“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Effect of number of
Possible configurations
(randomness) on reactions
What we’ve Learned So Far
Energy
Ea
Ea
endothermic
exothermic
A+B
C + heat
A + B + heat
C
rate  k  A B
Ea



RT   
rate   A exp  A B


Does not tell us if spontaneous or not
A messy room is more probable than an organized
one.
Can think of this as
stored energy.
spontaneous?
or probable?
Spontaneous Reactions:
heat
or
randomness
-heat
+randomness
reactants
-heat
+ randomness
products
-H
+S
OJO
Randomness = more possibilities = entropy (S)
What is the most probable
configuration for n tossed
quarters?
1 coin =2 sides
2 coins
1
2
1
Most probable configuration is least
organized
2 configurations =21
4 configurations =22
What will the pattern be for three coins?
1
3
3 coins
8 configurations = 23
4 coins
? configurations = 2?
#configurations  2 n
3
1
1. Most probable configuration
is least organized
2. Number of possible
configuration increases
exponentially
3. No. configurations contains
information about compound
(sides)
Probability of finding
An electron – everywhere!
Very random
How does this affect
the trends in entropy of
various elements?
1s electron
2p electron probability
Is more confined – less
Random, less entropy
Probability of
Finding an electron
For a d orbital –
More spatially constrained,
Less entropy
Entropy and Atomic Mass
Entropy of Pure Elements
row 2
90
3
4
80
Entropy (J/mol-K)
70
60
50
40
1.
30
20
2.
10
Entropy contraction across row
– related to organization of
electrons
Entropy increase down group –
related to volume occupied
0
0
10
20
30
40
50
Atomic Number
Half filled d
60
70
80
90
Entropy and Group
Group 1
Li
Na
K
Rb
Cs
S (J/mol-K)
29.09
51.45
64.67
76.78
85.5
Group 2 S (J/mol-K)
Be
9.44
Mg
32.51
Ca
41.4
Sr
54.392
Ba
63.2
Group 4 S (J/mol-K)
C
2.43
Si
18.7
Ge
42.42576
Sn
44.7688
Pb
68.85
What do you observe?
Entropy in a Single Group
90
Same observations!
Group 1 (Li to Cs)
Group 2 (Be to Ba)
Group 4A/14 (C to Pb)
80
70
o
S (J/mol-K)
60
50
40
30
20
10
0
0
1
2
3
4
Row of Periodic Table
5
6
7
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Reaction Entropy
The change in entropy in a reaction is the
difference between the summed entropy of the
products minus the summed entropy of the
reactants scaled by the number of moles
 Srx 
nS
i
products
i

nS
i
reac tan ts
i
Example Calculation 1: Calculate the change in
reaction entropy that occurs for each of the
following two phase changes given the data below:
1.
Css 
 Cs
2
Cs 
 Cs g
 Srx 
nS
i
products
i

Cs(s)
Cs(l)
Cs(g)
nS
i
reac tan ts
So (J/mol-K)
85.15
92.07
175.6
i




J
J
  1moleCss  8515

 Srx  1moleCs  92.07
.
molCs  K 
molCss  K 


J 
J 
J

 Srx    92.07    8515
.
    6.92 



K
K  
K



Example Calculation 1: Calculate the change in
reaction entropy that occurs for each of the
following two phase changes given the data below:
1.
Css 
 Cs
2
Cs 
 Cs g
 Srx 
nS
i
products

 Srx  1moleCsg
i

Cs(s)
Cs(l)
Cs(g)
nS
i
reac tan ts
So (J/mol-K)
85.15
92.07
175.6
i




J
J
 175.6
  1moleCs  92.07




molCsg  K 
molCs  K 



J 
J 
J

 Srx    175.6    92.07     8353
.

