CHEMISTRY
The Molecular Nature of Matter and Change
Third Edition
Chapter 20
Lecture Outlines*
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Chapter 20
Thermodynamics:
Entropy, Free Energy and the
Direction of Chemical Reactions
20-2
Thermodynamics: Entropy, Free Energy, and the
Direction of Chemical Reactions
20.1 The Second Law of Thermodynamics: Predicting
Spontaneous Change
20.2 Calculating the Change in Entropy of a Reaction
20.3 Entropy, Free Energy, and Work
20.4 Free Energy, Equilibrium, and Reaction Direction
20-3
Limitations of the First Law of Thermodynamics
Esystem = Ek + Ep
DEsystem = heat + work
DEsystem = q + w
Euniverse = Esystem + Esurroundings
DEsystem = -DEsurroundings
The total energy-mass of the universe is constant.
But, this does not tell us anything about the
direction of change in the universe.
20-4
Figure 20.1
A spontaneous endothermic chemical reaction
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water
Ba(OH)2.8H2O(s) + 2NH4NO3(s)
Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
DH0rxn = + 62.3 kJ
20-5
The Concept of Entropy (S)
Entropy refers to the state of order.
A change in order is a change in the number of ways
of arranging the particles, and it is a key factor in
determining the direction of a spontaneous process.
more order
solid
more order
crystal + liquid
more order
crystal + crystal
20-6
less order
liquid
gas
less order
ions in solution
less order
gases + ions in solution
Figure 20.2
20-7
The number of ways to arrange a deck of playing cards
Figure 20.3
Spontaneous expansion of a gas
stopcock
closed
1 atm
evacuated
stopcock
opened
0.5 atm
20-8
0.5 atm
1877 Ludwig Boltzman
S = k ln W
S = entropy
W = number of ways of arranging the components of a system
k = Boltzman constant
•A system with relatively few equivalent ways to arrange its
components (smaller W) has relatively less disorder and low entropy.
•A system with many equivalent ways to arrange its components
(larger W) has relatively more disorder and high entropy.
2nd law of thermodynamics
The universe is going to a higher state of entropy
DSuniverse = DSsystem + DSsurroundings > 0
20-9
Figure 20.4
3rd Law of
thermodynamics
A perfect crystal has
zero entropy at a
temperature of
absolute zero.
Ssystem = 0 at 0 K
20-10
Random motion in a crystal
Predicting Relative S0 Values of a System
1. Temperature changes
S0 increases as the temperature rises.
2. Physical states and phase changes
S0 increases as a more ordered phase changes to a less ordered phase.
3. Dissolution of a solid or liquid
S0 of a dissolved solid or liquid is usually greater than the S0 of
the pure solute. However, the extent depends upon the nature of
the solute and solvent.
4. Dissolution of a gas
A gas becomes more ordered when it dissolves in a liquid or solid.
5. Atomic size or molecular complexity
In similar substances, increases in mass relate directly to entropy.
In allotropic substances (e.g. O2 & O3), increases in complexity
(e.g. bond flexibility) relate directly to entropy.
20-11
Figure 20.5
20-12
The increase in entropy from solid to liquid to gas
Figure 20.6
The entropy change accompanying the dissolution of a salt
pure solid
MIX
pure liquid
solution
20-13
Figure 20.7
The small increase in entropy when ethanol dissolves in water
Ethanol
20-14
Water
Solution
of ethanol
and water
Figure 20.8
The large decrease in entropy when a gas dissolves in a liquid
O2 gas
20-15
Figure 20.9
Entropy and vibrational motion
NO
NO2
20-16
N 2 O4
Sample Problem 20.1:
Predicting Relative Entropy Values
PROBLEM: Choose the member with the higher entropy in each of the
following pairs, and justify your choice [assume constant
temperature, except in part (e)]:
(a) 1mol of SO2(g) or 1mol of SO3(g)
(b) 1mol of CO2(s) or 1mol of CO2(g)
(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)
(d) 1mol of KBr(s) or 1mol of KBr(aq)
(e) Seawater in midwinter at 20C or in midsummer at 230C
(f) 1mol of CF4(g) or 1mol of CCl4(g)
PLAN: In general less ordered systems have higher entropy than ordered
systems and entropy increases with an increase in temperature.
