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Spontaneity, Entropy (熵)and Free Energy (自由能)

insulation

idealgas vacuum

Question: Why a particular process occurs?Is it due to decrease of energy?

Model study 1: An adiabatic (絕熱) system

q = 0w = PV = 0 (note: expansion against zero P)E = q + w = 0

This is a spontaneous process but no change of E!!

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Model study 2KI(s) K+(aq) + I(aq) H = +21 kJ

Endothermic yet spontaneous!!

Model study 3H2O(s) H2O(l) H > 0

Spontaneous above 0 oC

What's in common?Increase randomness or disorder

The measurement of randomness: entropy (亂度 or 熵)

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◎ How to measure randomness?Probability

A model of positional probability

A B

Arrange two balls in two boxes: 22 = 4 possibilitiesPossible arrangements Probability

A,B -- ¼-- A,B ¼A B ¼B A ¼ ½(more probability)

Four balls: A, B, C, D (24 = 16 possibilities)

Possible arrangements ProbabilityABCD -- 1/16 ABC D 4/16 = 1/4 (4×3×2/3! = 4)AB CD 6/16 = 3/8 (4×3/2! = 6)

More probabilityBoltzmann defines:

S = kBlnΩ(kBNA = R; Ω: the number of microstates)

Ω = 1 S = 0Boltzmann constant

Avogadro’s number

S = S2 – S1 = kBlnΩ2 – kBlnΩ1 = kBln(Ω2/Ω1)

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Ex.1 mol N2 gas

at 1 atm, 25 oC1 mol N2 gas

at 1.0 × 12 atm, 25 oC

Larger V Higher S

※ Entropy and the second law of thermodynamics

The second law:In any spontaneous process there is always an increase in the entropy of the universe

Suniv = Ssys + Ssurr > 0 for spontaneous process

Suniv < 0 spontaneous for the reverse process

Suniv = 0 at equilibrium

Ex. In a living cell, large molecules are assembled from simple ones. Against the 2nd law?

Ssys is negative

Need to consider Ssurr also

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※ Reversible process

Model study: An ideal gas, isothermal (constant T)P

V

Pi

Pf

Vi Vf

Pextideal gas(P, V, T)

For ideal gasE = 3/2kT E = 3/2kTT = 0 for isothermal process E = 0 = q + w q = w

P

V

Pi

Pf

Vi Vf

Pextideal gas(P, V, T)

Case I:Pext changes to Pf suddenly Vi expands to Vf

w = Pf(Vf – Vi) q = Pf(Vf – Vi)

Impossible to go back through the same path irreversible

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Pextideal gas(P, V, T)

Pext decreases infinitesimal amount along the path until V = Vf Moves along the PV line A reversible process

Note:

P

V

Pi

Pf

Vi Vf

Conclusion: S = Tqrev (for a reversible process)

Ex. H2O(l) H2O(g) Hvaporization

This is an equilibrium process at 100 oCand reversible H = qrev

S = TH onvaporizati

For a reversible process at constant pressure:∵ qrev = qP = H

∴ S = TH

Tq

rev

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※ The effect of q and T

System SurroundingsHsys

When Hsys is negative:Increases randomness of atoms in the surroundings

Qualitatively consider the surroundings as a huge systemWith constant P, V and T

Ssurr is primarily determined by the heat flow

Heat surroundings Ssurr increasesHeat surroundings Ssurr decreases

Ssurr also depends on T

For the same amount of heat:Lower T more randomnessHigher T less randomness

System Surroundings

Exothermic Ssurr = +

Endothermic Ssurr =

heat (J)T (K)

heat (J)T (K)

We can define

Ssurr = Hsys

T(Note: Hsurr = Hsys)

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Ex. Sb2S3(s) + 3Fe(s) 2Sb(s) + 3FeS(s) H = 125 kJSsurr? At 25 oC, 1 atm

125 kJ

298 KSsurr = = 419 J/K

The unit of entropy

☆ A quick analysisSsys Ssurr Suniv

+ + + spontaneous non-spontaneous+ ? + ? depends on the

relative size

※ Free energy

Suniv = Ssys + Ssurr > 0 for spontaneous process

Suniv = Ssys – > 0 (at constant P, qP = H)Hsys

TTSuniv = TSsys – Hsys > 0

TSsys – Hsys = T(Sf – Si) – (Hf – Hi) = TSf – Hf – TSi + Hi

Now, define Gibbs free energy: G = H – TS

State functionTSsys – Hsys= – Gf + Gi = Gsys at constant T and P

★ For a spontaneous process: G > 0or G < 0 at constant T and P

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Ex. H2O(s) H2O(l)Ho = 6.03 × 103 Jmol1, So = 22.1 JK1mol1

T (oC) Ho (J/mol) So (JK1mol1) TSo (J/mol) Go = Ho – TSo(J/mol)

10 6.03 × 103 22.1 5.81 × 103 +2.2 × 102

0 6.03 × 103 22.1 6.03 × 103 0

10 6.03 × 103 22.1 6.25 × 103 2.2 × 102

(Assume Ho and So do not change at different T)

◎ Qualitatively:

H TS = G + spontaneous

(exothermic) (more random)

+ + non-spontaneous(endothermic) (less random)

spontaneous at low Tnon-spontaneous at high T

+ + spontaneous at high Tnon-spontaneous at low T

In general: S is more important at higher T

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Ex. Br2(l) Br2(g)Ho = 31.0 kJmol1, So = 93.0 JK1mol1At what T is the reaction spontaneous at 1 atm?

