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Chapter Sixteen:
SPONTANEITY,
ENTROPY, AND FREE
ENERGY
講義
Assignment
• 1-100題裡每10題中挑選1-2題完成。
Chapter 16 | Slide 2
Copyright © Houghton Mifflin Company. All rights reserved.
• 本章投影片做為課前課後參考。
• 在課堂講解的投影片主要採用Atkins書的第
17章：
http://140.117.34.2/faculty/phy/sw_ding/teaching/gchem/gc17.ppt
Chapter 16 | Slide 3
Copyright © Houghton Mifflin Company. All rights reserved.
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
o
dissolved
o in 100 mL of water; (b) 1 g of water at 25 C or a 1 g of water
at 50 C?
Ans:
•
•
1 g of NaCl dissolved
o in 100 mL of water.
1 g of water at 50 C.
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)
o
Which has the greater entropy:
a
sample
of
liquid
mercury
at
-15
C
o
or the same sample at 0 C?
o
Ans: The sample of liquid mercury at 0 C.
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
o
of 100.0 J of energy is transferred to it reversibly as heat at 20.0 C.
qrev
S 

T
100.0J
293 K
=+0.341 J/K
o
Calculate the change in entropy of a large iron block at 20.0 C when
1500.0 J of energy escapes as heat from the block to the surroundings.
qrev
S 

T
1500.0J
297 K
=-5.05 J/K
Classroom Exercise:
Calculating the Change in Entropy
o
Calculate the change in entropy of a large swimming pool at 28.0 C
when 240.0 J of escapes from the pool as heat to the surroundings.
qrev
S 

T
240.0J
299 K
=-0.803 J/K
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?
o
(a) oOne mole of CO2 at 25 C and 1 atm occupies
larger volume than one mole of CO2 at 25
o
C and 3 atm One mole of CO2 at 25 C 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 higherothe temperature, the higher the molar entropy
o
He at 100 C has the higher molar entropy than He at 25 C
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.
Reaction Entropy
ΔSr o = nSmo (products)   nSmo (reactants)
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
(2)
Exercise
N2O4 (g)  2NO2 (g)
Reactant: 1 mol, Product: 2 mol
 increase in entropy.
ΔSr o = 2Som (NO 2 ,g)  Som (N 2O 4 ,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 (C2 H6 ,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
o
solution spontaneous at 25 C?
+
NH4NO3 (s)  NH4 (aq) + NO3 (aq)
Reaction entropy:
o
sol
ΔH
= 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
S surr
(endothermic)
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 (CO2 ,g)  6Smo (6H 2O,g)  6 Smo (6O 2 ,g)  Smo (C6H12O6 ,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
Free energy is effectively the same
as the total entropy (except for the
negative sign and coefficient T)
Josiah Willard Gibbs (1839–1903).
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  H vap  Tb Svap  0
Tb 
H vap
Svap
98000J/mol
Tb  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
6410J/mol
Tm  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  H r  T Sr
ΔHr = nHf (products)   nHf (reactants)
o
o
o
ΔSr = nSm (products)   nSm (reactants)
o
o
o
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.
G f minimized
The most
stable form of
a compound is
the state with
lowest free
energy of
formation.
G f minimized
Standard Reaction Free Energies
Gr  H r  T Sr
o
Calculate the standard free energy of formation of HI(g) at 25 C 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  H f (HI,g)   H f (H 2 ,g)  H f (I 2 ,s) 
2
2

=H f (HI,g)=+26.48kJ/mol
1
1

Sr  Sm (HI,g)   Sm (H 2 ,g)  Sm (I 2 ,s) 
2
2

1
1
 206.59  130.68  116.14J/K/mol
2
2
Gr  H r  T Sr
=83.18J/K/mol
 26.48 kJ/mol -298 K  0.08318 kJ/K/mol
=1.69 kJ/mol = Gf (HI,g)
More Exercise
o
Calculate the standard free energy of formation of NH3(g) at 25 C.
1
3
N 2 (g)+ H 2 (g)  NH 3 (g)
2
2
1
3

H r  H f (NH3 ,g)   H f (H 2 ,g)  H f (N 2 ,g) 
2
2

=Hf (NH3 ,g)=-46.11 kJ/mol
1
3

Sr  Sm (NH 3 ,g)   Sm (H 2 ,g)  Sm (N 2 ,g) 
2
2

3
1
 192.45   130.68  191.61J/K/mol
2
2
Gr  H r  T Sr
=99.375 J/K/mol
 46.11 kJ/mol -298 K  0.099375 kJ/K/mol
=-16.5 kJ/mol = Gf (NH3 ,g)
Classroom Exercise
Calculate
o the standard free energy of formation of cyclopropane, C3H6(g)
at 25 C.
3C(s)+3H2 (g)  C3H6 (g)
Hr  Hf (C3H6 ,g)  3Hf (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)  3Sm (H 2 ,g)  3Sm (C,s)
 237.4  3 130.68  3 158.10J/K/mol
=-628.94 J/K/mol
Gr  H r  T Sr
 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,
thermodynamically
 stable.
, then it is
G f  0
If the standard free energy of formation of a compound is larger than 0,
thermodynamically
unstable.

