chapter 7 thermodynamics: the second …hosting01.snu.ac.kr/~shinkj/ch_07.pdf · 2010-05-17 ·...
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2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
CHAPTER 7 THERMODYNAMICS: THE SECOND AND THIRD LAWS
Direction of natural change : Increasing disorder of energy and matter
Measure of disorder : Entropy
Link between thermodynamics and chemical equilibrium : Gibbs free energy
Second law : Irreversibility (Spontaneity), △S, △G
Third law: Unavailability of 0 K Absolute entropy
◈ ENTROPY
7.1 Spontaneous Change
Spontaneous process : Natural tendency to occur
How fast? depends on the rate
Spontaneous⎯⎯⎯⎯⎯⎯→
Not spontaneous←⎯⎯⎯⎯⎯⎯⎯⎯
Fig. 7.1 Spontaneous cooling of a block.
Spontaneous⎯⎯⎯⎯⎯⎯→
Not spontaneous←⎯⎯⎯⎯⎯⎯⎯⎯
Fig. 7.2 Spontaneous filling of a gas into an empty flask.
7.2 Entropy and Disorder
Dispersion of energy and matter proceeds in a disorderly fashion.
◈ Thermodynamic definition of entropy (1865, Clausius):
rev qST
∆ =
Entropy A measure of degree of disorder, a state function
cf. Carnot cycle
Rudolf Clausius (獨,1882-1888)
The second law of thermodynamics
The entropy of an isolated system increases in the course of any spontaneous change.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
7.3 Change in Entropy
◈ Temperature dependence of entropy
revdqdST
=
revdq CdT= , where or P VC C C=
CdTdST
=
2 2
1 1
2
1
lnT T
T T
CdT dT TS C CT T
∆ = = =∫ ∫ T (2)
Fig. 7.3 Change of entropy with temperature
◈ Volume dependence of entropy at constant T
For an isothermal expansion of an ideal gas:
0U q w∆ = + =
2rev rev
1
lnVq w nRTV
= − =
rev rev 2
1
lnq wST T
−∆ = = =
VnRV
(3a)
◈ Pressure dependence of entropy at constant T
Fig. 7.4 Change of entropy with volume
2 1 1 2/ /V V P P= (Boyle’s law)
1
2
ln PS nRP
∆ = (3b)
Ex. 7.2 Calculating the entropy change due to an increase in temperature.
20.0 L of N2 gas at 5.00 kPa is heated from 20oC to 400oC at constant volume. ?S∆ =
. 1V,m 2(N ) 20.81 J KC −= ⋅
[Solution]
1 1 1
(5.00 kPa) (20.0 L)/ 0.0411 mol(8.3145 L kPa K mol ) (293 K)
n PV RT − −
×= = =
⋅ ⋅ ⋅ ×
2 1 V,m 2 1ln( / ) ln( / )S C T T nC T T∆ = =
1 1 673 K(0.0411 mol) (20.81 J K mol ) ln 0.710 J K293 K
− −= × ⋅ ⋅ × = + 1−⋅
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
7.4 Entropy Changes Accompanying Changes in Physical State
▶ Normal melting point, Tf, normal boiling point, Tb at 1 atm
◈ Phase transition
1. Transition temperature is kept constant as heat is supplied.
2. At the transition temperature, the transfer of heat is reversible.
3. Since transition takes place at constant pressure, the heat supplied is equal to H∆ .
◆ At the boiling point, vapvap
b
HS
T∆
∆ = revqST
⎛ ⎞← ∆ =⎜⎝ ⎠
⎟ entropy of vaporization (4)
◆ At the freezing point, fusfus
f
0HST
∆∆ = > entropy of fusion (5)
◈ Entropy of vaporization of water at 25oC and 1 bar
1. Entropy change of heating of liquid water from 25oC to Tb. Eq.(2)
2. Entropy of vaporization at Tb =100oC. Eq.(4)
3. Entropy change of cooling of water vapor from Tb to 25oC. Eq.(2)
o o ovap
o ovap
(25 C) (heating liquid from 25 C to 100 C)
(100 C) (cooling liquid from 100 C to 25 C)
S S
S S
∆ = ∆
+ ∆ + ∆ o
7.5 A Molecular Interpretation of Entropy
The third law of thermodynamics
At , the entropies of all perfect crystalline substances approach zero. as 0T = 0S → 0T →
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
Perfect crystal zero positional disorder
0 K zero thermal disorder
◈ Statistical definition of entropy (Boltzmann, 1877)
logS k W =
S : Absolute entropy k : Boltzmann constant
W : Number of available microstates of a system with the same energy
▶ Ensemble (앙상블) : a large number of replicas (with different microstates) of the system
Ex. 7.7 Calculate the entropy of a tiny solid made up of four randomly oriented CO molecules at 0 K.
