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Unformatted text preview: Chapter 11 Objectives
11.1. Define equilibrium and explain on a microscopic level why
equilibrium is observed. 11.2. Introduce the equilibrium constant as a way of defining the extent
of equilibrium. Include a discussion of the use of pressure,
concentration, and activity when expressing equilibrium. 11.3. Describe the relationships which exist among equilibrium
constants for heterogeneous reactions. 11.4. Provide examples of how to solving equilibrium problems.
11.5. Outline the fundamentals of Le Chatelier's Principle 10/6/2009 Zumdahl Chapter 2 H2O (g) + CO (g) 10/6/2009 1 H2 (g) + CO2 (g) Zumdahl Chapter 2 2 1 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. Four fundamental characteristics of equilibrium states in isolated
isolated
systems:
1. They display no macroscopic evidence of change.
2. They are reached through spontaneous processes.
3. They show a dynamic balance of forward and backward
processes.
4.
10/6/2009 Zumdahl Chapter 2 3 The Form of Equilibrium Expressions
Consider the equilibrium reaction where 2 molecules of NO2
combine reversibly to form N2O4 : 2 NO2 ↔ N2O4 (1) Rate of forward reaction (2 NO2 → N2O4) =
Rate of reverse reaction (N2O4 → 2 NO2) =
At equilibrium, the concentrations have adjusted to make the rate
rate
of the forward reaction equal to the rate of the reverse reaction:
reaction: (2)
Rearranging eqt. (2) allows us to define a Pseudo Equilibrium
eqt.
Constant (K′) as follows: K′ =
10/6/2009 Zumdahl Chapter 2 = cN2O4 / cNO22
4 2 “Legitimate” Equilibrium Constants are Unitless
2 NO2 ↔ N2O4
= K′ cref
K′ K = We can also define an equilibrium constant in terms of the partial pressures of
partial
reactants and products: KP =
Typically, the reference concentration (pressure) is chosen to be 1 of whatever
be
unit is being used, i.e., 1 mol L1, 1 atm, 1 bar, … It is often inferred.
atm,
inferred. An equilibrium constant is a unitless number that has no
meaning unless the balanced chemical equation and the
reference pressure or concentration are also specified 10/6/2009 Zumdahl Chapter 2 5 Law of Mass Action
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
if pressures if concentrations (PC )c (PD )d
(PA )a (PB )b [C]c [D]d
[A]a [B]b = KP =K Reference pressuresZumdahl Chapter 2
or concentrations are inferred 10/6/2009 6 3 From Text, p. 506:
“The law of mass action is widely applicable. It correctly describes
describes
equilibrium behavior of all chemical reaction systems whether they occur in
they
solution or in the gas phase. Although, as we will see later, corrections for nonnonideal behavior must be applied in certain cases, such as for concentrated
cases,
aqueous solutions and for gases at high pressures, the law of mass action
mass
provides a remarkably accurate description of all types of chemical equilibria.
chemical equilibria. When dealing with nonideal behavior, activity replaces concentration
nonor partial pressure. Gas phase activity is often referred to as fugacity.
fugacity. 10/6/2009 Zumdahl Chapter 2 7 Sample Problem
Write equilibrium expressions for the reactions defined by the
following equations. Reference pressures can be inferred.
