Chapter 15
Chemical
Equilibrium
15.1 The Concept of Equilibrium
Most chemical reactions are reversible.
reversible reaction = a reaction that proceeds
simultaneously in both directions
Examples:

 2 NH (g )
N2 (g )  3 H2 (g ) 

3

 CH OH(g )
CO(g )  2 H2 (g ) 

3

 2 NO (g )
N2O 4 (g ) 

2

 ) denote an equilibrium reaction.
Double arrows (

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Equilibrium

 2 NO (g )
Consider the reaction N2O 4 (g ) 

2
At equilibrium,
the forward reaction: N2O4(g)  2 NO2(g), and
the reverse reaction: 2 NO2(g)  N2O4(g)
proceed at equal rates.
Chemical equilibria are dynamic, not static – the
reactions do not stop.
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Equilibrium
Let’s use 2 experiments to study the reaction

 2 NO (g )
N2O 4 (g ) 

2
each starting with a different reactant(s).
Exp #1
Exp #2
pure N2O4
pure NO2
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Equilibrium
Experiment #1

 2 NO (g )
N2O 4 (g ) 

2
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Equilibrium
Experiment #2

 2 NO (g )
N2O 4 (g ) 

2
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Equilibrium

 2 NO (g )
N2O 4 (g ) 

2
Are the equilibrium pressures of NO2 and N2O4
related? Are they predictable?
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15.2 The Equilibrium Constant
At equilibrium,
rate forward  ratereverse

 2 NO (g )
N2O 4 (g ) 

2

2
kf [N2O4 ]eq  kr [NO2 ]eq
or
2
kf [NO 2 ]eq

 Kc
kr [N 2O 4 ]eq
where Kc is the equilibrium constant
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The Equilibrium Constant
This constant value is termed
the equilibrium constant,
Kc, for this reaction at 25°C.
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The Equilibrium Constant
For the NO2 / N2O4 system:

 2 NO (g )
N 2O 4 ( g ) 

2
[NO2 ]2
K 
 0.143
[N2O4 ]
at 25C
equilibrium constant
equilibrium constant expression
Note: at 100°C, K =
6.45
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The Equilibrium Constant
reaction quotient = Qc = the value of the
“equilibrium constant expression” under
any conditions.

 COCl (g )
CO( g )  Cl2 (g ) 

2
For,
Kc 
[COCl2 ]eq
[CO]eq [Cl 2 ]eq
Q > K
Q = K
Q < K
Qc 
[COCl2 ]
[CO][Cl2 ]
 reverse reaction favored
 equilibrium present
 forward reaction favored
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The Equilibrium Constant

 c C  d D
a A  bB 

For a reaction:
For gases:
For solutions:
KP
PCc  PDd

PA a  PBb
P in atm
Kc
[C]c [D]d

[A]a [B]b
[ ] = mol/L
The Law of Mass Action:
Cato Maximilian Guldberg & Peter Waage,
Forhandlinger: Videnskabs-Selskabet i
Christiana 1864, 35.
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The Equilibrium Constant
Note:
• The equilibrium constant expression has products
in the numerator, reactants in the denominator.
• Reaction coefficients become exponents.
• Equilibrium constants are temperature dependent.
• Equilibrium constants do not have units. (pg. 622)
• If K >>> 1, products favored (reaction goes
nearly to completion).
• If K <<< 1, reactants favored (reaction hardly
proceeds).
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15.3 Equilibrium Expressions
homogeneous equilibria = equilibria in which all
reactants and products are in the same phase.
heterogeneous equilibria = equilibria in which all
reactants and products are not in the same phase.
Ex:

 CaO(s )  CO (g )
CaCO3 (s ) 

2
The equilibrium constant expression is,
K = [CO2]
• [CaO] and [CaCO3] are solids.
• Pure solids and liquids are omitted from equilibrium
constant expressions.
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Exercise: Write the expressions for Kp for the
following reactions:

 N O(g )  2 H O(g )
(a) NH4NO3 (s ) 

2
2

 CuCl (s )
(b) Cu(s )  Cl (g ) 

2
2
Solution:

(a) K P  P N2O  PH2O
(b) K P 

2
1
PCl2
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Equilibrium Expressions
A. Reverse Equations

 2 NO (g )
For, N2O 4 (g ) 

2
K1 
For,
 PNO 
[1]
2
 0.143
2
PN2O4
at 25 C

 N O (g )
2 NO2 (g ) 

2 4
K2 
PN2O4
 PNO 
2
1

 6.99
0.143
2
Conclusion:
K2
1

K1
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[2]
at 25 C
Equilibrium Expressions
B. Coefficient Changes
For,
For,

 2 NO (g )
N2O 4 (g ) 
[1]

2
2
PNO2
K1 
 0.143
at 25 C
PN2O4
1

 NO (g )
N2O 4 (g ) 
[3]