K 
K  
K

Spontaneous Reactions:
heat
or
randomness
-heat
Reactions
+randomness
Increase the
Spontaneity
By an increase
reactants
products
In randomness
-heat
+ randomness
-H
+S
H2O(l)
H2O(g)
Hg(l)
Hg(g)
So (J/mol-K)
69.91
188.83
“Medicine is the Art of
Observation” (Al B. Benson)
What do you observe?
77.40
174.89
Entropy Changes of Cs With Phase
200
160
140
o
85.15
92.07
175.6
S (J/mol-K)
Cs(s)
Cs(l)
Cs(g)
180
120
100
80
60
40
20
0
solid
1. Large change liquid to gas
2. S depends on Temp!!!
liquid
gas
Some typical Standard S values
Element
Ag
Ag+
AgBr
AgCl
AgI
AgNO3
Ag2O
Al
Ba
Ba2+
BaCl2
BaCO3
BaO
Br2
BrC
CCl4
ChCl3
CH4
C2H2
C2H4
C3H8
CH3OH
C2H5OH
CO
CO2
CO32-
form
s
aq
s
s
s
s
s
s
s
aq
s
s
s
l
aq
s
l
l
g
g
g
g
l
l
g
g
aq
S kJ/K-mol
0.0426
0.0727
0.1071
0.0962
0.1155
0.1409
0.1213
0.0283
0.0628
0.0096
0.1237
0.1121
0.0704
0.1522
0.0824
0.0057
0.2164
0.2017
0.1862
0.2008
0.2195
0.2699
0.1268
0.1607
0.1976
0.2136
-0.0569
Element
Ca
CaCl2
CaCO3
CaO
Ca(OH)2
CaSO4
Cd
Cd2+
CdCl2
CdO
Cl2
ClClO3ClO4Cr
CrO42Cr2O3
Cr2O72Cu
Cu+
Cu2+
CuO
Cu2O
CuS
Cu2S
CuSO4
F2
form
s
s
s
s
s
s
s
aq
s
s
g
aq
aq
aq
s
aq
s
aq
s
aq
aq
s
s
s
s
s
g
S kJ/K-mol
0.0414
0.1046
0.0929
0.0398
0.0834
0.1067
0.0518
-0.0732
0.1153
0.0548
0.223
0.0565
0.1623
0.182
0.0238
0.0502
0.0812
0.2619
0.0332
0.0406
-0.0996
0.0426
0.0931
0.0665
0.1209
0.1076
0.2027
Element
FFe
Fe2+
Fe3+
Fe(OH)
Fe2O3
Fe3O4
H2
H+
HBr
HCl
HCO3HF
HI
HNO3
H2O
H2O
H2O2
H2PO4HPO42H2S
H2SO4
HSO4Hg
Hg2+
HgO
I2
form
aq
s
aq
aq
s
s
s
g
aq
g
g
aq
g
g
l
g
l
l
aq
aq
g
l
aq
l
aq
s
s
S kJ/K-mol
-0.0138
0.0273
-0.1377
-0.3159
0.1067
0.0874
0.14645
0.1306
0
0.1986
0.1868
0.0912
0.1737
0.2065
0.1556
0.1887
0.0699
0.196
0.0904
-0.0335
0.2057
0.1569
0.1318
0.075
-0.0322
0.0703
0.1161
Element
IK
K+
KBr
KCl
KClO3
KClO4
KNO3
Mg
Mg2+
“Medicine is the Art of Observation”
Let’s rearrange to make this a little easier
form
aq
s
aq
s
s
s
s
s
s
aq
S kJ/K-mol
0.1113
0.0642
0.0642
0.0959
0.0826
0.1431
0.151
0.133
0.0327
-0.1381
Element
C
Cr
Fe
Al
Mg
Cu
CaO
Ca
Ag
CuO
Cd
CdO
Ba
K
CuS
HgO
BaO
form
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
average
S kJ/K-mol
0.0057
0.0238
0.0273
0.0283
0.0327
0.0332
0.0398
0.0414
0.0426
0.0426
0.0518
0.0548
0.0628
0.0642
0.0665
0.0703
0.0704
0.0446
Element
Cr2O3
KCl
Ca(OH)2
Fe2O3
CaCO3
Cu2O
KBr
AgCl
CaCl2
CaSO4
Fe(OH)
AgBr
CuSO4
BaCO3
CdCl2
AgI
I2
Cu2S
Ag2O
BaCl2
KNO3
AgNO3
KClO3
Fe3O4
KClO4
form
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
average
S kJ/K-mol
0.0812
0.0826
0.0834
0.0874
0.0929
0.0931
0.0959
0.0962
0.1046
0.1067
0.1067
0.1071
0.1076
0.1121
0.1153
0.1155
0.1161
0.1209
0.1213
0.1237
0.133
0.1409
0.1431
0.14645
0.151
0.11139
Element
CCl4
CHCl3
H2O2
C2H5OH
H2SO4
HNO3
Br2
CH3OH
Hg
H2O
form
l
l
l
l
l
l
l
l
l
l
average
0.15112
Pure Liquid
Element
C3H8
Cl2
C2H4
CO2
HI
H2S
F2
C2H2
HBr
CO
H2O
HCl
CH4
HF
H2
form
g
g
g
g
g
g
g
g
g
g
g
g
g
g
g
Solids
Gas
Do we notice anything?
1. More atoms = more S
average
2. Pure liquids higher S than s
3. Aqueous species can have neg S! Why?
4. Gas highest S
S kJ/K-mol
0.2164
0.2017
0.196
0.1607
0.1569
0.1556
0.1522
0.1268
0.075
0.0699
S kJ/K-mol
0.2699
0.223
0.2195
0.2136
0.2065
0.2057
0.2027
0.2008
0.1986
0.1976
0.1887
0.1868
0.1862
0.1737
0.1306
0.20026
Element
Cr2O72ClO4ClO3HSO4IHCO3H2PO4BrAg+
K+
ClCrO42Cu+
Ba2+
H+
FHg2+
HPO42CO32Cd2+
Cu2+
Fe2+
Mg2+
Fe3+
form
S kJ/K-mol
0.2619
0.182
0.1623
0.1318
0.1113
0.0912
0.0904
0.0824
0.0727
0.0642
0.0565
0.0502
0.0406
0.0096
0
-0.0138
-0.0322
-0.0335
-0.0569
-0.0732
-0.0996
-0.1377
-0.1381
-0.3159
average
0.021092
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
aq
Aqueous
Prediction reactions
producing more
Gas phase species
will increase in
entropy
Possible configurations?
solid < glass, plastic, liquid < gas
Ssolid < Ssolid, plastic,liquid
< Sgas
S
Note: this
Refers to a single
Molecule or element
Water ice, liquid, gas
Absolute
Zero, no motion
T (K)
A closely related idea to change in entropy
In phase changes is the change in entropy in
Going from individual ligands to chelates
From Module 18 we considered the electrostatic
attraction between the electron pairs on ligand
functional groups and the positive nucleus of a
metal ion.
Module 18 review
 q1q2 

E el  k 
r

r
 1 2
Cu( NH ) 
3 4
z
x
y
Module 18 review
2
Krx  K f 1 K f 2 K f 3 K f 4 K f 5 K f 6 ....... K fn
We observed that multidentate ligands had very high
Kf values
Lead Complexation
Constants
Ligand
logK1
F1.4
Cl1.55
Br1.8
I1.9
OH6.3
Acetate
2.7
Oxalate
4.9
Citrate
5.7
EDTA
17.9
logK2
1.1
0.6
0.8
1.3
4.6
1.4
1.9
LogK3
logK4
-0.4
-0.1
0.7
2
-0.7
-0.3
0.6
Which are polydentate?
What do you observe?
Module 18 review
logKf
2.5
1.05
2.2
4.5
12.9
4.1
6.8
5.7
17.9
Why so large?
Example 2: Predict the entropy change in the following reaction by
considering volume occupied and number of possible configurations
between the reactants and products
Note that the electrostatic attraction which shows up in the enthalpy
is similar for both compounds
M  NH2 CH3  4  X  2  2en 
 M  en 2  X  2  4 NH2 CH3
NH 2 CH 3
Example 2: Predict the entropy change in the
following reaction by considering volume occupied
and number of possible configurations between the
reactants and products
M  NH2 CH3  4  X  2  2en 
 M  en 2  X  2  4 NH2 CH3
5
3
o
 S exp
Cd  NH 3 CH 3  4  2en  Cd  en
2
2
2
 4 NH 3 CH 3
J
79.5
mol  K
o
 S calc
J
58.5
mol  K
J. Chem. Ed. 61,12, 1984, Entropy Effects in Chelation Reactions, Chung-Sun Chung
Example Calculation 3: Calculate the reaction
entropy changes for the reaction shown below given:
So (J/mol-K)
S
rhombic
S
orthoclinic
Cu(s)
CuS(s)
Cu2S(s)
32
33
85.15
92.07
175.6
CuS s  Cus  Cu2 S s
First let’s think about this a bit
1. Internal Entropy
a.Why should CuS have
more entropy than Cu?
b. Why should Cu2S have
much more entropy than
CuS?
CuS s  Cus  Cu2 S s
Which one might have
More configurations?
http://www.haraldthielenredlich.onlinehome.de/k
kch/cu1+cu2s.jpg
http://images.google.com/imgres?imgurl=http://www.unisa.edu.au/synchrot
ron/res/projects/chalcociteCu2S.jpg&imgrefurl=http://www.unisa.edu.au/sy
nchrotron/res/projects/default.asp&h=242&w=180&sz=20&hl=en&start=7
&um=1&tbnid=UYBN7p1lBciHwM:&tbnh=110&tbnw=82&prev=/images
%3Fq%3DCu2S%2B%26svnum%3D10%26um%3D1%26hl%3Den%26cli
ent%3Dfirefox-a%26channel%3Ds%26rls%3Dorg.mozilla:enUS:official%26sa%3DG
A single, individual, H atom can occupy 4
Different locations – thus the compound will be
CH4
C2H4
So (J/mol-K)
186.
219.4
More random than one with three locations for
The hydrogen
This reflects internal entropy which often scales
with # of Atoms in the molecule.
Example Calculation 3: Calculate the reaction
entropy changes for the reaction shown below given:
So (J/mol-K)
CuS s  Cus  Cu2 S s
2
1
First let’s think about this a bit
2. Spatial Volume of rx
entropy
Which will be more imp in rx
entropy? Internal entropy
or spatial volume entropy?
We take two separate chemical units and make
Them into 1 chemical unit – implies decrease in
entropy
S
rhombic
S
orthoclinic
Cu(s)
CuS(s)
Cu2S(s)
32
33
85.15
92.07
175.6
Example Calculation 3: Calculate the reaction
entropy changes for the reaction shown below given:
So (J/mol-K)
S
rhombic
S
orthoclinic
Cu(s)
CuS(s)
Cu2S(s)
32
33
85.15
92.07
175.6
CuS s  Cus  Cu2 S s
 Srx 
nS
i
products


J

 Srx  1moleCus Ss  175.6
molCus Ss  K 




J
  1moleCuS
  1moleCus  8515
.
s
molCus  K 



i

nS
i
reac tan ts
i






J
 92.07
 
molCuSs  K  


J  
J 
J

 Srx   175.6     8515
.
   92.07  



K
K 
K 

Example Calculation 3: Calculate the reaction
entropy changes for the reaction shown below given:
So (J/mol-K)
S
rhombic
S
orthoclinic
Cu(s)
CuS(s)
Cu2S(s)
32
33
85.15
92.07
175.6
CuS s  Cus  Cu2 S s
2
J  
J 
J

 Srx   175.6     8515
.
   92.07  



K
K 
K 

J  
J 
J

 Srx   175.6     177.22      162
.






