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Chapter 17 The Direction of Chemical Change
(The Second Law of Thermodynamics)Life thrives on Earth because there is a constant supply of energy from the Sun. Some of this energy is stored in plants through photosynthesis. Although photosynthesis is only about 3% efficient, it supports nearly all plants on Earth and the animals that feed on them. The concepts in this chapter are critical for understanding the conversion of energy from one form to another and for research into the availability and deployment of energy.
Assignment for Chapter 17
Exercises:17.3, 17.8, 17.14, 17.19, 17.28, 17.35, 17.45, 17.53
Supplementary Exercises:17.64, 17.69, 17.78
Applied Exercises:
17.81
Integrated Exercises:17.90
The Big Question
• Energy (loss or gain) is required—obvious!• How energy is distributed also affects the
tendency of a change—not that obvious!• Both entropy and enthalpy are important to drive a
change although some processes are primarily driven by enthalpy while others by entropy. Overall, the Gibbs free energy is the parameter that controls the possibility of a change.
Why do some changes take place?What parameter(s) control or drive a change?
We’ll find how this conclusion is drawn.
Spontaneous Change
Fig.1 The direction of spontaneous change is for a hot block of metal (left) to cool to the temperature of its surroundings (right). A block at the same temperature as its surroundings does not spontaneously become hotter.
A change that tends to occur without needing to be driven by an external influence.
Fig.2 The direction of spontaneous change for a gas is toward filling its container. A gas that already fills its container does not collect spontaneously in a small region of the container. A glass cylinder containing a brown gas (upper piece of glassware in the left illustration) is attached to an empty flask. When the stopcock between them is opened, the brown gas fills both upper and lower vessels (right illustration). The brown gas is nitrogen dioxide.
NO2
Fig.3 We can understand the natural direction of the migration of heat from a hot region to a cold region by thinking about the jostling between the vigorously moving atoms in the hot region. Molecules jostle their neighbors, and the thermal motion spreads.
Spontaneity—the spreading of energy to more and more degrees of freedom. Entropy is the measure of the number of degrees of freedomaffected by thermal motion.
Fig.4 We can understand the natural direction of the migration of matter by visualizing how the random motion of molecules results in their spreading throughout the available space.
The second law of thermodynamics:The entropy of an isolated systemtends to increase.
Spontaneity—the spreading of energy to more and more degrees of freedom. Entropy is the measure of the number of degrees of freedomaffected by thermal motion.
Fig.5 A representation of the arrangement of molecules in (a) a solid and (b) a liquid. When the solid melts, there is an increase in the disorder of the system and hence a rise in entropy.
For a given sample and at the same temperature, the entropy of liquid state is larger than that of solid state.
Fig.6 The entropy of a solid increase as its temperature is raised. The entropy increase sharply when the solid melts to form the more disordered liquid and then gradually increases again up to the boiling point. A second, larger jump in entropy occurs when the liquid turns into a vapor.
For a given sample, the entropy is larger at higher temperatures than that at lower temperatures.
Example: predicting the relative entropies of two samples
Which has the greater entropy: (a) 1 g of pure solid NaCl or 1 g of NaCl dissolved in 100 mL of water; (b) 1 g of water at 25 oC or a 1 g of waterat 50 oC?
Ans:• 1 g of NaCl dissolved in 100 mL of water.• 1 g of water at 50 oC.
For a given sample and at the same temperature, the entropy of liquid state is larger than that of solid state.
For a given sample, the entropy is larger at higher temperatures than that at lower temperatures.
Exercise: predicting the relative entropies of two samples
Which has the greater entropy: 1 mol of CO2 (s) or 1 mol of CO2 (g)at the same temperature?
Ans: 1 mol of CO2 (g)
Which has the greater entropy: a sample of liquid mercury at -15 oC or the same sample at 0 oC?
Ans: The sample of liquid mercury at 0 oC.
Entropy: Macroscopic Definition and Calculation
heat supplied Change in entropy =
temperature at which
the transfer takes place
, unit: J/K
reversibly
rev
rev
q dqS
T T
The greater the energy transferred to the system as heat, the greater the increase in entropy.
If the transfer is made to a hot system, the increase in entropy is smaller than when the same amount of energy is transferred to a cool system.
A reversible process is one that can be reversed by an infinitesimal change in a variable. (when heat is transferred to a system, the sourcemust have the same temperature as the system itself.)
Calculating the Change in Entropy
Calculate the change in entropy of a large tank of water when a totalof 100.0 J of energy is transferred to it reversibly as heat at 20.0 oC.
100.0J293 K
= 0.341 + J/KrevqS
T
Calculate the change in entropy of a large iron block at 20.0 oC when1500.0 J of energy escapes as heat from the block to the surroundings.
1500.0J297 K
= 5.05 - J/KrevqS
T
Classroom Exercise:Calculating the Change in Entropy
Calculate the change in entropy of a large swimming pool at 28.0 oCwhen 240.0 J of escapes from the pool as heat to the surroundings.
240.0J299 K
= 0.803 - J/KrevqS
T
Fusion and vaporization entropies
The discontinuities correspondto phase transition.
The jump of entropy is the Signature of first order phase transition.
Calculating the Entropy of Vaporization
Calculate the change in molar entropy when water vaporizes at its boiling point.
40.7kJ/mol373 K
= 109 J/ /m l+ K ovaprevvap
b b
HqS
T T
Calculating the Entropy of Fusion
Calculate the change in molar entropy of ice at its melting point.(Look Table 6.2 for fusion enthalpy)
6.06kJ/mol273 K
= 22.0 J/ /mol+ Kfusrevfus
m m
HqS
T T
Classroom Exercise:Calculating the Entropy of Vaporization
Calculate the change in molar entropy when ammonia vaporizes at its boiling point (239.7 K). The vaporization enthalpy of ammoniais 23.4 kJ/mol.
23.4kJ/mol239.7 K
= 97.62 J/K/mo+ lvaprevvap
b b
HqS
T T
Entropy: Microscopic Definition and Calculation
Entropy = The number of microscopic states for a given macroscopic state
lnS k W k: Boltzmann constant
Entropy is a measure of how the energy of a system is storedfor a given macroscopic state.
Absolute Entropies
• The third law of thermodynamics: The entropy of a perfect crystal approaches 0 as the absolute temperature approaches 0.
• All absolute entropies are positive Standard molar entropy Smo (298 K).
More complicated compounds have higher entropies
Entropies are higher at higher temperatures
Fig.7 The entropy change due to heat transfer depends on both the amount of heat transferred and the temperature of the system. A lot of heat transferred to a cold system (upper left) results in a large increase in the entropy of the system. A small quantity of heat transferred to a hot system (lower right) results in a small increase in entropy of the system. You take more entropy after you drink a cup of
hot tea than a cup of iced tea.You take more entropy when you’re in fever than when you’re normal after you drink a cup of tea.
The entropy is higher for
• Higher temperature
• Larger volume
• More complex structures
• Larger sample size
• Heavier atoms (entropy is not disorder!)
• Vapor relative to liquid or solid
• Liquid relative to solid
Estimating the relative value of the molar entropy
Which substance in each pair has the higher molar entropy: (a) CO2 at 25 oC and 1 atm or CO2 at 25 oC and 3 atm; (b) Br2(l) or Br2(g) at the same temperature and pressure; (c) methane gas, CH4, or propane gas, CH3CH2CH3 at the same temperature and pressure?
(a) One mole of CO2 at 25 oC and 1 atm occupies larger volume than one mole of CO2 at 25 oC and 3 atm One mole of CO2 at 25 oC and 1 atm has the higher molar entropy.
(b) Gas has the higher molar entropy than liquid. Therefore, Br2(g) has the higher molar entropy at the same temperature and pressure.
(c) One mole of CH4 is lighter than one mole of CH3CH2CH3 . Therefore, CH3CH2CH3 has the higher molar entropy at the same temperature and pressure.
Exercise Which substance in each pair has the higher molar
entropy: (a) He at 25 oC or He at 100 oC in a container of the same volume; (b) Br(g) or Br2(g) at the same temperature and pressure?
The higher the temperature, the higher the molar entropyHe at 100 oC has the higher molar entropy than He at 25 oC in a container of the same volume.The heavier the molecule/atom, the higher the molar entropyBr2(g) has the higher molar entropy than Br(g) at the same temperature and pressure.
Classroom Exercise Which substance in each pair has the higher molar
entropy at the same temperature and pressure : (a) Pb(s) or Pb(l); (b) SbCl3(g) or SbCl5(g) ?
Liquid has the higher molar entropy than solid. Therefore, Pb(l) has the higher molar entropy than Pb(s) at the same temperature and pressure.
The heavier the molecule/atom, the higher the molar entropySbCl5(g) has the higher molar entropy than SbCl3(g) at the same temperature and pressure.
Reaction Entropyo o o
r m mΔS = nS (products) nS (reactants) (2)
Because the molar entropy of a gas is so much greater than that of Solids and liquids, a change in the amount of gas normally dominatesany other entropy change in a reaction. A net consumption of gas usually results in a negative reaction entropy. A net production ofgas usually results in a positive reaction entropy.
N2(g) + 3H2 (g) 2NH3 (g)
Reactant: 4 mol, Product: 2 mol decrease in entropy.o o o o
r m 3 m 2 m 2ΔS = 2S (NH ,g) [S (N ,g) 3S (H ,g)]
2 192.4 (191.6 3 130.7)
198.9 J/K/mol
The standard molar entropies of common compounds are listed in Appendix 2
Exercise
N2O4 (g) 2NO2 (g)
Reactant: 1 mol, Product: 2 mol increase in entropy.
o o or m 2 m 2 4ΔS = 2S (NO ,g) S (N O ,g)
2 240.06 304.29
175.83 J/K/mol
Classroom Exercise
C2H4 (g) +H2(g) C2H6 (g)
Reactant: 2 mol, Product: 1 mol decrease in entropy.
o o o or m 2 6 m 2 m 2 4ΔS = S (C H ,g) [S (H ,g) S (C H ,g)]
229.60 219.56 130.68
120.64 J/K/mol
Why does ice freeze spontaneously?Why exothermic reactions occur
spontaneously?
Total entropy change
=entropy change of system
+entropy change of surroundings
(3)tot surrS S S
A process is spontaneous as long as the total entropy change is positive. A spontaneous process does NOT require the increase of the entropy of the system.
Fig.8 (a) In an exothermic process, heat escapes into the surroundings and increases their entropy. (b) In an endothermic process, the entropy of the surroundings decreases. The blue-green arrows indicate the direction of entropy change in the surroundings.
Entropy change of surroundings
heat trandferred to surroundings=
temperature of surroundings
enthalpy change of system=
temperature of surroundings
(4)surr
HS
T
Exothermic reactions occur spontaneously because the increase of the entropy of the surroundings is more than the decrease of the system.
N2(g) + 3H2 (g) 2NH3 (g)
Standard reaction enthalpy = -92.22 kJ/mol < 0 exothermic The entropy of the surroundings increases.
