Thermodynamics

1. 20-1
Chapter 20
Thermodynamics
Entropy, Free Energy and the Direction
of Chemical Reactions

2. 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 (we will
consider later in the semester)

3. 20-3
The Larger Question
A B
What factors must be identified in order to predict
whether the reaction will proceed spontaneously
in the direction written (as opposed to in the opposite
direction)?
Enthalpy and entropy changes are combined to yield a
new term, the Gibbs free energy. The sign of the latter
allows predictions of reaction spontaneity.
These are thermodynamic, not kinetic, considerations!

4. 20-4
Limitations of the First Law of Thermodynamics
DE = q + w
Euniverse = Esystem + Esurroundings
DEsystem = -DEsurroundings
The total energy-mass of the universe is
constant.
However, these energy changes do not explain the
direction of change in the universe!
DEsystem + DEsurroundings = 0 = DEuniverse

5. 20-5
Figure 20.1
A spontaneous endothermic chemical reaction
water
Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3
-(aq) + 2NH3(aq) + 10H2O(l)
.
DHo
rxn = + 62.3 kJ
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The sign of the enthalpy change is insufficient to predict
reaction spontaneity!

6. 20-6
The Missing Factor: Entropy (S)
Entropy refers to the state of order/disorder of a system.
A change in order is a change in the number of ways of
arranging the particles. It is a key factor in determining the
direction of a spontaneous process.
solid liquid gas
more order less order
crystal + liquid ions in solution
more order less order
more order less order
crystal + crystal gases + ions in solution

7. 20-7
Figure 20.2
The number of ways to arrange a deck of playing cards
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8. 20-8
Figure 20.3
1 atm evacuated
Spontaneous expansion of a gas
stopcock
closed
stopcock
opened
0.5 atm 0.5 atm
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This process occurs
spontaneously
without a change in
the total internal
energy of the
system!
This is an entropy-
driven process!

9. 20-9
Ludwig Boltzmann, 1877
S = k ln W
S = entropy, W = the number of ways of arranging the components of
a system having equivalent energy, and k = the Boltzmann constant =
R/NA (R = universal gas constant, NA = Avogadro’s number) = 1.38 x
10-23 J/K.
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.
DSuniv = DSsys + DSsurr > 0
The Second Law of Thermodynamics
Entropy is a state function.

10. 20-10
All processes occur spontaneously
in the direction that increases
the entropy of the universe
(system + surroundings).

11. 20-11
Random motion in a crystal
The Third Law of
Thermodynamics
A perfect crystal has
zero entropy at a
temperature of
absolute zero.
Ssystem = 0 at 0 K
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Figure 20.4
(increased arrangements, greater S)

12. 20-12
Standard Molar Entropies (So)
Standard states: 1 atm for gases; 1 M for solutions; pure substance
in its most stable form for solids and liquids.
units: J/mol/K at 25 oC
The Third Law leads to absolute values for the entropies of substances!
This value equals the entropy increase in a substance
upon raising its temperature from 0 K to the specified temperature.

13. 20-13
Predicting relative So values of a system
1. Temperature changes
2. Physical states and phase changes
3. Dissolution of a solid or liquid
5. Atomic size or molecular complexity
4. Dissolution of a gas
So increases as temperature rises.
So increases as a more ordered phase changes to a less
ordered phase.
So of a dissolved solid or liquid is usually greater than So of the
pure solute. However, the extent depends on the nature of the
solute and solvent.
A gas becomes more ordered when it dissolves in a liquid or solid.
In similar substances, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g., bond
flexibility) relate directly to entropy.

14. 20-14
Figure 20.5
The increase in entropy from solid to liquid to gas
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Large changes in S occur
at phase transitions!

15. 20-15
Figure 20.6
The entropy change accompanying the dissolution of a salt
pure solid
pure liquid
solution
MIX
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salt becomes more disordered (dissociation): S increases
water becomes more ordered (ion-dipole interactions): S decreases

16. 20-16
Figure 20.7
ethanol water solution of
ethanol and
water
The small increase in entropy when ethanol dissolves in water
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Freedom of movement remains essentially unchanged;
S increase is due solely to random mixing effects.