Fig. 7.7 Array of small boxes containing 4 solid carbon monoxide molecules.
(a) Each of 4 CO molecules can take two orientations: CO or OC
Total number of possible orientations: 42 2 2 2 2 16W = × × × = =
23 1 23 1ln ln16 (1.3807 10 J K ) ln16 3.828 10 J KS k W k − − −= = = × ⋅ × = × ⋅ −
(b) 1 mol CO molecules: , 236.02 102W ×= 15.76 J KS −= ⋅
Experimental value: 14.6 J KS −= ⋅ at near T = 0 K residual entropy
due to small electric dipole moment of CO no preference in ordering in one direction
(c) For a perfect crystal, every member of the ensemble is identical:
1W = ⎯⎯→ ln ln1 0S k W k= = =
cf. close to for solid HCl large electric dipole moment 0S ≈ 0T =
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
Ex. 7.8 Resiudal entropy of 1.00 mol FClO3(s) is 10.1 J·K–1.
Find possible molecular disorder (orientations) in the shape of molecule.
From VSEPR theory, an FClO3(s) molecule has tetrahedral shape.
Four possible orientations of a tetrahedral FClO3 molecule in a solid.
Total number of microstates for a crystal containing N molecules:
factors(4 4 4 4) 4NNW = × × ×⋅⋅ ⋅× =
236.02 10 1ln ln 4 11.5 J KS k W k × −= = = ⋅
Fig. 7.6 Ludwig Boltzmann (墺,1844-1906)
Zentralfriedhof (Central Cemetery, Group 14C, Number 1), Vienna, Austria
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
7.6 The Equivalence of Statistical and Thermodynamic Entropies
▶ Qualitative explanation of the volume dependence of entropy change:
Increasing the width of the box at constant temperature (increasing the volume)
lowering of energy levels increasing the disorder in occupation of energy levels (increasing W)
Fig. 7.9 Energy levels of a particle in a box (a) small width and
(b) large width. The change from (a) to (b) corresponds to the
isothermal expansion of an ideal gas.
▶ Qualitative explanation of the temperature dependence of entropy change:
At low temperature, only a few lower levels are occupied.
At high temperature, a large number of levels are occupied. increasing W
Fig. 7.10 More energy levels are accessible as temperature increases.
◈ Quantitative equivalence of thermodynamic and statistical definitions of entropy
Isothermal expansion of an ideal gas
Assume that for a single molecule. constant W = V×
V
For N molecules, the number of microstates available is proportional to the N th power of the volume:
factors[(constant ) (constant ) (constant )] (constant )NNW V V V= × × × ×⋅ ⋅ ⋅× × = ×
ln ln(constant )NS k W k V= = ×
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
Isothermal expansion from V1 to V 2 :
2 22 1 A
1 1
ln(constant ) ln(constant ) ln ln lnN N V VS k V k V Nk nN k nRV V
∆ = × − × = = = 2
1
VV
7.7 Standard Molar Entropies
( ) (0) (heating from 0 to ) (heating from 0 to )S T S S T S T= + ∆ = ∆ (0) 0S =∵
2
11 2(heating from to )
T
T
CdTS T TT
∆ = ∫
Set and and heat the system at constant pressure: 1 0T = 2T T=
P0
(heating from 0 to )T C dTS T
T∆ = ∫
Invoking the third law, , (0) 0S = P0
( )T C dTS T
T= ∫
Fig. 7.11 Experimental determination of entropy.
(a) Measurements of Cp at various T. (b) Calculate the area under the curve Cp/T vs T. (c) Transfer the area as S.
▶ as P 0C → 0T →
Available energy is so small that there is not enough to stimulate transitions to higher energy states.
The sample can not take up energy, and its “capacity for heat” is zero.
