Chapter 17
1
Microstates and Entropy
The Austrian physicist Ludwig Boltzmann introduced a model
to relate the total number of microstates (the multiplicity,
W) to entropy (S).
Boltzmann constant (k)
S = k ln W
k = 1.38 x 1023J/K
units on S is J/K
R
k
NA
Gas constant
Avogadro constant
Boltzmann was the founding father of statistical mechanics,
a completely new way of thinking about theoretical physics.
His work was ridiculed by his fellow professors.
Standard Entropy
• Difficult to count the number of
microstates directly.
• Measured by calorimetery.
• Standard entropy (S) absolute entropy
of a substance at 1 atm (typically at 25C)
• Units of J K-1 mol-1
• All positive values.
• Absolute values (unlike DH which is
relative)
3
Why is there a –S?
S = k ln W
3+
Al + 6H2O
2+
Al(H2O)6 (aq)
4
Chapter 17
5
Spontaneity
Spontaneous: process that does occur
under a specific set of conditions.
Nonspontaneous: process that does not
occur under a specific set of conditions.
The reverse of a spontaneous process
is a nonspontaneous one.
Spontaneity depends on temperature.
6
Predicting Spontaneity
If we mix reactants and products together
will the reaction occur?
Enthalpy (H) is used to quantify the heat flow into or out of
a system in a process that occurs at constant pressure.
DHrxn = Hproducts – Hreactants
Hproducts < Hreactants
DH < 0
Exothermic
Enthalpically favorable
Hproducts > Hreactants
DH > 0
Endothermic
Enthalpically unfavorable
Not enough information. We need to know the change in
entropy to predict if a reaction will spontaneously occur.
7
Predicting Spontaneity
If we mix reactants and products together
will the reaction occur?
Entropy (S) Is a measure of the number of specific ways
(microstates) in which a thermodynamic system may be arranged.
DSrxn = Sproducts – Sreactants
Sproducts < Sreactants
DS < 0
Entropy decreases
Entropically unfavorable
Sproducts > Sreactants
DS > 0
Entropy increases
Entropically favorable
Still not enough information. We need to know total entropy
change to predict if a reaction will spontaneously occur.
8
Predicting Spontaneity
If we mix reactants and products together
will the reaction occur?
DSuniv = DSsys + DSsurr
Equilibrium : DSuniv = 0
Nonspontaneous process: DSuniv < 0
Spontaneous process: DSuniv > 0
DSuniv = DSsys + DSsurr
If we know DSsurr we
can calculate DSuniv
Can calculate from:
DS0rxn = S nS0(products) - S mS0(reactants)
9
Predicting Spontaneity
DSuniv = DSsys + DSsurr
If we know DSsurr we
can calculate DSuniv
Can calculate from:
DS0rxn = S nS0(products) - S mS0(reactants)
Surrounding = everything in the
universe except the system.
Very, very difficult to measure!
If not impossible.
10
Heat and Entropy
Exothermic Reaction
DHsys
0 < DHsys
Endothermic Process
+DHsys
0 > DHsys
Surroundings + heat = ↑ S
Surroundings - heat = ↓ S
DSsurr > 0
DSsurr < 0
DSsurr ∝ -DHsys
11
Heat and Entropy
Heat released by the system increases the disorder of the surroundings.
DSsurr ∝ -DHsys
The effect of -DHsys on the surroundings depends on temperature:
– At high temperature, where there is already considerable
disorder, the effect is muted
– At low temperature the effect is much more significant
– The difference between tossing a rock into a calm pool (low T)
and a storm-tossed ocean (high T)
DSsurr =
-DHsys
T
12
Predicting Spontaneity
DSuniv = DSsys + DSsurr
Substitution:
DSuniv = DSsys +
-DHsys
DSsurr =
-DHsys
T
T
Multiply by -T:
-TDSuniv = -TDSsys + DHsys
Rearrange:
-TDSuniv = DHsys - TDSsys
This equation relates DSuniv to DHsys and DSsys.
Both in terms of the system.
13
Gibbs Free Energy
-TDSuniv = DHsys - TDSsys
DG = DHsys - TDSsys
in K
Gibbs free energy (DG)• AKA Free energy.
• Relates S, H and T of a system.
• Can be used to predict
spontaneity.
• G is a state function.
Josiah Willard Gibbs (1839-1903)
First American Ph.D. in Engineering (Yale, 1863)
Praised by Albert Einstein as "the
greatest mind in American history"
14
Died at 64 from an acute intestinal obstruction.