K
K 
K

1
Example Calculation 4: Calculate the change in
entropy for the allotropic forms of elemental S
So (J/mol-K)
S
rhombic
S
orthoclinic
Cu(s)
CuS(s)
Cu2S(s)
32
33
85.15
92.07
175.6
Gr: allos = others
S s , r hom bic  S s,orthoclinic
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Entropy of the surroundings
Spontaneous Process: Universal entropy increases
(The universe is winding down.)
 Stotal ( universe)   S system ( chemical rx )   S surroundings
 S universe  0  spon tan eous
Example 1: Ssurroundings
Cs  O2, g  CO2, g  heat
Change in Ssurroundings?
Change in Sreaction?
More organized (fewer molecules)
implies less entropy, less random
Sreaction <0
heat will
increase
kinetic
energy of
gases =
Ssurround >0
Example 2: Ssurroundings
From surroundings, withdraw heat,
Less kinetic energy, less motion, less entropy
H2 O  heat  H2 Og
H2 Og
H2 O
+ heat
Although this process requires
heat, it is spontaneous, driven
by entropy of chemical reaction
Reaction less random
rx more random
The two reactions (the system)
Cs  O2, g  CO2, g  heat
H2 O  heat  H2 Og
Interact with the surroundings by exchange of heat
Heat of reaction must be related to entropy of
surroundings
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Randomness of the
“surroundings” affected
By enthalpy
 Stotal ( universe)   S system ( chemical rx )   S surroundings
 S surroundings   Hreaction
related to enthalpy
or heat of reaction
proportional
Where will impact on Ssurroundings be greatest?
a.
1 J at 600oC
b. 1 J at 25oC
Predict entropy change is largest at low temperatures
 Ssurroundings  as T 
 S surroundings
 S surroundings
 H reaction

T
 H reaction
 
T
sign change accounts for the
fact that entropy increases with
exothermic reactions
Context Slide for a calculation on entropy of the
surroundings
Historically Ag was mined as Ag2S found in the presence of PbS, galena.
Part of the process of releasing the silver required oxidizing the galena.
The lead oxide recovered was used in glass making. The
fumes often killed animals near by and have left a permanent record in
the artic ice. Large regions near silver mines were deforested.
One reason that this process was discovered so early in history was
The low temperature at which it could be carried out.
Lead in Artic Ice
Who has the “honor” of most contaminating
Medicine is the art
The artic ice?
of observation
Calculating Ssurrounding Example 1
2 PbS s  3O2, g  2 PbOs  2SO2, g
Compare the change in entropy of the surroundings for
this reaction at room temperature and at the
temperature of a campfire (~600 oC).
Know:
reaction
o
T  25 C
o
T  600 C
 S surroundings
HO 
Don’t know
entropy
 H reaction
 
T
o
n

H

f , products 
o
n

H

f ,reac tan ts
red herrings?
none
2 PbS s  3O2, g  2 PbOs  2SO2, g
Substance
O2(gas)
PbS
PbO
SO2(gas)
HO 
 n H
Hf0 (kJ/mole)
0
-100
-219
-297
o
f , products

 n H
o
f ,reac tan ts

  297 kJ  
  219 kJ  
 
  2moleSO 
 H   2mole PbOs 
2,g 

mole

PbOs  

 moleSO2 , g  

O
 0kJ  

  100kJ  

  3moleO 
 2mole PbSs 
2,g 

 mole PbSs  

 moleO2 , g  
 H O    438kJ   594kJ    200kJ
 H O   1032  200   832kJ
 S surroundings
 H reaction
 
T
 S surroundings  
  832kJ 
T
Calculating Ssurrounding Example 1
2 PbS s  3O2, g  2 PbOs  2SO2, g
Compare the change in entropy of the surroundings for
this reaction at room temperature and at the
temperature of a campfire (~600 oC).
 S surroundings  
T  25o C
SSsurroundings
surroundings  
298
298
T
T  600 C
T  873K
o
T  298K
832
832kJ
kJ
  832kJ 
  2.792kJ
SSsurroundings
surroundings  
 832
832kJ
kJ
873
873
  0.953kJ
Our prediction was right!  S surroundings Larger at low T
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Total Entropy change
With reaction enthalpy
 Stotal ( universe)   S system ( chemical rx )   S surroundings
 S surroundings
 H reaction
 
T
 Stotal ( universe)   S system ( chemical rx ) 
 Hchemical rx
T
Reaction Entropy Example Calculation 3 Compare
the total entropy change for the following reaction at
25oC and 600oC
2 PbS s  3O2, g  2 PbOs  2SO2, g
Substance
O2(gas)
PbS(solid)
PbO(solid)
SO2(gas)
S0 (J/K-mole)
205  Srx   ni S o i , products   ni S o i ,reac tan ts
91
66.5
248


 


J
J
 
  2moleSO2  248
 S rx   2mole PbO  66.5
K  mol PbO 
K  molSO2  





 


J
J
 
  3 MoleO2  205
  2mole PbS  91
K  moleO2  
 K  mole PbS 


Reaction Entropy Example Calculation 3 Compare
the total entropy change for the following reaction at
25oC and 600oC
2 PbS s  3O2, g  2 PbOs  2SO2, g
 Srx 
o
n
S
 i i , products 
o
n
S
 i i ,reac tan ts


 


J
J
 
  2moleSO2  248
 S rx   2mole PbO  66.5
K  mol PbO 
K  molSO2  





 


J
J
 
  3 MoleO2  205
  2mole PbS  91
K  moleO2  
 K  mole PbS 


J 
J

 Srx    143    496   
K 
K 

J  
J 

  182    615  
K
K 

J
 S rx   168
K
 Stotal ( universe )   S system ( chemical rx ) 
 S reaction
J
  168
K
  832kJ 
T
  832kJ 
T
T  600 o C
T  25 C
o
J
J
 S   168  2,791
K
K
J
 S  2623
K
Spontaneous
T
 S surroundings  
 Stotal ( universe )   168 J 
J   832kJ 
 Stotal ( universe)   168 
K
298K
J
kJ
 S total )   168  2.791
K
K
 Hchemical x
 Stotal ( universe)
 S total )
J   832kJ 
  168 
K
873K
J
kJ
  168  0.953
K
K
 S total
 S total
J
J
  168  953
K
K
J
 785
K
Less spontaneous
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
“Free energy” is a
Way of accounting
For contribution of randomness
Hchemical
H

chemicalr xx 
Ssystem
 SStotal
T
total( (universe
universe))  S
system((chemical
chemicalrx
rx)) 
T


 T Stotal ( universe)   T S system ( chemical rx )   Hchemical rx
 T Stotal ( universe)   Hchemical rx  T S system ( chemical rx )
Define
 T S total ( universe )   G free energy rx
 G free energy   Hr x  T Srx
Gibb’s free energy
 G free energy  0
a) enthalpy of bonds
b) organization of atoms Spontaneous reaction
c) randomness of surroundings
Galen, 170
Marie the Jewess, 300
Charles Augustin
James Watt
Coulomb 1735-1806 1736-1819
Justus von
Thomas Graham
Liebig (1803-1873 1805-1869
Ludwig Boltzman
1844-1906
Gilbert N
Lewis
1875-1946
Henri Louis
LeChatlier
1850-1936
Johannes
Bronsted
1879-1947
Jabir ibn
Hawan, 721-815
Luigi Galvani
1737-1798
Richard AC E
Erlenmeyer
1825-1909
An alchemist
Count Alessandro G
A A Volta, 1747-1827
James Joule
(1818-1889)
Henri Bequerel
1852-1908
Lawrence Henderson
1878-1942
Galileo Galili Evangelista Torricelli
1564-1642
1608-1647
Amedeo Avogadro
1756-1856
Rudolph Clausius
1822-1888
Jacobus van’t Hoff
1852-1911
Niels Bohr
1885-1962
John Dalton
1766-1844
William Thompson
Lord Kelvin,
1824-1907
Johannes Rydberg
1854-1919
William Henry
1775-1836
Johann Balmer
1825-1898
J. J. Thomson
1856-1940
Erwin Schodinger Louis de Broglie
1887-1961
(1892-1987)
Fitch Rule G3: Science is Referential
Jean Picard
1620-1682
Jacques Charles
1778-1850
Francois-Marie
Raoult
1830-1901
Heinrich R. Hertz,
1857-1894
Friedrich H. Hund
1896-1997
Daniel Fahrenheit
1686-1737
Max Planck
1858-1947
Rolf Sievert,
1896-1966
Blaise Pascal
1623-1662
Georg Simon Ohm
1789-1854
James Maxwell
1831-1879
Robert Boyle,
1627-1691
Isaac Newton
1643-1727
Michael Faraday
1791-1867
B. P. Emile
Clapeyron
1799-1864
Dmitri Mendeleev
1834-1907
Svante Arrehenius
Walther Nernst
1859-1927
1864-1941
Fritz London
1900-1954
Wolfgang Pauli
1900-1958
Johannes D.
Van der Waals
1837-1923
Marie Curie
1867-1934
Anders Celsius
1701-1744
Germain Henri Hess
1802-1850
J. Willard Gibbs
1839-1903
Fritz Haber
1868-1934
Thomas M Lowry
1874-1936
Werner Karl Linus Pauling Louis Harold Gray
1905-1965
Heisenberg 1901-1994
1901-1976
Conceptually:
 G free energy   Hr x  T Srx
H reaction
+
+
Sreaction
+
+
-
 G free energy  0
Spontaneous?
always
at high T, 2nd term lg.
at lowT, 2nd term sm
never
Gibbs Free Energy Example 1
When will this reaction be spontaneous, hi or lo T?
2 PbS s  3O2, g  2 PbOs  2SO2, g
 G free energy   Hr x  T Srx
 H rxO 
 832 kJ
H reaction
+
+
 Srx
  168 J
Sreaction
+
+
-
Spontaneous?
always
at high T, 2nd term lg.
at lowT, 2nd term sm
never
At LowT!!!
200
Non Spontaneous Reaction
Free Energy >0
Free Energy (kJ)
0
-200
-400
More
spontaneous
-600
Spontaneous Reactions
Free Energy <0
Can we figure
Out exactly at what T
this reaction becomes
Spontaneous?
-800
-1000
0
1000
2000
3000
4000
5000
6000
7000
KC
o
2 PbS s  3O2, g  2 PbOs  2SO2, g
 G free energy
J

  832 kJ  T   168 

K
To find when a reaction will just go
Spontaneous (or not)
1. Use the equation:
 G free energy   Hr x  T Srx
2. Set Go to zero (equilibrium)
0   Hr x  T Srx
3. Solve for T.
T Srx   Hr x
Tbecomes spon tan eous 
 Hr x
 S rx
4. Depending upon sign of enthalpy entropy
determine if temperature decrease/increase
causes Go to go negative
Gibbs Free Energy Example 2
At what T will this reaction become change between
Spontaneous and non-spontaneous?
2 PbS s  3O2, g  2 PbOs  2SO2, g
 G free energy   Hr x  T Srx
 G free energy
J

