Transcript No Slide Title

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
Why do some changes take place?
What parameter(s) control or drive a change?
• 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.
We’ll find how this conclusion is drawn.
Spontaneous Change
A change that tends to occur without needing to be driven by
an external influence.
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.
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 freedom
affected 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.
Spontaneity—the spreading of
energy to more and more degrees
of freedom. Entropy is the measure
of the number of degrees of freedom
affected by thermal motion.
The second law of thermodynamics:
The entropy of an isolated system
tends to increase.
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 water
at 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 reversibly
Change in entropy =
temperature at which
the transfer takes place
qrev
dq
S 
  , unit: J/K
T
T
rev
A reversible process is one that can be reversed by an infinitesimal
change in a variable. (when heat is transferred to a system, the source
must have the same temperature as the system itself.)
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.
Calculating the Change in Entropy
Calculate the change in entropy of a large tank of water when a total
of 100.0 J of energy is transferred to it reversibly as heat at 20.0 oC.
qrev 100.0J
S 
 293 K =+0.341 J/K
T
Calculate the change in entropy of a large iron block at 20.0 oC when
1500.0 J of energy escapes as heat from the block to the surroundings.
qrev 1500.0J
S 
 297 K =-5.05 J/K
T
Classroom Exercise:
Calculating the Change in Entropy
Calculate the change in entropy of a large swimming pool at 28.0 oC
when 240.0 J of escapes from the pool as heat to the surroundings.
qrev 240.0J
S 
 299 K =-0.803 J/K
T
Fusion and vaporization
entropies
The discontinuities correspond
to 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.
Svap
qrev H vap



Tb
Tb
40.7kJ/mol
373 K
=+109 J/K/mol
Calculating the Entropy of Fusion
Calculate the change in molar entropy of ice at its melting point.
(Look Table 6.2 for fusion enthalpy)
S fus
qrev H fus



Tm
Tm
6.06kJ/mol
273 K
=+22.0 J/K/mol
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 ammonia
is 23.4 kJ/mol.
Svap
qrev H vap



Tb
Tb
23.4kJ/mol
239.7 K
=+97.62 J/K/mol
Entropy:
Microscopic Definition and Calculation
Entropy = The number of microscopic states for a given macroscopic state
S  k ln W
k: Boltzmann constant
Entropy is a measure of how
the energy of a system is stored
for 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
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.
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.
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 entropy
He 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 entropy
Br2(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 entropy
SbCl5(g) has the higher molar entropy than SbCl3(g) at the same
temperature and pressure.
ΔSr o
Reaction Entropy
=nS (products)  nS (reactants)
o
m
o
m
(2)
The standard molar entropies of common compounds are listed in
Appendix 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 dominates
any other entropy change in a reaction. A net consumption of gas
usually results in a negative reaction entropy. A net production of
gas usually results in a positive reaction entropy.
N2(g) + 3H2 (g)  2NH3 (g)
Reactant: 4 mol, Product: 2 mol
 decrease in entropy.
ΔSr o = 2Som (NH3 ,g)  [Som (N 2 ,g)  3Som (H 2 ,g)]
 2 192.4  (191.6  3 130.7)
 198.9 J/K/mol
Exercise
N2O4 (g)  2NO2 (g)
Reactant: 1 mol, Product: 2 mol
 increase in entropy.
ΔSr o = 2Som (NO2 ,g)  Som (N 2O4 ,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.
ΔSr o = Som (C2H6 ,g)  [Som (H 2 ,g)  Som (C2H 4 ,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
Stot  S  S surr
(3)
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
S surr  
H
T
(4)
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)
ΔH r o = 2Hfo (NH3 ,g)  [ H fo (N 2 ,g)  3H fo (H 2 ,g)]
 2  (46.11)  (0  3  0) kJ/mol
 92.22 kJ/mol
Standard reaction enthalpy = -92.22 kJ/mol < 0
 exothermic
 The entropy of the surroundings increases.
S surr
H


T
92220J/mol
298K
 309J/K/mol
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.
ΔS > 0,ΔSsurr > 0
ΔS  0,ΔSsurr > 0, ΔSsurr >| ΔS |,
ΔStot = ΔS + ΔSsurr > 0
ΔStot = ΔS + ΔSsurr > 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.
ΔS  0,ΔSsurr  0,| ΔSsurr | ΔS,
ΔStot = ΔS + ΔSsurr > 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)
Reaction entropy:
ΔH sol o = H fo (NH +4 ,aq)  H fo (NO3- ,aq)  H fo (NH 4 NO3 ,s)
 135.21  205.0  (365.56)kJ/mol
 28.0kJ/mol (endothermic)
S surr
H