92220J/mol298K 309J/K/molsurr
HS
T
o o o or f 3 f 2 f 2ΔH = 2H (NH ,g) [ (N ,g) 3 (H ,g)]
2 ( 46.11) (0 3 0) kJ/mol
92.22 kJ/mol
H H
Fig.9 In an exothermic reaction, (a) the overall entropy change is certainly positive when the entropy of the system increases. (b) The overall entropy change is positive even when the entropy of the system decreases, provided that the entropy increase in the surroundings is greater. The reaction is spontaneous in both cases.
surr
tot surr
ΔS > 0,ΔS > 0
ΔS = ΔS + ΔS > 0
, surr
tot surr
surrΔS >|ΔS 0,ΔS > 0 ,
ΔS = Δ
ΔS |
S + ΔS > 0
Fig.10 An endothermic reaction is spontaneous only when the entropy of the system increases enough to overcome the decrease in entropy of the surroundings, as it does here.
| |, surr
tot
sur
surr
rΔS ΔSΔS 0,ΔS 0 ,
ΔS = ΔS + ΔS > 0
A process is spontaneous if the change of total entropy is positive
Is the dissolution of ammonium nitrate to form a dilute aqueous solution spontaneous at 25 oC?
NH4NO3 (s) NH4+ (aq) + NO3
-(aq)
o o + o - osol f 4 f 3 f 4 3Δ = (NH ,aq) (NO ,aq) (NH NO ,s)
135.21 205.0 ( 365.56)kJ/mol
28.0kJ/mol
H H H H
Reaction entropy:
(endothermic)
28000J/mol298K 94J/K/molsurr
HS
T
o o + o - osol m 4 m 3 m 4 3
o o otot sol surr
Δ = (NH ,aq) (NO ,aq) (NH NO ,s)
113.4 146.4 151.08 J/K/mol 108.7 J/K/mol
Δ = Δ Δ = 108.7-94.0 J/K/mol = 14.7 J/K/mol+
S S S S
S S S
A process is spontaneous if the change of total entropy is positiveA model for the combustion of wood:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) +6H2O(g)
o o o o osol f 2 f 2 f 6 12 6 f 2Δ = 6 (CO ,g) 6 (H O,g) (C H O ,s) 6 (O ,g)
6 393.51 6 241.82 ( 1268) 6 0kJ/mol
2543.98kJ/mol
H H H H H
Reaction entropy:
(exothermic)
2543980J/mol298K 8536.85J/K/molsurr
HS
T
o o o o ocomb m 2 m 2 m 2 m 6 12 6
o o otot comb surr
Δ = 6 (CO ,g) 6 (6H O,g) 6 (6O ,g) (C H O ,s)
6 213.74 6 188.83 212 J/K/mol 2203.42 J/K/mol
Δ = Δ Δ = 8536.85+2203.42 J/K/mol = 10.7403 kJ/K/mo+ l
S S S S S
S S S
Gibbs free energy must decrease for a spontaneous
change: ΔStot>0ΔG = -TΔStot<0
t
tot surr
tot
ot
S S S
HS S
G TT
H T SS
Josiah Willard Gibbs (1839–1903).
Free energy is effectively the sameas the total entropy (except for thenegative sign and coefficient T)
ΔG < 0 the process is spontaneousΔG > 0 the reverse of the process is spontaneousΔG = 0 both the process and its reverse are spontaneous equilibrium.
Free energy must decrease for a spontaneous
change: ΔG<0
Free energy and equilibrium
3
3
6.00 kJ/mol
21.97 J/k mol= 21.97 10 kJ/k mol
( 6.00kJ/mol) (273.15k) ( 21.97 10 kJ/k mol)
=0.00kJ/mol
freeze fus
freeze fus
freeze freeze freeze
freeze
H H
S S
G H T S
G
0totG T S
At equilibrium, both directions are equally spontaneous, then
This condition applies to any phase change and any chemical reactionat equilibrium at constant temperature and pressure.
Example: water-ice equilibrium----
Predicting the boiling point of a substance
Liquid metals, such as mixtures of sodium and potassium, are used as coolants in some nuclear reactors. Predict the normal boiling point of liquid sodium, given that the standard entropy of vaporization of liquid sodium is 84.8 J/K/mol and that its standard enthalpy of vaporization is 98.0 kJ/mol
0 0tot vap b vapG T S H T S At equilibrium, both directions are equally spontaneous, then
vap
vap
H
b ST 98000J/mol
84.8J/K/mol 1160KbT
Predicting the melting point of a substance
Predict the normal melting point of solid chlorine, given that the standard entropy of fusion is 837.3 J/K/mol and that its standard enthalpy of fusion is 6.41 kJ/mol.