  832 kJ  T   168 

K
J

0   832 kJ  T   168 

K
J

832 kJ  T  168 

K
832 kJ
T
kJ
0168
.
K
4952K  T
Rx spontaneous at T<4952K
Gibbs Free Energy Example 2: The only good substitute for PbCO3 for
white paint is TiO2. To manufacture this paint need to be able to process
titanium ore TiO2. (Different allotrope). At what temperature does the
following reaction become spontaneous?
TiOs  2Cs  Ti s  2COg
Substance
Tisolid
CO(gas)
TiO2(solid)
O2(gas)
SO2(gas)
Csolid
Hf0 (kJ/mole) S0(J/K-mole)
485
179.45
-110.5
198
-945
50
0
205
-297
248
0
0
 G free energy   Hr x  T Srx
Substance
Tisolid
CO(gas)
TiO2(solid)
O2(gas)
SO2(gas)
Csolid
Hf0 (kJ/mole)
485
-110.5
-945
0
-297
0
S0(J/K-mole)
179.45
198
50
205
248
0
TiOs  2Cs  Ti s  2COg
 G free energy   Hr x  T Srx
kJ
..55kJ
kJ kJ    0kJ  

485
kJ 
110
.5kJ
 945
485
  110
       945
485
kJ
485
kJ
110
kJ
 G free energy  111mole
mole
 
  1mole
 
2

mole
mole
 mole
1mole
 
mole mole 222
mole
mole









mole
mole
mole
mole
mole
mole
 mole  
 mole     


.45.45
J J  
198
J J    
J J 
 
 179
 179
 198
  5050
 0J  
 T T1mole
1
mole

2

mole
2
mole


1
mole
1
mole

2
mole
















   













KK
 mole
 mole 
KK
 mole
 mole   
KK
 mole
 mole 
K  mole  
 
 G free energy   485  221    945 kJ
 T 575  50
 G free energy
J
K
J
 1652 kJ  T  525
K
Substance
TiO2solid
Ti
O2(gas)
SO2(gas)
Csolid
CO(gas)
Hf0 (kJ/mole)
-945
485
0
-297
0
-110.5
S0(J/K-mole)
50
179.45
205
248
0
198
TiOs  2Cs  Ti s  2COg
 G free energy   Hr x  T Srx
 G free energy
J
 1652 kJ  T  525
K
When is this reaction spontaneous:
at high or low temp?
T  3144 K
0  1652  T  0.525
Rx spontaneous > 3144K
T  0.525  1652
Context Slide 1
WWII
titanium was not routinely processed until
after WWII (jet engine technology). So TiO2 purified
not available cheaply for paint until after WWII
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Reference states for
Free Energy
As for enthalpy and entropy, there are tables
Of values obtained via Hess’s Law
 G,orx 
o
n

G
 i f ,i , products 
o
n

G
 i fi ,reac tan ts
f means formation at standard state 25 oC!!!!!
Properties and Measurements
Property
Size
Volume
Weight
Temperature
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
1.66053873x10-24g
quantity
mole
Pressure
Energy, General
Animal hp
heat
BTU
calorie
Kinetic J
Electrostatic
electronic states in atom
Electronegativity F
horse on tread mill
1 lb water 1 oF
1 g water 1 oC
m, kg, s
1 electrical charge against 1 V
Energy of electron in vacuum
As for enthalpy and entropy, there are tables
Of values obtained via Hess’s Law
 G,orx 
o
n

G
 i f ,i , products 
o
n

G
 i fi ,reac tan ts
f means formation at standard state 25 oC!!!!!
State of Matter
Standard (Reference) State
Solid
Liquid
Gas
Solution
Elements
Pure solid
Pure liquid
1 atm pressure
1 M concentration
Gfo0
Gibbs Standard Free Energy Example Calc. 1:
What Is the standard free energy change of the
following Reaction? 2 PbS s  3O2, g  2 PbOs  2SO2, g
Substance
PbO
SO2(gas)
PbS
O2(gas)
 Grxo 
Gf 0 (kJ/mole)
-188.9
-300
-99
0
o
n

G
 i f ,i , products 
o
n

G
 i fi ,reac tan ts



kJ 
kJ  
 
 G   2moles PbO   188.9
  2molesSO2 , g   300
moles PbO 
moleSO2  



o
rx


 

kJ 
kJ

  3moleO  0
  2moles PbSs   99
2,g 

mole PbSs 

 moleO2 , g  

Gibbs Standard Free Energy Example Calc. 1:
What Is the standard free energy change of the
following Reaction? 2 PbS s  3O2, g  2 PbOs  2SO2, g



kJ 
kJ  
 G   2moles PbO   188.9 moles   2molesSO2 , g   300 mole  


PbO
SO2  



 

kJ 
kJ

  3moleO  0
  2moles PbSs   99
2,g 

mole PbSs 

 moleO2 , g  

o
rx
 Grxo    377.8kJ   600kJ 
   198kJ 
 Grxo   779.8kJ
Gibbs Standard Free Energy Example Calc. 1:
What Is the standard free energy change of the
following Reaction? 2 PbS s  3O2, g  2 PbOs  2SO2, g
 Grxo 
o
n

