Chapter 11 Chemical Equilibrium

1
Chapter 11Chemical 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.
Temperature (oC) Vapor Pressure (atm)
15.0 0.01683 17.0 0.01912 19.0 0.02
168 21.0 0.02454 23.0 0.02772 25.0 0.03126
30.0 0.04187 50.0 0.1217
H2O(l) H2O(g) _at_ 25oC
K 0.03126
H2O(l) H2O(g) _at_ 30oC
K 0.04187
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 dD
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
10
Activities
The concept of Activity (i-th component) ai
Pi / P reference
H2O (l) H2O (g) Kp P H2O
_at_ 25oC
Kp 0.03126 atm
Pref is numerically equal to 1
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)
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 Solids
CaCO3(s) CaO(s) CO2(g) KPCO2
K is independent of the amounts of CaCO3(s) or
CaO(s)
15
Heterogeneous Equilibrium
Liquids Solutions
H2O(l) H2O(g) KPH2O
I2(s) I2(aq) KI2
16
Relationships Among the Ks of Related Reactions
1 The equilibrium constant for a reverse
reaction is always the reciprocal of the
equilibrium constant for the corresponding
forward reaction.
aA bB cC dD
cC dD aA bB
versus
2 H2 (g) O2 (g) 2 H2O (g)
1
K1 1/K2
2 H2O (g) 2 H2 (g) O2 (g)
2
17
Relationships Among the Ks 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.
2 H2 (g) O2 (g) 2 H2O (g) Rxn 1
1
18
Relationships Among the Ks 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)
wrong arrow
Br2 (g) I2 (g) ? 2 IBr (g)
wrong arrow
2 BrCl (g) I2 (g) ? 2 IBr (g) Cl2(g)
K1K2
wrong arrow
(0.45)(0.051) 0.023 _at_ 25oC
X
K1K2 K3
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.
22

1. determine the equilibrium partial pressures of
   GeO and W2O6, and
2. determine the equilibrium constant for the
   reaction.

0 2x 0.980 atm GeWO4 x 0.490 atm
1.000 2(0.490) 0.020 atm GeO
1.000 0.490 0.510 atm W2O6
(PGeWO4)2
K

(PGeO)2(PW2O6)
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.
Ni (s) 4CO (g) ? Ni(CO)4 (g)
P CO (atm) P Ni(CO)4 (atm)
Construct an ICE table
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
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
28
Non-Equilibrium Conditions The Reaction
Quotient (1)
wrong arrow
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)
wrong arrow
If Q lt K, reaction proceeds in a forward
direction (toward products)
If Q gt 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.
K 1.39 _at_ 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)(273.15400oC)/(25.0L)
6.08 atm
PP2(init) nP2(init)RT/V
(1.08mol)(0.08206 atm L mol-1
K-1)(273.15400oC)/(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
Châteliers Principle
A system in equilibrium that is subjected to a
stress reacts in a way that counteracts the
stress.
Le Châteliers 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 _at_ 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
11 2 moles
Lets decrease the volume of the reaction
container
Less room less amount (fewer moles)
Shifts reaction to restore
equilibrium
Lets 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)
V reactants gt V products Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants)
V reactants lt V products Equilibrium Shifts left (toward reactants) Equilibrium shift right (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
Lets increase the temperature of the reaction,
what direction does the equilibrium reaction shift
Lets decrease the temperature of the reaction
38
Effects of Changing the Temperature
Exothermic heat liberated by a
reaction Reactants ? Products heat
Lets increase the temperature of the reaction
Lets 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 Reaction (absorb heat) Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants)
Exothermic Reaction (liberate heat) Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products)
If a forward reaction is exothermic, Then the
reverse reaction must be endothermic
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 gt V products Equilibrium shift right (toward products) Equilibrium Shifts left (toward reactants)
  Temperature Raised Temperature Lowered
Exothermic Reaction (liberate heat) Equilibrium Shifts left (toward reactants) Equilibrium shift right (toward products)