SOLUTION:
(a) 1mol of SO3(g) - more atoms
(d) 1mol of KBr(aq) - solution > solid
(b) 1mol of CO2(g) - gas > solid
(e) 230C - higher temperature
(c) 3mol of O2(g) - larger #mols
(f) CCl4 - larger mass
20-17
Sample Problem 20.4: Calculating DG0 from Enthalpy and Entropy Values ( 1of 2)
PROBLEM:
Potassium chlorate, one of the common oxidizing agents in
explosives, fireworks, and matchheads, undergoes a solidstate redox reaction when heated. In this reaction, note that
the oxidation number of Cl in the reactant is higher in one of
the products and lower in the other (disproportionation):
+5
+7
-1
D
4KClO3(s)
3KClO4(s) + KCl(s)
Use DH0f and S0 values to calculate DG0sys (DG0rxn) at 250C for this reaction.
PLAN:
Use Appendix B values for thermodynamic entities; place them into
the Gibbs Free Energy equation and solve.
SOLUTION:
DH0rxn = S mDHf0products - S nDHf0reactants
DH0rxn = (3mol)(-432.8kJ/mol) + (1mol)(-436.7kJ/mol) (4mol)(-397.7kJ/mol)
DH0rxn = -144kJ
20-18
Sample Problem 20.4:
Calculating DG0 from Enthalpy and Entropy Values
continued
DS0rxn = S mS0products - S nS0reactants
DS0rxn = (3mol)(151J/mol*K) + (1mol)(82.6J/mol*K) (4mol)(143.1J/mol*K)
DS0rxn = -36.8J/K
DG0rxn = DH0rxn - T DS0rxn
DG0rxn = -144kJ - (298K)(-36.8J/K)(kJ/103J)
DG0rxn = -133kJ
20-19
Sample Problem 20.2:
PROBLEM:
Calculating the Standard Entropy of Reaction, DS0rxn
Calculate DS0rxn for the combustion of 1mol of propane at 250C.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
PLAN: Use summation equations. It is obvious that entropy is being lost
because the reaction goes from 6mols of gas to 3mols of gas.
SOLUTION:
Find standard entropy values in the Appendix or other table.
DS = [(3mol)(S0 CO2) + (4mol)(S0 H2O)] - [(1mol)(S0 C3H8) + (5mol)(S0 O2)]
DS = [(3mol)(213.7J/mol*K) + (4mol)(69.9J/mol*K)] - [(1mol)(269.9J/mol*K) +
(5mol)(205.0J/mol*K)]
DS = - 374 J/K
20-20
Figure B20.1
Lavoisier studying human respiration as a form of combustion
20-21
Figure B20.2
20-22
A whole-body calorimeter
Figure 20.10
Components of DS0universe for spontaneous reactions
DSuniverse = DSsystem + DSsurroundings
endothermic
exothermic
system becomes more disordered
system becomes more disordered
exothermic
system becomes more ordered
20-23
Sample Problem 20.3:
PROBLEM:
Determining Reaction Spontaneity
At 298K, the formation of ammonia has a negative DS0sys;
N2(g) + 3H2(g)
2NH3(g)
DS0sys = -197J/K
Calculate DS0rxn, and state whether the reaction occurs
spontaneously at this temperature.
PLAN: DS0universe must be > 0 in order for this reaction to be spontaneous, so
DS0surroundings must be > 197J/K. To find DS0surr, first find DHsys; DHsys =
DHrxn which can be calculated using DH0f values from tables. DS0universe
= DS0surr + DS0sys.
