Transcript CHEM1310 Lecture

Chapter 15 - Spontaneity, Entropy, and Free Energy
1 Spontaneous Processes
2 The Isothermal Expansion and Compression of an Ideal Gas
3 The Definition of Entropy
4 Entropy and Physical Changes
5 Entropy and the Second Law of Thermodynamics
6 The Effect of Temperature on Spontaneity
7 Free Energy
8 Entropy Changes in Chemical Reactions
9 Free Energy and Chemical Reactions
10 The Dependence of Free Energy on Pressure
11 Free Energy and Equilibrium
12 Free Energy and Work (skip)
13 Reversible and Irreversible Processes: A Summary (skip)
14 Adiabatic Processes (skip)
• Kinetics vs Thermo
– Thermodynamics predicts
direction and “driving force”.
– Kinetics predicts speed
(rate).
• Spontaneous Processes
– Occur on some timescale
(maybe slowly) without
outside intervention
(examples: a battery will
discharge, a hot cup of coffee
will cool to ambient
temperature).
– All spontaneous processes
proceed toward “states”
(macrostates) with the
greatest number of
accessible microstates.
Microstates and Macrostates:
An available microstate describes a specific detailed microscopic
configuration (molecular rotations, translations, vibrations, electronic
configuration) that a system can visit in the course of its fluctuations.
A macrostate describes macroscopic properties such as temperature
and pressure.
For a gas at constant T: the number of available microstates increases
with volume.
For gas, liquid or solid, the number of available microstates increases
with T (the number of available vibrational microstates, electronic microstates,
etc. increases with T). When you heat anything, you increase the number of
available microstates.
When a liquid vaporizes, the number of available microstates
increases.
When a liquid freezes, the number of available microstates decreases.
What is a spontaneous process?
probable
not probable
This is not a spontaneous process.
The reverse process (going from right to left) is spontaneous.
a) A gas will spontaneously expand to fill the available space.
b) There is a ‘driving’ force that causes a gas to spontaneously expand to fill
a vacuum.
c) The entropy of the universe increases with a gas expands to fill a
vacuum.
a=b=c
Why is this not a spontaneous process?
probable
not probable
There are more available microstates on the right hand side than on the left
hand side.
If a system gains degrees of freedom (more constituents, more room to
move, more available quantum states, more available rotational, vibrational,
translational or electronic states), then it gains entropy.
A spontaneous process increases entropy (but you must consider both the
system and the surroundings)
probable
not probable
The probability of finding both molecules on the left side is ¼
(this is one available microstate out of four possible
microstates that will give this arrangement)
The probability of finding one molecule on the each side is ½.
(this are two possible microstates out of four possible
microstates that will give this arrangement)
The probability of finding both molecules on the right side is ¼
(this is one microstate out of four possible microstates)
The probability of occurrence of a particular arrangement (state)
depends on the number of ways (microstates) in which that
arrangement can be achieved. All microstates are equally
probable.
There are three
possible arrangements
of four molecules in
two chambers.
The arrangement with
the greatest number of
microstates is most
probable. Label the
molecules a,b,c,d and
count the microstates.
You will see that
arrangement III is most
probable.
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2
1 1 1
1
x =  =
2 2 2
4
n
1
 
2
2 molecules
Probability of
finding 2
molecules on the
same side is 1/4
Definitions of Entropy
“S”
• Entropy is related to probability
• If a system has several available macrostates, it will
spontaneously proceed to the one with the largest
number of available microstates.
• The macrostate with the greatest probability (largest
number of available microstates) has the highest
entropy.
• When you heat something you increase its entropy.
S = kB ln Ω Joules/Kelvin
Kb = Boltzmann’s constant, the gas constant per molecule (R/NA)
Ω = the number of available microstates of a given state
∆S = q/T J / mol-K
Ludwig Boltzmann (1844-1906)
Highlights
– Established the logarithmic connection
between entropy and probability in his
kinetic theory of gases.
– The Boltzmann constant (k or kB) is the
physical constant relating temperature
to energy.
Moments in a Life
– Suffered from bipolar disorder and
depression
– Ironically, in Max Planck’s Nobel Prize
speech in 1918, it was pointed out that
Boltzmann never introduced the
constant k, Planck did.
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Zumdahl Chapter 10
One He in the
gas phase
expands from
volume V1 to 2V1
Ω2 = 2Ω1
Twice the number of
microstates
S  S 2  S1  k B ln  2  k B ln 1
 2 
 21 
n
 k B ln 

