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1
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!!
17
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)
2
◎ 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)
3
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
4
※ 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
5
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
6
※ 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)
7
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
8
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
9
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:
10
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
11
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
12
◎ 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
13
※ 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
14
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
16
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
17
※ 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↓
18
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
19
※ 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
20
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
21
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