0 0tot fus m fusG T S H T S
At equilibrium, both directions are equally spontaneous, then
fus
fus
H
m ST 6410J/mol
37.3 J/K/mol =172KmT
Classroom Exercise:Predicting the boiling point of methanol
Predict the normal boiling point of methanol, CH3OH, given that the standard entropy of vaporization is 104.7 J/K/mol and that its standard enthalpy of vaporization is 35.3 kJ/mol.
vap
vap
H
b ST
o35300J/mol104.7J/K/mol 337K=64 CbT
Case Study 17 (a)The bubbles on the leaves of this underwater plant are oxygen produced by photosynthesis. Molecules such as the chlorophyll that colors the leaves green capture sunlight to begin the transformation of carbon dioxide and water to glucose and oxygen.
Case Study 17 (b)A weight with a small mass can be lifted into the air by another weight of the same or greater mass. What would appear unnatural if we saw it by itself (a weight rising) is actually part of a spontaneous event overall. The “natural” fall of the heavier weight causes the “unnatural” rise of the smaller weight.
r r rHG T S
Standard Reaction Free Energies
o o or f fΔH = n H (products) n H (reactants)
o o or m mΔS = nS (products) nS (reactants)
o o or f fΔG = n G (products) n G (reactants)
Standard free energies of formation
The most stable form of an element is the state with lowest free energy of formation.
minimizedfG
The most stable form of a compound is the state with lowest free energy of formation.
minimizedfG
2 2
1 1H (g) I (s) HI(g)
2 2
r r rHG T S
Standard Reaction Free Energies
Calculate the standard free energy of formation of HI(g) at 25 oC fromIts standard entropy and standard enthalpy of formation.
r f f 2 f 2
f
1 1H H (HI,g) H (H ,g) H (I ,s)
2 2
= H (HI,g)=+26.48kJ/mol
r m m 2 m 2
1 1S S (HI,g) (H ,g) (I ,s)
2 2
1 1206.59 130.68 116.14J/K/mol
2 2=83.18J/K/mol
S S
r r r
f
H
26.48 kJ/mol -298 K 0.08318 kJ/K/mol
=1.69 kJ/mol = (HI,g)
S
G
G T
2 2 3
1 3N (g)+ H (g) NH (g)
2 2
More Exercise
Calculate the standard free energy of formation of NH3(g) at 25 oC.
r f 3 f 2 f 2
f 3
3 1H H (NH ,g) H (H ,g) H (N ,g)
2 2
= H (NH ,g)=-46.11 kJ/mol
r m 3 m 2 m 2
3 1S S (NH ,g) (H ,g) (N ,g)
2 2
3 1192.45 130.68 191.61J/K/mol
2 2=99.375 J/K/mol
S S
r
3
r
f
rH
46.11 kJ/mol -298 K 0.099375 kJ/K/mol
=-16.5 kJ/mol = (NH ,g)
G S
G
T
2 3 63C(s)+3H (g) C H (g)
Classroom ExerciseCalculate the standard free energy of formation of cyclopropane, C3H6(g) at 25 oC.
r f 3 6 f 2 fH H (C H ,g) 3 H (H ,g) 3 H (C,s)
=20.42-3 0-3 716.68 kJ/mol=-2129.62 kJ/mol
r m 3 6 m 2 mS S (C H ,g) 3 (H ,g) 3 (C,s)
237.4 3 130.68 3 158.10J/K/mol
=-628.94 J/K/mol
S S
r
6
r
f 3
rH
2129.62 kJ/mol -298 K ( 0.62894) kJ/K/mol
=-1942.1959 kJ/mol = (C H ,g)
G T S
G
Figure 17.12 The standard free energies of formation of compounds are defined as the standard reaction free energy for their formation from the elements. They represent a thermodynamic “altitude” with respect to the elements at “sea level.” The numerical values are in kilojoules per mole.
