1. Chapter 11
Chemical Equilibrium
11.1 The Equilibrium Condition
11.2 The Equilibrium Constant
11.3 Equilibrium Expressions Involving Pressures
11.4 The Concept of Activity
11.5 Heterogeneous Equilibria
11.6 Applications of the Equilibrium Constant
11.7 Solving Equilibrium Problems
11.8 Le Chatelier's Principle
11.9 Equilibria Involving Real Gases

2. The Equilibrium Condition (General)
Thermal equilibrium indicates two systems in thermal contact
with each do not exchange energy by heat. If two bricks are in
thermal equilibrium their temperatures are the same.
Chemical equilibrium indicates no unbalanced potentials (or
driving force). A system in equilibrium experiences no change
over time, even infinite time.
The opposite of equilibrium systems are non-equilibrium
systems that are off balance and change with time.
Example 1 atm O2 + 2 atm H2 at 298K

3. The Equilibrium Condition (Chem Rxn)
aA + bB cC + dD
The same equilibrium state
is achieved whether
starting with pure reactants
or pure products.
The equilibrium state can
change with temperature.

4. The Equilibrium State (Chem Rxn)
H2O (g) + CO (g) H2 (g) + CO2 (g)
Change
[CO] to PCO
[H2O] to PH2O
etc

5. Chemical Reactions and Equilibrium
As the equilibrium state is approached, the
forward and backward rates of reaction
approach equality. At equilibrium the rates
are equal, and no further net change occurs
in the partial pressures of reactants or
products.
Fundamental characteristics of equilibrium states:
1. No macroscopic evidence of change.
2. Reached through spontaneous processes.
3. Show a dynamic balance of forward and backward processes.
4. Same regardless of the direction from which they are approached.
5. No change over time.

6. Arrows: Chemical Symbolism
Use this in an
equilibrium expression.
↔ Use this to indicate
resonance.

7. Chemical Reactions and Equilibrium
The equilibrium condition for every reaction can be
described in a single equation in which a number, the
equilibrium constant (K) of the reaction, equals an
equilibrium expression, a function of properties of the
reactants and products.
H2O(l) H2O(g) @ 25oC Temperature (oC) Vapor Pressure (atm)
15.0 0.01683
17.0 0.01912
K = 0.03126 19.0 0.02168
H2O(l) H2O(g) @ 30oC 21.0 0.02454
23.0 0.02772
K = 0.04187 25.0 0.03126
30.0 0.04187
50.0 0.1217

8. Law of Mass Action (1)
Partial pressures and concentrations of products appear in the
numerator and those of the reactants in the denominator.
Each is raised to a power equal to its coefficient in the
balanced chemical equation.
aA + bB cC +
if gases dD if concentrations
c d c d
( PC ) ( PD ) [C] [D]
=K
a b a b =K
( PA ) ( PB ) [ A] [B]

9. Law of Mass Action (2)
1. Gases enter equilibrium expressions as partial pressures, in
atmospheres. E.g., PCO2
2. Dissolved species enter as concentrations, in molarity (M)
moles per liter. E.g., [Na+]
3. Pure solids and pure liquids are represented in equilibrium
expressions by the number 1 (unity); a solvent taking part in a
chemical reaction is represented by unity, provided that the
solution is dilute. E.g., I2(s) ↔ I2(aq) [I2 (aq) ] = K
I 2 ( s ) ↔ I 2 ( aq )
[ I 2 (aq )] [ I 2 ( aq )]
K= = = [ I 2 ( aq )]
[ I 2 ( s )] 1

10. Activities
The concept of Activity (i-th component)
= ai = Pi / P reference
H2O (l) H2O (g) Kp = P H2O @ 25oC Kp = 0.03126 atm
PH2O
=K Pref is numerically equal to 1
Pref K = 0.03126
The convention is to express all pressures in atmospheres
and to omit factors of Pref because their value is unity. An
equilibrium constant K is a pure number.

