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

2. Entropy, Free Energy, and the Direction of Chemical Reactions
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. 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. Limitations of the First Law of Thermodynamics
DE = q + w
Euniverse = Esystem + Esurroundings
DEsystem = -DEsurroundings
DEsystem + DEsurroundings = 0 = DEuniverse
The total energy-mass of the universe is constant.
However, these energy changes do not explain the direction of change in the universe!

5. The sign of the enthalpy change is insufficient to predict
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) .
DHorxn = kJ
The sign of the enthalpy change is insufficient to predict
reaction spontaneity!
Figure 20.1

6. Entropy refers to the state of order/disorder of a system.
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.
more order
less order
solid liquid gas
more order
less order
crystal + liquid ions in solution
more order
less order
crystal + crystal gases + ions in solution

7. The number of ways to arrange a deck of playing cards
Figure 20.2
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8. Spontaneous expansion of a gas
This process occurs
spontaneously without a change in the total internal energy of the system!
stopcock closed
This is an entropy-
driven process!
evacuated
1 atm
stopcock opened
0.5 atm
Figure 20.3
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9. S = k ln W The Second Law of Thermodynamics
Ludwig Boltzmann, 1877
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 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.
The Second Law of Thermodynamics
DSuniv = DSsys + DSsurr > 0
Entropy is a state function.

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

11. Random motion in a crystal
(increased arrangements, greater S)
The Third Law of Thermodynamics
A perfect crystal has zero entropy at a temperature of absolute zero.
Ssystem = 0 at 0 K
Figure 20.4
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12. Standard Molar Entropies (So)
The Third Law leads to absolute values for the entropies of substances!
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
This value equals the entropy increase in a substance
upon raising its temperature from 0 K to the specified temperature.

13. So increases as temperature rises.
Predicting relative So values of a system
1. Temperature changes
So increases as temperature rises.
2. Physical states and phase changes
So increases as a more ordered phase changes to a less ordered phase.
3. Dissolution of a solid or liquid
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.
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, increases in complexity (e.g., bond flexibility) relate directly to entropy.

14. Large changes in S occur
The increase in entropy from solid to liquid to gas
Large changes in S occur
at phase transitions!
Figure 20.5
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15. The entropy change accompanying the dissolution of a salt
pure solid
MIX
pure liquid
salt becomes more disordered (dissociation): S increases
water becomes more ordered (ion-dipole interactions): S decreases
Figure 20.6
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16. The small increase in entropy when ethanol dissolves in water
solution of ethanol and water
water
ethanol
Freedom of movement remains essentially unchanged;
S increase is due solely to random mixing effects.
Figure 20.7
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17. The large decrease in entropy when a gas dissolves in a liquid
O2 gas
O2 dissolved
Figure 20.8
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18. Entropy and vibrational motion
more motions, greater S
N2O4 NO
NO2
Internal molecular motions influence entropy!
Figure 20.9

19. 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) 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.
SOLUTION:
(a) 1 mol of SO3(g) - more atoms
(d) 1 mol of KBr(aq) - solution > solid
(b) 1 mol of CO2(g) - gas > solid
(e) 23 oC - higher temperature
(c) 3 mol of O2(g) - larger # mols
(f) CCl4 - larger mass

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

21. DSorxn = - 374 J/K Sample Problem 20.2
Calculating the standard entropy of reaction, DSorxn
PROBLEM:
Calculate DSorxn for the combustion of 1 mol of propane at 25 oC.
C3H8(g) + 5O2(g) CO2(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.
DSorxn = [(3 mol)(So CO2) + (4 mol)(So H2O)] - [(1 mol)(So C3H8) + (5 mol)(So O2)]
DSorxn = [(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)]
DSorxn = J/K

22. The initial temperature of the surroundings affects the
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
and
DSsurr a 1/T

23. DSsurr = -qsys/T DSsurr = -DHsys/T Combining these ideas......
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.

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

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

26. DSouniv = DSosurr + DSosys = 0
What happens when equilibrium is reached?
DSouniv = DSosurr + DSosys = 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. Components of DSouniv for spontaneous reactions
exothermic
endothermic
system becomes more disordered
exothermic
system becomes more disordered
system becomes more ordered
Figure 20.10

28. The Gibbs free energy (G) accomplishes this goal, where
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. 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. spontaneous in the opposite direction (DG < 0).
Some Key Concepts
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.