SOLUTION:
DH0rx = [(2mol)(DH0fNH3)] - [(1mol)(DH0fN2) + (3mol)(DH0fH2)]
DH0rx = -91.8kJ
DS0surr = -DH0sys/T = -(-91.8x103J/298K) = 308J/K
DS0universe = DS0surr + DS0sys = 308J/K + (-197J/K) = 111J/K
DS0universe > 0 so the reaction is spontaneous.
20-24
Useful Thermodynamic Relationships
Different Ways to Calculate DG0rxn
0 = Std conds: 25oC, 1 atm, 1 M solutions
(Eqn 1)
DG0rxn = DH0rxn - TDS0rxn
(Eqn 2)
DG0rxn = S mDGf0products - S nDGf0reactants
(Eqn 3) DG0 = - RT lnKeq
R = 8.314 J/mol.K
When not at standard conditions
(Eqn 4)
DGrxn = DG0rxn + RT ln Q
(Q = Reaction Quotient)
Use these equations with equation #1
DS0rxn = S mS0products - S nS0reactants
DH0rxn = S mDHf0products - S nDHf0reactants
20-25
Sample Problem 20.6:
PROBLEM:
Determining the Effect of Temperature on DG0 (1 of 2)
An important reaction in the production of sulfuric acid is
the oxidation of SO2(g) to SO3(g):
2SO2(g) + O2(g)
2SO3(g)
At 298K, DG0 = -141.6kJ; DH0 = -198.4kJ; and DS0 = -187.9J/K
(a) Use the data to decide if this reaction is spontaneous at 250C, and predict
how DG0 will change with increasing T.
(b) Assuming DH0 and DS0 are constant with increasing T, is the reaction
spontaneous at 900.0C?
PLAN:
The sign of DG0 tells us whether the reaction is spontaneous
and the signs of DH0 and DS0 will be indicative of the T effect.
Use the Gibbs free energy equation for part (b).
SOLUTION: (a) The reaction is spontaneous at 250C because DG0 is (-).
Since DH0 is (-) but DS0 is also (-), DG0 will become less
spontaneous as the temperature increases.
20-26
Sample Problem 20.6:
Determining the Effect of Temperature on DG0 (2 of 2)
continued
(b) DG0rxn = DH0rxn - T DS0rxn
DG0rxn = -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/103J)
DG0rxn = 22.0 kJ; the reaction will be nonspontaneous at 900.0C
20-27
DG and the Work a System Can Do
For a spontaneous process, DG is the maximum work obtainable from the
system as the process takes place: DG = workmax
For a nonspontaneous process, DG is the maximum work that must be done to the
system as the process takes place: DG = workmax
An example
20-28
Table 20.1 Reaction Spontaneity and the Signs of DH0, DS0, and DG0
DH0
DS0
-TDS0
DG0
-
+
-
-
Spontaneous at all T
+
-
+
+
Nonspontaneous at all T
+
+
-
+ or -
Spontaneous at higher T;
nonspontaneous at lower T
-
-
+
+ or -
Spontaneous at lower T;
nonspontaneous at higher T
20-29
Description
Sample Problem 20.5:
PROBLEM:
PLAN:
Calculating DG0rxn from DG0f Values
Use DG0f values to calculate DGrxn for the reaction in
Sample Problem 20.4:
D
4KClO3(s)
3KClO4(s) + KCl(s)
Use the DG summation equation.
SOLUTION:
DG0rxn = S mDGf0products - S nDGf0reactants
DG0rxn = (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) - (4mol)(-296.3kJ/mol)
DG0rxn = -134kJ
20-30
Figure 20.11
20-31
The effect of temperature on reaction spontaneity
The coupling of a nonspontaneous reaction to the hydrolysis of ATP.