k
ln

k
ln
2
 k B ln 2
 B 

B
 1 
 1 
 
If 1 mole of He (instead of 2 He)
2
6 x1023
2
1

S  k B ln 2
6 x1023

and the gas expands from V1 to V2
 (6 x1023 )(k B ln 2)  N A k B ln 2  R ln 2
 V2 
SV1 V 2  nR ln  
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 V1 
The change in entropy of a gas is
dependent
Zumdahl Chapter 10on the change in
12
volume of the gas
The isothermal expansion of an ideal gas.
• Isothermal – system and surroundings maintain constant
temperature.
ΔE = 0 = q + w
then q = – w
• Consider only reversible and irreversible processes
– For a reversible, cyclic process both the system and
the surroundings are returned exactly to their original
positions.
• Cyclic expansion-compression process “work is
converted to heat”
Work →
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Heat
Zumdahl Chapter 10
13
The isothermal expansion of an ideal gas.
∆E=0 (energy of a perfect gas depends only on T)
∆E= w + q
w = -q
 V2 
wrev  nRT ln    qrev
 V1 
 V2 
qrev  nRT ln  
 V1 
qrev
S 
T
 V2 
S  nR ln  
 V1 
This important relationship entropy
(determined by number of available
microscopic states) is related to a
macroscopic properties of heat and
temperature.
Brick A (warm)
Brick B (warm)
w(A)
q(A)
ΔE(A)
ΔH(A)
ΔS(A)
W(B)
Q(B)
ΔE(B)
ΔH(B)
ΔS(B)
↓
Brick A (cold)
Brick B (hot)
ΔS(A) < 0 (cools)
ΔS(B) > 0 (heats)
|ΔS(A)| > |ΔS(B)|
ΔS(uni) = ΔS(A) + ΔS(B)
ΔS(uni) < 0
This is not a spontaneous process.
Entropy and Physical Change
Temperature Dependence of Entropy:
ST1 T2
 T2 
 nC p ln   at constant Pressure
 T1 
ST1 T2
 T2 
 nCV ln   at constant Volume
 T1 
Cp and Cv are is the heat capacities of the system.
ΔS(T1 to T2) here should be written ΔSsys(T1 to T2)
Example
ΔST1
→
Calculate the change in entropy that occurs when a sample
containing 1.00 mol of water (ice) is heated from – 20 °C to
+20°C at 1 atm pressure. The molar heat capacities of H2O (s)
and H2O (l) are 38.1 J K-1mol-1 and 75.3 J K-1mol-1 respectively
and the enthalpy of fusion (melting) is 6.01 kJ mol-1 at 0°C.
T2
qrev H fus
S 

T
T
 T2 
= nCpln   at const. P
 T1 
Solution
1. ΔS from 253K to 273K = n Cp ln(T2/T1) = (1.00)(38.1)ln (273/253) = 2.90 J/K
2. ΔS phase change from liq to gas = qrev/T = ΔHfus/T = (6010/273) = 22.0 J/K
3. Δfrom 273K to 293K = n Cp ln(T2/T1) = (1.00)(75.3)ln (293/273) = 5.3 J/K
4. Total ΔS = ΔS1 + ΔS2 + ΔS3
Entropy Change (J/K)
Entropy Change from -20C to +20C
ice to water
35
30
25
20
15
10
5
0
250
260
270
280
Temperature (K)
290
300
17
First Law of Thermodynamics
The change in the internal energy of a system is equal to
the work done on it plus the heat transferred to it. The
Law of Conservation of Energy
E = q + w
Second Law of Thermodynamics
For a spontaneous process the Entropy of the universe
(meaning the system plus its surroundings) increases.
Suniverse > 0
Third Law of Thermodynamics
In any thermodynamic process involving only pure
phases at equilibrium, the entropy change,  S,
approaches zero at absolute zero temperature; also the
entropy of a crystalline substance approaches zero at
0K.
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Zumdahl Chapter 10
18
S
= 0 at 0 K
• 1st Law of Thermodynamics
– In any process, the total energy of the universe remains
unchanged: energy is conserved
– A process and its reverse are equally allowed by the first law
0 =ΔEforward + ΔEreverse
(Energy is conserved in both directions)
• 2nd Law of Thermodynamics
– Processes that increase ΔSuniverse are spontaneous.
ΔSuniv > 0
Spontaneous Forward
ΔSuniv = 0
At Equilibrium
ΔS
univ < 0
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Spontaneous
Reverse
Zumdahl Chapter
10
19
Suniverse = Ssystem + Ssurroundings
• The sign of ΔSsur
depends on the
direction of the heat
flow.
• The magnitude of
ΔSsur depends on the
temperature
S surr
H