If the standard free energy of formation of a compound is smaller than 0, , then it is thermodynamically stable.
Standard Free Energy of Formation Decides Thermodynamic Stability
0fG
2 6 6 f
6 6 2 f
6C(s)+3H (g) C H (l) G 124 kJ/mol >0
C H (l) 6C(s)+3H (g) G 124 kJ/mol<0
Thermodynamically unstable
Thermodynamically stable
Example:
If the standard free energy of formation of a compound is larger than 0, then it is thermodynamically unstable.0,fG
Standard Free Energy of Formation Decides Thermodynamic Stability
Is glucose stable relative to its elements at 25 oC and under standard conditions?
C6H12O6 (s) 6C (s) + 6H2(g) + 3O2(g)
o o o of f 6 12 6 f f 2 f 2Δ = (C H O ,s) 6 (C,s) 6 (H ,g) 3 (O ,g)
910.0 0 0 0kJ/mol
910kJ/mol<0
G G G G G
Standard free energy of formation (From Appendix 2A—Page A13):
thermodynamically stable.
Is methylamine, CH3NH2, stable relative to its elements at 25 oC and under standard conditions?
of 3 2(CH NH ,g) 32.16kJ/mol > 0G thermodynamically unstable.
Standard Free Energy of Formation Decides Thermodynamic Stability
Is methylamine, CH3NH2, stable relative to its elements at 25 oC and under standard conditions?
of 3 2(CH NH ,g) 32.16kJ/mol > 0G thermodynamically unstable.
Look up Appendix 2A, page A13, and we find the standard free energy of formation of methylamine is 32.16 kJ/mol.
o o or f fΔG = nΔG (products) nΔG (reactants)
The Chemical Reaction Proceeds So That The Free Energy of Reaction
< 0. orΔG
3 2 24NH (g)+5O (g) 4NO(g)+6H O(g)
o o o o or f f 2 f 3 f 2ΔG = 4ΔG (NO,g)+6ΔG (H O,g) 4ΔG (NH ,g)+5ΔG (O ,g)
= 4 86.55+6 ( 228.57) 4 ( 16.45)+0 kJ/mol
= 959.42 kJ/mol
spontaneous reaction.
r f 2 f f 22 (CO , ) 2 (CO, ) (O , )G G g G g G g
Negative Free Energy of Reaction Means Spontaneous Reaction
2 22CO(g)+O (g) 2CO (g)
Very negative spontaneous reaction.
2 ( 394.36) 2 ( 137.17) 0
514.38kJ/mol
Look up Appendix 2A, we have
r f 3 f 2 f 22 (SO , ) 2 ( , ) (O , )G G g G SO g G g
Negative Free Energy of Reaction Means Spontaneous Reaction
2 2 32SO (g)+O (g) 2SO (g)
Negative spontaneous reaction.
2 ( 371.06) 2 ( 300.19) 0
141.74kJ/mol
Look up Appendix 2A, we have
r f 6 12 6 f 2 f 2 f 2(C H O , ) 6 (O , ) 6 (CO , ) 6 (H O, )G G s G g G g G l
Negative Free Energy of Reaction Means Spontaneous Reaction
2 2 6 12 6 26CO (g)+6H O(l) C H O ( ) 6O (g)s
Very positive not spontaneous reaction (the reverse is).
910 6 ( 394.36) 6 ( 237.13)
2878.94kJ/mol
Look up Appendix 2A, we have
In biological systems, glucose is synthesized by assistance of a special bioenzyme.
or rΔG =ΔG +RT 1n Q
RT=(8.314 51J/k mol) (298.15 k)=2.4790 kJ/mol at 298.15K
Reaction Free Energy Varies With Temperature
Reaction free energy can be determined from reaction quotient.