17. 20-17
Figure 20.8
The large decrease in entropy when a gas dissolves in a liquid
O2 gas
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O2 dissolved

18. 20-18
Figure 20.9
NO
NO2
N2O4
Entropy and vibrational motion
Internal molecular motions influence entropy!
more motions, greater S

19. 20-19
Sample Problem 20.1
SOLUTION:
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) 1 mol of SO2(g) or 1 mol of SO3(g)
(b) 1 mol of CO2(s) or 1 mol of CO2(g)
(c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3)
(d) 1 mol of KBr(s) or 1 mol of KBr(aq)
(e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC
(f) 1 mol of CF4(g) or 1 mol of CCl4(g)
PLAN: In general less ordered systems have higher entropy than ordered
systems and entropy increases with increasing temperature.
(a) 1 mol of SO3(g) - more atoms
(b) 1 mol of CO2(g) - gas > solid
(c) 3 mol of O2(g) - larger # mols
(d) 1 mol of KBr(aq) - solution > solid
(e) 23 oC - higher temperature
(f) CCl4 - larger mass

20. 20-20
Standard Entropies of Reaction, DSo
rxn
By analogy to calculating standard heats of reaction: DHo
rxn
DHo
rxn = S mDHo
f (products) - S nDHo
f
(reactants)
DSo
rxn = S mSo (products) - S nSo (reactants)
From Chapter 6
As before, m and n are the appropriate coefficients
in the balanced chemical equation.

21. 20-21
Sample Problem 20.2 Calculating the standard entropy of reaction, DSo
rxn
PROBLEM: Calculate DSo
rxn for the combustion of 1 mol of propane at 25 oC.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
PLAN: Use summation equations. Entropy is expected to decrease
because the reaction goes from 6 moles of gas to 3 moles of gas.
SOLUTION: Find standard entropy values in an appropriate data table.
DSo
rxn = [(3 mol)(So CO2) + (4 mol)(So H2O)] - [(1 mol)(So C3H8) + (5 mol)(So O2)]
DSo
rxn = [(3 mol)(213.7 J/mol.K) + (4 mol)(69.9 J/mol.K)] - [(1 mol)(269.9
J/mol.K) + (5 mol)(205.0 J/mol.K)]
DSo
rxn = - 374 J/K

22. 20-22
We can’t forget entropy changes in the surroundings!
For spontaneous reactions in which a decrease in the entropy of
the system occurs: must be outweighed by a concomitant increase
in the entropy of the surroundings (DSuniv > 0)
For an exothermic process: qsys < 0, qsurr > 0, DSsurr > 0
For an endothermic process: qsys > 0, qsurr < 0, DSsurr < 0
The initial temperature of the surroundings affects the
magnitude of DSsurr.
DSsurr a -qsys DSsurr a 1/T
and

23. 20-23
DSsurr = -qsys/T
For a process of constant pressure, where qp = DH:
DSsurr = -DHsys/T
Implication: DSsurr can be calculated by measuring
DHsys and the temperature at which the change occurs.
Combining these ideas......

24. 20-24
Sample Problem 20.3
SOLUTION:
Determining reaction spontaneity
PROBLEM: At 298 K, the formation of ammonia has a negative DSo
sys.
Calculate DSo
univ and state whether the reaction
occurs spontaneously at this temperature.
N2(g) + 3H2(g) 2NH3(g) DSo
sys = -197 J/K
PLAN: DSo
univ must be > 0 in order for the reaction to be spontaneous, so
DSo
surr must be > +197 J/K. To find DS o
surr, first find DHsys. DHo
sys =
DHo
rxn which can be calculated using DHo
f values from data tables.
Then apply DSo
univ = DSo
surr + DSo
sys.
DHo
rxn = [(2 mol)(DHo
f NH3)] - [(1 mol)(DHo
f N2) + (3 mol)(DHo
f H2)]
DHo
rxn = -91.8 kJ
DSo
surr = -DHo
sys/T = -(-91.8 x 103J/298 K) = +308 J/K (>+197 J/K)
DSo
univ = DSo
surr + DSo
sys = 308 J/K + (-197 J/K) = +111 J/K
DSo
univ > 0: the reaction is spontaneous!

25. 20-25
Figure B20.2
A whole-body calorimeter
Living systems do not violate the Second Law
DSuniv = DSsys + DSsurr > 0

26. 20-26
What happens when equilibrium is reached?
DSo
univ = DSo
surr + DSo
sys = 0
Examples: phase changes (fusion, vaporization)
When a system reaches equilibrium, neither the forward
nor the reverse reaction is spontaneous; neither proceeds further
because there is no driving force.

27. 20-27
Figure 20.10
Components of DSo
univ for spontaneous reactions
exothermic
system becomes more disordered
exothermic
system becomes more ordered
endothermic
system becomes more disordered

28. 20-28
But measuring DSsys and DSsurr is inconvenient
to determine reaction spontaneity
More convenient to use parameters that apply only to the system!
The Gibbs free energy (G) accomplishes this goal, where
G = H - TS or DG = DH - TDS
G combines the system’s enthalpy and entropy; thus
G is a state function.
Where does the equation come from?

