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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, high-
energy 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
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
Chapter Outline
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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):
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Ho = -2200 kJ
1. propane combustion:
2. water melting:
H2O(s) H2O(l) Ho = 6.01 kJ
3. gas expansion:
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
Chapter Outline
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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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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
Chapter Outline
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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 < So
liquid < Sogas
Trends in S: Phase Changes Ssolid < Sliquid < Sgas
S = Sfinal - Sinitial
Units: J/molK
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The Effect of Molecular Structure on S
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 S
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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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
Chapter Outline
Changes in Entropy
S = Sf - Si
S > 0
S < 0
Suniverse
Spontaneous Nonspontaneous
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The Second Law of Thermodynamics: The total entropy of the universe increases in any spontaneous process
Suniv = Ssys + Ssurr 0 Spontaneous process:
sys = system
surr = surroundings 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:
Ssurr -Hsys
if Hsys < 0 (exothermic), then
Ssurr > 0 (entropy of the
surroundings increases)
if Hsys > 0 (endothermic), then
Ssurr < 0 (entropy of the
surroundings decreases)
Ssurr 1
Tsurr
If the temperature of the
surroundings is already high,
then pumping heat in or out
causes less change in disorder
than at lower temperatures
Combining the two:
and
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.
e.g. N2(g) + 3H2(g) 2NH3(g) Ssys = -198.3 J/K
Hsys = -92.6 kJ*
*from Ch 9: H0 rxn nH0 (products) f = S mH0 (reactants) f S -
Ssurr -Hsys Ssurr 1
Tsurr
so Ssurr = -Hsys
T
(Tsurr usually = Tsys)
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Suniv = Ssys + Ssurr
Is the reaction spontaneous at 25 oC?
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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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
Chapter Outline
Standard Entropy of Reaction (Srxn)
The standard entropy of reaction (S0 ) is the entropy
change for a reaction carried out at 1 atm and 250C. rxn
aA + bB cC + dD
S0 rxn nS0(products) = S mS0(reactants) S -
S0 rxn dS0(D) cS0(C) = [ + ] - bS0(B) aS0(A) [ + ]
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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/molK] 151.1 113.4 146.4
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
Chapter Outline
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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:
G = