Transcript Chemistry: McMurry and Fay, 5th Edition

John E. McMurry • Robert C. Fay
C H E M I S T R Y
Fifth Edition
Chapter 16
Thermodynamics: Entropy, Free Energy, and
Equilibrium
Lecture Notes
Alan D. Earhart
Southeast Community College • Lincoln, NE
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Spontaneous Processes
Spontaneous Process: A process that, once
started, proceeds on its own without a continuous
external influence.
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Chapter 16/2
Spontaneous Processes
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Chapter 16/3
Enthalpy, Entropy, and
Spontaneous Processes
State Function: A function or property whose value
depends only on the present state, or condition, of the
system, not on the path used to arrive at that state.
Enthalpy Change (DH): The heat change in a reaction
or process at constant pressure; DH = DE + PDV.
Entropy (S): The amount of molecular randomness in a
system.
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Chapter 16/4
Enthalpy, Entropy, and
Spontaneous Processes
Exothermic:
CH4(g) + 2O2(g)
DH° = -890.3 kJ
CO2(g) + 2H2O(l)
Endothermic:
H2O(s)
H2O(l)
H2O(l)
H2O(g)
N2O4(g)
NaCl(s)
H2O
DHfusion = +6.01 kJ
DHvap = +40.7 kJ
2NO2(g)
DH° = +57.1 kJ
Na1+(aq) + Cl1-(aq)
DH° = +3.88 kJ
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Chapter 16/5
Enthalpy, Entropy, and
Spontaneous Processes
DS = Sfinal - Sinitial
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Chapter 16/6
Enthalpy, Entropy, and
Spontaneous Processes
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Chapter 16/7
Enthalpy, Entropy, and
Spontaneous Processes
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Chapter 16/8
Entropy and Probability
S = k ln W
k = Boltzmann’s constant
= 1.38 x 10-23 J/K
W = The number of ways that the
state can be achieved.
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Chapter 16/9
Entropy and Temperature
Third Law of Thermodynamics: The entropy of a
perfectly ordered crystalline substance at 0 K is zero.
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Chapter 16/10
Entropy and Temperature
Standard Molar Entropies and
Standard Entropies of Reaction
Standard Molar Entropy (S°): The entropy of 1 mole
of a pure substance at 1 atm pressure and a specified
temperature.
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Chapter 16/12
Standard Molar Entropies and
Standard Entropies of Reaction
DS° = S°(products) - S°(reactants)
aA + bB
cC + dD
DS° = [cS°(C) + dS°(D)] - [aS°(A) + bS°(B)]
Products
Reactants
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Chapter 16/13
Standard Molar Entropies and
Standard Entropies of Reaction
Using standard entropies, calculate the standard entropy
change for the decomposition of N2O4.
N2O4(g)
2NO2(g)
DS° = 2 S°(NO2(g)) - S°(N2O4(g))
J
J
- (1 mol) 304.2
DS° = (2 mol) 240.0
K mol
K mol
DS° = 175.8 J/K
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Chapter 16/14
Entropy and the Second Law of
Thermodynamics
First Law of Thermodynamics: In any process, spontaneous
or nonspontaneous, the total energy of a system and its
surroundings is constant.
Second Law of Thermodynamics: In any spontaneous
process, the total entropy of a system and its surroundings
always increases.
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Chapter 16/15
Entropy and the Second Law of
Thermodynamics
DStotal = DSsystem + DSsurroundings
or
DStotal = DSsys + DSsurr
DStotal > 0
The reaction is spontaneous.
DStotal < 0
The reaction is nonspontaneous.
DStotal = 0
The reaction mixture is at equilibrium.
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Chapter 16/16
Entropy and the Second Law of
Thermodynamics
Dssurr a - DH
Dssurr a
1
T
Dssurr =
- DH
T
Free Energy
Free Energy: G = H - TS
DG = DH - TDS
Using: DStotal = DSsys + DSsurr
Dssurr =
- DH
T
DS = DSsys
DG = DH - TDS = -TDStotal
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Chapter 16/18
Free Energy
Using the second law and DG = DH - TDS = -TDStotal
DG < 0
The reaction is spontaneous.
DG > 0
The reaction is nonspontaneous.
DG = 0
The reaction mixture is at equilibrium.