Glucose-6-phosphate
Enzyme: hexose kinase
Glucose
Enzyme
1st Rxn of Glycolysis catalyzed by Hexose Kinase:
Glucose + HPO42-  Glucose-6-phosphate + H2O
ATP4- + H2O  ADP3- + HPO42Glucose + HPO42-  Glucose-6-phosphate + H2O
Glucose
20-32
+ ATP  Glucose-6-phosphate + ADP3-
DG0 = +13.8 kJ
DG0 = -30.5 kJ
DG0 = +13.8 kJ
DG0 = -16.7 kJ
Figure B20.4
20-33
The cycling of metabolic free energy through ATP
Figure B20.5
20-34
Why is ATP a high-energy molecule?
Free Energy, Equilibrium and Reaction Direction
•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (DG < 0)
•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0)
•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0)
DG = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and lnQ = 0
so...
DG0 = - RT lnK
20-35
Table 20.2 The Relationship Between DG0 and K at 250C
DG0(kJ)
100
3x10-18
50
2x10-9
10
2x10-2
1
7x10-1
0
1
Essentially no forward reaction;
reverse reaction goes to completion
Forward and reverse reactions
proceed to same extent
1.5
-10
5x101
-50
6x108
-100
3x1017
-200
1x1035
Forward reaction goes to
completion; essentially no reverse
reaction
REVERSE REACTION
9x10-36
Significance
FORWARD REACTION
200
-1
20-36
K
Sample Problem 20.7:
PROBLEM:
Calculating DG at Nonstandard Conditions
The oxidation of SO2, which we considered in Sample Problem 20.6
2SO2(g) + O2(g)
2SO3(g)
is too slow at 298K to be useful in the manufacture of sulfuric acid. To
overcome this low rate, the process is conducted at an elevated temperature.
(a) Calculate K at 298K and at 973K. (DG0298 = -141.6kJ/mol of reaction as
written using DH0 and DS0 values at 973K. DG0973 = -12.12kJ/mol of reaction
as written.)
(b) In experiments to determine the effect of temperature on reaction
spontaneity, two sealed containers are filled with 0.500atm of SO2,
0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In
which direction, if any, will the reaction proceed to reach equilibrium at each
temperature?
(c) Calculate DG for the system in part (b) at each temperature.
PLAN: Use the equations and conditions found on slide
20-37
.
Sample Problem 20.7:
Calculating DG at Nonstandard Conditions
continued (2 of 3)
SOLUTION: (a) Calculating K at the two temperatures:
DG0
= -RTlnK so K  e
(DG / RT )
0
(-141.6kJ/mol)(103J/kJ)
At 298, the exponent is -DG0/RT = -
= 57.2
(8.314J/mol*K)(298K)
K e
(DG / RT )
0
= e57.2 = 7x1024
(-12.12kJ/mol)(103J/kJ)
At 973, the exponent is -DG0/RT
= 1.50
(8.314J/mol*K)(973K)
K e
20-38
(DG / RT )
0
= e1.50 = 4.5
Calculating DG at Nonstandard Conditions
Sample Problem 20.7:
continued (3 of 3)
pSO32
(b) The value of Q =
(pSO2)2(pO2)
(0.100)2
=
= 4.00
(0.500)2(0.0100)
Since Q is < K at both temperatures the reaction will shift right; for 298K
there will be a dramatic shift while at 973K the shift will be slight.
(c) The nonstandard DG is calculated using DG = DG0 + RTlnQ
DG298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(ln4.00)
DG298 = -138.2kJ/mol
DG973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(ln4.00)
DG298 = -0.9kJ/mol
20-39
Figure 20.12
The relation between free energy and the extent of reaction
DG0 < 0
K >1
20-40
DG0 > 0
K <1
Useful Thermodynamic Relationships
Different Ways to Calculate DGrxn
(Eqn 1)
DG0rxn = DH0rxn - TDS0rxn
(Eqn 2)
DG0rxn = S mDGf0products - S nDGf0reactants
(Eqn 3)
DG0 = - RT lnKeq
(Eqn 4)
DG = DG0 + RT lnQ
R = 8.314 J/mol.K
Q = Rxn Quotient
Use these with equation #1
DS0rxn = S mS0products - S nS0reactants
DH0rxn = S mDHf0products - S nDHf0reactants
20-41