Soln. Find the equilibrium point first Go = 0 = Ho – TSo Ho = TSo

T = Ho

So=

31.0 × 103

93.0= 333 K

Equilibrium T = bp(bp: boiling point)

When T > 333 K TSo > Ho Go < 0 Spontaneous

※ Entropy changes in chemical reactions

N2(g) + 3H2(g) 2NH3(g) volume decreasesS:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) volume increasesS: +

CaCO3(s) CaO(s) + CO2(g) S: +

2SO2(g) + O2(g) 2SO3(g) volume decreasesS:

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In general:for homogeneous reactionA + B C S: A B + C S: +A + B C + D S: difficult to estimate

※ Absolute entropies

E and H can not be determinedBut S can be determined

The third law of thermodynamics:Entropy of a perfect crystal at 0 K is zero

0 K TS = 0 S ST

From experiment Can be determined

Standard entropy: So (at 298 K, 1 atm)

Sorxn = npSoproducts – nrS

oreactants

reaction coefficient

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Ex. Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)SoAl2O3(s) = 51 JK

1mol1 SoH2(g) = 131SoAl(s) = 28 SoH2O(g) = 189

Soln.Sorxn = 2SoAl + 3SoH2O – 3S

oH2 – S

oAl2O3

= 2(28) + 3(189) – 3(131) – 51= 179 JK1mol1

Note: has more rotational and vibrational freedom than H2

OHH

自由度

In general: the more complex the molecule the higher the entropy

※ Free energy and chemical reactions

G is measured indirectly

◎ Method 1 Go = Ho – TSo

Ex. 2SO2(g) + O2(g) 2SO3(g)

Hfo(SO2) = 297 kJ/mol So(SO2) = 248 JK1mol1

Hfo(SO3) = 396 So(SO3) = 257

So(O2) = 205

Soln: Horxn = 2(396) – 2(297)= 198 kJ

Sorxn = 2(257) – 2(248) – 1(205)= 187 J/K = 187 × 103 kJ/K

Gorxn = 198 – (298)(187 x 103) = 142 kJ

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◎ Method 2 Free energy is a state functionThe change is independent of the pathway

Derive from known reactions

Ex.2CO(g) + 4H2O(g) 2CH4(g) + 3O2(g) Go = 1088 kJ2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) Go = 1602 kJ

2CO(g) + O2(g) 2CO2(g) Go = 1088 – 1602 = 514 kJ

◎ Method 3 From standard free energy of formation(same as enthalpy treatment)

Gorxn = npGoproducts – nrG

oreactants

Ex. 2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)Gorxn = ?

Known Gof forCH3OH = 163 kJ/molCO2 = 394 kJ/molH2O = 229 kJ/mol

Soln.Gorxn = 2(394) + 4(229) – 2(163) = 1378 kJ

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※ The dependence of G on P

From definition: G = H – TS H = E + PV

G = E + PV – TS One can show that

dG = RTd(lnP) (1 mol and constant T)Set a reference state with G = Go and P = 1 atmIntegrate from this state to any state of interest

G – Go = RTln(P/1) = RTlnP G = Go + RTlnP

free energy of interest

When P = 1 atm G = Go (free energy of the gas at 1 atm)a function of T

Ex. N2(g) + 3H2(g) 2NH3(g)

Grxn = npGproducts – nrGreactants= 2GNH3 – GN2 – 3GH2

From the relationship of G and P:GNH3 = G

oNH3 + RTlnPNH3

GN2 = GoN2 + RTlnPN2

GH2 = GoH2 + RTlnPH2

Grxn = 2GoNH3 + 2RTlnPNH3 – GoN2 – RTlnPN2

– 3GoH2 – 3RTlnPH2= (2GoNH3 – G

oN2 – 3G

oH2) + RT(2lnPNH3–lnPN2 – 3lnPH2)

= Gorxn + RTlnP2NH3

PN2P3H2

Quotient of the reaction = Q

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G = Go + RTlnQReaction quotient

Standard free energy change(P = 1 atm for all gases)

Ex. CO(g) + 2H2(g) CH3OH(l)G? At 25 oC start from PCO = 5.0 atm, PH2 = 3.0 atm

Known: Gfo (CH3OH) = 166 kJ

Gfo (H2) = 0 kJ

Gfo (CO) = 137 kJ

Go = (166) – (137) – 2(0) = 29 kJ

G = Go + RTlnQ= 2.9 × 104 + (8.31)(298)ln

= 3.8 × 104 J per mol of CO

1(5.0)(3.0)2

Soln.

