CHAPTER 7 THERMODYNAMICS: THE SECOND …hosting01.snu.ac.kr/~shinkj/Ch_07.pdf · 2010-05-17 ·...

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.


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−⋅


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 →


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 =


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


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= = ×


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.


▶ 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


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.”


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.


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.


◈ 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∆


(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.


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)


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−


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


▶ 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

− −= × ⋅ − × ⋅= +


▶ 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


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


(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

∆ = ∆ − ∆

= ×∆ − ×∆= − − − = +

∑ ∑


{ }{ }

{ } { }

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)


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 −= − = = −


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

CHAPTER 7 THERMODYNAMICS: THE SECOND …hosting01.snu.ac.kr/~shinkj/Ch_07.pdf · 2010-05-17 ·...

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