inferred. 3 H2(g) + SO2(g) ↔ H2S(g) + 2 H2O(g)
= KP
2 C2F5Cl(g) + 4 O2(g) ↔ Cl2(g) + 4 CO2(g) + 5 F2(g)
= KP
10/6/2009 Zumdahl Chapter 2 8 4 Heterogeneous
Equilibria
Solids CaCO3(s) ↔ CaO(s) + CO2(g) Liquids H2O(l) ↔ H2O(g) KP is independent of the
amounts of CaCO3(s) or CaO(s)
CaO(s) Dissolved
I2(s) ↔ I2(aq)
species
10/6/2009 Zumdahl Chapter 2 9 Law of Mass Action Extended to Heterogeneous Equilibria
1. Gases enter equilibrium expressions as partial pressures,
e.g., PCO2
2. Dissolved species enter as concentrations, 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
aq)
aq) I 2 ( s ) ↔ I 2 ( aq )
K =
10/6/2009 [ I 2 ( aq )] [ I 2 ( aq )]
=
= [ I 2 ( aq )]
[ I 2 ( s )]
1
Zumdahl Chapter 2 10 5 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
forward
reaction. #1 2 H2(g) + O2(g) ↔ 2 H2O(g)
O(g (PH2O)2
(PH2)2(PO2) = K1 #2 2 H2O(g) ↔ 2 H2(g) + O2(g)
O(g (PH2)2(PO2)
(PH2O)2 = K2 10/6/2009 Zumdahl Chapter 2 11 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. #1
#3 2 H2(g) + O2(g) ↔ 2 H2O(g) Rxn 1
O(g (PH2O)2
(PH2)2(PO2) = K1 H2(g) + ½ O2(g) ↔ H2O(g) Rxn 3 = Rxn 1 times 1/2
O(g
(PH2O)
= K3
(PH2)(PO2)½ 10/6/2009 Zumdahl Chapter 2 12 6 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
constant
associated with the new equation.
(PBr2)(PCl2) 2 BrCl (g) ↔ Br2 (g) + Cl2 (g)
2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g) (PBr2)(PCl2)
(PBrCl)2 X = K1 = 0.45 @ 25oC (PBrCl)2 Br2 (g) + I2 (g) ↔ 2 IBr (g) (PIBr)2
o
(PBr2) (PI2) = K2 = 0.051 @ 25 C (PIBr)2
(PBr2) (PI2) =
= (0.45)(0.051) = 0.023 @ 25oC 10/6/2009 Zumdahl Chapter 2 13 The Reaction Quotient
aA + bB (PC )c (PD )d
(PA )a (PB )b forward
⎯⎯ ⎯→
cC + dD
reverse
←⎯ ⎯
⎯ (PC )c (PD )d
(PA )a (PB )b =K
P = QP Note that KP (the Equilibrium Constant) uses
equilibrium partial pressures
Note that QP (the reaction quotient) uses prevailing
partial pressures, not necessarily at equilibrium
10/6/2009 Zumdahl Chapter 2 14 7 The Reaction Quotient (con’t)
forward
⎯⎯ ⎯→
aA + bB reverse cC + dD
←⎯ ⎯
⎯ (PC )c (PD )d
(PA )a (PB )b =Q P (PC )c (PD )d
(PA )a (PB )b =K
P If Q < K, reaction proceeds in a
K,
If Q > K, reaction proceeds in a
K,
If Q = K, the reaction is at equilibrium
K,
10/6/2009 Zumdahl Chapter 2 15 The equilibrium constant for the reaction P4(g) ↔ 2 P2(g) is 1.39 at 400
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.
oC . ( PP 2 )2
= K = 1.39
( PP 4 )1 P PV = nRT KP = 1.39 @ 400oC
nP4t=0 = 2.75 mol
nP2t=0 = 1.08 mol PP4t=0 = nP4t=0 RT/V = [(2.75 mol)(0.0821 atm L mol1 K1)(673 K)] ÷ (25.0 L) = 6.08 atm
PP2t=0 = nP2t=0 RT/V = [(1.08 mol)(0.0821 atm L mol1 K1)(673 K)] ÷ (25.0 L) = 2.39 atm QP = (2.39)2 / (6.08) = 0.939 (assuming Pref = 1 atm)
atm) 10/6/2009 Zumdahl Chapter 2 16 8 Sample Problem
Consider the equilibrium 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
O(g
The three gases are introduced into a container at partial pressures of 3.6 atm
pressures
(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. Evaluate the equilibrium constant of the reaction at this
temperature, assuming that no competing reactions occur.
4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
O(g
initial partial pressure (atm)
change in partial pressure (atm)
equilibrium partial pressure (atm) 10/6/2009 Zumdahl Chapter 2 17 Evaluate the equilibrium constant of the reaction at this temperature, assuming
temperature,
that no competing reactions occur.