◆ Inclusion of phase transitions (solid liquid vapor)
P0
(solid)( ) fT C dS TT
= ∫T
fus
f
HT
∆+ P(liquid)b
f
T
T
C dT
+∫T
vap
b
HT
∆+ P (vapor)
b
T
T
C dT
+∫T
Fig. 7.12 Temperature dependence of entropy.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
▶ Standard molar entropy, , at 1 bar and at 25oSmoC
Ex. 273.15o P,m
m 0
(ice)(water)
C dTS
T= ∫
ofus,m
f ( 273.15 K)H
T∆
+=
298.15 P,m
273.15
(water)C dT
+∫T
Diamond ( =2.4 J·KomS –1·mol–1 ) : rigid bonds
Lead ( =64.8 J·KomS –1·mol–1 ): more jiggly
H2 ( =130.7 J·KomS –1·mol–1 )
N2 ( =191.6 J·KomS –1·mol–1 )
Heavier atom(or molecule) has more energy levels
CaCO3 ( =92.9 J·KomS o
mS–1·mol–1 ) vs. CaO ( =39.8 J·K–1·mol–1 ):
Higher molar entropy for large, complex molecule
Fig. 7.13 Energy levels of a particle in a box for (a) light and (b) heavy molecules.
Fig. 7.14 White tin ( =51.55 J·KomS –1·mol–1 ) Gray tin ( =44.15 J·Ko
mS –1·mol–1 ) at 25oC
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
7.8 Standard Reaction Entropies
▶ Standard reaction entropy, oS∆
o oom m(products) (reactants)S nS nS∆ = −∑ ∑
◈ GLOBAL CHANGES IN ENTROPY
Freezing of water to ice at low temperature Spontaneous process but decrease in entropy !
Entropy increases in a spontaneous process in an isolated system.
Isolated system (universe) = system + surroundings
Spontaneity determined by the entropy changes for both the system and the surroundings.
7.9 The Surroundings Entropy change Entropy change Total entropy of system of surroundings change
tot sys surrS S S∆ = ∆ + ∆
o o 1(water,0 C) 65.2 J K molS − −= ⋅ ⋅ 1m
1m
H
o o 1(ice,0 C) 43.2 J K molS − −= ⋅ ⋅
∴ upon freezing. 1 1sys 22.0 J K mol 0S − −∆ = − ⋅ ⋅ <
syssurrq = −∆ (at const T and P)
syssurr sys,freeze0 0
HS H
T∆
∆ = − > ∆ <∵ ( = –6.0 kJ·mol–1)
Fig. 7.15 The global isolated system.
Fig. 7.16 Flow of heat into (a) hot and (b) cool surroundings. is larger for (b). surrS∆“Sneeze in (a) the street and (b) the library.”
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
Ex. 7.10 Calculate the change in entropy of the surroundings when water freezes at –10oC.
1 ofus 2(H O) 6.0 kJ mol at 10 CH −∆ = ⋅ −
1freeze fus 6.0 kJ molH H −∆ = −∆ = − ⋅
( 10 273) K 263 KT = − + =
3 1sys 1 1
surr( 6.0 10 J mol ) 23 J K mol
263 KH
ST
−− −∆ − × ⋅
∆ = − = − = + ⋅ ⋅
o o 1 1sys sys( 10 C) (0 C) 22.0 J K molS S − −∆ − ≈ ∆ = − ⋅ ⋅
1 1 1 1tot sys surr ( 22.0 23.0) J K mol 1.0 J K mol 0S S S − − − −∆ = ∆ + ∆ = − + ⋅ ⋅ = + ⋅ ⋅ >
Fig. 7.17 Flow of heats in (a) an exothermic (△Ssurr>0) and (b) an endothermic (△Ssurr<0) processes.
7.10 The Overall Change in Entropy
If is positive (an increase), the process is spontaneous. totS∆
If is negative (a decrease), the reverse process is spontaneous. totS∆
If , the process has no tendency to proceed in either direction. tot 0S∆ =
Fig. 7.18 Exothermic reaction: Positive total entropy change when (a) △Ssys > 0 or (b) △Ssys< 0.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
Ex. 7.11 Calculating the total change in entropy accompanying a reaction.
