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Entropy and Free Energy
The Basis for Thermodynamics
First law of thermodynamics:
The change in the energy of a system
∆U = q + w
is the sum of the heat and the work
done by or on the system.
the first law does not, however, account for the observation
that natural processes have a direction of progress.
q
∆S 
T
Second Law of Thermodynamics:
The Second Law of Thermodynamics states that the state of
entropy of the entire universe, as a closed isolated system,
will always increase over duration.
Entropy always increases in the direction of duration.
The second law also states that the changes in the entropy in
the universe can never be negative
Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 0 0C and ice melts above 0 0C
• Heat flows from a hotter object to a colder object
• A gas expands in an evacuated bulb
• Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2
spontaneous
nonspontaneous
18.2
Entropy (S): A measure of the degrees of freedom of the system.
(A measure of the disorder of the system.)
Second Law of Thermodynamics:
All processes occur spontaneously in the direction that increases the entropy of the
universe.
∆Suniv = ∆Ssys+ ∆Ssurr > 0
k – Boltzmann constant 1.38 x 10-23 J/K
R/NA
S = k ln W W – ways of arranging the components of a system without
changing its energy.
(microstates)
Third Law of Thermodynamics:
A perfect crystal has zero entropy at a temperature of absolute zero.
Random Motion in a Crystal
[Not at 0 K]
∆Suniv = 0
at 0 K
Richard Feynman on Entropy :
“So we now have to talk about what we mean by disorder and
what we mean by order. ... Suppose we divide the space into little
volume elements. If we have black and white molecules, how many
ways could we distribute them among the volume elements so that
white is on one side and black is on the other? On the other hand,
how many ways could we distribute them with no restriction on
which goes where? Clearly, there are many more ways to arrange
them in the latter case. We measure "disorder" by the number of
ways that the insides can be arranged, so that from the outside it
looks the same. The logarithm of that number of ways is the
entropy. The number of ways in the separated case is less, so the
entropy is less, or the "disorder" is less.”
Adding energy to a system always changes
the entropy of the system.
The definition of entropy derived from the second law of thermodynamics has no
molecular interpretation.
The Austrian physicist Ludwig Boltzmann (1844-1907) showed in 1877 that entropy
has a fundamental molecular significance: It is a measure of the disorder of a system.
This disorder can be asymmetry or displacement.
This significance is relevant in the
sub-atomic nature of the physical universe as well.
Boltzmann proposed that entropy is related to the number of different microscopic
views of obtaining a specified macroscopically definable and observable situation.
If the number of equivalent ways of constructing a system is W, then the entropy is
proportional to the logarithm of W:
“In the admission of synthetic expertise among chemists, it is embarrassingly
admitted also that it is impossible to re-create even the simplest enzyme or
structural protein found in biotic materials. The process that requires entropy
reduction to increase complexity also requires heat energy to form chemical bonds.
This produces an obstacle for our un-hindered production of amazingly complex molecules.”
Pharmaceuticals, polymers, enzymes and structural and nutrient proteins all require
the original biotic material to begin with, as it contains an entropy reducing structure:
catalysts and replication directing chemical structures containing information (RNA,DNA)
as well as sophisticated electronic energy work. The smallest DNA/RNA strand contains
more information than the whole of the Encyclopedia Britannica
Information is capable of reducing entropy-but information is also susceptible to entropy.
An example is the polymerization of amino acids to form protenoids:
a Amino acid A + b Amino acid B  c polymer + d H2O
La Chateliers principle indicates that this equilibrium would lie far to the left and entropy
requirements will grow rapidly with molecular weight. Some small protenoids have been
formed in the laboratory, but they are small molecular weight species. The free energy
to incorporate many high M.W. amino acids into larger polymers has a limiting factor we
shall derive later.
Binomial Distribution
Only four states
Are solid crystals
The rest are gases
Counter examples exist:
A new state function can be defined, the free energy, G:
G = H - TS
For reactions at constant temperature and pressure, any reaction is
spontaneous when G decreases.
In a reversible process in which no external work is involved the free energy of a
reacting system does not change and G=0. If the system does electric, magnetic
or gravitational work then:
where all other work is wext
so that:
When a system does work on its surroundings in a reversible way, the decrease
In the free energy exactly balances the work done other than pressure-volume
Work.
Thermodynamics
State functions are properties that are determined by the state of the system, regardless of
how that condition was achieved.
energy, enthalpy, pressure, volume, temperature, entropy.
Potential energy of hiker 1 and hiker 2 is the same even
though they took different paths.
6.7
Entropy Changes in the System (∆Ssys)
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at
rxn
1 atm and 250C.
aA + bB
∆S0
[
=
rxn
∆S0
rxn
cS0(C) +
= Σ nS0(products)
cC + dD
dS0(D) ] - [
aS0(A) +
bS0(B) ]
- Σ mS0(reactants)
What is the standard entropy change for the following reaction at 250C?
