CHAPTER 14
Thermodynamics: Spontaneous Processes,
Entropy, and Free Energy
Useful energy is being "degraded" in
the form of unusable heat, light, etc.
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•
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A tiny fraction of the sun's
energy is used to produce
complicated, ordered, highenergy systems such as life
Our observation is that natural processes proceed from ordered,
high-energy systems to disordered, lower energy states.
In addition, once the energy has been "degraded", it is no longer
available to perform useful work.
It may not appear to be so locally (earth), but globally it is true (sun,
universe as a whole).
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Thermodynamics - quantitative description
of the factors that drive chemical reactions, i.e.
temperature, enthalpy, entropy, free energy.
Answers questions such as-
• will two or more substances react when they are mixed
under specified conditions?
• if a reaction occurs, what energy changes are associated
with it?
• to what extent does a reaction occur to?
Thermodynamics does NOT tell us the RATE of a reaction
Spontaneous Processes
A spontaneous process is one that is capable of proceeding in
a given direction without an external driving force
• 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
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Spontaneous chemical and physical changes are frequently
accompanied by a release of heat (exothermic H < 0) C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
Ho = -2200 kJ
Although many spontaneous processes are
exothermic (e.g., the previous combustion
reaction), not true for all spontaneous reactions:
Endothermic (ΔH > 0),
but reaction is
spontaneous.
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Some processes are accompanied by no change in enthalpy
at all (Ho = 0.0), as is the case for an ideal gas
spontaneously expanding:
spontaneous
nonspontaneous
There's another factor promoting spontaneity in these processes,
and that's the increasing randomness or disorder of the system:
1. propane combustion:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
2. water melting:
H2O(s)  H2O(l)
Ho = -2200 kJ
Ho = 6.01 kJ
3. gas expansion:
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Thermodynamics: Entropy
 Second Law of Thermodynamics:
• The total entropy of the universe increases
in any spontaneous process
 Entropy (S):
• A measure of the amount of disorder
(qualitative), or unusable energy in a system
at a specific temperature (quantitative).
• Entropy is affected by molecular motion, or
disorder from volume changes (e.g. the
previous gas expansion example).
Types of Molecular Motion
 Three types of motion:
• Translational—movement
through space.
• Rotational—spinning motion
around axis  to bond.
• Vibrational—movement of
atoms toward/away from each
other.
 As temperature increases, the
amount of motion increases.
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Trends in Entropies
Ssolid < Sliquid < Sgas
Entropy is expected
to INCREASE for
these types of
processes (S > 0) :
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Changes in Entropy
order
S
disorder
S
S = Sf - Si
S > 0
S < 0
Suniverse
Example - Predict whether the entropy change is greater than
or less than zero for each of the following processes:
(a) freezing ethanol
(b) evaporating a beaker of liquid bromine at room temperature
(c) dissolving sucrose in water
(d) cooling nitrogen gas from 80 oC to 20 oC.
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Entropy Changes in the System (Ssys)
The standard entropy of reaction (S0rxn ) is the entropy
change for a reaction carried out at 1 atm and 25 0C.
aA + bB
cC + dD
from Appendix
S0rxn = [cS0(C) + dS0(D)] - [aS0(A) + bS0(B) ]
S0rxn = S nS0(products) - S mS0(reactants)
The Second Law of Thermodynamics:
The total entropy of the universe increases in any spontaneous process
Spontaneous process:
sys = system
surr = surroundings
Equilibrium process:
Suniv = Ssys + Ssurr > 0
One can be negative but the
other will be even more positive
Suniv = Ssys + Ssurr = 0
At equilibrium the forward and reverse
rates are equal, e.g vapor pressure
(l)
=
(g)
So there is no net change in entropy
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Entropy Changes in the Surroundings (Ssurr)
Exothermic Process
Ssurr > 0
Endothermic Process
Ssurr < 0
The change in entropy of the surroundings
can be calculated:
Ssurr  -Hsys
Ssurr  1
Tsurr
if Hsys < 0 (exothermic), then
Ssurr > 0 (entropy of the
surroundings increases)
if Hsys > 0 (endothermic), then
Ssurr < 0 (entropy of the
surroundings decreases)
If the temperature of the
surroundings is already high,
then pumping heat in or out
causes less change in disorder
than at lower temperatures
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Combining the two:
Ssurr  -Hsys
and
Ssurr  1
Tsurr
so
Ssurr = -Hsys
T
(Tsurr usually = Tsys)
e.g. N2(g) + 3H2(g)  2NH3(g)
Ssys = -198.3 J/K
Hsys = -92.6 kJ*
The two main driving forces are in opposition to each other the release of heat favors a spontaneous reaction while the
decrease in entropy does not. Calculating Suniv will decide
the issue (next slide). Remember: for a spontaneous
reaction the entropy of the universe increases.
*from Ch 5: H0rxn = S nH0f (products) - S mHf0 (reactants)
Is the reaction spontaneous at 25 oC?
Suniv = Ssys + Ssurr
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The previous example with ammonia illustrated that
maybe entropy will decrease in the system, but this
will always be accompanied by a greater increase in
the entropy of the surroundings such that Suniv > 0.
Suniv = Ssys + Ssurr > 0
Another way of stating the 2nd Law is that "You Can't Win!"
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