Figure 17.13 At constant temperature and pressure, the direction of spontaneous change is toward lower free energy. The equilibrium composition of a reaction mixture corresponds to the lowest point on the curve. In this example, substantial quantities of both reactants and products are present at equilibrium, and K is close to 1.
Figure 17.14 In this reaction, the free energy is a minimum when products are much more abundant than reactants. The equilibrium lies in favor of the products, and K 1. This reaction effectively goes almost to completion.
Figure 17.15 In this reaction, the free energy is a minimum when the reactants are much more abundant than the products. The equilibrium lies in favor of the reactants, and K 1. This reaction “does not go.”
Free Energy of Reaction Gives The Temperature Dependence of The Equilibrium Constant
or
r
o
or
-o
r
0=ΔG + lnK
ΔG = lnK
K<1 when ΔG >0
K>1 when ΔG <0
oGrRTK e
RT
RT
2
2 4
2NO
2 4 2 pN O
N O (g) 2NO (g) KP
P
2 4 2N O (g) 2NO (g)
Equilibrium Constant Can Be Found From Free Energies
Calculate Kp at 25 oC for the equilibrium
r f 2 f 2 4
r 4730J/molp 8.3145J/K/mol×298K
p
2 ( , ) ( , )
ln 1.91RT
0.15
G G NO g G N O g
GK
K
Figure 17.16 A negative value of the standard reaction free energy corresponds to an equilibrium constant greater than 1 and to products (yellow) favored over reactants (purple) at equilibrium. A positive value of the standard reaction free energy corresponds to an equilibrium constant of less than 1 and to reactants favored over products at equilibrium.
Estimate the minimum temperature at which K>1
K>1 when the reaction free energy becomes negative
0
r
r
r r r
H
S
G H T S
T
Objectives (1)
Skills You Should Have Mastered• Conceptual 1. State and explain the implications of the second law of
thermodynamics, Section 17.2. 2. Explain how temperature, volume, and state of matter affect the
entropy of a substance, Sections 17.2 and 17.3.
3. Show how ΔSsurr is related to ΔH for a change at constant temperature and pressure and justify the relationship, Section 17.5.
4. Show how the free energy change accompanying a process is related to the direction of spontaneous reaction and the position of equilibrium, Sections 17.7 and 17.10.
• Descriptive 1. Describe the criteria for spontaneity of a reaction, Sections 17.5
and 17.6. 2. Identify thermodynamically unstable compounds from their
standard free energies of formation, Section 17.8.
Objectives (2)• Problem-Solving1. Predict which of two systems has the greater entropy, given their compositions
and conditions, Toolbox 17.1 and Example 17.1.2. Calculate the change in entropy of a system due to heat transfer and phase
changes, Toolbox 17.1 and Examples 17.2 and 17.3.3. Estimate the relative molar entropies of two substances, Toolbox 17.1 and
Example 17.4.4. Calculate the standard reaction entropy from standard molar entropies,
Example 17.5.5. Judge the spontaneity of a reaction from its standard reaction enthalpy and
standard reaction entropy, Example 17.6.6. Predict the boiling point and melting point of a substance from the changes in
entropy and enthalpy of the substance, Example 17.7.7. Calculate a standard free energy of formation from the standard enthalpy of
formation and standard molar entropies, Example 17.8.8. Calculate the standard reaction free energy from free energies of formation,
Example 17.9.9. Calculate the reaction free energy from ΔGr° and the reaction quotient,
Example 17.10.10. Calculate an equilibrium constant from ΔGr° at a given temperature, Toolbox
17.2 and Example 17.11.11. Predict the temperature at which a process with known DH and DS becomes
spontaneous, Example 17.12.
Assignment for Chapter 17
Exercises:17.3, 17.8, 17.14, 17.19, 17.28, 17.35, 17.45, 17.53
Supplementary Exercises:17.64, 17.69, 17.78
Applied Exercises:
17.81
Integrated Exercises:17.90