Assess whether the combustion of magnesium is spontaneous at 25oC under standard conditions:
2 Mg(s) + O2(g) 2 MgO(s) ⎯⎯→ o 1sys 217 J KS −∆ = − ⋅ , o
sys 1202 kJH∆ = −
Determine the change in
entropy of the system:
o 1sys 217 J KS −∆ = − ⋅
Determine the change in
entropy of the surroundings:
oo sys
surr
6
3 1
( 1.202 10 J) 298 K
4.03 10 J K
HS
T
−
∆∆ = −
− ×= −
= + × ⋅
Determine the total change in
entropy:
o otot sys surr
1 3 1
3 1
217 J K (4.03 10 J K ) 3.81 10 J K
S S S− −
−
∆ = ∆ + ∆
= − ⋅ + × ⋅
0= × ⋅+ >
Since , the reaction is spontaneous. otot 0S∆ >
▶ Spontaneous endothermic reaction
Spontaneous reaction
direction of decreasing energy (X)
Spontaneous endothermic reaction
reactants are rising spontaneously to higher energies (X)
Spontaneous reaction
increase in the total entropy of the system and the surroundings
Spontaneous endothermic reaction
Entropy of the surroundings decreases
Entropy of the system must increase to overcome
the decrease in the entropy of the surroundings
Fig. 7.19 Spontaneous endothermic
reaction.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
◈ Clausius inequality
rev irrev− > −w w rev irrev<w w
rev rev irrev irrevU q q∆ = + = +w w
∴ rev irrevq>q
rev irrevq qS∆ = >T T
qST
∆ ≥
For an isolated system, . 0q = 0 S∆ ≥
▶ Reversible and irreversible paths between two given states of the system leave the surroundings in
different states.
total sys,rev surr,rev 0S S S∆ = ∆ + ∆ =
total sys,irrev surr,irrev 0S S S∆ = ∆ + ∆ >
sys,rev sys,irrevS S∆ = ∆
But surr,rev surr,irrevS S−∆ > −∆ surr,rev surr,irrevS S∆ < ∆
Fig. 7.20 Total entropy change in (a) a reversible and (b) an irreversible systems.
Ex. 7.12 Calculating the total entropy change for the expansion of an ideal gas.
Calculate , , and for (a) the isothermal, reversible expansion and (b) the isothermal,
free expansion of 1.00 mol of ideal gas molecules from 8.00 L to 20.00 L at 292 K.
sysS∆ surrS∆ totalS∆
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
(a) For the reversible path,
( ) 1 1sys 2 1
20.00 Lln / (1.00 mol) (8.3145 J K mol ) ln8.00 L
S nR V V − −∆ = = × ⋅ ⋅ ×
17.6 J K−= + ⋅
0U∆ = for an isothermal expansion of an ideal gas. q = −w
( )surr sys sys 2 1ln /q q nRT V V= − = = −w
( ) 1surr surr 2 1 sys/ ln / 7.6 JS q T nR V V S K−∆ = = − = −∆ = − ⋅
1 1tot sys surr 7.6 J K 7.6 J K 0S S S − −∆ = ∆ + ∆ = ⋅ − ⋅ =
reversible process
(b) For the irreversible path (free expansion),
1sys 7.6 J KS −∆ = + ⋅ the same as in (a)
sys 0=w and 0∆ =U sys 0q = surr 0S∆ =
∴ 1
total 7.6 J KS −∆ = + ⋅
spontaneous process
Fig. 7.21 Total entropy change in (a) an isothermal reversible and (b) an isothermal irreversible expansions.
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
7.11 Equilibrium
Dynamic equilibrium : forward and reverse processes undergo at the same rate total 0S∆ =
▶ Thermal equilibrium: A block of metal in contact with its surroundings at the same temperature
No net flow of heat
Fig. 7.22 Heat flow from a high-temp region to a low-temp region through a conducting wall.
▶ Mechanical equilibrium: Pressure of a gas inside the cylinder confined by a piston is the same as that of
the surroundings No net motion of the piston
▶ Ice in contact with liquid water at 0oC and 1 atm
▶ Chemical reactions in equilibrium
◈ GIBBS FREE ENERGY
▶ Criterion for spontaneity: Entropy changes for both the system and the surroundings
▶ Gibbs free energy:
Focusing on the system at constant P and T
non pV-work
7.12 Focusing on the System
At constant T and P, . surr sys /S H∆ = −∆ T
sys
systot sys surr sys
/H T
HS S S S
T−∆
∆∆ = ∆ + ∆ = ∆ − (13)
Define a state function, the Gibbs free energy, G, as
G H TS= − (14)
Fig. 7.23 Josiah Willard Gibbs(美,1839-1903)
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
G H T∆ = ∆ − ∆S at constant T (15)
sys syssys
G HS
T T∆ ∆
= − ∆
Using Eq. (13),
systotal
GS
T∆
= −∆ at constant T and P (16)
As increases, totalS∆ sysG∆ decreases
Direction on spontaneity is the direction of decreasing sysG∆ .