then it is
G f  0,
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
o
Is glucose stable relative to its elements at 25 C 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
 910kJ/mol<0
thermodynamically stable.
o
Is methylamine, CH3NH2, stable relative to its elements at 25 C and
under standard conditions?
Gfo (CH3 NH 2 ,g)  32.16kJ/mol > 0
thermodynamically unstable.
Standard Free Energy of
Formation Decides
Thermodynamic Stability
o
Is methylamine, CH3NH2, stable relative to its elements at 25 C 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 (CH3 NH 2 ,g)  32.16kJ/mol > 0
thermodynamically unstable.
The Chemical Reaction Proceeds
So That The Free Energy of
o
Reaction 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 (CO2 ,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 +RT 1n Q
o
RT=(8.314 51J/K  mol)  (298.15 K)=2.4790 kJ/mol at 298.15K
At T  0 K, Q  0 (no reaction occurs), RTlnQ
is generally not zero so ΔG r (0K)  ΔG r o
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
o
Calculate Kp at 25 C for the equilibrium
N 2O4 (g)
N 2O4 (g)
2NO2 (g)
2NO2 (g) K p 
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.
•
Objectives
(2)
Problem-Solving
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.
Thermodynamics
vs. Kinetics
Chapter 16 | Slide 71
Copyright © Houghton Mifflin Company. All rights reserved.
Spontaneous Processes and Entropy
• Thermodynamics lets us predict whether a
process will occur but gives no
information about the amount of time
required for the process.
• A spontaneous process is one that occurs
without outside intervention.
16.1
Chapter 16 | Slide 72
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• Consider 2.4 moles of a gas contained in a 4.0
L bulb at a constant temperature of 32°C. This
bulb is connected by a valve to an evacuated
20.0 L bulb. Assume the temperature is
constant.
a) What should happen to the gas when you
open the valve?
16.1
Chapter 16 | Slide 73
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• Consider 2.4 moles of a gas contained in a 4.0
L bulb at a constant temperature of 32°C. This
bulb is connected by a valve to an evacuated
20.0 L bulb. Assume the temperature is
constant.
b) Calculate H, E, q, and w for the process
you described above.
16.1
Chapter 16 | Slide 74
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• Consider 2.4 moles of a gas contained in a
4.0 L bulb at a constant temperature of 32°C.
This bulb is connected by a valve to an
evacuated 20.0 L bulb. Assume the
temperature is constant.
c) Given your answer to part b, what is the
driving force for the process?
16.1
Chapter 16 | Slide 75
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The Expansion of An Ideal Gas Into an
Evacuated Bulb
16.1
Chapter 16 | Slide 76
Copyright © Houghton Mifflin Company. All rights reserved.
Entropy
• The driving force for a spontaneous
process is an increase in the entropy of
the universe.
• A measure of molecular randomness or
disorder.
16.1
Chapter 16 | Slide 77
Copyright © Houghton Mifflin Company. All rights reserved.
Entropy
• Thermodynamic function that describes
the number of arrangements that are
available to a system existing in a given
state.
• Nature spontaneously proceeds toward
the states that have the highest
probabilities of existing.
16.1
Chapter 16 | Slide 78
Copyright © Houghton Mifflin Company. All rights reserved.
The Microstates That Give a Particular
Arrangement (State)
16.1
Chapter 16 | Slide 79
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Positional Entropy
• A gas expands into a vacuum because
the expanded state has the highest
positional probability of states available
to the system.
• Therefore: Ssolid < Sliquid << Sgas
16.1
Chapter 16 | Slide 80
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• Predict the sign of S for each of the following,
and explain:
a) The evaporation of alcohol
b) The freezing of water
c) Compressing an ideal gas at constant
temperature
d) Heating an ideal gas at constant pressure
e) Dissolving NaCl in water
16.1
Chapter 16 | Slide 81
Copyright © Houghton Mifflin Company. All rights reserved.
Second Law of Thermodynamics
• The entropy of the universe is increasing.
• The total energy of the universe is
constant, but the entropy is increasing.
Suniverse = ΔSsystem + ΔSsurroundings
16.2
Chapter 16 | Slide 82
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• For the process A(l)  A(s), which direction
involves an increase in energy randomness?
Positional randomness? Explain your answer.
• As temperature increases/decreases (answer
for both), which takes precedence? Why?
• At what temperature is there a balance
between energy randomness and positional
randomness?
16.3
Chapter 16 | Slide 83
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
Describe the following as spontaneous/nonspontaneous/cannot tell, and explain.
A reaction that is:
a) Exothermic and becomes more positionally
random
b) Exothermic and becomes less positionally random
c) Endothermic and becomes more positionally
random
d) Endothermic and becomes less positionally
random
Explain how temperature affects your answers.
16.3
Chapter 16 | Slide 84
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
Use the ideas of energy randomness and
positional randomness in your discussion of the
following:
a) Under what conditions is the freezing of
water spontaneous? Why?
b) Under what conditions is the melting of
ice spontaneous? Why?
c) Under what conditions is the freezing of
water as likely as the melting of ice? Why?
16.3
Chapter 16 | Slide 85
Copyright © Houghton Mifflin Company. All rights reserved.
ΔSsurr
• The sign of ΔSsurr depends on the
direction of the heat flow.
• The magnitude of ΔSsurr depends on the
temperature.
ΔSsurr = –ΔH/T
16.3
Chapter 16 | Slide 86
Copyright © Houghton Mifflin Company. All rights reserved.
Interplay of Ssys and Ssurr in
Determining the Sign of Suniv
16.3
Chapter 16 | Slide 87
Copyright © Houghton Mifflin Company. All rights reserved.
Free Energy (G)
• ΔSuniv = –ΔG/T (at constant T and P)
• A process (at constant T and P) is
spontaneous in the direction in which
the free energy decreases.
– Negative ΔG means positive ΔSuniv.
16.4
Chapter 16 | Slide 88
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Free Energy (G)
• ΔG = ΔH – TΔS (at constant T and P)
16.4
Chapter 16 | Slide 89
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• A liquid is vaporized at its boiling point.
Predict the signs of w, q, H, S, Ssurr and
G.
• Explain your answers.
16.4
Chapter 16 | Slide 90
Copyright © Houghton Mifflin Company. All rights reserved.
Exercise
• The value of Hvaporization of substance X is 45.7
kJ/mol, and its normal boiling point is 72.5°C.
• Calculate S, Ssurr, and G for the
vaporization of one mole of this substance at
72.5°C and 1 atm.
16.4
Chapter 16 | Slide 91
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Spontaneous Reactions
16.4
Chapter 16 | Slide 92
Copyright © Houghton Mifflin Company. All rights reserved.
Effect of H and S on Spontaneity
H
S
Result