Gibbs Free Energy
For a constant temperature and constant pressure process:
-TDSuniv = DHsys - TDSsys
DG = DHsys - TDSsys
Gibbs free energy (DG)- Can be used to predict spontaneity.
DG < 0
The reaction is spontaneous in the forward direction.
DG = -T(+DSuniv)
DG > 0
DSuniv > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
+DG = -T(-DSuniv)
DS < 0
univ
DG = 0
The reaction is at equilibrium.
DG = -T(DSuniv) = 0
DSuniv = 0
15
Gibbs Free Energy
DG = DH - TDS
• If you know DG for reactants and products then you can
calculate if a reaction is spontaneous.
• If you know DG for two reaction then you can calculate if the
sum is spontaneous.
• If you know DS, DH and T then you can calculate spontaneity.
• Can predict the temperature when a reaction becomes
spontaneous.
• If you have DHvap or DHfus and DS you can predict boiling and
freezing points.
• If you have DHvap or DHfus and T you can predict the entropy
change during a phase change.
• Can predict equilibrium shifts.
16
Standard Free Energy Changes
0 ) is the free-energy
The standard free-energy of reaction (DGrxn
change for a reaction when it occurs under standard-state
conditions.
aA + bB
cC + dD
0
DGrxn
= S nDG0f (products) - S mDG0f (reactants)
Standard free energy of formation (DGf0) is the free-energy change
that occurs when 1 mole of the compound is formed from its
elements in their standard states.
0
DGrxn
= [ cDG0f (C) + dDG0f (D) ] - [aDG0f (A) + bDG0f (B) ]
DG0f of any element in its stable form is zero.
17
Standard Free Energy Changes
DG is a state function so free energy can be calculated from the
table of standard values just as enthalpy and entropy changes.
aA + bB
cC + dD
0
DGrxn
= S nDG0f (products)
- S mDG0f (reactants)
Standard free energy of formation
(DGf0) is the free-energy change that
occurs when 1 mole of the
compound is formed from its
elements in their standard states.
18
Standard Free Energy Changes
Calculate the standard free-energy change for the following reaction:
2KClO3(s)  2KCl(s) + 3O2(g)
0
DGrxn
= S nDG0f (products) - S mDG0f (reactants)
DG0rxn = [2(408.3 kJ/mol) + 3(0)]  [2(289.9 kJ/mol)]
DG0rxn = 816.6  (579.8)
DG0rxn = 236.8 kJ/mol
Is the reaction spontaneous?
DG0rxn < 0
From appendix 3:
KClO3(s) DGf = -289.9 kJ/mol
KCl(s) DGf = -408.3 kJ/mol
O2(g)
DGf =
0 kJ/mol
Yes!
19
Example 17.4
Calculate the standard free-energy changes for the following
reactions at 25°C.
(a) CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔG°rxn = [ΔG°f (CO2) + 2ΔG°f (H2O)] - [ΔG°f (CH4) + 2ΔG°f (O2)]
ΔG°rxn =[(-394.4 kJ/mol) + (2)(-237.2 kJ/mol)] - [(-50.8 kJ/mol) + (2) (0 kJ/mol)]
ΔG°rxn = -818.0 kJ/mol
Spontaneous?
Yes.
From appendix 3:
CH4(g)
O2(g)
CO2(g)
H2O (l)
DGf = -50.8 kJ/mol
DGf =
0 kJ/mol
DGf = -394.4 kJ/mol
DGf = -237.2 kJ/mol
20
Example 17.4
Calculate the standard free-energy changes for the following
reactions at 25°C.
(b) 2MgO(s)
2Mg(s) + O2(g)
ΔG°rxn = [2ΔG°f (Mg) + ΔG°f (O2)] - [2ΔG°f (MgO)]
ΔG°rxn = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-569.6 kJ/mol)]
ΔG°rxn = 1139 kJ/mol
From appendix 3:
Spontaneous?
MgO(s) DGf = -569.6 kJ/mol
O2(g)
DGf =
0 kJ/mol
Mg(s) DGf =
0 kJ/mol
No.
But the reverse is…
2Mg(s) + O2(g)
2MgO(s)
21
More DG° Calculations
Appendix 3
DG0rxn = S nDG0f (products)
- S mDGf0 (reactants)
To predict spontaneity of any rxn:
1) Pick any reactants and products.
2) Write a balanced equation.
3) Calculate DG0rxn.
4) Is the reaction spontaneous?