11. The Equilibrium State
H2O (g) + CO (g) H2 (g) + CO2 (g)
PH 2 PCO 2
Kp =
PH 2 O PCO

12. The Equilibrium Expressions
aA + bB cC +
dD
In a chemical reaction in which a moles of species A and b moles of
species B react to form c moles of species C and d moles of species D,
The partial pressures at equilibrium are related through
K = PcCPdD/PaAPbB

13. Write equilibrium expressions for the
following reactions
3 H2(g) + SO2(g) H2S(g) + 2 H2O(g)
2 C2F5Cl(g) + 4 O2(g) Cl2(g) + 4 CO2(g) + 5 F2(g)

14. Heterogeneous Equilibrium
Gases and CaCO3(s) CaO(s) + CO2(g)
Solids
K=PCO2
K is independent of the amounts
of CaCO3(s) or CaO(s)

15. Heterogeneous Equilibrium
Liquids H2O(l) H2O(g)
K=PH2O
I2(s) I2(aq)
Solutions
K=[I2]

16. Relationships Among the K’s of Related Reactions
#1: The equilibrium constant for a reverse reaction is always
the reciprocal of the equilibrium constant for the
corresponding forward reaction.
1
K for = or K for K rev = 1
K rev
aA + bB cC + dD versus cC + dD aA + bB
(PH2O)2
#1 2 H2 (g) + O2 (g) 2 H2O (g)
(PH2)2(PO2) = K1
K1 = 1/K2
2 H2 (g) + O2 (g) (PH2) (PO22)
2
#2 2 H2O (g) = K2
(PH2O)

17. Relationships Among the K’s of Related Reactions
# 2: When the coefficients in a balanced chemical
equation are all multiplied by a constant factor, the
corresponding equilibrium constant is raised to a power
equal to that factor.
(PH2O)2
#1 2 H2 (g) + O2 (g) 2 H2O (g) Rxn 1 (PH2)2(PO2) = K1
#3 H2 (g) + ½ O2 (g) H2O (g) Rxn 3 = Rxn 1 times 1/2
(PH2O)
= K3 K3 = K1½
(PH2)(PO2)½

18. Relationships Among the K’s of Related Reactions
# 3: when chemical equations are added to give a new
equation, their equilibrium constants are multiplied to
give the equilibrium constant associated with the new
equation.
2 BrCl (g) ↔ Br2 (g) + Cl2 (g) (PBr2)(PCl2)
= K1 = 0.45 @ 25oC
wrong arrow
(PBrCl)2
Br2 (g) + I2 (g) ↔ 2 IBr (g) (PIBr)2
wrong arrow = K2 = 0.051 @ 25oC
(PBr2) (PI2)
2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g)
wrong arrow
= K1K2
= (0.45)(0.051)
(PBr2)(PCl2) (PIBr)2
X (P ) (P ) = K1K2 = K3
(PBrCl)2 Br2 I2 =0.023 @ 25oC

19. Calculating Equilibrium Constants
Consider the equilibrium
4 NO2(g)↔ 2 N2O(g) + 3 O2(g)
The three gases are introduced into a container at partial pressures of 3.6
atm (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reach
equilibrium at a fixed temperature. The equilibrium partial pressure of the
NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the
reaction at this temperature, assuming that no competing reactions occur.
4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
initial partial pressure (atm)
change in partial pressure (atm)
equilibrium partial pressure (atm)

20. Calculate the equilibrium constant of the reaction at this temperature,
assuming that no competing reactions occur.
4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
initial partial pressure (atm) 3.6 5.1 8.0
change in partial pressure (atm) – 4x +2x +3x
equilibrium partial pressure (atm) 2.4 5.1 + 2x 8.0 + 3x
5.1 + 2(0.3 atm) = 5.7 atm N2O
3.6 – 4x = 2.4 atm NO2;
x = 0.3 atm 8.0 + 3(0.3 atm) = 8.9 atm O2
(PN2O)2(PO2)3
K= =
(PNO2)4

21. The compound GeWO4(g) forms at high temperature in the reaction
2 GeO (g) + W2O6(g) ↔ 2 GeWO4(g)
Some GeO (g) and W2O6 (g) are mixed. Before they start to react,
their partial pressures both equal 1.000 atm. After their reaction at
constant temperature and volume, the equilibrium partial pressure of
GeWO4(g) is 0.980 atm. Assuming that this is the only reaction that
takes place, (a) determine the equilibrium partial pressures of
GeO and W2O6, and (b) determine the equilibrium constant for
the reaction.
2 GeO (g) + W2O6 (g) ↔ 2 GeWO4(g)
initial partial pressure (atm) 1.000 1.000 0
change in partial pressure (atm) – 2x –x +2x
equilibrium partial pressure (atm)

22. (a) determine the equilibrium partial pressures of GeO and W2O6, and
(b) determine the equilibrium constant for the reaction.
2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g)
initial partial pressure (atm) 1.000 1.000 0
change in partial pressure (atm) –2x –x +2x
equilibrium partial pressure (atm) 1.000 – 2x 1.000 – x 0.980
0 + 2x = 0.980 atm GeWO4; 1.000 – 2(0.490) = 0.020 atm GeO
x = 0.490 atm 1.000 – 0.490 = 0.510 atm W2O6
(PGeWO4)2
K = (P )2(P
GeO W2O6 ) =

23. • Skip Solving quadratic equations
• Will utilize approximation method
– Systems that have small equilibrium
constants.
– Assume “x” (the change in concentration)
is small (less than 5%) of the initial
concentration.