  28000J/mol
 94J/K/mol
298K
T
ΔSsol o = Smo (NH +4 ,aq)  Smo (NO3- ,aq)  Smo (NH 4 NO3 ,s)
 113.4  146.4  151.08 J/K/mol  108.7 J/K/mol
ΔS tot o = ΔSsol o  ΔSsurr o = 108.7-94.0 J/K/mol = +14.7 J/K/mol
A process is spontaneous if the
change of total entropy is positive
A model for the combustion of wood:
C6H12O6 (s) + 6O2 (g)  6CO2 (g) +6H2O(g)
Reaction entropy:
ΔH sol o = 6H fo (CO2 ,g)  6H fo (H 2O,g)  H fo (C6H12O6 ,s)  6H fo (O 2 ,g)
 6  393.51  6  241.82  (1268)  6  0kJ/mol
 2543.98kJ/mol(exothermic)
S surr
H

  2543980J/mol
 8536.85J/K/mol
298K
T
ΔScomb o = 6Smo (CO 2 ,g)  6 Smo (6H 2O,g)  6 Smo (6O 2 ,g)  Smo (C 6H12O 6 ,s)
 6  213.74  6 188.83  212 J/K/mol  2203.42 J/K/mol
ΔS tot o = ΔScomb o  ΔSsurr o = 8536.85+2203.42 J/K/mol = +10.7403 kJ/K/mol
Gibbs free energy must decrease for a spontaneous change:
ΔStot>0ΔG = -TΔStot<0
Stot  S  S surr
H
Stot  S 
T
G  T Stot  H  T S
Josiah Willard Gibbs (1839–1903).
Free energy is effectively the same
as the total entropy (except for the
negative sign and coefficient T)
ΔG < 0  the process is spontaneous
ΔG > 0  the reverse of the process is spontaneous
ΔG = 0  both the process and its reverse are spontaneous
 equilibrium.
Free energy must decrease for a spontaneous change:
ΔG<0
Free energy and equilibrium
At equilibrium, both directions are equally spontaneous, then
G  T Stot  0
This condition applies to any phase change and any chemical reaction
at equilibrium at constant temperature and pressure.
Example: water-ice equilibrium----
H freeze  H fus  6.00 kJ/mol
S freeze  S fus  21.97 J/k  mol=  21.97 103kJ/k  mol
G freeze  H freeze  T S freeze
G freeze  (6.00kJ/mol)  (273.15k)  (  21.97 103kJ/k mol)
=0.00kJ/mol
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
At equilibrium, both directions are equally spontaneous, then
G  T Stot  0  Hvap  TbSvap  0
Tb 
H vap
Svap
Tb 
98000J/mol
84.8J/K/mol
 1160K
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.
At equilibrium, both directions are equally spontaneous, then
G  T Stot  0  H fus  TmS fus  0
Tm 
H fus
S fus
Tm 
6410J/mol
37.3 J/K/mol
=172K
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.
Tb 
Tb 
35300J/mol
104.7J/K/mol
H vap
Svap
 337K=64 C
o
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.
Standard Reaction Free Energies
Gr  Hr  T Sr
ΔHr =nHf (products)  nHf (reactants)
o
o
o
ΔSr =nS (products)  nS (reactants)
o
o
m
o
m
ΔGr o =nGf o (products)  nGf o (reactants)
Standard free energies of formation
The most stable form of an element is the
state with lowest free energy of formation.
Gf minimized
The most stable
form of a
compound is the
state with lowest
free energy of
formation.
Gf minimized
Standard Reaction Free Energies
Gr  Hr  T Sr
Calculate the standard free energy of formation of HI(g) at 25 oC from
Its standard entropy and standard enthalpy of formation.
1
1
H 2 (g)  I 2 (s)  HI(g)
2
2
1
1

H r  Hf (HI,g)   Hf (H 2 ,g)  Hf (I2 ,s) 
2
2

=Hf (HI,g)=+26.48kJ/mol
1
1

Sr  Sm (HI,g)   S m (H 2 ,g)  S m (I 2 ,s) 
2
2

1
1
 206.59  130.68  116.14J/K/mol
2
2
Gr  H r  T S r
=83.18J/K/mol
 26.48 kJ/mol -298 K  0.08318 kJ/K/mol
=1.69 kJ/mol = Gf (HI,g)
More Exercise
Calculate the standard free energy of formation of NH3(g) at 25 oC.
1
3
N 2 (g)+ H 2 (g)  NH3 (g)
2
2
1
3