22
Another Example
C(s, diamond) + O2(g)
CO2(g)
ΔG°rxn = [ΔG°f (CO2)] - [ΔG°f (C, diamond) + ΔG°f (O2)]
From appendix 3:
DG°rxn = -396.4 kJ
C, diamond(s) DGf =
2.9 kJ/mol
O2(g)
DGf =
0 kJ/mol
CO2(g)
DGf = -394.4 kJ/mol
+ O2
Therefore, diamonds are contributing to global warming!
very slowly
23
More DG° Calculations
Similar to DH°, one can use the DG° for various
reactions to determine DG° for the reaction of
interest (a “Hess’ Law” for DG°)
Hess’ Law- states that regardless
of the multiple stages or steps of a
reaction, the total enthalpy change
for the reaction is the sum of all
changes.
What is the DG° for this reaction:
Given:
C(s, diamond) + O2(g)
CO2(g) DG° = -397 kJ
C(s, graphite) + O2(g)
CO2(g) DG° = -394 kJ
24
More DG° Calculations
What is the DG° for this reaction:
Given:
C(s, diamond) + O2(g)
C(s, graphite) + O2(g)
CO2(g)
C(s, diamond) + O2(g)
CO2(g)
C(s, diamond)
CO2(g)
CO2(g)
DG° = -397 kJ
DG° = -394 kJ
C(s, graphite) + O2(g)
DG° = +394 kJ
CO2(g)
DG° = -397 kJ
C(s, graphite) + O2(g)
DG° = +394 kJ
C(s, graphite)
DG° = -3 kJ
DG°rxn < 0…..rxn is spontaneous
25
Alternative DG Calculation
Is the following reaction spontaneous at 298 K?
(assume standard conditions)
4KClO 3 (s) 
 3KClO 4 (s) + KCl(s)
Given:
KClO3(s)
KClO4(s)
KCl (s)
DH°f (kJ/mol)
-397.7
-432.8
-436.7
S° (J/mol.K)
143.1
151.0
82.6
Given
DG°rxn = DH°rxn - TDS°rxn
Then Find
First Calculate
26
Alternative DG Calculation
4KClO 3 (s) 
 3KClO 4 (s) + KCl(s)
KClO3(s)
KClO4(s)
DH°f (kJ/mol)
-397.7
-432.8
S° (J/mol.K)
143.1
151.0
KCl (s)
-436.7
82.6
DH rxn  3DH  f KClO4  + DH  f KCl   4DH  f KClO3 
 3(432.8kJ ) + (436.7kJ )  4(397.7kJ )
 144kJ
DS rxn  3S KClO4  + S KCl   4S KClO3 
 3(151.0 J
 36.8 J
K
K
) + (82.6 J
K
)  4(143.1 J
K
)
27
Alternative DG Calculation
4KClO 3 (s) 
 3KClO 4 (s) + KCl(s)
KClO3(s)
KClO4(s)
DH°f (kJ/mol)
-397.7
-432.8
S° (J/mol.K)
143.1
151.0
KCl (s)
-436.7
82.6
DH°rxn = -144 kJ
DS°rxn = -36.8 J/K
Enthalpically favorable
Entropically unfavorable
DGrxn  DHrxn  TDSrxn
 1kJ 
J
 144kJ  298K  38.6

K 
1000J 
 133kJ


DG°rxn < 0…..rxn is spontaneous at 298 K
What about at 5000 K?
(Assuming the same DH and S)
DG°rxn = 50 kJ
DG°rxn > 0…rxn is
nonspontaneous at 5000
K
28
DG Temperature Dependence
4KClO 3 (s) 
 3KClO 4 (s) + KCl(s)
DG°rxn = -133 kJ at 298 K
Spontaneous
DG°rxn = 50 kJ at 5000 K Nonspontaneous
Reaction spontaneity is a temperature dependent phenomenon!
DGrxn = DHrxn -
TDSrxn
29
DG Temperature Dependence
Enthalpy:
Entropy:
DHrxn < 0
The reaction is enthalpically favorable.
DSrxn > 0
The reaction is entropically favorable.
Need both to predict spontaneity. And sometimes temperature!
DG = DH - TDS
DG < 0
DG > 0
DG = 0
Spontaneous
Nonspontaneous
Equilibrium
1) If DH < 0 and DS > 0, then DG is negative at all T
2) If DH > 0 and DS > 0, then DG depends on T
3) If DH < 0 and DS < 0, then DG depends on T
4) If DH > 0 and DS < 0, then DG is positive at all T
30
DG Temperature Dependence
DG = DH - TDS
3) DH < 0 and DS < 0
DG < 0
DG > 0
DG = 0
Spontaneous
Nonspontaneous
Equilibrium
(Enthalpically favorable, entropically unfavorable)
If DH < TDS, then DG is positive.