24. A vessel holds pure CO (g) at a pressure of 1.282 atm and a temperature of
354K. A quantity of nickel is added, and the partial pressure of CO (g)
drops to an equilibrium value of 0.709 atm because of the reaction
Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)
Compute the equilibrium constant for this reaction at 354K.
PNi(CO)4 PNi(CO)4
K= 4
=
(PCO ) [Ni(s)] (PCO ) 4 (1)
Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)
Construct an “ICE” table P CO (atm) P Ni(CO)4 (atm)
initial partial pressure (atm) 1.282 0
change in partial pressure (atm) -4x +1x
equilibrium partial pressure (atm) 0.709 x
At equil. Pco= x = PNi(CO)4 =

25. Equilibrium Calculations
At a particular temperature, K = 2.0 x 10-6 mol/L for the
reaction
2CO2 (g) 2CO (g) + O2 (g)
If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate
the equilibrium concentrations of all species.
2CO2 (g) 2CO (g) + O2 (g)
initial partial pressure (mol/L) 0.4 0 0
change in partial pressure (mol/L) – 2x +2x +1x
equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x

26. At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction
2CO2 (g) 2CO (g) + O2 (g)
If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium
concentrations of all species.
2CO2 (g) 2CO (g) + O2 (g)
initial partial pressure (mol/L) 0.4 0 0
change in partial pressure (mol/L) – 2x +2x +1x
equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x
[CO]2 [O 2 ] (2 x) 2 ( x)
Kp = = 2.0x10-6 =
[CO 2 ]2 (0.40 − 2 x) 2
check assumption
assume 2x << 0.40
2x 2(4.3x10-3)
= x 100 = 2.2%
0.40 0.40
(2x) 2 (x) assumpiton valid, less than 5%
2.0x10-6 = ⇒ x = 4.3x10-3 M
(0.40) 2

27. At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction
2CO2 (g) 2CO (g) + O2 (g)
If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium
concentrations of all species.
2CO2 (g) 2CO (g) + O2 (g)
initial partial pressure (mol/L) 0.4 0 0
change in partial pressure (mol/L) – 2x +2x +1x
equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x
[CO]2 [O 2 ]
Kp = = 2.0x10-6
[CO 2 ]2 [CO 2 ] = 0.40 - 2x =
(2 x) 2 ( x)
=
(0.40 − 2 x) 2 = 0.40 - 2(4.3x10-3 ) = 0.39M
x = 4.3x10-3M
-3
[CO] = 2x = 2(4.3 x 10 )
-3
[O 2 ] = x = 4.3 x 10 M = 8.6 x 10-3 M

28. Non-Equilibrium Conditions:
The Reaction Quotient (1)
φορωαρδ
 →
aA + bB ρεϖερσε cC + dD wrong arrow
← 

(PC ) (PD )
c d
(PC ) (PD )
c d
=Q
?
=K Q= K
(PA ) (PB )
a b
(PA ) (PB )
a b
K (the Equilibrium Constant) uses equilibrium
partial pressures
Q (the reaction quotient) uses prevailing partial
pressures, not necessarily at equilibrium

29. The Reaction Quotient (2)
φορωαρδ
arrow→

aA + bB ρεϖερσε cC + dD
wrong
←  
(PC )c (PD )d =Q
?
Q= K
(PC )c (PD )d =K
(PA )a (PB )b (PA )a (PB )b
If Q < K, reaction proceeds in a
forward direction (toward
products);
If Q > K, reaction proceeds in a backward direction
(toward reactants);
If Q = K, the reaction is in equilibrium.

30. The equilibrium constant for the reaction P4(g) ↔ 2 P2(g) is
1.39 at 400oC. Suppose that 2.75 mol of P4(g) and 1.08 mol of
P2(g) are mixed in a closed 25.0 L container at 400oC. Compute
Q(init) (the Q at the moment of mixing) and state the direction in
which the reaction proceeds.
?
(PP 2 )2
= K = 1.39 Q= K PV = nRT
(PP 4 )1
K = 1.39 @ 400oC; nP4(init) = 2.75 mol; nP2(init) = 1.08 mol
PP4(init) = nP4(init)RT/V = [(2.75mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L)
= 6.08 atm
PP2(init) = nP2(init)RT/V = [(1.08mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L)
= 2.39 atm
Q=

31. Henri Louis Le Châtelier (1850-1936)
Highlights
– 1884 Le Chatelier's Principle: A system in
equilibrium that is subjected to a stress
reacts in a way that counteracts the stress
– If a chemical system at equilibrium
experiences a change in concentration,
temperature or total pressure the
equilibrium will shift in order to minimize
that change.
– Industrial chemist involved with industrial
efficiency and labor-management
relations
Moments in a Life
– Le Chatelier was named "chevalier"
(knight) of the Légion d'honneur in 1887,
decoration established by Napoléon
Bonaparte in 1802.