H r  Hf (NH3 ,g)   Hf (H 2 ,g)  Hf (N 2 ,g) 
2
2

=Hf (NH3 ,g)=-46.11 kJ/mol
1
3

Sr  Sm (NH 3 ,g)   S m (H 2 ,g)  S m (N 2 ,g) 
2
2

3
1
 192.45  130.68   191.61J/K/mol
2
2
Gr  H r  T S r
=99.375 J/K/mol
 46.11 kJ/mol -298 K  0.099375 kJ/K/mol
=-16.5 kJ/mol = Gf (NH 3 ,g)
Classroom Exercise
Calculate the standard free energy of formation of cyclopropane, C3H6(g)
at 25 oC.
3C(s)+3H2 (g)  C3H6 (g)
H r  Hf (C3H6 ,g)  3H f (H 2 ,g)  3H f (C,s)
=20.42-3  0-3  716.68 kJ/mol=-2129.62 kJ/mol
Sr  Sm (C3H 6 ,g)  3S m (H 2 ,g)  3S m (C,s)
 237.4  3  130.68  3 158.10J/K/mol
=-628.94 J/K/mol
Gr  H r  T S r
 2129.62 kJ/mol -298 K  (0.62894) kJ/K/mol
=-1942.1959 kJ/mol = Gf (C3H 6 ,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.
Standard Free Energy of Formation
Decides Thermodynamic Stability
If the standard free energy of formation of a compound is smaller
than 0, G f  0 , then it is thermodynamically stable.
If the standard free energy of formation of a compound is larger than
0, G f  0, then it is thermodynamically unstable.
Example:
6C(s)+3H 2 (g)  C6H6 (l) G f  124 kJ/mol >0 Thermodynamically unstable
C6H6 (l)  6C(s)+3H 2 (g) G f  124 kJ/mol<0 Thermodynamically stable
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)
Standard free energy of formation (From Appendix 2A—Page A13):
ΔGf = Gfo (C6H12O6 ,s)  6Gfo (C,s)  6Gfo (H 2 ,g)  3Gfo (O2 ,g)
 910.0  0  0  0kJ/mol
thermodynamically stable.
 910kJ/mol<0
Is methylamine, CH3NH2, stable relative to its elements at 25 oC and
under standard conditions?
Gfo (CH3NH2 ,g)  32.16kJ/mol > 0 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?
Look up Appendix 2A, page A13, and we find the standard free
energy of formation of methylamine is 32.16 kJ/mol.
Gfo (CH3NH2 ,g)  32.16kJ/mol > 0 thermodynamically unstable.
The Chemical Reaction Proceeds So
That The Free Energy of Reaction
o
ΔG r < 0.
ΔGr =nΔGf (products) nΔGf (reactants)
o
ΔG r o
o
o
4NH3 (g)+5O2 (g)  4NO(g)+6H2O(g)
= 4ΔG (NO,g)+6ΔG (H O,g)  4ΔG (NH ,g)+5ΔG (O ,g)
o
f
o
f
o
2
f
o
3
= 4  86.55+6  (  228.57)  4  (  16.45)+0 kJ/mol
=  959.42 kJ/mol  spontaneous reaction.
f
2
Negative Free Energy of Reaction Means
Spontaneous Reaction
2CO(g)+O2 (g)  2CO2 (g)
Gr  2Gf (CO 2 , g )  2Gf (CO, g )  Gf (O 2 , g )
Look up Appendix 2A, we have
 2  (394.36)  2  (137.17)  0
 514.38kJ/mol
Very negative  spontaneous reaction.
Negative Free Energy of Reaction Means
Spontaneous Reaction
2SO2 (g)+O2 (g)  2SO3 (g)
Gr  2Gf (SO3 , g )  2Gf ( SO2 , g )  Gf (O 2 , g )
Look up Appendix 2A, we have
 2  (371.06)  2  (300.19)  0
 141.74kJ/mol
Negative  spontaneous reaction.
Negative Free Energy of Reaction Means
Spontaneous Reaction
6CO2 (g)+6H2O(l)  C6H12O6 (s)  6O2 (g)
Gr  Gf (C6H12O6 , s)  6Gf (O2 , g )  6Gf (CO 2 ,g )  6Gf (H 2O, l )
Look up Appendix 2A, we have
 910  6  (394.36)  6  (237.13)
 2878.94kJ/mol
Very positive  not spontaneous reaction (the reverse is).
In biological systems, glucose is synthesized by assistance of
a special bioenzyme.
Reaction Free Energy Varies With Temperature
ΔG r =ΔG r o +RT 1n Q
RT=(8.314 51J/k  mol)  (298.15 k)=2.4790 kJ/mol at 298.15K
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
0=ΔG r +RT lnK
o
ΔG r =  RT lnK  K  e
o
K<1 when ΔG r >0
o
K>1 when ΔG r <0
o
Gr o
- RT
Equilibrium Constant Can Be Found From Free Energies
Calculate Kp at 25 oC for the equilibrium
N2O4 (g)
N 2O 4 (g)
2NO2 (g)
2NO 2 (g)
Kp 
PNO2 2
PN2O4
Gr  2Gf ( NO2 , g )  Gf ( N 2O4 , g )
Gr
4730J/mol
ln K p  
  8.3145J/K/mol×298K
 1.91
RT
K p  0.15
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

r

r

r
G  H  T S  0
T
H r
Sr
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.
• Problem-Solving
Objectives (2)
1. 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