+
spontaneous at all temps
+
+
spontaneous at high temps


spontaneous at low temps
+

not spontaneous at any temp
16.4
Chapter 16 | Slide 93
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• Gas A2 reacts with gas B2 to form gas AB at
constant temperature and pressure. The bond
energy of AB is much greater than that of
either reactant.
• Predict the signs of:
H
Ssurr
S
Suniv
• Explain.
16.5
Chapter 16 | Slide 94
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Third Law of Thermodynamics
• The entropy of a perfect crystal at 0 K is
zero.
• The entropy of a substance increases
with temperature.
16.5
Chapter 16 | Slide 95
Copyright © Houghton Mifflin Company. All rights reserved.
Standard Entropy Values (S°)
• Represent the increase in entropy that
occurs when a substance is heated from
0 K to 298 K at 1 atm pressure.
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
16.5
Chapter 16 | Slide 96
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Standard Free Energy Change (ΔG°)
• The change in free energy that will occur
if the reactants in their standard states
are converted to the products in their
standard states.
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
16.6
Chapter 16 | Slide 97
Copyright © Houghton Mifflin Company. All rights reserved.
Concept Check
• A stable diatomic molecule spontaneously
forms from its atoms.
• Predict the signs of:
H°
S°
G°
• Explain.
16.6
Chapter 16 | Slide 98
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Concept Check
• Consider the following system at equilibrium at
25°C.
PCl3(g) + Cl2(g)
PCl5(g)
G° = −92.50 kJ
• What will happen to the ratio of partial pressure
of PCl5 to partial pressure of PCl3 if the
temperature is raised?
• Explain.
16.6
Chapter 16 | Slide 99
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Free Energy and Pressure
G = G° + RT ln(P)
or
ΔG = ΔG° + RT ln(Q)
16.7
Chapter 16 | Slide 100
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Concept Check
• Sketch graphs of:
1. G vs. P
2. H vs. P
3. ln(K) vs. 1/T (for both endothermic
and exothermic cases)
16.7
Chapter 16 | Slide 101
Copyright © Houghton Mifflin Company. All rights reserved.
Free Energy and Equilibrium
• The equilibrium point occurs at the
lowest value of free energy available to
the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
16.8
Chapter 16 | Slide 102
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Change in Free Energy to Reach
Equilibrium
16.8
Chapter 16 | Slide 103
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Qualitative Relationship Between the
Change in Standard Free Energy and
the Equilibrium Constant for a Given
Reaction
16.8
Chapter 16 | Slide 104
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Free Energy and Work
• Maximum possible useful work
obtainable from a process at constant
temperature and pressure is equal to the
change in free energy.
wmax = ΔG
16.9
Chapter 16 | Slide 105
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Free Energy and Work
• Achieving the maximum work available
from a spontaneous process can occur
only via a hypothetical pathway. Any
real pathway wastes energy.
• All real processes are irreversible.
16.9
Chapter 16 | Slide 106
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