Fig. 7.24 Direction of spontaneous change
toward lower G.
sys 0G∆ < when is large negative. Large increase in sysH∆ surrS∆
sys 0G∆ < even if is positive provided sysH∆ sysT S∆ is large positive. Large increase in sysS∆
Criterion for equilibrium: total 0S∆ = sys 0G∆ = at constant T and P (17)
Ex. 7.13 Deciding whether a process is spontaneous
2 2H O( ) H O( )s l⎯⎯→ at 1 atm and (a) at 10m ?G∆ = oC and (b) 0oC
m mG H T S∆ = ∆ − ∆ m
1−
At 10oC,
m m
1 1m
(273.15+10.) K
1 1
1
6.01 kJ mol (283 K) (22.0 J K mol )
6.01 kJ mol 6.23 kJ mol 0.22 kJ mol
H S
G − −
∆ ∆
− −
−
∆ = ⋅ − × ⋅ ⋅
= ⋅ − ⋅
= ⋅−
At 0oC,
m m
1 1m
1 1
6.01 kJ mol (273 K) (22.0 J K mol )
= 6.01 kJ mol 6.01 kJ mol 0
H T S
G − −
∆ ∆
− −
∆ = ⋅ − × ⋅ ⋅
⋅ − ⋅ =
1−
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
G H TS= −
G decreases as T increases.
G decreases more sharply for vapor than liquid as T increases. ∵ m m( ) ( )S g S l>
Fig. 7.25 Temperature dependence of G. Fig. 7.26 Unusual temperature dependence of G. ex. CO2
Spontaneous freezing at low temperature
m m( ) ( )G s G l<
Spontaneous melting between Tf and Tb Spontaneous condensation between Tf and Tb
m m( ) ( )G l G s< m m( ) ( )G l G g<
Spontaneous vaporization above Tb Spontaneous sublimation above Ts
m m( ) ( )G g G l< m m( ) ( )G g G s<
7.13 Gibbs Free Energy of Reaction
▶ Gibbs free energy of reaction, G∆
m m(products) (reactants)G nG nG∆ = −∑ ∑ (18)
Spontaneous reaction when 0G∆ <
▶ Standard Gibbs free energy of reaction, oG∆ , in terms of the standard molar Gibbs free energies:
o oom m(products) (reactants)G nG nG∆ = −∑ ∑ (19)
oG∆ is fixed for a given reaction and temperature
G∆ varies and might even change sign as the reaction proceeds
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
▶ Standard Gibbs free energy of formation, oG∆ f
Standard Gibbs free energy of reaction per mole for the formation of a compound from its elements
in their most stable from.
1 12 22 2H ( ) I ( ) HI( )g s g+ ⎯⎯→ o 1
f (HI, ) 1.70 kJ molG g −∆ = + ⋅
f 2 = f 2
o (I , ) 0G s∆ (most stable form), o (I , ) 0G g∆ ≠
Ex. 7.14 Calculating a standard Gibbs free energy of formation at 25oC from enthalpy and entropy data.
1 12 22 2H ( ) I ( ) HI( )g s+ ⎯⎯→ g o
f (HI, ) ?G g∆ =
oo (1 mol) (HI, ) 26.48 kJH H g∆ = ×∆ = +f
o oom m(products) (reactants)S nS nS∆ = −∑ ∑
{ }{ }
( ) ( ){ }
o o o1 1m m 2 m 22 2
1 1
1 11 12 2
1
(HI, ) (H , ) (I , )
(1 mol) (206.6 J K mol )
mol (130.7) mol (116.1) J K mol
0.0832 kJ K
S g S g S s− −
− −
−
= − +
= × ⋅ ⋅
− × + × ⋅ ⋅
= + ⋅
o oG H T S∆ = ∆ − ∆ o
1 1(1 mol) (+26.48 kJ mol ) (298 K) (0.0832 kJ K )1.69 kJ
− −= × ⋅ − × ⋅= +
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
▶ Standard Gibbs free energy of formation, oG∆ f
f
f
“Thermodynamic altitude” w. r. t. the elements at
“sea level”.
