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CHAPTER 12: Thermodynamics
Why Chemical Reactions Happen
Useful energy is being "degraded" in
the form of unusable heat, light, etc.
•
•
•
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
Chapter Outline
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12.1 Spontaneous Processes
12.2 Entropy
12.3 Absolute Entropy and Molecular Structure
12.4 Applications of the Second Law
12.5 Calculating Entropy Changes
12.6 Free Energy
12.7 Temperature and Spontaneity
12.8 Coupled Reactions
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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
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
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But sometimes a spontaneous process
can be endothermic H > 0 -
Some processes are accompanied by no change in enthalpy
at all (Ho = 0), as is the case for an ideal gas spontaneously
expanding:
spontaneous
nonspontaneous
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There's another factor promoting spontaneity in these processes,
and that's the increasing randomness or disorder of the system
(this is a qualitative description only – quantitative coming up):
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:
Chapter Outline
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
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




12.1 Spontaneous Processes
12.2 Entropy
12.3 Absolute Entropy and Molecular Structure
12.4 Applications of the Second Law
12.5 Calculating Entropy Changes
12.6 Free Energy
12.7 Temperature and Spontaneity
12.8 Coupled Reactions
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Thermodynamics: Entropy
 Second Law of Thermodynamics:
• The entropy of the universe increases in any
spontaneous process
• Suniv = Ssys + Ssurr > 0
 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 perpendicular to
bond
– Vibrational: Movement of
atoms toward/away from
each other
• As temperature increases,
the amount of motion
increases.
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Third Law of Thermodynamics
 The entropy of a perfect crystal is
zero at absolute zero
 Provides a point of reference or
baseline for quantitating entropy
(placing a numerical value on it)
 Heat plays a role in the amount of
entropy a system has
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Chapter Outline
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





12.1 Spontaneous Processes
12.2 Entropy
12.3 Absolute Entropy and Molecular Structure
12.4 Applications of the Second Law
12.5 Calculating Entropy Changes
12.6 Free Energy
12.7 Temperature and Spontaneity
12.8 Coupled Reactions
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Standard Molar Entropy, So
(absolute entropy content, J/mol K
The entropy of one mole of
a substance in its standard
state at 1 atm and 298 K.
Sosolid < Soliquid < Sogas
Trends in Entropies of Phase Changes
Ssolid <  Sliquid <  Sgas
S = Sfinal - Sinitial
Units: J/molK
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The Effect of Molecular Structure on Entropy
Summary:
S is expected to
INCREASE for these
types of processes
(S > 0) :
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Sample Exercise 12.1:
Predicting the Sign of an Entropy Change
Predict whether or not an increase or decrease in entropy
accompanies each of these processes when they occur at
constant temperature:
(a)H2O(l)  H2O(g)
(b)NH3(g) + HCl(g)  NH4Cl(s)
(c)C12H22O11(s)  C12H22O11(aq)
Sample Exercise 12.2:
Comparing Standard Molar Entropy Changes
Without consulting any standard reference sources, select the
component in each of the following pairs that has the greatest
standard molar entropy at 298 K. Assume that there is one mole
of each component in its standard state (the pressure of each
gas is 1 bar and the concentration of each solution is 1 M).
(a)HCl(g), HCl(aq)
(b)CH3OH(l), CH3CH2OH(l)
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Chapter Outline








12.1 Spontaneous Processes
12.2 Entropy
12.3 Absolute Entropy and Molecular Structure
12.4 Applications of the Second Law
12.5 Calculating Entropy Changes
12.6 Free Energy
12.7 Temperature and Spontaneity
12.8 Coupled Reactions
Changes in Entropy
Spontaneous
Nonspontaneous
S = Sf - Si
S > 0
S < 0
Suniverse
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The Second Law of Thermodynamics:
The total entropy of the universe increases in any spontaneous process
Spontaneous process:
sys = system
surr = surroundings
Suniv = Ssys + Ssurr > 0
One can be negative but the
other will be even more positive
Entropy Changes in the Surroundings (Ssurr)
Exothermic Process
Ssurr > 0
Endothermic Process
Ssurr < 0
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The change in entropy of the surroundings
can be calculated:
if Hsys < 0 (exothermic), then
Ssurr > 0 (entropy of the
surroundings increases)
Ssurr  -Hsys
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
Ssurr  1
Tsurr
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 9: H0rxn = S nH0f (products) - S mHf0 (reactants)
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Is the reaction spontaneous at 25 oC?
Suniv = Ssys + Ssurr
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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Chapter Outline








12.1 Spontaneous Processes
12.2 Entropy
12.3 Absolute Entropy and Molecular Structure
12.4 Applications of the Second Law
12.5 Calculating Entropy Changes
12.6 Free Energy
12.7 Temperature and Spontaneity
12.8 Coupled Reactions
Standard Entropy of Reaction (Srxn)
The standard entropy of reaction (S0rxn ) is the entropy
change for a reaction carried out at 1 atm and 250C.
aA + bB
cC + dD
S0rxn = [cS0(C) + dS0(D)] - [aS0(A) + bS0(B) ]
S0rxn = S nS0(products) - S mS0(reactants)
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Sample Exercise 12.3:
Calculating So Values
Given the following standard molar entropy values at 298 K (found in
Appendix 4, Table A4.3), what is the of Sorxn for the dissolution of
ammonium nitrate under standard conditions?
NH4NO3(s)  NH4+(aq) + NO3-(aq)
So [J/molK]
151.1
113.4
146.4
Chapter Outline