2
2
PNO2
K3 
0.143  0.378
1/2 
PN2O4




Conclusion:
K3 
K1
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at 25 C
Equilibrium Expressions
C. Reaction Sum (related to Hess’ Law)

 2 NO (g )
For, N O (g ) 

2 4
K1 
PNO2 2
2
PN2O4

 2 NO(g )  O (g )
For, 2 NO2 (g ) 

2
 PNO 2 PO2
K4 
2
 PNO2 
Add [1] + [4],

 2 NO(g )  O (g )
N O (g ) 

2 4
 PNO  PO2
2
2
K5 
PN2O4
 K1  K 4
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[1]
[4]
[5]
Equilibrium Expressions
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Exercise: At 500ºC, KP = 2.5  1010 for,

 2 SO (g )
2 SO2 (g )  O2 (g ) 

3
(a) At 500ºC, which is more stable, SO2 or SO3?
Compute KP for each of the following:
(b) SO 2 (g) 
1

 SO (g)
O2 (g) 

3
2
3

 3 SO (g)
(c) 3 SO 2 (g) 
O2 (g) 

3
2

 SO (g)  1 O (g)
(d) SO3 (g) 

2
2
2
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15.4 Using Equilibrium
Expressions to Solve Problems
Predicting the direction of a reaction
Compare the computed value of Q to K
Q > K
Q = K
Q < K
 reverse reaction favored
 equilibrium present
 forward reaction favored
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Exercise #1: At 448°C, K = 51 for the reaction,

 2 HI(g )
H (g )  I (g ) 

2
2
Predict the direction the reaction will proceed, if at
448°C the pressures of HI, H2, and I2 are 1.3,
2.1 and 1.7 atm, respectively.
Solution:
2
PHI 


Q 
PH2  PI2
(1.3)2
 0.47
(2.1)  (1.7)
0.47 < 51  system not at equilibrium
Numerator must increase and denominator must
decrease.
Consequently the reaction must shift to the right.
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Exercise #2: At 1130°C, K = 2.59  102
for

 2 H (g )  S (g )
2 H2S(g ) 

2
2
At equilibrium, PH S = 0.557 atm and PH = 0.173 atm,
2
2
calculate PS at 1130°C.
2
 PH   PS
2
 PH S 
2
Solution:
K 
2
2
 2.59 102
2

(0.173) 2  PS2
(0.557) 2
 2.59 102
PS = 0.268 atm
2
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Exercise #3: K = 82.2 at 25°C for,

 2 ICl(g )
I (g )  Cl (g ) 

2
2
Initially, PI2 = PCl2 = 2.00 atm and PICl = 0.00 atm.
What are the equilibrium pressures of I2, Cl2, and ICl?
Solution:
I2 (g ) 
Initial
2.00 atm
Change
x
Equilibrium (2.00 – x)

 2 ICl(g )
Cl2 (g ) 

2.00 atm
0.00 atm
x
+2x
(2.00 – x)
2x
(2 x) 2
PICl 2

 82.2
K 
(2.00  x)(2.00  x)
PI2  PCl2
perfect square
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Exercise #3: (cont.)
(2 x) 2
 82.2
(2.00  x)(2.00  x)
square root 
(2 x)
 9.066
(2.00  x)
2 x = 18.132 – 9.066 x
11.066 x = 18.132
x = 18.132 / 11.066 = 1.639
PI2 = PCl2 = 2.00 – x = 2.00 – 1.639 = 0.36 atm
PICl = 2x = (2)(1.639) = 3.28 atm
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Exercise #4: At 1280°C, Kc = 1.1  103 for

 2 Br(g )
Br (g ) 

2
Initially, [Br2] = 6.3  102 M and [Br] = 1.2  102
M. What are the equilibrium concentrations of Br2
and Br at 1280°C?

 2 Br(g )
Br (g ) 
Solution:

2
Initial
6.3  102 M 1.2  102 M
Change
-x
+2x
Equilibrium (6.3  102) - x (1.2  102) + 2x
Kc
[Br]2
[(1.2 102 )  2 x]2


[Br2 ]
(6.3 102 )  x
 1.1103
4x2 + 0.0491x + (7.47  105) = 0
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4x2 + 0.0491x + (7.47  10-5) =
0
quadratic equation: ax2 + bx + c = 0
b 
b 2  4ac
x 
solution:
2a
x = 1.779  103 and 1.050  102
Q: Two answers? Both negative? What’s happening?
Equilibrium Conc.
x = 1.779  103
[Br2] = (6.3  102) – x =
0.0648 M
[Br] = (1.2  102) + 2x =
0.00844 M
[Br2] = 6.5  102 M
[Br] = 8.4  103 M
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 1.050  102
0.0735 M
 0.00900 M
impossible
Exercise #5: A pure NO2 sample reacts at 1000 K,

 2 NO(g )  O (g )
2 NO (g ) 

2
2
KP is 158. If at 1000 K the equilibrium partial
pressure of O2 is 0.25 atm, what are the equilibrium
partial pressures of NO and NO2.

