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Lecture 3, Entropy and Free Energy
Part b
Text: Chapter 7, Sections 5 through end
Preliminary: reminder of results from macroscopic (thermodynamic) view.
Reversible work wrev = -nRT∙ ln (V2/V1)
Heat Capacity C = q / ΔT
Entropy change ΔS = qrev/ T = ΔH / T
ΔS = C ln (T2/T1)
ΔS = nR∙ ln (V2/V1) = nR∙ ln (P2/P1)
2) Microscopic Viewpoint
The second description for entropy is called the microscopic or molecular
interpretation. It applies the ideas from Probability Theory to the idea of
“microstates”. According to this viewpoint,
entropy is a measure of available states.
Boltzman
S = k lnW
W-number of states
Probability: One coin toss: either heads or tails: two available states, 2 1
Two coin tosses: 2 heads, 2 tails, head-tail, tail-heal: 4 available states, 22
Three coin tosses: 23 = 8 available states
Let’s take the example of four heteronuclear diatomic molecules as our sample.
The figure below shows the 24 =16 states available of positioning the molecules
head-to-head or head-to-tail. We call these 16 microstates. The microstates will be
at different energy levels depending on the number of intermolecular attractions or
repulsions. At T=0K there is no energy available to overcome the forces of
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attraction for a molecule such as HCl. Therefore, there is only one state available,
W=1, ln1=0. S=0
This is an example of the Third Law of Thermodynamics:
The entropies of all perfect crystals approach zero as the absolute temperature
approaches zero.
However, if the crystal cannot arrive at a “perfect” structure, then there will be
what is termed “residual entropy” at 0K.
In the case of the CO molecule, the dipole-dipole force of attraction of attraction is
very low. (You can demonstrate this by drawing formal charges.) So, the 16
microstates are available at 0K. (Self-tests 7.9) The entropy per 4 molecules at 0K
would be S=k∙ln (24) = (1.4x10-23J/K)∙4∙ln2 = 3.9x10-23J/K
The entropy for 1 mole is calculated using Avogadro’s number N=6.02x10 23.
S = k∙ln (2N) = (1.4x10-23J/K)∙ 6.02x1023∙ln2 = 5.7 J/K
Let’s look at more complicated molecules than diatomics, FClO3 for example.
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There are four different ways the molecule can position itself in the crystal. So for
1 mole at 0K, there is a residual entropy of
S = k∙ln (4N) = (1.4x10-23J/K)∙ 6.02x1023∙ln4 = 11. J/K
There is already a trend appearing. As molecules
become more complex, their residual entropy at 0K
seems to increase.
Now let’s see what happens to the entropy content
when T is raised far above 0K or when a molecule
interacts with (captures) electromagnetic radiation
and turns it into internal motion.
Energy Diagram
molecule 1
molecule 2
excited state
²E
excited state
²E
ground
state
S2
> S1
ground
state
because excited state is more available
1) All energy is quantized, i.e. there are only certain energy states.
2) All motion involves energy. Therefore, forms of motion are quantized.
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Energy States
ETotal = Eelectronic + Evibrational + Erotational + Etranslational
•
Eelectronic
Electronic transitions (orbitals)
Ultra-violet and visible ranges of spectrum
Energy
E
•
Evibrational
Vibrations
O
H
O
H
“Angle bend”
H
O
H
“Asymmetric stretch”
H
H
“Symmetric stretch”
Vibrational energies involve infra-red radiation
Each motion has a ground state and excited states.
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Energy
2
E
1
•
Erotational
Rotational Motion (microwave region)
Whole molecule spins on an axis
O
H
•
Etranslational
H
Translational Motion
Whole molecule vibrates in place; or gas molecules move through space..
Predicting entropies
S (for a particular substance) depends on physical state
Sgas >> Sliquid > Ssolid
• More complex (floppier) molecules have larger entropies
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1) Higher MW, usually higher S
CH4
<
186.3
<
C2H6
C3H8
269.9 (J/mol•K)
229.6
2) Similar MW, but more complex, always higher S
Ar
154.7
<
<
CO2
C3H8
269.9 (J/mol•K)
213.7
• Ionic Solids: entropies are larger as attractive forces become smaller (because
there are more vibrational states attainable at a particular T).
CaO <
LiF
<
CsI
ΔS (for reactions)
• Increases when solids dissolve in liquids
∆Sgas >> ∆Sliquid > ∆Ssolid
• Increases when gas escapes solution
• If a reaction produces more molecules, entropy increases
CaCO3 (s)
CaO (s) + CO2 (g)
Ag+ (aq) + Br– (aq)
Entropy Increases
AgBr (s)
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Entropy Decreases
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Since S is a state function:
1) Hess’s Law applies; 2) can use standard entropies of formation Sof
Srxn   nS f ( prod.)   mS f (react.)
N2 g  3H2 g   2NH3 g 



Srxn  2S NH3   S N 2   3S H2 
Further Discussion
The term, “Entropy is a measure of disorder” is misleading and can lead to
incorrect conclusions. (So, don’t use it!) For instance, answer the question.
“Which room has more entropy; the one with a desk with 4 drawers and papers in
each drawer, or another room with a desk with 6 drawers and the same number of
papers neatly stacked in one drawer?”
If we use this analogy for the entropy of substances, it is the room with the desk
with six drawers. How the papers are distributed at any one given moment (a snap
shot view) is irrelevant since things are always moving around at the molecular
level. More drawers means more available states, which means more entropy. If
you want to continue the analogy to the molecular level, place a fan in each room.