29. 20-29
Deriving the Gibbs free energy equation
DSuniv = DSsys + DSsurr
DSsurr = -DHsys/T
DSuniv = DSsys - DHsys/T
-TDSuniv = DHsys - TDSsys
DGsys = DHsys - TDSsys
DGsys = -TDSuniv
DSuniv > 0 or DGsys < 0 for a spontaneous process
DSuniv < 0 or DGsys > 0 for a nonspontaneous process
DSuniv = 0 or DGsys = 0 for a process at equilibrium

30. 20-30
If a process is nonspontaneous in one direction (DG > 0), then it is
spontaneous in the opposite direction (DG < 0).
The magnitude and sign of DG are unrelated to the rate
(speed) of the reaction.
Some Key Concepts

31. 20-31
DGo
sys = DHo
sys - TDSo
sys
Standard Free Energy Changes
All components are in their standards states.
A reference state; similar to DHo and So

32. 20-32
Sample Problem 20.4
SOLUTION:
Calculating DGo from enthalpy and entropy values
PROBLEM: Potassium chlorate, one of the common oxidizing agents in
explosives, fireworks and match heads, undergoes a solid-state
redox reaction when heated. In this reaction, the oxidation
number of Cl in the reactant is higher in one of the products and
lower in the other (a disproportionation reaction).
4KClO3(s) 3KClO4(s) + KCl(s)
D
+7 -1
+5
Use DHo
f and So values to calculate DGo
sys (DGo
rxn) at 25 oC for this reaction.
PLAN: Obtain appropriate thermodynamic data from a data table; insert
them into the Gibbs free energy equation and solve.
DHo
rxn = S mDHo
f (roducts) - S nDHo
f (reactants)
DHo
rxn = [(3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol)] -
[(4 mol)(-397.7 kJ/mol)]
DHo
rxn = -144 kJ

33. 20-33
Sample Problem 20.4 (continued)
DSo
rxn = S mSo
products - S nSo
reactants
DSo
rxn = [(3 mol)(151 J/mol.K) + (1 mol)(82.6 J/mol.K)] -
[(4 mol)(143.1 J/mol.K)]
DSo
rxn = -36.8 J/K
DGo
rxn = DHo
rxn - TDSo
rxn
DGo
rxn = -144 kJ - (298 K)(-36.8 J/K)(kJ/103J)
DG0
rxn = -133 kJ
The reaction is spontaneous.

34. 20-34
DGo
rxn = S mDGo
f (products) - S nDGo
f (reactants)
Another way to calculate DGo
rxn
From standard free energies of formation, DGo
f
DGo
f = the free energy change that occurs when 1 mol of compound
is made from its elements, with all components in their standard states.
Thus,
DGo
f values have properties similar to DHo
f values.

35. 20-35
Sample Problem 20.5
PROBLEM:
Calculating DGo
rxn from DGo
f values
Use DGo
f values to calculate DGo
rxn for the following reaction:
4KClO3(s) 3KClO4(s) + KCl(s)
D
PLAN: Use the DGo
rxn summation equation.
SOLUTION: DGo
rxn = S mDGo
f (products) - S nDGo
f (reactants)
DGo
rxn = [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] -
[(4 mol)(-296.3 kJ/mol)]
DGo
rxn = -134 kJ

36. 20-36
DG and Work
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 minimum work that must be
done to the system to make the process take place.
An example of a
gas doing work

37. 20-37
A reversible process: one that can be changed in either direction by an
infinitesimal change in a variable. The maximum work from a spontaneous
process is obtained only if the work is carried out reversibly.
In most real processes, work is performed irreversibly; thus,
maximum work is not obtained; some free energy is lost to
the surroundings as heat and is thus unavailable to do work.
A reaction at equilibrium cannot do work.
More Key Concepts

38. 20-38
Effect of Temperature on Reaction Spontaneity
DGo
sys = DHo
sys - TDSo
sys
The sign of DGo
sys is T-independent when DHo
sys and
DSo
sys have opposite signs.
The sign of DGo
sys is T-dependent when DHo
sys and
DSo
sys have the same signs.

39. 20-39
Table 20.1 Reaction Spontaneity and the Signs of
DHo, DSo and DGo
DHo DSo -TDSo DGo description
- + - -
+ - + +
+ + - + or -
- - + + or -
spontaneous at all T
nonspontaneous at all T
spontaneous at higher T;
nonspontaneous at lower T
spontaneous at lower T;
nonspontaneous at higher T

40. 20-40
Sample Problem 20.6
PROBLEM:
PLAN:
SOLUTION:
Determining the effect of temperature on DGo
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 298 K, DGo = -141.6 kJ, DHo = -198.4 kJ, and DSo = -187.9 J/K.
(a) Use these data to decide if this reaction is spontaneous at 25 oC, and
predict how DGo will change with increasing T.
(b) Assuming DHo and DSo are constant with increasing T, is the reaction
spontaneous at 900. oC?
The sign of DG o tells us whether the reaction is spontaneous
and the signs of DHo and DSo will be indicative of the T effect.
Use the Gibbs free energy equation for part (b).
(a) The reaction is spontaneous at 25 oC because DGo is (-).
Since DHo is (-) and DSo is (-), DGo will become less negative and
the reaction less spontaneous as temperature increases.