If DH > TDS, then DG is negative.
Nonspontaneous at high T
Spontaneous at low T
2) DH > 0 and DS > 0
(Enthalpically unfavorable, entropically favorable)
If DH < TDS, then DG is negative.
If DH > TDS, then DG is positive.
Spontaneous at high T
Nonspontaneous at low T
31
DG Temperature Dependence
DG = DH - TDS
32
DG Temperature Dependence
Four possible scenarios:
(1) DH < 0, DS > 0
Nonspontaneous
(2) DH > 0, DS > 0
(3) DH < 0, DS < 0
Spontaneous
(4) DH > 0, DS < 0
33
DSuniv and Spontaneous Reactions
DG = DH - TDS
Class 1:
Class 3:
endothermic
exothermic
system becomes more disordered
Class 2:
system becomes more disordered
exothermic
34
system becomes more ordered
Predicting T from Gibbs Equation
4KClO 3 (s) 
 3KClO 4 (s) + KCl(s)
DG°rxn = -133 kJ at 298 K
Spontaneous
DG°rxn = 50 kJ at 5000 K Nonspontaneous
Reaction spontaneity is a temperature dependent phenomenon!
At what temperature will the reaction become spontaneous?
Find T where DG changes from positive to negative.
I.e. when DG =0.
DG = DH – TDS = 0
T = DH/DS
35
Predicting T from Gibbs Equation
At what T is the following reaction spontaneous?
Br2(l)
Br2(g)
Given: DH°= 30.91 kJ/mol
DS°= 93.2 J/mol.K
DG = DH – TDS = 0
T = DH/DS
T = (30.91 kJ/mol) /(93.2 J/mol.K)
T = 331.7 K
DH > 0, DS > 0
The reaction will be
spontaneous when
T > 331.7 K
36
Predicting T from Gibbs Equation
At what T is the following reaction spontaneous?
CaCO3 (s)
DG0 = DH0 – TDS0 = 0
CaO (s) + CO2 (g)
Equilibrium Pressure of CO2
T = D0H/DS0
DH0 = 177.8 kJ/mol
DS0 = 160.5 J/K·mol
DG0 = 0 at 835 oC
37
Chapter 11
38
Liquid ↔ Gas
Liquid
At time = 0
At time > 0
At time = ∞
At equilibrium!
DG = DH - TDS
DG < 0
DG > 0
DG = 0
Spontaneous
Nonspontaneous
Equilibrium
# of molecules in =
# of molecules out
rate of evaporation =
rate of condensation
39
Gibbs Equation and Phase Change
H2O (l)
H2O (g)
Molar heat of vaporization (DHvap) is the energy required
to vaporize 1 mole of a liquid at its boiling point.
If we know DHvap and boiling point we can calculate DS!
DG = DH – TDS = 0
DHvap = 40.79 kJ/mol
BP(H2O) = 373 K
DS =
DH
T
40.79 kJ/mol
DS =
373 K
40
Example 17.5
The molar heats of fusion and vaporization of benzene are
10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy
changes for the solid → liquid and liquid → vapor transitions for
benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at
80.1°C.
vapor
liquid
solid
41
Example 17.5
The molar heats of fusion and vaporization of benzene are
10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy
changes for the solid → liquid and liquid → vapor transitions for
benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at
80.1°C.
solid → liquid → vapor
DHfus
DHvap
Molar heat of vaporization (DHvap) is the
energy required to vaporize 1 mole of liquid
to gas.
Molar heat of fusion (DHfus) is the energy
required to melt 1 mole of solid.
42
Example 17.5
The molar heats of fusion and vaporization of benzene are
10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy
changes for the solid → liquid and liquid → vapor transitions for
benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at
80.1°C.
solid → liquid → vapor
DHfus
DG = DH – TDS = 0
ΔSfus
ΔH fus
=
Tf
DHvap
DH
DS =
T
ΔSvap =
ΔH vap
Tb
(10.9 kJ/mol)(1000 J/1 kJ)
=
(5.5 + 273)K
=
= 39.1 J / K  mol
= 87.8 J / K  mol
(31.0 kJ/mol)(1000 J/1 kJ)
(80.1 + 273)K
43
Chapter 17
45