1. Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Chemistry 140 Fall 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 19: Spontaneous Change: Entropy and Gibbs Energy
Philip Dutton
University of Windsor, Canada
N9B 3P4
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2. General Chemistry: Chapter 20
Chemistry 140 Fall 2002
Contents
19-1 Spontaneity: The Meaning of Spontaneous Change
19-2 The Concept of Entropy
19-3 Evaluating Entropy and Entropy Changes
19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics
19-5 Standard Gibbs Energy Change, ΔG°
19-6 Gibbs Energy Change and Equilibrium
19-7 ΔG° and Keq as Functions of Temperature
19-8 Coupled Reactions
Focus On Coupled Reactions in Biological Systems
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General Chemistry: Chapter 20

3. 19-1 Spontaneity: The Meaning of Spontaneous Change
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General Chemistry: Chapter 20

4. General Chemistry: Chapter 20
Spontaneous Process
A process that occurs in a system left to itself.
Once started, no external actions is necessary to make the process continue.
A non-spontaneous process will not occur without external action continuously applied.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
H2O(s)  H2O(l)
2 Fe2O3(s) → 4 Fe(s) + 3 O2(g)
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General Chemistry: Chapter 20

5. General Chemistry: Chapter 20
Spontaneous Process
Potential energy decreases.
For chemical systems the internal energy U is equivalent to potential energy.
Berthelot and Thomsen 1870’s
Spontaneous change occurs in the direction in which the enthalpy of a system decreases.
Mainly true but there are exceptions.
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6. 19-2 The Concept of Entropy
Entropy, S.
The greater the number of configurations of the microscopic particles among the energy levels in a particular system, the greater the entropy of the system.
ΔS > 0 spontaneous
ΔU = ΔH = 0
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7. The Boltzmann Equation for Entropy
Chemistry 140 Fall 2002
The Boltzmann Equation for Entropy
S = k lnW
S is the entropy
k = R/NA: Boltzmann constant
W = number of microstates.
The particular way in which particles are distributed amongst the microscopic energy levels called states available in a system with the same total energy.
Ludwig Boltzmann.
Associated the number of energy levels in a system with the number of ways of arranging the particles in these levels.
General Chemistry: Chapter 20

8. Boltzmann Distribution FIG. 19-3
Chemistry 140 Fall 2002
Boltzmann Distribution FIG. 19-3
FIGURE 19-3
Energy levels for a particle in a one-dimensional box
(a) The energy levels of a particle in a box become more numerous and closer
together as the length of the box increases. The range of thermally accessible levels
is indicated by the tinted band. The solid circles signify a system consisting of 15
particles. Each drawing represents a single microstate of the system. Can you see that
as the box length increases there are many more microstates available to the particles
for a given amount of thermal energy? As the number of possible microstates for a
given total energy increases, so does the entropy. (b) More energy levels become
accessible in a box of fixed length as the temperature is raised. Because the average
energy of the particles also increases, the internal energy and entropy both increase as
the temperature is raised.
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General Chemistry: Chapter 20

9. General Chemistry: Chapter 20
Chemistry 140 Fall 2002
Entropy Change FIG. 19-4
ΔS =
qrev T
FIGURE 19-4
Three processes in which entropy increases
Each of the processes pictured (a) the melting of a
solid, (b) the evaporation of a liquid, and (c) the dissolving
of a solute results in an increase in entropy. For part (c),
the generalization works best for nonelectrolyte solutions,
in which ion dipole forces do not exist.
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10. 19-3 Evaluating Entropy and Entropy Changes
Chemistry 140 Fall 2002
19-3 Evaluating Entropy and Entropy Changes
Phase transitions.
Exchange of heat can be carried out reversibly.
ΔS = ΔH
Ttr
H2O(s, 1 atm)  H2O(l, 1 atm) ΔHfus° = 6.02 kJ at K
ΔSfus =
ΔHfus
Ttr =
6.02 kJ mol-1 K
= 22 J mol-1 K-1
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General Chemistry: Chapter 20

11. General Chemistry: Chapter 20
Chemistry 140 Fall 2002
Trouton’s Rule
The molar boiling entropy changes are similar for many materials (e.g. toluene, 87.3, benzene, 89.5, chloroform, 87.9,…):
ΔS =
ΔHvap
Tbp
= 87 kJ mol-1 K-1
Failures of this rule are understandable:
Water and hydrogen bonding. Sliquid is lower than expected so ΔS is larger than 87.
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General Chemistry: Chapter 20

12. General Chemistry: Chapter 20
Chemistry 140 Fall 2002
Absolute Entropies
Third law of thermodynamics.
The entropy of a pure perfect crystal at 0 K is zero.
Standard molar entropy.
Tabulated in Table 17.
Extensive state function, entropy can be summed in a multistep reaction
ΔS = [ ∑pS°(products) - ∑rS°(reactants)]
 are the stoichiometric coefficients.
General Chemistry: Chapter 20

13. Molar entropies for the standard conditions
Compound
Smo, J K-1mol-1
Solids
C (diamond)
2.38
C (graphite)
5.74 Na
51.2 K
64.2 Rb
69.5 Cs
85.2
NaCl
72.1
KCl
82.6 I2
116.1
General Chemistry: Chapter 20

14. Molar entropies for the standard conditions
Compound
Smo, J K-1mol-1
Liquids Hg
76.0
Br2
152.2
H2O
69.9
CH3OH
126.8
C2H5OH
160.7
C6H6
172.8
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15. Molar entropies for the standard conditions
Compound
Smo, J K-1mol-1
Triatomic Gases
H2O
188.8
NO2
240.1
H2S
205.8
CO2
213.7
SO2
248.2
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16. Calculate DSº for the complete combustion of ethane (C2H6) with oxygen (O2).
Equation: C2H6 + 7/2 O2  → 2 CO2 + 3 H2O
Vapor
S°, JK-1mol-1
C2H6
229.6 O2
205.1
CO2
213.7
H2O
188.8
DSº= [2(213.7) + 3(188.8)] - [ /2(205.1)] = JK-1mol-1
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17. Entropy as a Function of Temperature
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18. Vibrational Energy and Entropy
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19. General Chemistry: Chapter 20
19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics.
ΔStotal = ΔSuniverse = ΔSsystem + ΔSsurroundings
The Second Law of Thermodynamics (Clausius):
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
All spontaneous processes produce an increase in the entropy of the universe.
(Macroscopic, non equilibrium, isolated systems – spontaneous process: the entropy increases; it is maximal in the equilibrium.)
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20. Gibbs Energy and Gibbs Energy Change
Chemistry 140 Fall 2002
Gibbs Energy and Gibbs Energy Change
Hypothetical process:
only pressure-volume work, at constant T and P.
qsurroundings = -qp = -ΔHsys
Make the enthalpy change reversible.
large surroundings, infinitesimal change in temperature.
Under these conditions we can calculate entropy.
Want to evaluate the system only, not the surroundings.
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21. Ice Melting Tsys= 273 K < Tsurr= 298 K and qsys = DHsys>0.
P=const.
Tsys= 273 K < Tsurr= 298 K and qsys = DHsys>0.
Egyensúly

22. Ice Melting: Entalpy and Entropy
The entropy change of the surroundings and system:
The ice melts because the total entropy change is positive: the entropy of the universe increases.
Egyensúly

23. Gibbs Energy and Energy Change
For the universe (P and Tsurr = const.) reversible:
Tsurr ΔSuniv. = Tsurr ΔSsys – ΔHsys = -(ΔHsys – TsurrΔSsys)
-TΔSuniv. = ΔHsys – TΔSsys
For the system (P and T = const.):
G = H - TS
ΔG = ΔH - TΔS
ΔGsys = - TΔSuniverse
General Chemistry: Chapter 20

24. Criteria for Spontaneous Change
ΔGsys < 0 (negative), the process is spontaneous.
ΔGsys = 0 (zero), the process is at equilibrium.
ΔGsys > 0 (positive), the process is non-spontaneous.
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25. 19-5 Standard Gibbs Energy Change, ΔG°
The standard Gibbs energy of formation, ΔfG°.
The Gibbs energy change for a reaction in which a substance in its standard state is formed from its elements in reference forms in their standard states.
The standard Gibbs energy of reaction, ΔG°.
where i s are the stoichiometric factors
ΔG° = [ ∑p ΔfG°(products) - ∑r ΔfG°(reactants)]
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26. Standard changes of extensive quantities
Reaction Gibbs energy: ΔG° = [ ∑p ΔfG°(products) - ∑r ΔfG°(reactants)]
Reaction entropy change: ΔS° = [ ∑p S°(products) - ∑r S°(reactants)]
Reaction enthalpy change: ΔH° = [ ∑p ΔfH°(products) - ∑r ΔfH°(reactants)]
where i s are the stoichiometric factors.
General Chemistry: Chapter 20

27. Általános Kémia, 2008 tavasz
Termodinamic table
P = 1 atm, T =298 K
DfH° S°
DfG°
kJ mol-1
J mol-1 K-1
H2O(l)
-285.8
69.9
-237.1
H2O(g)
-241.8
188.8
-228.6
H2O(l) → H2O(g)
44.0
118.9
8.5
The enthalpy change is positive, endothermic.
DG° positive, the reaction is non spontaneous at atmospheric pressure
DG° = 0 if P = mmHg - equilibrium.
Equilibrium

28. Gibbs Energy Change and Equilibrium
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29. 19-6 Gibbs Energy Change and Equilibrium
DG(T) = DH –TDS
DS negative
General Chemistry: Chapter 20

30. Temperature dependence of the Gibbs free energy
DH and DS are constant for K.
At the boiling point: (DG = 0)
𝑇= Δ𝐻 Δ𝑆
Tboil = /118.9 = 370 K ( 373 K)
Above this temperature the DG < 0, and the water boils, evaporates completely.
Δ𝐺=Δ𝐻−𝑇⋅Δ𝑆
0=Δ𝐻−𝑇⋅Δ𝑆
Egyensúly

31. Water boiling, P = 1 atm
Egyensúly

32. + Spontneous reactions DG(T) = DH –TDS DH DS DG(T) Examples -
H2O(l) → H2O(g)
High T helps
H2O(g) → H2O(l)
Low T helps
4Fe(s) +3O2(g) →2Fe2O3(s) always
2Fe2O3(s)→ 4Fe(s) +3O2(g) never T T 32
11/18/2018
Professor Gabor I. Csonka

33. Relationship of ΔG° to ΔG for Nonstandard Conditions
N2(g) + 3 H2(g)  2 NH3(g)
ΔG = ΔH - TΔS
ΔG° = ΔH° - TΔS°
For ideal gases ΔH = ΔH°
ΔG = ΔH° - TΔS
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34. Relationship Between S and S°Vf
qrev = -w = RT ln Vi
ΔS =
qrev T
= R ln Vf Vi
ΔS = Sf – Si
= R ln Vf Vi Pi Pf
= -R ln
S = S°
- R ln P P°
= S° 1
= S° - R ln P
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General Chemistry: Chapter 20