T
This is ΔH of the system.
If the reaction is exothermic, ΔH has a
negative sign and ΔSsurr is positive
If the reaction is endothermic, ΔH has a
positive sign and ΔSsurr is negative
Ssystem + Ssurroundings = Suniverse
Summary of Entropy
• Entropy is a quantitative measure of the number of
microstates available to the molecules in a system. It is
a measure of the number of ways in which energy or
molecules can be arranged.
• Entropy is the degree of randomness or disorder in a
system
• The Entropy of all substances is positive
Ssolid < S liquid < Sgas
• ΔSsys is the Entropy Change of the system
• ΔSsur is the Entropy Change of the surroundings
• ΔSuni is the Entropy Change of the universe
• S has the units J K-1mol-1
Josiah Willard Gibbs (1839-1903)
Highlights
– Devised much of the theoretical
foundation for chemical
thermodynamics.
– Established the concept free energy
Moments in a Life
– 1863 Yale awarded him the first
American Ph.D. in engineering
– Book: Equilibrium of Heterogeneous
Substances, deemed one of the
greatest scientific achievements of
the 19th century.
– Will never be famous like Michael
Jackson.
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Zumdahl Chapter 10
23
Gibbs Free Energy
ΔG = ΔHsys - TΔSsys
•
Allows us to focus on the system only, without
considering the surroundings.
•
G is called
1.
2.
3.
Gibbs Function, or
Gibbs Free Energy, or
Free Energy.
Free Energy
ΔG < 0
Spontaneous
ΔG = 0
Equilibrium
ΔG > 0
Spontaneous Reverse
Entropy
ΔSuniv > 0
Spontaneous Forward
ΔSuniv = 0
Equilibrium
ΔSuniv < 0
Spontaneous Reverse
Gibbs Free energy
Benzene, C6H6, boils at 80.1°C (at 1
atm) and ΔHvap = 30.8 kJ
– a) Calculate ΔSvap for 1 mole of benzene at
60°C and pressure = 1 atm.
Benzene, C6H6, boils at 80°C at 1 atm.
ΔHovap = 30.8 kJ
– a) Calculate ΔSvap for 1 mole of benzene.
Start with ΔGvap=ΔHvap-TΔSvap
at the boiling point, ΔGvap = 0
so ΔHvap = TbΔSvap
Δ Hv ap
30.8 x 103 J
Δ Sv ap 