2CO2 (g)
2CO (g) + O2 (g)
S0(CO) = 197.9 J/K•mol
S0(CO2) = 213.6 J/K•mol
S0(O2) = 205.0 J/K•mol
∆S0
∆S0
rxn
rxn
= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
= 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
18.3
3 Factors Involved in Determining Whether a
Reaction is Spontaneous or Not
 ∆H; a negative value (exothermic reactions)
favors a spontaneous reaction
 ∆S; a positive value (disorder increases) favors
a spontaneous reaction
 Temperature; If the above factors conflict;
temperature determines if reaction is
spontaneous or not
∆S
-∆H
+∆S
Spontaneous at
all Temperatures
∆G=∆H-T∆S
+∆H
+∆S
Spontaneous at
High Temperatures
∆H
-∆H
-∆S
Spontaneous at
Low Temperatures
+∆H
-∆S
NonSpontaneous at
all Temperatures
Conditions for Spontaneity of a Reaction
∆S osurroundings = -
Recall:
∆H osystem
T
∆Suniv = ∆Ssys + ∆Ssurr > 0
Josiah Willard Gibbs (1878)
Introduced the spontaneity test as the “Free Energy”
term for a reaction.
Gibbs Equation:
∆Gsys = ∆Hsys - T∆Ssys
“∆G” is the “Free Energy Change”
Gibbs Free Energy and Spontaneity
∆G < 0
Spontaneous
∆G > 0
Nonspontaneous
∆G = 0
Equilibrium
The value of ∆G indicates the amount of
energy which is available to do useful work!
Gibbs Free Energy
For a constant-temperature process:
Gibbs free energy (G)
∆G = ∆Hsys -T∆Ssys
∆G < 0
The reaction is spontaneous in the forward direction.
∆G > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
∆G = 0
The reaction is at equilibrium.
18.4
∆G; Change in Gibb’s Free Energy
∆G = ∆Hsys -T∆Ssys
1. Determine if reaction is spontaneous or
not.
2. wmax; free energy available to do useful
work. ∆G = wmaximum
∆Hsys; represents energy available
for useful work
-T∆Ssys; represents energy that can’t
Be harnessed to do useful work
Ways to Obtain ∆Greaction
1. From ∆Gfo Values
2. From Gibb’s Free Energy Equation
∆G = ∆Hsys -T∆Ssys
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction
rxn
when it occurs under standard-state conditions.
aA + bB
∆G0
∆G0
rxn
=
[
c∆G0 (C) +
f
= Σ n∆G0 (products)
rxn
f
d∆G0 (D) ] - [
f
cC + dD
a∆G0 (A) +
f
b∆G0 (B) ]
f
- Σ m∆G0 (reactants)
f
Standard free energy of formation (∆G0) is the
free-energy change that occurs when 1 mole of the
f
compound is formed from
its elements in their
standard states.
∆G0 of any element in its stable form is zero.
f
18.4
What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
∆G0
∆G0
∆G0
rxn
rxn
12CO2 (g) + 6H2O (l)
= Σ n∆G0 (products)
rxn
f
12∆G0 (CO2) +
f
- Σ m∆G0 (reactants)
f
6∆G0 (H2O)] - [
f
=
[
=
[ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
2∆G0 (C6H6)]
f
Is the reaction spontaneous at 25 0C?
∆G0 = -6405 kJ
<0
spontaneous
18.4
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s)
CaO (s) + CO2 (g)
∆H0 = 177.8 kJ; Does not favor spontaneous reaction
∆S0 = 160.5 J/K; Favors spontaneous reaction
∆G0 = ∆H0 – T∆S0
Determine the temperature range where the
above reaction will be spontaneous.
-∆G; spontaneous
∆G=0;equilibrium
+∆G;nonspontaneous
0 = ∆H0 – T∆S0
∆H0 = T∆S0
T=∆Η0/ ∆S0
T = 177.8 kJ/ 0.1605 kJ/K =1108K
Entropy and States of Matter
[Note entropy and temperature changes.]
Predicting Relative Standard
Molar Entropy Values [So] of
a System
1 atm
1M
298 K (25oC)
Most Stable Form
for Solids or
Liquids
Fig. 20.5
Gibbs Free Energy and Phase Transitions
Ice melting;
Liquid vaporizing
Solid subliming
H2O (l)
H2O (g)
Phase changes happen at equilibrium
∆G0 = ∆H0 – T∆S0
∆G0 = 0
T=∆H0/ ∆S0
Determine the boiling point of water given that
∆Svap = 109.3 J/mole K and ∆Hvap = 40.62 kJ/mole.
T = 40.6 kJ/mole
= 373 K = 100 °C
0.1093 kJ/mole K
18.4