At one speed of the fan, the papers will be blowing around and distributing
themselves in the drawers. At high speeds (higher T) the papers will be
distributing themselves among more drawers. Electrons and atoms and molecules
are always moving and redistributing themselves among the states.
It would be more appropriate to say that “disorder” is measured by T∙ΔS
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Tying together the Macroscopic (Thermodynamic) and Microscopic (Molecular)
Descriptions
Since energy is quantized into allowed states, we can use our simplest of models
for quantum systems as a guide: a particle in a box model. Recall what happens to
the energy levels when you increase the length of the box… What? You don’t
remember? Well, get out your guitar, put your finger at a fret way down (short
string) and pluck: twing,twing, twing (high frequency). Now take you finger off
the frets and pluck: twung twung, twung (low frequency). E=h∙
So, energy levels come down when you increase the length of the box, or the
volume of a 3D box.(See figure above, on right.) Lowering the energy levels
means making them more available, which means more entropy content.
W depends on Volume
W = constant ∙ V for each particle
W = (constant ∙ V)N for N particles
S = k∙lnW = k∙ln(constant ∙ V)N = kN∙ln(constant ∙ V)
ΔS = kN∙ln(constant ∙ Vf) - kN∙ln(constant ∙ Vi)
ΔS = kN∙ln(constant ∙ Vf / constant ∙ Vi)
ΔS = kN∙ln(Vf / Vi) = nR∙ln(Vf / Vi)
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Global Changes in Entropy
Remember: The Second Law states that the entropy of an isolated system will
increase to a maximum.
To answer the question how can a system become
more comlex, we must consider any system to be
part of a larger isolated system that includes the
surroundings of the system of interest.
ΔStotal = ΔS + ΔSsurroundings
(ΔS ≡ ΔSsystem )
and ΔHsystem = - ΔHsurroundings
so ΔSsurroundings = - ΔH / T
Self-test 7.14
When heat flows out of a system into hot
surroundings it produces less disorder (increase in
entropy is smaller) than when the heat flows into
cooler surroundings, where more disorder will be
produced by that amount of heat.
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Overall Change in Entropy
When using entropy as a judge of the direction of a spontaneous reaction, the total
entropy must be used.
For the process “A goes to B”,
If ΔStotal > 0, the process is spontaneous.
If ΔStotal < 0, the reverse process is spontaneous. “B goes to A”.
If ΔStotal = 0, “equilibrium”, no tendency to proceed in either direction.
Self-tests 7.15
Chemical Equilibrium
The necessary condition for chemical equilibrium is that the rate of the forward
reaction must equal the rate of the reverse reaction. This is a dynamic equilibrium
(akin to juggling) rather a static equilibrium (such as a see-saw or balance)
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Reversible and Irreversible Processes
Reversible processes are at equilibrium at
every infinitesimal change along the pathway, and the ΔS of the system will equal
the ΔS of the surroundings
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Self-tests 7.16
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Gibbs Free Energy
Two Tendencies for Spontaneity
1.
Lose heat
2.
“Disorder”
∆H < 0
"exothermic"
- T∆S
• as T rises, T∆S rises: more disorder
• want to give "free energy" a negative sign when spontaneous

Free Energy Change
amount of energy available to do chemical work
G  H  TS
Sum of two tendencies
- heat to evolve
- randomness
Spontaneous reaction
G  0
balance between
H & S
Standard Free Energy Changes
Standard conditions: 1 atm for gases, pure solids, pure liquids, solutions at 1M (p
243)
Greaction   G f ( products)   G f (reactants)
State Functions: H, S, and G are state functions, whose changed values (∆H, ∆S,
∆G) are determined only by the initial and final values.
Grxn   G f ( prod.)   G f (react.)
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Gsys  Hsys  TSsys
Self-tests 7.19-22
A
B
For A  B
if ∆G < 0
spontaneous for A  B
if ∆G > 0
not spontaneous (but B  A is)
if ∆G = 0
equilibrium
Gibbs Free Energy is the “maximum non-expansion work”, which is the chemical
generation of electrical power and has significance in everything from
electrochemistry to bioenergetics.
The Effect of Temperature:
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G  H TS
∆H ∆S ∆G
- +
Always neg.
+ -
always pos.
Reaction characteristics
reactions is spontaneous at all
temperatures
reaction is non spontaneous at all
temperatures: reverse reaction is always
spontaneous
-
-
neg. at low temps:
Reactions is spontaneous at low
pos at high temps
temperatures but becomes
nonspontaneous at high temperatures
+ +
Pos. at low temps:
Reaction is nonspontaneous at low
neg at high temps
temperatures but becomes spontaneous at
high temperatures.
For the 3rd and 4th cases, the point at which a non-spontaneous reaction becomes
spontaneous, and vice-versa, is at equilibrium: ΔG = 0 = ΔH – TΔS
or
T = ΔH / ΔS
Self-tests 7.23
Bonus: Given that the heat of atomization of a diatomic molecule is always
negative, prove that there is a temperature above which the molecule will
spontaneously decompose.
XY (g) -> X(g) + Y(g)
The Laws of Thermodynamics
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First Law of Thermodynamics
The total energy of the universe is a constant
E  q  w
E  Esurr  0
Second Law of Thermodynamics
The entropy of an isolated process increases in the course of a spontaneous
process
Third Law of Thermodynamics
The entropy of a pure, perfectly formed crystalline substance at absolute zero is
zero.
S = k∙lnW
at absolute zero, there is only one possible state available to the
crystal, W = 1,
S = k∙ln(1) = 0
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