 87.2J K1
Tb (273.15  80.1)
Gibbs Free energy
Benzene, C6H6, boils at 80.1°C (at 1
atm) and ΔHvap = 30.8 kJ
– b) Does benzene spontaneously boil at
60°C?
G
G
G
vap
vap
vap
 H
vap
- TS
vap
o
-1
 30,800J  (273K 60 C)(87.2 J K )
 1749 J = 1.7 kJ
Since G
vap
o
> 0, benzene does not boil at 60 C, 1 atm.
Effects of Temperature on ΔG°
• Typically ΔH and ΔS are almost constant over a broad
range
3NO (g) → N2O (g) + NO2 (g)
• For the reaction above, as Temperature increases ΔG
becomes more positive, i.e., less negative.
Effects of Temperature on ΔG
For temperatures other than 298K or 25C
ΔG = ΔH - T·ΔS
B
A
D
C
For temperatures other than 298K or 25C
ΔG = ΔH - T·ΔS
Case C
ΔH° > 0
B
ΔS° < 0
A
ΔG = ΔH - T·ΔS
ΔG = (+) - T·(-) = positive
 ΔG > 0 or non-spontaneous at all
Temp.
Case B
ΔH° < 0
ΔS° > 0
ΔG = ΔH - T·ΔS
ΔG = (-D
) - T·(+) = negative
 ΔG < 0 or spontaneous at all
temp.
C
For temperatures other than 298K or 25C
ΔG = ΔH - T·ΔS
Case D
ΔH° < 0
ΔS° < 0B
A
ΔG = ΔH - T·ΔS at a low Temp
ΔG = (-) - T·(-) = negative
 ΔG < 0 or spontaneous at low
Temp.
Case A
ΔH° > 0
ΔS° > 0
D
ΔG = ΔH - T·ΔS
C
ΔG = (+) - T·(+) at a High
Temp
 ΔG < 0 or spontaneous
at high Temp.
Entropies of Reaction
ΔSrxn° = ΣS°products – ΣS°reactants
ΔSrxn° is the sum of products minus the sum of the
reactants, for one mole of reaction (that is what °
means)
For a general reaction
aA+bB→cC+dD
ΔS  cS (C)  dS (D)  aS (A)  bS (B)
o
o
o
o
Appendix 4 tabulates standard molar entropy
values, S° in units JK-1mol-1
o
Example
(a) Calculate ΔSr° at 298.15 K for the reaction
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
(b) Calculate ΔS° of the system when 26.71 g of
H2S(g) reacts with excess O2(g) to give
SO2(g) and H2O(g) and no other products at
298.15K
(a) Calculate ΔSr° at 298.15 K for the reaction
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
Solution
(a) Look up each S° of formation [Note this is for “one
mole of the reaction”
as written: i.e. 2 moles of H2S, 3 moles of O2, etc]
ΔSrxn°= 2S°(SO2(g) ) + 2S°(H2O(g)) -2S°(H2S(g) ) 3S°(O2(g))
ΔSrxn°= 2(248) + 2(189) -2(206) - 3(205)
= – 153 JK-1mol-1
2H2S(g) + 3O2(g) → 2SO2(g) +2H2O(g)
(a) ΔSrxn= -153 JK-1mol-1
(b) Calculate ΔS° when 26.7 g of H2S(g)
reacts with excess O2(g) to give SO2(g)
and H2O(g) and no other products at
298.15K
Solution:
1
1 mol reaction (153 J K1 )
S  26.7 g H2S 


34 g H2S
2 mol H2S
1 mol reaction
mol
S  60.0 J K1
Free Energy and Chemical Reactions
ΔG = ΔH - T·ΔS
• ΔGf° is the standard molar Gibbs
function of formation
• Because G is a State Property, for a
general reaction
aA+bB→cC+dD
ΔG  cΔG (C)  dΔG (D)  aΔG (A)  bΔG (B)
o
r
o
f
o
f
o
f
o
f
Calculate ΔG° for the following reaction at
298.15K. Use Appendix 4 for additional
information needed.
3NO(g) → N2O(g) + NO2(g)
Solution From Appendix 4
ΔGf°(N2O) = 104 kJ mol-1
ΔGf°(NO2) = 52
ΔGf° (NO) = 87
ΔG°= 1(104) + 1(52) – 3(87)
ΔG°= − 105 kJ therefore,
Zumdahlspontaneous
Chapter 10
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40
The Dependence of Free Energy on Pressure
ΔG = ΔG° + RT ln Q
Where Q is the reaction quotient
aA+bB↔cC+dD
• If Q > K the rxn shifts towards the reactant side
– The amount of products are too high relative to the amounts of
reactants present, and the reaction shifts in reverse (to the left) to
achieve equilibrium
• If Q = K equilibrium
• If Q < K the rxn shifts toward the product side
– The amounts of reactants are too high relative to the amounts of
products present, and the reaction proceeds in the forward
direction (to the right) toward equilibrium
compare
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 PCc PDd 
 PCc PDd 
Q   a b 
K   a b 
 PA PB any conditions
Zumdahl Chapter 10
 PA PB equilibriu m
41
ΔG = ΔG° + RT ln Q
– Where Q is the reaction quotient
aA+bB↔cC+dD
• If Q < K the rxn shifts towards the product side
• If Q = K equilibrium
• If Q > K the rxn shifts toward the reactant side
At Equilibrium conditions, ΔG = 0
ΔG° = -RT ln K
NOTE: we can now calculate equilibrium constants
(K) for reactions from standard ΔGf functions of
formation
c
d
[C] [D]
K
a
b
[A] [B]
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ΔG  cΔG (C)  dΔG (D)
o
r
o
f
o
f
 aΔG (A)  bΔG (B)
o
f
Zumdahl Chapter 10
o
f
42
Calculate the equilibrium constant for
this reaction at 25C.
3NO(g) ↔ N2O(g) + NO2(g)
• Strategy
Use - ΔG ° = RT ln K
Use ΔG°= - 105 kJ mol -1 (from
previous)
3NO(g) ↔ N2O(g) + NO2(g)
• Solution
K
[ PN2O ]1[ PNO2 ]1
[ PNO ]3
 1.8x1018
Use
– ΔG ° = RT ln K
Rearrange
 G 
ln K 
RT
ΔGrxn°= – 105 kJ mol-1
(105,0 00 J mol1 )
ln K 
 42
1
1
(8.3145 J K mol )(298.15 K)
K  e 42  2 x 1018
ΔG = ΔG° + RT ln Q
Where Q is the reaction quotient
aA+bB↔cC+dD
Criteria for Spontaneity in a Chemical Reaction
Spontaneous
Processes
Equilibrium
Processes
Nonspontaneous
Processes
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGf < 0 ΔGf = 0 ΔGf > 0
Q<K
Q=K
Q>K
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Zumdahl Chapter 10
Conditions
All conditions
Constant P and T
Constant P and T
45
The Temperature Dependence of
Equilibrium Constants
 G   H  S 
ln K 