32. Effects of External Stresses on Equilibria:
Le Châtelier’s Principle
A system in equilibrium that is subjected to a
stress reacts in a way that counteracts the stress.
Le Châtelier’s Principle provides a way to predict the
response of an equilibrium system to an external
perturbation, such as…
1. Effects of Adding or Removing Reactants or Products
2. Effects of Changing the Volume (or Pressure) of the System
3. Effects of Changing the Temperature

33. Effects of Adding or Removing Reactants or Products
PCl5(g) PCl3(g) + Cl2(g) K = 11.5 @ 300oC = Q
add extra PCl5(g)
add extra PCl3(g)
remove some PCl5(g)
remove some PCl3(g)
A system in equilibrium that is subjected to a stress reacts in a
way that counteracts the stress. In this case adding or
removing reactants or products

34. Effects of Changing the Volume of the System
PCl5(g) PCl3(g) + Cl2(g)
1 mole 1+1 = 2 moles
Let’s decrease the volume of the reaction container
Less room :: less amount (fewer moles)
Shifts reaction to restore equilibrium
Let’s increase the volume of the reaction container
More room :: more amount (greater moles)
Shifts reaction to restore equilibrium
A system in equilibrium that is subjected to a
stress reacts in a way that counteracts the stress.
In this case a change in volume

35. Volume Decreased Volume Increased
(Pressure Increased) (Pressure Decreased)
Equilibrium shift right Equilibrium Shifts left
V reactants > V products
(toward products) (toward reactants)
Equilibrium Shifts left Equilibrium shift right
V reactants < V products
(toward reactants) (toward products)
V reactants = V products Equilibrium not affected Equilibrium not affected
2 P2(g) P4 (g)
PCl5(g) PCl3(g) + Cl2(g)
CO (g) +H2O (g) CO2 (g) + H2 (g)
Boyles Law:
PV = Constant
A system in equilibrium that is subjected to a stress reacts in a way
that counteracts the stress. In this case a change in volume (or
pressure)

36. Effects of Changing the Temperature
Endothermic: heat is aborbed by a reaction
Reactants + heat gives Products
Exothermic: heat is liberated by a reaction
Reactants gives Products + heat
A system in equilibrium that is subjected to a stress reacts
in a way that counteracts the stress. In this case a change
in temperature

37. Effects of Changing the Temperature
Endothermic: absorption of heat by a reaction
Reactants + heat gives Products
Let’s increase the temperature of the reaction, what
direction does the equilibrium reaction shift
Let’s decrease the temperature of the reaction

38. Effects of Changing the Temperature
Exothermic: heat liberated by a reaction
Reactants ↔ Products + heat
Let’s increase the temperature of the reaction
Let’s decrease the temperature of the reaction
A system in equilibrium that is subjected to a stress reacts
in a way that counteracts the stress. In this case a change
in temperature

39. Temperature Raised Temperature Lowered
Endothermic
Equilibrium shift right Equilibrium Shifts left
Reaction
(toward products) (toward reactants)
(absorb heat)
Exothermic
Equilibrium Shifts left Equilibrium shift right
Reaction
(toward reactants) (toward products)
(liberate heat)
If a forward reaction is exothermic,
forward
aA + bB → cC + dD + Heat
Then the reverse reaction must be endothermic
reverse
cC + dD + Heat  aA + bB
→
φορωαρδ
 →
aA + bB ρεϖερσε cC + dD
← 


40. Driving Reactions to Completion/ Increasing Yield
Industrial Synthesis of Ammonia (Haber)
N2 (g) + 3H2 (g) ↔ 2NH3 (g)
Forward reaction is exothermic
What conditions do we need to increase the yield, i.e.,
produce more ammonia?
Volume Decreased Volume Increased
(Pressure Increased) (Pressure Decreased)
V reactants > Equilibrium shift Equilibrium Shifts left
right (toward
V products (toward products) reactants)
Temperature Raised Temperature Lowered
Exothermic Equilibrium Shifts left Equilibrium shift right
Reaction (toward (toward
(liberate heat) reactants) products)