G
 i f ,i , products 
o
n

G
 i fi ,reac tan ts
 Grxo   779.8kJ
For comparison, we calculated from before:
 Hrx   832 kJ
 G free energy
 G free energy
J
 S rx   168
K
J

  832 kJ  T   168 

K
J

  832 kJ   298 K    168 

K
 G free energy   782kJ
Not too bad of
Agreement!
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Summing Reactions
Rx#
1
A+B
2
nC + D
nC
E
3
E
A+B+D
Greaction
Greaction 1
Greaction 2
Greaction 1 + Greaction 2
Summing Free Energy Example Calculation Why
was lead one of the first elements first processed by
man? A. Calculate the standard free energy of the
Combined reactions. B. Calculate the free energy of
the reaction at 600 oC (campfire temp).
2 PbS s  3O2, g  2 PbOs  2SO2, g
2 PbOs  2Cs  2 Pbs  2COg
2 PbS s  3O2, g  2Cs  2 Pbs  2SO2 , g  2COg
Summing Free Energy Example Calculation Why
was lead one of the first elements first processed by
man? A. Calculate the standard free energy of the
Combined reactions. B. Calculate the free energy of
the reaction at 600 oC (campfire temp).
2 PbS s  3O2, g  2 PbOs  2SO2, g
 H   832kJ
O
rx
J
 S   168
K
0
rx
 Grxo   779.8kJ
2 PbOs  2Cs  2 Pbs  2COg
Need standard free energy to solve A
But! Will also need standard enthalpy and S
To solve B – so solve for those
Substance
Pb
CO(gas)
PbS
PbO
O2(gas)
SO2(gas)
Csolid
Hf0 (kJ/mole) S0(J/K-mole)
0
0
-110.5
198
-100
91
-219
66.5
0
205
-297
248
0
0
2PbOsolid + 2Csolid
2Pbsolid + 2CO(gas)
ΔH= [{(2(0)+2(-110.5)}-{2(-219)+2(0)}]=+217kJ
ΔS=[{2(0)+2(198)}-{2(66.5)+2(0)}]=263J/K
?
2 PbOs  2Cs  2 Pbs  2COg
 H  217kJ
O
rx
J
 S  263
K
0
rx
 G free energy   Hr x  T Srx
 G free energy
J   kJ 

 217 kJ  T  263   3 

K   10 J 
At standard conditions
kJ 

 G  217 kJ  T  0.263 

K
o
kJ 

 G  217 kJ  (25  273) K  0.263    138.6

K
o
At campfire conditions
kJ 

 G  217 kJ  (873) K  0.263    12.56

K
o
 Grxo   138.6kJ
Net reaction at 25 oC
Grx
2PbS(solid) + 3O2(gas)
PbO(solid) + 2SO2(gas)
2PbOsoloid + 2Csolid
2Pbsolid + 2CO(gas)
2PbS + 3O2(gas) + 2Csolid
-779.8 kJ
+138.6kJ
2Pbsolid + 2SO2gas + 2COgas
sum = -641kJ
The net standard free energy for the coupled two
reactions is -641 kJ, spontaneous
Net reaction at 600 oC
2PbS(solid) + 3O2(gas)
PbO(solid) + 2SO2(gas)
2PbOsoloid + 2Csolid
2Pbsolid + 2CO(gas)
2PbS + 3O2(gas) + 2Csolid
Grx
-685 kJ
-12.6kJ
2Pbsolid + 2SO2gas + 2COgas
sum = -697kJ
The reduction of Pb in PbS to metal and oxidation
of S in PbS to sulfur dioxide gas is spontaneous at
campfire temperatures of 600oC
Context point: can manufacture pure lead in a campfire
Context Slide
Where did all the lead go?
Decade
1914-23
1920-29
1930-39
1940-49
1950-59
1960-69
1970-1979
Context Slide
Estimate lbs white lead/housing unit
110
87
TiO2 makes inroads
42
particularly in Europe
22
7
3
White lead restricted
1
2 2
2 2
2 22
5 5
300gPb
gPb
1lbPb
lbPb 12
12
.gsoil
gsoil 228
228
.44
mi
gsoil
1609
.
km




300
1
.
12
.
.
44
mi
1609
.
km
10
10
cm
cm




x
x
x
x
6
3
x
x
x
x
x
x
x3cm  14,790,239lbPb
6 gsoil






3 3
3


45359
gPb cm
10
cm
Chicago
45359
.. gPb
mi
106 gsoil
cm
Chicago   mi     km
km  
300ppm = “background” level of Chicago soil lead
Depth: does
Not move down
Because of Oh Card me PleaSe
Percent of Children with
Elevated Blood Lead Levels
Chicago, 1999
ugPb/gsoil
<250
250-500
500-1000
>1000
N
W
E
S
Percent of Children
with Elevated Blood
Lead Levels
< 5%
5% - 15%
16% - 25%
> 25%
5
0
5 Miles
Context Slide
Relates to
a) Age of Housing
b) “Gentrification”
Relevant Chem 102 Concepts:
1. Temperature dependence
of spontaneous reactions
2. Stability of soil lead form
(Oh Card me PleaSe)
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #20
Spontaneity
Relating Free Energy
To Concentrations
The free energy of the reaction related to
a) standard free energy change
b) and the ratio of concentrations of
products to reactants, Q
 G   G  RT ln Q
o
In this equation you can use (simultaneously)
Pressures
Concentrations
The ln(Q) is treated as unitless
Free Energy and Conc. Example Calc. Calculate the free
energy of the reaction if the partial pressures of the gases are each
0.1 atm, 298 K. Remember, we calculated ΔGrx to be -641 kJ at
298K (25 oC)
2 PbS( s)  3O2 s  2C( s)  2 Pb( s)  2SO2( g )  2CO( g )
 G   G  RT ln Q
o
 