35. N2(g) + 3 H2(g)  2 NH3(g) SN2 = S°N2 – R·ln PN2 SH2 = S°H2 – R·ln PH2
SNH3 = S°NH3 – R·ln PNH3
ΔSrxn = 2·(S°NH3 – R·ln PNH3) – (S°N2 – R·ln PN2) – 3· (S°H2 – R·ln PH2)
PN2PH2 3
ΔSrxn = 2 S°NH3 –S°N2 –3S°H2+ R·ln
PNH3 2
PN2PH2 3
ΔSrxn = ΔS°rxn + R·ln
PNH3 2
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36. ΔG Under Non-standard Conditions
PN2PH2 3
ΔG = ΔH° - TΔS
ΔSrxn = ΔS°rxn + Rln
PNH3 2
PN2PH2 3
ΔG = ΔH° - TΔS°rxn – TR ln
PNH3 2
PNH3 2
ΔG = ΔG° + RT ln
PN2PH2 3
ΔG = ΔG° + RT ln Q
Q reaction quotient
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37. ΔG and the Equilibrium Constant Keq
ΔG = ΔG° + RT ln Q
ΔG = ΔG° + RT ln Keq= 0
If the reaction is at equilibrium then:
ΔG° = -RT ln Keq
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38. Criteria for Spontaneous Change
Every chemical reaction consists of both a forward and a reverse reaction.
The direction of spontaneous change is the direction in which the Gibbs energy decreases.
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39. Significance of the Magnitude of ΔG0
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40. The Thermodynamic Equilibrium Constant: Activities
Chemistry 140 Fall 2002
The Thermodynamic Equilibrium Constant: Activities
For ideal gases at 1.0 bar:
S = S°
- R ln P P°
= S° 1
PV=nRT or P=(n/V)RT, pressure is an effective concentration
Therefore, in solution:
S = S°
- R ln c c°
= S°
- R ln a
Activity is the effective concentration divided by the effective concentration in the standard state.
The effective concentration in the standard state for an ideal solution is c° = 1 M.
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General Chemistry: Chapter 20

41. General Chemistry: Chapter 20
Activities
For pure solids and liquids:
a = 1
For ideal gases:
a = P (in bars, 1 bar = atm)
For ideal solutes in aqueous solution:
a = c (in mol L-1)
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General Chemistry: Chapter 20

42. The Thermodynamic Equilibrium Constant, Keq
A dimensionless equilibrium constant expressed in terms of activities.
Often Keq = Kc
Must be used to determine ΔG.
ΔG = ΔG° + RT ln
agah…
aaab…
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43. 19-7 ΔG° and Keq as Functions of Temperature
ΔG° = ΔH° -TΔS°
ΔG° = -RT ln Keq
ln Keq =
-ΔG° RT =
-ΔH°
TΔS° +
ln Keq =
-ΔH° RT
ΔS° R +
ln =
-ΔH°
RT2
ΔS° R +
RT1 - = 1 T2 T1
Keq1
Keq2
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44. Temperature Dependence of Keq
ln Keq =
-ΔH° RT
ΔS° R +
slope =
-ΔH° R
-ΔH° = R · slope
= J mol-1 K-1 · 2.2·104 K
= -1.8·102 kJ mol-1
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General Chemistry: Chapter 20

45. General Chemistry: Chapter 20
19-8 Coupled Reactions
In order to drive a non-spontaneous reactions we changed the conditions (i.e. temperature or electrolysis)
Another method is to couple two reactions.
One with a positive ΔG and one with a negative ΔG.
Overall spontaneous process.
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General Chemistry: Chapter 20

46. General Chemistry: Chapter 20
Smelting Copper Ore
Cu2O(s) → 2 Cu(s) + ½ O2(g) ΔG°673K = +125 kJ Δ
Cu2O(s) → 2 Cu(s) + ½ O2(g)
Non-spontaneous reaction:
+125 kJ
C(s) + ½ O2(g) → CO(g)
Spontaneous reaction:
-175 kJ
Cu2O(s) + C(s) → 2 Cu(s) + CO(g)
-50 kJ
Spontaneous reaction!
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47. Focus On Coupled Reactions in Biological Systems
ADP3- + HPO42- + H+ → ATP4- + H2O
ΔG° = -9.2 kJ mol-1
ΔG = ΔG° + RT ln
aATPaH2O
aADPaPiaH3O+
But [H3O+] = 10-7 M not 1.0 M.
ΔG = -9.2 kJ mol kJ mol-1
= kJ mol-1 = ΔG°'
The biological standard state:
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48. Focus On Coupled Reactions in Biological Systems
Glucose → 2 lactate + 2 H+
-218 kJ
2 ADP HPO H+ → 2 ATP H2O
+64 kJ
2 ADP HPO42- + glucose → 2 ATP H2O + 2 lactate
-153 kJ
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