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Chapter 16/19
Free Energy
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Chapter 16/20
Standard Free-Energy
Changes for Reactions
Thermodynamic Standard State: Most stable form of a
substance at 1 atm pressure and at a specified
temperature, usually 25 °C; 1 M concentration for all
substances in solution.
DG° = DH° - TDS°
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Chapter 16/21
Standard Free-Energy
Changes for Reactions
Calculate the standard free-energy change at 25 °C for
the Haber synthesis of ammonia using the given values
for the standard enthalpy and standard entropy changes:
N2(g) + 3H2(g)
2NH3(g)
DH° = -92.2 kJ
DS° = -198.7 J/K
DG° = DH° - TDS°
=
-92.2 kJ - 298 K
x
-198.7 J
K
x
1 kJ
1000 J
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= -33.0 kJ
Chapter 16/22
Standard Free Energies of
Formation
DG° = DG°f (products) - DG°f (reactants)
aA + bB
cC + dD
DG° = [cDG°f (C) + dDG°f (D)] - [aDG°f (A) + bDG°f (B)]
Products
Reactants
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Chapter 16/23
Standard Free Energies of
Formation
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Chapter 16/24
Standard Free Energies of
Formation
Using table values, calculate the standard free-energy
change at 25 °C for the reduction of iron(III) oxide with
carbon monoxide:
Fe2O3(s) + 3CO(g)
2Fe(s) + 3CO2(g)
DG° = [2 DG°f (Fe(s)) + 3 DG°f (CO2(g))]
- [1 DG°f (Fe2O3(s)) + 3 DG°f (CO(g))]
DG° = [(2 mol)(0 kJ/mol) + (3 mol)(-394.4 kJ/mol)]
- [(1 mol)(-742.2 kJ/mol) + (3 mol)(-137.2 kJ/mol)]
DG° = -29.4 kJ
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Chapter 16/25
Free Energy Changes and the
Reaction Mixture
DG = DG° + RT ln Q
DG = Free-energy change under nonstandard conditions.
For the Haber synthesis of ammonia:
2
P NH
N2(g) + 3H2(g)
2NH3(g)
Qp =
3
3
PN
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2
PH
2
Chapter 16/26
Free Energy Changes and the
Reaction Mixture
Calculate DG for the formation of ethylene (C2H4) from
carbon and hydrogen at 25 °C when the partial
pressures are 100 atm H2 and 0.10 atm C2H4.
2C(s) + 2H2(g)
C2H4(g)
Qp =
PC H
2 4
2
PH
2
Calculate ln Q:
0.10
ln
1002
= -11.51
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Chapter 16/27
Free Energy Changes and the
Reaction Mixture
Calculate DG:
DG = DG° + RT ln Q
= 68.1 kJ/mol + 8.314
J
K mol
(298 K)(-11.51)
1 kJ
1000 J
DG = 39.6 kJ/mol
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Chapter 16/28
Free Energy and Chemical
Equilibrium
DG = DG° + RT ln Q
•
When the reaction mixture is mostly reactants:
Q << 1
RT ln Q << 0
DG < 0
The total free energy decreases as the reaction
proceeds spontaneously in the forward direction.
•
When the reaction mixture is mostly products:
Q >> 1
RT ln Q >> 0
DG > 0
The total free energy decreases as the reaction
proceeds spontaneously in the reverse direction.
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Chapter 16/29
Free Energy and Chemical
Equilibrium
DG = DG° + RT ln Q
At equilibrium, DG = 0 and Q = K.
DG° = -RT ln K
Free Energy and Chemical
Equilibrium
Calculate Kp at 25 °C for the following reaction:
CaCO3(s)
CaO(s) + CO2(g)
Calculate DG°:
DG° = [DG°f (CaO(s)) + DG°f (CO2(g))] - [DG°f (CaCO3(s))]
= [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]
- [(1 mol)(-1128.8.0 kJ/mol)]
DG° = +130.4 kJ/mol
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Chapter 16/32
Free Energy and Chemical
Equilibrium
Calculate ln K:
DG° = -RT ln K
-130.4 kJ/mol
-DG°
ln K =
=
RT
8.314
J
K mol
(298 K)
1 kJ
1000 J
ln K = -52.63
Calculate K:
-52.63
K=e
= 1.4 x 10-23
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Chapter 16/33