[More negative than Go (at 1 atm): agrees with Le Chatelier’s principle]

※ Note about significant figures

ln(1/45) = 2.303㏒(1/45) = 2.303㏒45= 2.303㏒(4.5 × 10)= 2.303[(㏒ 4.5) + 1]

= 2.303(1.65)= 3.80

exact number∥0.65 two significant figures

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※ Free energy and equilibrium

Grxn = npGproducts – nrGreactants= Hrxn – TSrxnHA

H

A B

H

HB

reactant product1 0

mole of A (or nA)10

mole of B (or nB)

SA

S

A B

S

SB

A B

G = H TSGA

GB

G

H = XAHA + XBHB

Randomness due to mixing A and B

A B

The equilibrium point:The point at which there is no driving force to go towards either direction

G = Go + RTlnQ

At equilibrium: G = 0Q = K (equilibrium constant)

0 = Go + RTlnK Go = RTlnK

The standard free energy change with all reactants and products at 1 atm partial pressure

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G = Go + RTlnQ= RTlnK + RTlnQ= RTln(Q/K)

When Q = K, G = RTln1 = 0 at equilibrium

When Q > K, G > 0 backward

When Q < K, G < 0 forward

Agrees with Le Châtelier’s principle!!

Ex. 4Fe(s) + 3O2(g) 2Fe2O3(s) K?

Hfo (kJ/mol) So (JK1mol1)

Fe2O3(s) 826 90Fe(s) 0 27O2(g) 0 205

Soln. Ho = 2(826) = 1652 kJSo = 2(90) – 4(27) – 3(205) = 543 JK1

= 0.543 kJK1

Go = Ho – TSo= 1652 – (298)(0.543)= 1.490 × 103 kJ = 1.490 × 106 J= RTlnK = (8.31)(298)lnK

lnK = 601 2.303㏒K = 601 ㏒K = 261K = 10261 a highly favorable process

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※ The temperature dependence of K

Go = RTlnK = Ho – TSo

lnK =

=

Ho

RTSo

R+

Ho

RSo

R+1T

( )

Assume Ho, So are T independent (not quite so)A linear relationship of lnK and 1/T

with slope = (Ho/R)intercept = So/R

★ Experimentally this is very useful

When Ho > 0 (endothermic) slope: ()lnK

1/T

lnK

1/T

When Ho < 0 (exothermic) slope: ()

T↑ 1/T↓ lnK↑

T↑ 1/T↓ lnK↓

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Consider T1 T2K1 K2

lnK2 =Ho

RT2

So

R+lnK1 =

Ho

RT1

So

R+

lnK2 – lnK1 =Ho

R1T2

1T1

–

ln(K2/K1) =Ho

R1T2

1T1

–

The van’t Hoff equation:

Ex. N2(g) + 3H2(g) 2NH3(g)K = 3.7 × 106 at 900 KHo = 92 kJK at 550 K?

Soln.

K5503.7 x 10-6

(92000)8.31

1550

1900

ln = –

lnK550 = 4.82.3㏒K550 = 4.8㏒K550 = 2.1 = 0.9 – 3

K550 = 8 × 103

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※ Free energy and work

QualitativelyG: to know whether a reaction will occur or not

QuantitativelyG = wmax (maximum possible work obtainable)

For a spontaneous reactionG is the energy that is free to do work

For a non-spontaneous reactionG is the minimum amount of work that must be expended

Ex.

Battery Motor

Useful work is wasted as heathigher current more heatlower current less heatzero current zero heat

Maximum amount of work can be obtained (no waste) Only hypothetical

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Ex.

Batterycharged

Batterydischarged

w1

w2

w2 > w1unless the current ≈ 0 (a reversible process)

All real processes are irreversibleWork is always wasted as heat

Released into the surroundings S goes up

※ Biological relevance

◎ 自由能 (G) 的變化是反應的驅動力自由能 越低越好: G 愈負，驅動力愈強

生命體系最重要的能量轉換系統:ATP + H2O ADP + HPO42 + H+ G’

o = 30.5 kJ/mol

N

NN

N

NH2

O

OHOH

OPO

OOP

O

OOP

O

OHO

ATP

+ OPO

OHO

N

NN

N

NH2

O

OHOH

OPO

OOP

O

OHO

ADP

+ H2O

*G’: at pH 7 and based on water concentration of 55.5 M

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H OHCH2OH

CHO

OHHHHOOHH

+ HOPO32

H OHCH2OPO32

CHO

OHHHHOOHH

+ H2O G'o = 13.8 kJ/mol

Glucose Glucose 6-phosphate 正值: 無法進行

解決之道：

+ HOPO32 + H2O G'o = 13.8 kJ/molGlucose Glucose 6-phosphate

ATP + H2O ADP + HOPO32 + H+ G'o = 30.5 kJ/mol

Glucose + ATP Glucose 6-phosphate + ADP + H+ G'o = 16.7 kJ/mol

透過ATP的幫助，葡萄糖的磷酯化現在可行了！

例子