Transcript Chem1310 McKelvy Lecture - Georgia Institute of Technology

OFB Chapter 7
Chemical Equilibrium
7-1 Chemical Reactions in Equilibrium
7-2 Calculating Equilibrium Constants
7-3 The Reaction Quotient
7-4 Calculation of Gas-Phase Equilibrium
7-5 The effect of External Stresses on Equilibria:
Le Châtelier’s Principle
7-6 Heterogeneous Equilibrium
7-7 Extraction and Separation Processes
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OFB Chapter 7
1
aA  bB
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forward



reverse
cC  dD
OFB Chapter 7
2
aA  bB
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
forward

reverse
OFB Chapter 7
cC  dD
~220 K (-53oC)
3
7-1 Chemical Reactions and Equilibrium
The equilibrium condition for every reaction can be
summed up 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
K = 0.03126
H2O(l) ↔ H2O(g) @ 30oC
K = 0.04187
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Temperature (oC)
15.0
17.0
19.0
21.0
23.0
25.0
30.0
OFB Chapter 7
50.0
Vapor Pressure (atm)
0.01683
0.01912
0.02168
0.02454
0.02772
0.03126
0.04187
4
0.1217
Chemical Reactions and Equilibrium
H2O (l) ↔ H2O (g)
PH2O
Pref
=K
Pref is numerically equal to 1
The convention in the book 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.
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OFB Chapter 7
5
Chemical Reactions and Equilibrium
2 NO2(g)  N2O4(g)
N2O4(g)  2 NO2 (g)
An equilibrium reaction
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OFB Chapter 7
6
Chemical Reactions and Equilibrium
2 NO2(g)  N2O4(g)
N2O4(g)  2 NO2 (g)
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T
T
P
P
oC
2 NO2(g)
↔
N
O
(g);
K
=
8.8
@
25
2
4
OFB Chapter 7
7
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 systems:
1.
2.
3.
4.
They display no macroscopic evidence of change.
They are reached through spontaneous processes.
They show a dynamic balance of forward and backward processes.
They are the same regardless of the direction from which they are approached.

aA  bB reverse cC  dD

forward
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OFB Chapter 7
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The Form of Equilibrium Expressions
aA  bB

forward

reverse
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
provided that all species are present as low-pressure gases.
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OFB Chapter 7
9
Exercise 7-1
Write equilibrium expressions for the
reactions defined by the following equations:
3 H2(g) + SO2(g)  H2S(g) + 2 H2O(g)
2 C2F5Cl(g) + 4 O2(g)  Cl2(g) + 4 CO2(g) + 5 F2(g)
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OFB Chapter 7
10
7-2 Calculating Equilibrium Constants
Example 7-2
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)
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4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
initial partial pressure (atm)
change in partial pressure (atm)
equilibrium partial pressure (atm)
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
K=
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(PN2O)2(PO2)3
(PNO2)4
(5.7)2(8.9)3
=
=
(2.4)4
OFB Chapter 7
12
Exercise 7-2
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)
change in partial pressure (atm)
equilibrium partial pressure (atm)
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2 GeO(g) + W2O6(g)  2 GeWO4(g)
initial partial pressure (atm)
change in partial pressure (atm)
equilibrium partial pressure (atm)
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
(PGeWO2)2
(0.980)2
K=
=
(PGeO)2(PW2O6) (0.020)2(0.510)
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Relationships Among the K’s of Related Reactions
Rule 1: The equilibrium constant for a reverse reaction is
always the reciprocal of the equilibrium constant for the
corresponding forward reaction.
#1
#2
2 H2(g) + O2(g) ↔2 H2O(g)
(PH2O)2
(PH2)2(PO2) = K1
2 H2O(g) ↔ 2 H2(g) + O2(g)
(PH2)2(PO2)
(PH2O)2
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OFB Chapter 7
K1 = 1/K2
= K2
15
Relationships Among the K’s of Related Reactions
Rule 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
H2(g) + ½ O2(g) ↔ H2O(g)
(PH2O)
(PH2)(PO2)½ = K3
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Rxn 3
K3 = K1½ = K1
OFB Chapter 7
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Relationships Among the K’s of Related Reactions
Rule 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)
Br2(g) + I2(g) ↔ 2 IBr(g)
(PBr2)(PCl2)
oC
=
K
=
0.45
@
25
2
1
(PBrCl)
(PIBr)2
oC
=
K
=
0.051
@
25
2
(PBr2) (PI2)
= K1K2
= (0.45)(0.051)
=0.023 @ 25oC
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OFB Chapter 7
17
7-3 The Reaction Quotient
aA  bB
PC  PD 
a
b
PA  PB 
c
d
K

forward

reverse
PC  PD 
a
b
PA  PB 
c
cC  dD
d
Q
Note that K (the Equilibrium Constant) uses
equilibrium partial pressures
Note that Q (the reaction quotient) uses prevailing
partial pressures, not necessarily at equilibrium
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OFB Chapter 7
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The Reaction Quotient
aA  bB
forward