41. 20-41
Sample Problem 20.6 (continued)
(b) DGo
rxn = DHo
rxn - TDSo
rxn
DGo
rxn = -198.4 kJ - [(1173 K)(-187.9 J/mol.K)(kJ/103J)]
DGo
rxn = +22.0 kJ; the reaction will be nonspontaneous at 900.oC

42. 20-42
Determining the Temperature at which a
Reaction Becomes Spontaneous
(only when DH and DS have the same sign)
DGo = DHo - TDSo = 0 (solve for this condition)
DHo = TDSo
T = DHo/DSo

43. 20-43
Figure 20.11
The effect of temperature on reaction spontaneity
DHo: relatively
insensitive to T
DSo: increases
with increasing T
Cu2O(s) + C(s)
2Cu(s) + CO(g)
DHo = +58.1 kJ
DSo = +165 J/K

44. 20-44
Figure B20.4
The cycling of metabolic free energy through ATP

45. 20-45
Figure B20.3
The coupling of a nonspontaneous reaction to
the hydrolysis of ATP
D-glucose + ATP D-glucose 6P + ADP
(catalyzed by the enzyme, hexokinase)

46. 20-46
Figure B20.5
Why is ATP a high-energy molecule?
The DGo of ATP hydrolysis is -31 kJ/mol;
considerably more (-) under in vivo conditions!

47. 20-47
End of Assigned Material

48. 20-48
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 ln Q - RT ln K
Under standard conditions (1 M concentrations, 1 atm for gases), Q =
1 and ln Q = 0. Therefore:
DGo = - RT ln K

49. 20-49
FORWARD
REACTION
REVERSE
REACTION
Table 20.2 The relationship between DGo and K at 25 oC
DGo (kJ) K significance
200
100
50
10
1
0
-1
-10
-50
-100
-200
9 x 10-36
3 x 10-18
2 x 10-9
2 x 10-2
7 x 10-1
1
1.5
5 x 101
6 x 108
3 x 1017
1 x 1035
Essentially no forward reaction; reverse
reaction goes to completion
Forward and reverse reactions
proceed to the same extent
Forward reaction goes to
completion; essentially no reverse
reaction

50. 20-50
Sample Problem 20.7
PROBLEM:
Calculating DG at non-standard conditions
The oxidation of SO2,
2SO2(g) + O2(g) 2SO3(g)
is too slow at 298 K 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 298 K and at 973 K. (DGo
298 = -141.6 kJ/mol for the
reaction as written using DHo and DSo values. At 973 K, DGo
973 =
12.12 kJ/mol for the reaction as written.)
(b) In experiments to determine the effect of temperature on reaction
spontaneity, two sealed containers are filled with 0.500 atm of SO2,
0.0100 atm of O2, and 0.100 atm of SO3 and kept at 25 oC and at
700. oC. 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 previous slides.

51. 20-51
SOLUTION:
Sample Problem 20.7 (continued)
(a) Calculating K at the two temperatures:
DGo = -RTln K so K  e(DG0
/RT)
At 298 K, the exponent is -DGo/RT = -
(-141.6 kJ/mol)(103 J/kJ)
(8.314 J/mol.K)(298 K)
= 57.2
K  e(DG0
/RT)
= e57.2 = 7 x 1024
At 973 K, the exponent is -DGo/RT = -
(-12.12 kJ/mol)(103 J/kJ)
(8.314 J/mol.K)(973 K)
= 1.50
K  e(DG0
/RT)
= e1.50 = 4.5

52. 20-52
Sample Problem 20.7 (continued)
(b) The value of Q =
pSO3
2
(pSO2)2(pO2)
=
(0.100)2
(0.500)2(0.0100)
= 4.00
Since Q is < K at both temperatures the reaction will shift right; for 298 K
there will be a dramatic shift while at 973 K the shift will be slight.
(c) The non-standard DG is calculated using DG = DGo + RTlnQ
DG298 = -141.6 kJ/mol + (8.314 J/mol.K)(kJ/103J)(298 K)(ln 4.00)
DG298 = -138.2 kJ/mol
DG973 = -12.12 kJ/mol + (8.314 J/mol.K)(kJ/103J)(973 K)(ln 4.00)
DG298 = -0.9 kJ/mol

53. 20-53
Figure 20.12
The relation between free energy and the extent of reaction
DGo < 0
K >1
DGo > 0
K <1
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