If at a certain temperature, the compound is o 0G∆ <
more stable under standard conditions than the elements.
“Thermodynamically stable compound” ex. Water
If at a certain temperature, the compound is o 0G∆ >
less stable under standard conditions than the elements.
“Thermodynamically unstable compound” ex. Benzene
“Labile (레이빌,활성,活性) ” molecules decompose rapidly.
“Nonlabile” or “inert” compounds
the rate of decomposition is very slow
Fig. 7.27 Standard Gibbs free energy of
formation of several compounds.
▶ Standard Gibbs free energy of reaction, oG∆ , in terms of standard Gibbs free energies of formation:
o oof f(product) (reactants)G n G n G∆ = ∆ − ∆∑ ∑ (20)
Ex. 7.15 Calculating the standard Gibbs free energy of reaction.
3 2 24 NH (g) 5 O (g) 4 NO(g) 6 H O(g)+ ⎯⎯→ + o ?G∆ =
( ) ( ){ }( ) ( ){ }
[ ] [ ]
o oof f 2
o of 3 f 2
4 mol (NO,g) 6 mol (H O,g)
4 mol (NH ,g) 5 mol (O ,g)
4(86.55) 6( 228.57) 4( 16.45) kJ 959.42 kJ
G G G
G G
∆ = ×∆ + ×∆
− ×∆ + ×∆
= + − − − + = −
0
7.14 The Gibbs Free Energy and Nonexpansion Work
Prediction of the maximum nonexpansion work done at constant P and T
Gibbs free energy is a measure of the energy free to do nonexpansion work, ew
dG dH TdS= − at const T
dH dU PdV= + at const P
dG dU PdV TdS= + − at const T & P
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
dG d dq PdV TdS= + + −w at const T & P
Reversible process, maximum work
rev revdG d dq PdV TdS= + + −w at const T & P
rev revdG d TdS PdV TdS d PdV= + + − = +w w
dV
V
at const T & P
rev rev,e rev,expansiond d d= +w w w
rev,expansiond P= −w
rev rev,ed d Pd= −w w
rev,edG d= w at const T & P
e,maxdG d= w at const T & P
∴ at const T & P e,maxG∆ =w
Ex. Oxidation of glucose
6 12 6 2 2 2C H O ( ) 6 O ( ) 6 CO ( ) 6 H O( )s g g+ ⎯⎯→ + l
o
o 2879 kJG∆ = −
A peptide bond formation requires 17 kJ.
Oxidation of 1 mol of glucose Formation of 170(=2879/17) peptide bonds, theoretically.
Due to energy loss in biosynthesis, only 10 such links formed.
7.15 The Effect of Temperature
o oG H T S∆ = ∆ − ∆
(a) Exothermic reaction ( ), o 0H∆ < o 0S∆ < o 0T S∆ <
o 0G∆ < at low temp Spontaneous
(b) Endothermic reaction ( ), o 0H∆ > o 0S∆ > o 0T S∆ >
o 0G∆ < at high temp Spontaneous
(c) Endothermic reaction ( ), o 0H∆ > o 0S∆ < o 0T S∆ <
o 0G∆ > at all temp Not spontaneous
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
(d) Exothermic reaction ( ), o 0H∆ < o 0S∆ > o 0T S∆ >
o 0G∆ < at all temp Spontaneous
Fig. 7.28 Effect of temperature increase on the spontaneity of a reaction.
Ex. 7.16 Calculate the temperature at which an endothermic reaction becomes spontaneous.
Reduction of iron(III) oxide by carbon under standard condition endothermically:
2 3 22 Fe O (s) 3 C(s) 4 Fe(s) 3 CO (g)+ ⎯⎯→ + o 0H∆ > , o 0S∆ > case (b)
o 0G∆ < at the temperature for which o oH T S∆ < ∆
o oof f
o of 2 f 2 3
(products) (reactants)
(3 mol) (CO ,g) (2 mol) (Fe O ,s) 3( 393.5) 2( 824.2) kJ 467.9 kJ
H n H n H
H H
∆ = ∆ − ∆
= ×∆ − ×∆= − − − = +
∑ ∑
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
{ }{ }
{ } { }
o oom m
o om m 2
o om 2 3 m
1 1
(products) (reactants)
(4 mol) (Fe,s) (3 mol) (CO ,g)
(2 mol) (Fe O ,s) (3 mol) (C,s)
4(27.3) 3(213.7) 2(87.4) 3(5.7) J K 558.4 J K
S nS nS
S S
S S− −
∆ = −
= × + ×
− × + ×
= + − + ⋅ = + ⋅
∑ ∑
o 5
o 1
467.9 kJ
4.679 10 J 838 K558.4 J K
HTS −
∆ ×= = =∆ ⋅
∴ Minimum temperature at which reduction occurs at 1 bar is
838 K (or 565oC).
7.16 Impact on Biology: Gibbs Free Energy Change in Biological Systems
Non-spontaneous biological reactions can be driven by coupling to spontaneous reactions.