31. DGosys = DHosys - TDSosys
Standard Free Energy Changes
A reference state; similar to DHo and So
All components are in their standards states.
DGosys = DHosys - TDSosys

32. Sample Problem 20.4
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). +5 +7 -1
4KClO3(s) KClO4(s) KCl(s) D
Use DHof and So values to calculate DGosys (DGorxn) 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.
SOLUTION:
DHorxn = S mDHof (roducts) - S nDHof (reactants)
DHorxn = [(3 mol)( kJ/mol) + (1 mol)( kJ/mol)] - [(4 mol)( kJ/mol)]
DHorxn = kJ

33. The reaction is spontaneous.
Sample Problem 20.4
(continued)
DSorxn = S mSoproducts - S nSoreactants
DSorxn = [(3 mol)(151 J/mol.K) + (1 mol)(82.6 J/mol.K)] - [(4 mol)(143.1 J/mol.K)]
DSorxn = J/K
DGorxn = DHorxn - TDSorxn
DGorxn = kJ - (298 K)(-36.8 J/K)(kJ/103J)
DG0rxn = -133 kJ
The reaction is spontaneous.

34. DGorxn = S mDGof (products) - S nDGof (reactants)
Another way to calculate DGorxn
From standard free energies of formation, DGof
DGof = the free energy change that occurs when 1 mol of compound
is made from its elements, with all components in their standard states.
Thus,
DGorxn = S mDGof (products) - S nDGof (reactants)
DGof values have properties similar to DHof values.

35. Sample Problem 20.5
Calculating DGorxn from DGof values
PROBLEM:
Use DGof values to calculate DGorxn for the following reaction:
4KClO3(s) KClO4(s) + KCl(s) D
PLAN:
Use the DGorxn summation equation.
SOLUTION:
DGorxn = S mDGof (products) - S nDGof (reactants)
DGorxn = [(3 mol)( kJ/mol) + (1 mol)( kJ/mol)] - [(4 mol)( kJ/mol)]
DGorxn = -134 kJ

36. An example of a gas doing work
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. A reaction at equilibrium cannot do work.
More Key Concepts
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.

38. DGosys = DHosys - TDSosys
Effect of Temperature on Reaction Spontaneity
DGosys = DHosys - TDSosys
The sign of DGosys is T-independent when DHosys and
DSosys have opposite signs.
The sign of DGosys is T-dependent when DHosys and
DSosys have the same signs.

39. Table 20.1 Reaction Spontaneity and the Signs of DHo, DSo and DGo
-TDSo
DGo
description - +
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

40. Sample Problem 20.6
Determining the effect of temperature on DGo
PROBLEM:
An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g):
2SO2(g) + O2(g) SO3(g)
At 298 K, DGo = kJ, DHo = kJ, and DSo = 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?
PLAN:
The sign of DGo 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).
SOLUTION:
(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. Sample Problem 20.6
(continued)
(b) DGorxn = DHorxn - TDSorxn
DGorxn = kJ - [(1173 K)( J/mol.K)(kJ/103J)]
DGorxn = kJ; the reaction will be nonspontaneous at 900.oC

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

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

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

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

47. End of Assigned Material

48. Free energy, equilibrium and reaction direction
DG = RT ln Q/K = RT ln Q - RT ln K
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)
Under standard conditions (1 M concentrations, 1 atm for gases), Q = 1 and ln Q = 0. Therefore:
DGo = - RT ln K

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

50. Calculating DG at non-standard conditions
Sample Problem 20.7
Calculating DG at non-standard conditions
The oxidation of SO2,
PROBLEM:
2SO2(g) + O2(g) SO3(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. (DGo298 = kJ/mol for the reaction as written using DHo and DSo values. At 973 K, DGo973 = 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 atm of SO2, atm of O2, and 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. Sample Problem 20.7
(continued)
SOLUTION:
(a) Calculating K at the two temperatures:
DGo = -RTln K so
( kJ/mol)(103 J/kJ)
(8.314 J/mol.K)(298 K)
At 298 K, the exponent is -DGo/RT
= -
= 57.2
= e57.2
= 7 x 1024
( kJ/mol)(103 J/kJ)
(8.314 J/mol.K)(973 K)
At 973 K, the exponent is -DGo/RT = -
= 1.50
= e1.50
= 4.5

52. Sample Problem 20.7
(continued)
pSO32
(pSO2)2(pO2) =
(0.100)2
(0.500)2(0.0100)
(b) The value of Q =
= 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 = kJ/mol + (8.314 J/mol.K)(kJ/103J)(298 K)(ln 4.00)
DG298 = kJ/mol
DG973 = kJ/mol + (8.314 J/mol.K)(kJ/103J)(973 K)(ln 4.00)
DG298 = -0.9 kJ/mol

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