. What is the eq constant for the following reaction? 2 SO

O 2 SO
K p

P
2
SO
3


0.0166
atm

2
(0.0018
atm ) (0.0032
atm )
  4
SO
2
O
2
________________________________________________________________________
Another Example : What is the K p
of the reaction below at 325

C given K c
= 5.0?
Cl

CO COCl
K c

[ COCl
2
]

K c
units are
[ Cl
2
][ CO ]
K p

K RT c
)
 n 
5.0
L mol

0.08206
L mol
L atm mol K

K p

0.10

K
  
________________________________________________________________________
Application of Mass Action Rules I : Calculate the eq const for 2 D A

2 B given the info below.
A

2 B C
C 2 D
K c
K c


3.3
0.041
- we need to look from the reverse direction for both of these reactions
C A

2 B K c

1

3.3
0.303
2
2
D
D
C
A

2 B
K c
K c

1
0.041

24.4

0.303 24.4

7.39
________________________________________________________________________
Application of Mass Action Rules II : If the eq. constant at a given temperature is 2.4 x
10 -3 for 2 SO

O 2 SO what is the eq. constant for the reactions below? a.) b.)
SO
2 SO

1
2
O
2 SO
SO

O
K = (2.4 x 10 ) c
1
2

0.049
c
-3 -1
K = (2.4 x 10 )

417
-3
K = (2.4 x 10 )
1
2 20.4
c
  c.) SO SO

1
2
O
________________________________________________________________________
Example 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
Q

[ ]

2.0

[ ] 0.1
20 < 22 = K therefore it will go in the forward direction
________________________________________________________________________
Calculating Equilibrium Concentrations using "ICE" table, "I"nitial, "C"hange,
"E"q
ICE Example I : The value of K c
= 0.0900 at 298K for the reaction below, determine the
1

eq concentrations if initially [H
2
O] = 0.00432 M and [Cl
2
O] = 0.00442 M.
H O
2 ( )

Cl O
2 ( )
2 HOCl
K c
Initial
H
2
O
0.00432
Cl
2
O
0.00442
Change -x -x +2x
Eq

[ HOCl
[
2
]
0.00432-x 0.00442-x
]
2
 


0.00432

  x

2
0.00442

+2x x


4 x
2
1.9094 10

5 
8.74 10

2 x
 x
2
0.0900 (1.9094 10

5 
8.74 10

2 x
 x
2
)

4 x
2
HOCl
0.0

0.0900
1.1785 10

6  
0

3.9100
x
2  

4 x

0.0900
x
2 
4 x
2

4 x

Recall the quadratic equation:

6 for ax
2   
0 x

  b 2 
4 ac
2 a

4

6 x

  
4 


4

2
4 3.9100


 
6
 x

  
4 
6.187 10

7   
5
7.82
x

  
4 
0.005244

4 or x
7.
82 x must be greater than zero therefore x = 5.70x10
-4
 
7.71 10

4
Using this value we can determine what the concentrations are at eq:
[H
2
O] = 0.00432 - x = 0.00432 - 0.00057 = 0.00375 M
[Cl
2
O] = 0.00442 - x = 0.00385 M
[HOCl] = 2x = 0.00114 M
________________________________________________________________________
ICE Example II : The value of K c
for the thermal decomposition of hydrogen sulfide
2 H S
( )
2 H

S is 2.2 x 10
-6
at 1400K. A sample of gas in which [H
2
S] = 0.600M is heated to
1400K in a sealed vessel. After chemical eq has been achieved, what is the value of [H
2
S]? Assume no H
2
and S
2
was present in the original sample.
Initial
Change
Eq
H
2
S
0.600
-2x
0.600-2x
H
2
0.0
+2x
+2x
S
2
0.0
+x
+x
2


K c

[ H
2
][ S
2
]
 
2

 x

4 x 3 assume 0.600 >> 2 x
  x

0.600
4 x
3 x

2

2.2 10

6
4 x
3

0.36
x

2

2.2 10

6
0.00583
M check the assumption :

100% 1.94%

5%
 valid
0.600
Using x = 0.00734 M we can determine what the concentrations are at eq:
[H
2
S] = 0.600 - 2x = 0.600 - 2*0.00583 = 0.588 M
[H
2
] = 2x = 0.0117 M and [S
2
] = 0.00583 M
________________________________________________________________________
ICE Example III : What are the eq concentrations of each of the species in the following reaction, given the K c
= 5.1 at 700K and the initial concentration of all species is 0.050 M? CO

H O
2 ( )
CO

H
Initial
Change
Eq
CO
0.050
-x
H
2
O
0.050
-x
CO
+x
2
0.050
H
2
0.050
+x
K c
0.050-x

0.050-x 0.050+x 0.050+x
[ CO
2
][ H
2
[ ][
2
]
]

(0.050
(0.050

 x x
)(0.050
)(0.050

 x x
)
)

(0.050
(0.050

 x x
)
)
2
2
(0.050
 x )
2
(0.050
 x )
2

5.1

(0.050
 x )
(0.050
 x )
2.258(0.050
 x )

0.050

 x

0.050
x x x

2.258
0.0193
M

5.1
using this value we determine the concentrations:
[CO] = [H
[CO
2
] = [H
2
2
O] = (0.050-0.0193)M = 0.031 M
] = (0.050+0.0193)M = 0.069 M
________________________________________________________________________
LCP Example I : For each scenario predict the direction the reaction goes to attain eq:
CO
(g)

2H
2(g)
CH OH
3 (g) a.)
CO is added ……. reaction goes toward product (forward) b.) CH
3
OH is added ……. reaction goes toward reactants (reverse) c.) Pressure is reduced …….  n reactants
= 3,
 n products
= 1 reaction goes toward reactants (reverse) d.)
Volume is increased ……. reaction goes toward reactant (forward)
________________________________________________________________________
LCP Example II : In what direction will the eq shift when each of the following changes are made to the system at eq?
N O
4
 
2 NO
2
 

H

58.0
kJ
(a) N O
2 4
 
is added ……. reaction goes toward product
3

(b) NO
2
 
is removed ……. reaction goes toward product
(c) the total pressure is increased by adding N
2
 
……. reaction remains unchanged since the partial pressures of the reacting species is constant at constant volume
(d) the volume is decreased ……. reaction goes toward reactant since
 n reactants
= 1,
 n products
= 2
(e) the temperature is decreased ……. reaction goes toward reactant since the process is endothermic or N O
2 4
 

58.0
kJ 2 NO
2
 
4