 
 Pb 2 P 2 P 2 
J
s
SO2 CO 


 Grx   641kJ   8.314   298 K  ln
 PbS 2 P 3 C 2 

K


s
O2
s
2
2
 1 2 PSO

P
CO
2
 Grx   641kJ   2477.57 J  ln 2 3 2 
 1 Po2 1 
2
2
 PSO

P
CO
2

 Grx   641kJ   2.47757 kJ  ln
3
 PO2 
 01
. 2SO2 01
. 2CO 

 Grx   641kJ   2.47757 kJ  ln
3
. O2 
 01
 
Grx   641kJ   2.47757kJ  ln 01
.
 Grx   641kJ   2.47757kJ  2.302)
Grx   641kJ    5703
. kJ 
 Grx   647
kJ
mol
When Q = K (equilibrium):
0   G o  RT ln K
 RT ln K   G o
 G   RT ln K
o
 G   G o  RT ln Q
0
K
At equilbrium no
Energy to drive
Rx one way or other
K=1
-RT ln (1) = 0
K >1
-RT ln (>1) = -(+) < 0
K<1
-RT ln (<1) = -(-) > 0
Example Problem 2 Free Energy and Equilibrium:
What is the equilibrium constant for the reaction
at a campfire temperature?
2 PbS s  3O2, g  2Cs  2 Pbs  2SO2 , g  2COg
 G   RT ln K
o
G
 ln K
 RT
o
e
 G o
RT
Go = -697kJ/mol rx
kJ 

   697


mol 
Ke
 K

kJ 
 8.314 x103
 298 K
mol
K


K  e 281  10100
Example 3 Free Energy and Equilibrium:
The corrosion of Fe at 298 K is K = 10261 .
What is the equilibrium constant for corrosion
of lead?
2Pbsolid + O2gas
2PbOsolid
We don’t have any K values so we need
To go to appendix for various enthalpy and
Entropies to come at K from the backside
Substance
PbS
PbO
Pb
O2(gas)
SO2(gas)
Csolid
CO(gas)
Hf0 (kJ/mole) S0(J/K-mole)
-100
91
-219
66.5
0
0
0
205
-297
248
0
0
-110.5
198
2 Pbs  O2( g )  2 PbO
Go = Ho - TSo
Ho = 2(-219) - {2(0) + 2(0)} = -438kJ
So = 2(.0665) - {2(0) + 2(.205)} = -0.277kJ/K
Go = -438 - T(-.277) = -438 - (298)(-0.277) = -355kJ
RT ln K  - Go
 G   355kJ
kJ 

   355


mol 
Ke
 G 0
RT
e

kH 
 8.314 x103
 298 K
mol K 

K  e143  127 x1062
K for rusting of Fe = 10261
K for rusting of Pb = 1.27x1062
so: even though the reaction is favorable
it is less so than for iron.
Lead rusts less than iron = used for plumbing
“A” students work
(without solutions manual)
~7 problems/night.
Module #20
Spontaneity
What you need to
know
1. Be able to rank the entropy of various phases of
materials, including allotropes
2. Be able to rank the entropy of various compounds
3. Explain entropy concepts as related to chemical
geometry
4. Calc. standard entropy change for a reaction
5. Relate surrounding entropy to reaction enthalpy
6. Calc. temperature at which a reaction becomes
spontaneous
7. Explain why TiO2 was relatively late in replacing
PbCO3 as a white pigment; why lead was one of
first pure metals obtained by humanity
8. Convert standard free energy to equilibrium
constant
“A” students work
(without solutions manual)
~7 problems/night.
Module #20
Spontaneity
END
Entropy and Molecular Structure:
O2(g)
O3(g)
S,J/mol-K
161
205
237.6
C (g)
CO (g)
CO 2(g)
158.O
197.9
213.6
Cl (g)
Cl2(g)
165.2
2232.96
z
Pb
PbO
PbO2
Pb3O4
68.85
68.70
76.98
209.2
z
CH4
C2H4
186.4
219.4
C2H2
C2H4
C2H6
200.8
219.4
229.5
O(g)
z
#configurations  z
n
“z” = 2
z
Have we
Convinced
Ourselves
Yet?
“z” = 4
“z” =6
“z” = 8
Internal Entropy and Molecular Structure:
#configurations  z
n
Internal Entropy is generally increasing
With number of atoms in the molecule
Because the number of locations within the molecule
Where an atom could be found is increasing and
Because the possible orientations
of the molecule increases
Entropy of various Solids
14
1, average=38.26
12
Number observed
10
8
2, average=62.15
6
3, average = 95.42
4, average=137.9
4
5, average 103
Reliability of average
decreases with number
of averaged data points
2
6, average- 124
0
0
50
100
150
200
Entropy (J/mol-K)
250
300
350