reverse
PC c PD d
PA a PB b
cC  dD
Q
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 reactions stops because the reaction is in equilibrium.
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OFB Chapter 7
19
Exercise 7-4
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 
 K  1.39
1
 PP 4 
2
?
QK
PV  nRT
K = 1.39 @ 400oC; nP4(init) = 2.75 mol; nP2(init) = 1.08 mol
PP4(init) = nP4(init)RT/V
PP2(init) = nP2(init)RT/V
Q = (2.39)2/(6.08) = 0.939 0.939 < 1.39; Q < K
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7-4 Calculations of Gas-Phase Equilibria
Exercise 7-5
Carbon monoxide reacts with water to give hydrogen:
CO (g) +H2O (g) ↔ CO2 (g) + H2 (g)
At 900 K, the equilibrium constant for this reaction, the socalled shift reaction, equals 0.64. suppose the partial
pressures of three gases at equilibrium at 900 K are
PCO = 2.00 atm, PCO2 = 0.80 atm, and PH2 =0.48 atm
Calculate the partial pressure of water under these conditions.
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OFB Chapter 7
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7-4 Calculations of Gas-Phase Equilibria
Exercise 7-5
Carbon monoxide reacts with water to give hydrogen:
CO (g) +H2O (g) ↔ CO2 (g) + H2 (g)
At 900 K, the equilibrium constant for this reaction, the so-called shift reaction, equals 0.64.
suppose the partial pressures of three gases at equilibrium at 900 K are
PCO = 2.00 atm, PCO2 = 0.80 atm, and PH2 =0.48 atm
Calculate the partial pressure of water under these conditions.
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OFB Chapter 7
22
• Solving quadratic equations
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OFB Chapter 7
23
7-5 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…
Effects of Adding or Removing Reactants or Products
Effects of Changing the Volume of the System
Effects of Changing the Temperature
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Effects of Adding or Removing Reactants or Products
PCl5(g) ↔ PCl3(g) + Cl2(g)
add extra PCl5(g)
add extra PCl3(g)
remove some PCl5(g)
remove some PCl3(g)
P  P 
P 
1
PCl3
1
Cl 2
1
Q
PCl5
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Effects of Changing the Volume of the System
PCl5(g) ↔ PCl3(g) + Cl2(g)
Let’s decrease the volume of the reaction container
Let’s increase the volume of the reaction container
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Volume Decreased
Volume Increased
(Pressure Increased)
(Pressure Decreased)
V reactants > V products
Equilibrium shift right
(toward products)
Equilibrium Shifts left
(toward reactants)
V reactants < 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)
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Effects of Changing the Temperature
PCl5(g)  PCl3(g) + Cl2(g)
Let’s increase the temperature of the reaction
Let’s decrease the temperature of the reaction
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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
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Effects of Changing the Temperature
PCl5(g)  PCl3(g) + Cl2(g)
K = 11.5 @ 300oC = Q
Let’s increase the temperature of the reaction
Let’s decrease the temperature of the reaction
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Driving Reactions to Completion
Industrial Synthesis of Ammonia
N2 (g) + 3H2 (g) ↔ 2NH3 (g)
Forward reaction exothermic
V reactants > V products
Exothermic Reaction
(liberate heat)
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Volume Decreased
Volume Increased
(Pressure Increased)
(Pressure Decreased)
Equilibrium shift right
(toward products)
Equilibrium Shifts left
(toward reactants)
Temperature Raised
Temperature Lowered
Equilibrium Shifts left
(toward reactants)
Equilibrium shift right
(toward products)
OFB Chapter 7
31
Review: Chemical Equilibrium
aA  bB
PC  PD 
a
b
PA  PB 
c
d
K

forward

reverse
PC  PD 
a
b
PA  PB 
c
cC  dD
d
Q
?
QK
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
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OFB Chapter 7
32
Exercise 7-10
State the effect of an increase in temperature and
also of a decrease in volume on the equilibrium
yield of the products in each of the following
reactions.
a) CH3OCH3 (g) + H2O (g) ↔ 2 CH4 (g) + O2 (g)
Increase Temp?
Decrease Volume?
b) H2O (g) + CO (g) ↔ HCOOH (g)
Increase Temp?
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Decrease Volume?
OFB Chapter 7
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Heterogeneous Equilibrium
Solids
CaCO3(s) ↔ CaO(s) + CO2(g)
Liquids
H2O(l) ↔ H2O(g)
Dissolved
species
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I2(s) ↔ I2(aq)
OFB Chapter 7
34
Law of Mass Action
1. Gases enter equilibrium expressions as partial pressures, in
atmospheres. E.g., PCO2
2. Dissolved species enter as concentrations, in 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.,
I 2 ( s)  I 2 (aq)
[ I 2 (aq)] [ I 2 (aq)]
K

 [I2 ]
[ I 2 ( s)]
1
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OFB Chapter 7
35
Law of Mass Action
4. 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
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OFB Chapter 7
36
Exercise 7-11
Write equilibrium-constant equations for the following
Equilibria:
(a) Si3N4(s) + 4 O2(g)  3 SiO2(s) + 2 N2O(g)
(b) O2(g) + 2 H2O(l)  2 H2O2(aq)
(c) CaH2(s) + 2 C2H5OH(l)  Ca(OC2H5)2(s) + 2 H2(g)
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OFB Chapter 7
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Exercise 7-12
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.
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OFB Chapter 7
38
Exercise 7-12
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.
K
PNi(CO) 4
4

PNi(CO) 4
(PCO ) [Ni(s)] (PCO ) 4 (1)
P CO (atm) P Ni(CO)4 (atm)
initial partial pressure (atm)
change in partial pressure (atm)
equilibrium partial pressure (atm)
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OFB Chapter 7
39
7-7 Extraction and Separation Processes
• Extraction
– Partitioning of a solute between two immiscible
solvents
• Chromatography
– Solute is partitioned between a mobile phase
and a stationary phase
• Column Chromatography
• Gas-liquid Chromatography
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OFB Chapter 7
40