 2 NO( g )  O (g)
Solution: 2 NO2 (g ) 

2
Initial
?
0 atm
0 atm
0.50
+0.25
Change
+0.50
+0.50 atm 0.25 atm
Equilibrium PNO2
 PNO   PO
2
 PNO 
2
KP 
2
2

(0.50) 2 (0.25)
 PNO 
2
 158
2
rearrange and solve
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Exercise #5: (cont.)
(0.50)2 (0.25)
 PNO 
2
 158
2
 PNO 
2
2
(0.50)2 (0.25)

= 3.956  104
158
PNO2 
3.956  104  0.01989
PNO2 = 0.020 atm
PNO = 0.50 atm
see ICE table
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Exercise #6: The total pressure of an equilibrium
mixture of N2O4 and NO2 at 25°C is 1.30 atm.
For the reaction:

 2 NO (g )
N2O 4 (g ) 

2
KP = 0.143 at 25°C. Calculate the equilibrium
partial pressures of N2O4 and NO2.
KP 
 PNO 
2
2
PN2O4
PNO2 + PN
2O4
 0.143
= 1.30 atm
two equations and two unknowns – BINGO!
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Exercise #6: (cont.)
KP 
 PNO 
2
 0.143
2
PN2O4
PN
2O4
PNO2 + PN
2O4
= 1.30 atm
= 1.30 atm - PNO2
 PNO 
2
2
(1.30  PNO2 )
 0.143
PNO22 + 0.143 PNO2  0.1859 =
0
Use the quadratic formula,
PNO2 = +0.366 atm and 0.509 atm
PN
2O4
= 1.30 atm - PNO2 = 1.30  0.366 = 0.934 atm
P
= 0.93 atm
Copyright McGraw-Hill 2009 N O
2 4
15.5 Factors That Affect
Chemical Equilibrium
Le Châtelier’s Principle
“If an equilibrium system variable is changed, the
equilibrium will shift in the direction (right or left)
that tends to reduce the change.”
Example: N2, H2, and NH3 are at equilibrium in a
container at 500°C.
o


N2 ( g )  3 H2 ( g )  2 NH3 (g ) DHrxn  92kJ
(continued on next 5 slides)
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Case I:
Change: N2 is added
the right
Shift: to???
Q:
Why?
Ans: [N2] has increased. Which direction
will decrease [N2]?

 2 NH (g )
N2 (g )  3 H2 (g ) 

3
N2 decreases
N2 increases
right
left
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Case II:
Change: compress the system
Shift: to???
the right
Q:
N2 H2
NH3
Why?
Ans: Total pressure has increased. Which
direction will decrease the total pressure?
Recall: P  n

 2 NH (g )
N (g )  3 H (g ) 

2
2
(4 moles gas)
3
(2 moles gas)
more gas
more pressure
Copyright McGraw-Hill 2009
less gas
less pressure
Case III:
Change: increase the temperature
Shift: to???
the left
Q:
Why?
Ans: Temperature has increased. Which
direction decreases the temperature?
Recall, the reaction is exothermic.
ΔH rxn   92 kJ
right
endothermic
heat absorbed
left
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exothermic
heat evolved
Case IV:
Change: add helium at constant volume
Shift: none
???
Q:
Why?
Ans: Helium is not a reactant or product.
Adding helium (at constant V) does not
change PN2, PH2 or PNH3. Hence the
equilibrium will not shift.
Copyright McGraw-Hill 2009
Case V:
Change: add helium at constant
total pressure
Shift: to
???
the left
Q:
Why?
Ans: If the total pressure is constant, PN2 +
PH2 + PNH3 must decrease. Which
direction increases this sum?
Recall: P  n

 2 NH (g )
N2 (g )  3 H2 (g ) 

3
(4 moles gas)
(2 moles gas)
more gas
more pressure
less gas
less pressure
Copyright McGraw-Hill 2009
Exercise: Hydrogen (used in ammonia production)
is produced by the endothermic reaction,
Ni

 CO(g )  3H2 (g )
CH4 (g )  H2O(g ) 

750C
Assuming the reaction is initially at equilibrium,
indicate the direction of the shift (L, R, none) if
(a) H2O(g) is removed.
(b) The temperature is increased.
(c) The quantity of Ni catalyst is increased.
Left
Right
None
(d) An inert gas (e.g., He) is added.
(e) H2(g) is removed.
(f) The volume of the container is tripled.
None
Right
Right
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