RT
RT
R
• Where does this come from?
• Recall ΔG = ΔH - T·ΔS
• Divide by RT, then multiply by -1
ΔG  ΔH  ΔS 


RT
RT
R
 G 
 ΔG   ΔH  ΔS 


 lnK ln K 
RT
RT
RT
R
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Zumdahl Chapter 10
46
• Notice that this is y = mx + b the equation for a
straight line
 ΔH  ΔS 
lnK 

RT
R
• A plot of y = mx + b or
• ln K vs. 1/T
 ΔH 
slope 
R
ΔS
Y - intercept 
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R Zumdahl Chapter 10
47
• If we have two different Temperatures and K’s
(equilibrium constants)
 1 1
  
 T2 T1 
van't Hoff Equation
K2 - ΔH
ln

K1
R

or

K2 ΔH  1 1 
  
ln

K1
R  T1 T2 
van't Hoff Equat ion
• Now given ΔH and T at one temperature, we
can calculate K at another temperature, assuming
that ΔH and ΔS are constant over the temperature
range
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48
The Person Behind the Science
J.H. van’t Hoff (1852-1901)
Highlights
– Discovery of the laws of chemical dynamics and
osmotic pressure in solutions
– his work led to Arrhenius's theory of electrolytic
dissociation or ionization
– The Van't Hoff equation in chemical
thermodynamics relates the change in
temperature to the change in the equilibrium
constant given the enthalpy change.
 1 1
  
 T2 T1 
van't Hoff Equation
K2 - ΔH
ln

K1
R

Moments in a Life
van’t Hoff Factor (i)
moles of particles in solution
i
moles of solute dissolved
– 1901 awarded first Noble Prize in Chemistry
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ΔT = − i m K
49
The reaction
2 Al3Cl9 (g) → 3 Al2Cl6 (g)
Has an equilibrium constant of 8.8X103 at 443K and a
ΔHr°= 39.8 kJmol-1 at 443K. Estimate the equilibrium
constant at a temperature of 600K.
 1 1
   van't Hoff Eq.
 T2 T1 
- 39,800Jmol-1  1
1 



-1
-1 
8.315JK mol  600K 443K
K 2 - ΔH
ln

K1
R
K 600
ln
K 443
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
K 600
ln(
)  2.8
3
8.8x10
K 600
2.8

e
 16
3
8.8x10
5 Zumdahl Chapter 10
K 600  1.5x10
50
w   Pext V
P = mg/A
V
Work  
V2
V1
wrev
2 step expansion
6 step expansion
Pext dV
 V2 
 nRT ln    qrev
 V1 
Infinite-step expansion
Expansion (V2 > V1): Work flows out of the system and
the Work sign is negative
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51
P = mg/A
V
w   Pext V
Work  
V2
V1
wrev
Pext dV
 V2 
 nRT ln    qrev
 V1 
Compression (V2 < V1): Work is put into system,
ln(V2/V1) is negative and the Work is positive