12.1 Spontaneous Processes
12.2 Entropy
12.3 Absolute Entropy and Molecular Structure
12.4 Applications of the Second Law
12.5 Calculating Entropy Changes
12.6 Free Energy
12.7 Temperature and Spontaneity
12.8 Coupled Reactions
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Gibbs Free Energy
Suniv = Ssys + Ssurr
Could use this relation to calculate reaction spontaneity, but
not always easy to calculate Ssurr - so the expression is
rearranged to only include terms relating to the system:
Under constant temperature and pressure:
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.
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Calculating Free-Energy Changes Using
G = Hsys -TSsys
aA + bB
cC + dD
G = Hsys -TSsys
𝑜
∆𝐻𝑟𝑥𝑛
= Σ𝑛𝑝𝑟𝑜𝑑 ∙ ∆𝐻𝑓𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∙ ∆𝐻𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
𝑜
∆𝑆𝑟𝑥𝑛
= Σ𝑛𝑝𝑟𝑜𝑑 ∙ 𝑆𝑓𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∙ 𝑆𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
Sample Exercise 12.4: Predicting Reaction
Spontaneity under Standard Conditions
Consider the reaction of nitrogen gas and hydrogen gas at 298 K to
make ammonia at the same temperature:
N2(g) + 3 H2(g)  2 NH3(g)
𝑜
(a) Before doing any calculations, predict the sign of ∆𝑆𝑟𝑥𝑛
𝑜
(b) What is the actual value of ∆𝑆𝑟𝑥𝑛 ?
𝑜 ?
(c) What is the value of ∆𝐻𝑟𝑥𝑛
𝑜
(d) What is the value of ∆𝐺𝑟𝑥𝑛 ?
(e) Is the reaction spontaneous at 298 K and 1 atm?
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Calculating Free-Energy Changes Using
𝑜
∆𝐺𝑟𝑥𝑛
= Σ𝑛𝑝𝑟𝑜𝑑 ∆𝐺𝑓𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∆𝐺𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
Standard free energy of
formation (G0f ) is the free-energy
change that occurs when 1 mole
of the compound is formed from its
elements in their standard states.
G0f of any element in its most
stable allotropic form is zero, e.g.
graphite and not diamond
𝑜
Sample Exercise 12.5: Calculating ∆𝐺𝑟𝑥𝑛
Using
𝑜
Appropriate ∆𝐺𝑓 Values
Use the appropriate standard free energy of formation values in App. 4 to
calculate the change in free energy as ethanol burns under standard
conditions. Assume the reaction proceeds as described by the following
chemical equation:
CH3CH2OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
𝑜
∆𝐺𝑟𝑥𝑛
= Σ𝑛𝑝𝑟𝑜𝑑 ∆𝐺𝑓𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∆𝐺𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
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Chapter Outline








12.1 Spontaneous Processes
12.2 Entropy
12.3 Absolute Entropy and Molecular Structure
12.4 Applications of the Second Law
12.5 Calculating Entropy Changes
12.6 Free Energy
12.7 Temperature and Spontaneity
12.8 Coupled Reactions
Temperature and Spontaneity
G = H –TS
G = –TS + H
rearranging terms to match
the general formula for a
straight line -
G = –ST + H
y=
m x + b
m = –S
b = H
At what temperature
does the reaction
spontaneity change?
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At what temperature does the reaction
spontaneity change?
G = H –TS
When the reaction spontaneity changes, the sign of G changes from
_______ to ______, passing through _________ on the way. Therefore
G = 0 at the temperature that spontaneity changes, so -
 = H –TS
and
H = TS
and therefore
T=
G = H - T S
b = H (negative)
G 
Always
Spontaneous:
H < 0 and S > 0
m = - S (overall
negative)
G < 0 spontaneous at
all temperatures
0.0
b = H 
m = - S
T
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G = H - T S
b = H (positive)
Never
Spontaneous:
H > 0 and S < 0
m = - S (overall positive)
G 
G > 0 not spontaneous
at all temperatures
m = - S
b = H 
0.0
T
G = H - T S
Enthalpy-driven:
spontaneous at low T
H < 0 and S < 0
b = H 
m = - S (overall positive)
G < 0 spontaneous only
at low T
G 
0.0
b = H (negative)
m = - S
G < 0
T = H
S
T
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G = H - T S
Entropy-driven:
spontaneous at high T
H > 0 and S > 0
b = H (positive)
m = - S (overall negative)
G < 0 spontaneous only
at high T
b = H 
G 
m = - S
0.0
G < 0
T = H
S
T
G = H - T S
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Homework Problem #80
G 
The element H2 is not abundant in nature, but it is a useful reagent in, for
example, the potential synthesis of the liquid fuel methanol from gaseous
carbon monoxide. Under what temperature conditions if this reaction
spontaneous?
2 H2(g) + CO(g)  CH3OH(l)
G = 0
T
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