Fig. 7.29 A natural process represented by a falling weight.
◈ Hydrolysis of ATP (Adenosine triphosphate) to ADP(Adenosine diphosphate)
ATP + water ADP + inorganic phosphate, o 30 kJG∆ = −
6 12 6 2 2 2C H O ( ) 6 O ( ) 6 CO ( ) 6 H O( )s g g+ ⎯⎯→ + l o 2879 kJG∆ = −
Drives ~ 90 ADPs to restore to ATPs
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7 From Oxtoby, Gillis, Nachtrieb, “Principles of Modern Chemistry,” 5th ed.
8.3 ENTROPY AND HEAT: EXPERIMENTAL BASIS OF THE SECOND LAW OF THERMODYNAMICS
Background of the Second Law
Efficiency of heat engines (steam, internal combustion, ···)
heat ⇒ work
Thermodynamic efficiency of the Carnot cycle
~ fundamental limit of an engine
Equivalent formulations of the 2nd law of thermodynamics:
(Rudolf Clausius) There is no device that can transfer heat from a
colder to warmer reservoir without net expenditure of work.
(Lord Kelvin) There is no device that can transfer heat withdrawn from
a reservoir completely into work with no other effect.
(Thermodynamic) Definition of Entropy
Heat engine analysis
general thermodynamic processes
Carnot’s analysis
Efficiency for a reversible heat engine cycle
+ =q qT T
h l
h l
0 qT is a state function
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7 Clausius
∫dq
Trev : independent of path in any reversible process (state function)
revf
i
dqST
∆ ≡ ∫ Unit: J K−1
A DEEPER LOOK . . .
8.4 CARNOT CYCLES, EFFICIENCY, AND ENTROPY
The Carnot Cycle
◆ Carnot’s cycle (reversible)
Path AB : Isothermal expansion (at Th = TA = TB)
WAB = –qAB = –nRTh ln (VB/VA)
Path BC : Adiabatic expansion
qBC = 0, WBC =△UBC =qBC,V = ncV△T = –ncV(Th –Tl)
Path CD : Isothermal compression (at Tl = TC = TD)
WCD = –qCD = nRTl ln (VC/VD)
Path DA : Adiabatic compression
qDA = 0, WDA = ncV(Th –Tl)
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
Wnet = WAB + WBC + WCD + WDA
= –nRTh ln (VB/VA) + nRTl ln (VC/VD)
For an adiabatic process, T2/T1 = (V1/V2)γ−1
Th /Tl = (VC /VB)γ−1 for path BC
Th /Tl = (VD /VA)γ−1 for path DA
∴ (VC /VB)γ−1 = (VD /VA)γ−1
or VC /VB = VD /VA VB /VA = VC /VD
W net = –nR (Th –Tl ) ln (VB/VA)
Heat Engine
Efficiency of an engine
ε = – Wnet /qAB
For the ideal gas Carnot cycle,
1 1 net h l l
AB h h
w T T Tq T T
ε −= − = − <=
Efficiency of General Carnot Engines
For all Carnot engines operating reversibly between reservoirs of Th
and Tl (regardless of working fluids),
1 h
net h l l
h h
w T T Tq T T
ε −= − = = −
2010년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 7
Further Discussion of Efficiency
net h l h l
h h h
w q q T Tq q T
ε + −= − = =
1 1l l
h h
q Tq T
+ = − or 0 l h
l h
q qT T+ =
qT : state function
For a general reversible cyclic process,
, ,
, ,
0l i h i
l i h i
q qT T
+ = , , ,
, ,
0 l i h i
i l i h i
q qT T⎛ ⎞
+ =⎜ ⎟⎜ ⎟⎝ ⎠
∑
0 revdqT
=∫ rev
f
i
dqST
∆ ≡ ∫ : Clausius’s definition