4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
O(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 KP = 10/6/2009 (PN2O)2 (PO2)3
(PNO2)4 = Zumdahl Chapter 2 5.1 + 2x 8.0 + 3x = 690 = KP 18 9 Sample Problem
The compound GeWO4(g) forms at high temperature in the reaction 2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g)
GeO(g
Some GeO(g) and W2O6(g) are mixed. Before they start to react, their partial
GeO(g
pressures both equal 1.000 atm. After their reaction at constant temperature
constant
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.
reaction.
2 GeO (g) + W2O6 (g) ↔ 2 GeWO4(g)
(g
1.000
1.000
0 initial partial pressure (atm)
change in partial pressure (atm) – 2x –x equilibrium partial pressure (atm) 1.000 – 2x 1.000 – x 10/6/2009 +2x
+2x
0.980 Zumdahl Chapter 2 19 (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)
GeO(g
1.000
1.000
0 initial partial pressure (atm)
change in partial pressure (atm) –2x equilibrium partial pressure (atm) 1.000 – 2x 0 + 2x = 0.980 atm GeWO4
x = 0.490 atm KP = (PGeWO4)2
(PGeO)2 (PW2O6) 10/6/2009 = –x
1.000 – x +2x
0.980 1.000 – 2(0.490) = 0.020 atm GeO
1.000 – 0.490 = 0.510 atm W2O6
(0.980)2
= 4700 = KP
(0.020)2 (0.510)
Zumdahl Chapter 2 20 10 Sample Problem
A vessel holds pure CO(g) at a pressure of 1.282 atm and a temperature of
CO(g)
354K. A quantity of nickel is added, and the partial pressure of CO(g) drops to
of CO(g)
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.
354K. Construct an “ICE” table
ICE” PCO (atm) PNi(CO)4 (atm) initial partial pressure (atm)
change in partial pressure (atm)
equilibrium partial pressure (atm) => At equil., Pco =
equil., KP = PNi(CO) 1.282
4x
0.709 4 (P CO ) 4 [Ni(s)] 10/6/2009 = PNi(CO) 4 (P CO ) 4 (1) 0
+1x
+1x
x x = = KP = 0.143 / (0.709)4 = Zumdahl Chapter 2 21 Sample Problem
At a particular temperature, K = 2.0 x 106 mol/L for the reaction 2CO2(g) 2CO(g) + O2(g)
2CO(g If 2.0 mol CO2 is initially placed into a 5.0L vessel, calculate the equilibrium
5.0concentrations of all species.
2CO2(g)
2CO(g) + O2(g)
2CO(g
initial partial pressure (mol/L)
0.4
change in partial pressure (mol/L)
– 2x
equilibrium partial pressure (mol/L) 0.4 2x
Kp =
K 0
+2x
+2x
2x 0
+1x
1x [CO]2 [O 2 ]
(2 x)2 ( x)
= 2.0x106 =
2
[CO 2 ]
(0.40 − 2 x) 2 check assumption
103
assume 2x << 0.40 <――――――――> 2x = 2(4.3x103)) x 100 = 2.2%
――――――――>
0.40
0.40
assumpiton valid, less than 5%
assumption
(2x) 2 (x)
2.0x106 =
⇒ x = 4.3x103M
(0.40) 2
10/6/2009 Zumdahl Chapter 2 (cref = 1 M) 22 11 At a particular temperature, K = 2.0 x 106 mol/L for the reaction 2CO2 (g) 2CO (g) + O2 (g)
(g If 2.0 mol CO2 is initially placed into a 5.0L vessel, calculate the equilibrium
5.0concentrations of all species. 2CO2 (g) initial partial pressure (mol/L)
0.4
change in partial pressure (mol/L)
– 2x
equilibrium partial pressure (mol/L) 0.4 2x
Kp =
= 2CO (g) + O2 (g)
0
+2x
+2x
2x 0
+1x
1x [CO]2 [O 2 ]
= 2.0x106
[CO 2 ]2 (2 x) 2 ( x)
(0.40 − 2 x) 2 x = 4.3x103 M [O 2 ] = x = 4.3 x 103M
10/6/2009 Zumdahl Chapter 2 23 The Person Behind the Science Henri Louis Le Châtelier (18501936)
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 labormanagement relations
labor Moments in a Life
– Le Chatelier was named "chevalier"
(knight) of the Légion d'honneur in 1887,
decoration established by Napoléon
Napolé
Bonaparte in 1802. Zumdahl Chapter 2
10/6/2009 24 12 Effects of External Stresses on Equilibria:
Le Châtelier’s Principle Le Châtelier’s Principle provides a way to predict the response of
Châtelier’
an equilibrium system to an external perturbation, such as…
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
10/6/2009 Zumdahl Chapter 2 25 Effect of Adding or Removing Reactants or Products
PCl5(g) ↔ PCl3(g) + Cl2(g)
Q = [PCl3] [Cl2] / [PCl5]
K = [PCl3]eq [Cl2]eq / [PCl5]eq = Qeq
Suppose that a state of equilibrium initially exists, which is then perturbed by
then
addition or removal of a reactant or product. How does the system respond? add extra PCl5(g)
add extra PCl3(g)
remove some Cl2(g)
10/6/2009 Zumdahl Chapter 2 26 13 Effect of Changing the System Pressure/Volume
Volume Decreased Volume Increased (Pressure Increased) (Pressure Decreased) Preactants > Pproducts Equilibrium shift right
(toward products) Equilibrium Shifts left
(toward reactants) Preactants < Pproducts Equilibrium Shifts left
(toward reactants) Equilibrium shift right
(toward products) Preactants = Pproducts Equilibrium not affected Equilibrium not affected Boyles Law:
PV = Const. 2 P2(g) ↔ P4(g)
PCl5(g) ↔ PCl3(g) + Cl2(g)
CO(g) + H2O(g) ↔ CO2(g) + H2(g) 10/6/2009 Zumdahl Chapter 2 27 Effect of Changing the Temperature If a reaction is endothermic in the forward direction and we increase the
temperature, the equilibrium will shift toward products because some heat is used
temperature,
up to convert reactants to products, thus partially counteracting the temperature
counteracting
increase.
If a reaction is endothermic in the forward direction and we decrease the
temperature, the equilibrium will shift toward reactants because some heat is
temperature,
liberated when products are converted to reactants, thus partially counteracting
partially
the temperature decrease. 10/6/2009 Zumdahl Chapter 2 28 14 Driving Reactions to Completion
Industrial Synthesis of Ammonia (Haber Process) N2(g) + 3H2(g) ↔ 2NH3(g)
(Forward reaction is exothermic)
Forward
exothermic
What conditions should be adopted to increase
the yield of the product ammonia?
Volume Decreased
(Pressure Increased) Volume Increased
(Pressure Decreased)
Equilibrium Shifts left
(toward reactants) Temperature Raised Temperature Lowered Equilibrium Shifts left
(toward reactants)
10/6/2009 Zumdahl Chapter 2 29 Equilibria Involving Real Gases
N2(g) + 3 H2(g) ↔ 2 NH3(g) KP = PNH32 / (PN2 PH23) KP is an accurate equilibrium constant only in the limit where all gases behave
ideally. At high pressures, such as those employed in the industrial synthesis
industrial
of ammonia, nonideal behavior of gases must be taken into account:
non The φ’s are called fugacity coefficients and represent a measure of deviations
from ideal gas behavior for each gas. They are readily determinable
determinable
experimentally for pure gases, but are very difficult to determine in gas
determine
mixtures because they must be evaluated at the equilibrium composition which,
typically is not known if the φ’s are not known. The liquid phase analog of φ is
known.
the activity coefficient, γ.
coefficient,
10/6/2009 Zumdahl Chapter 2 30 15 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 10/6/2009 Zumdahl Chapter 2 31 16 ...
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This note was uploaded on 10/06/2009 for the course CHEM 1101 taught by Professor Bottomley during the Fall '08 term at Georgia Tech.
 Fall '08
 Bottomley
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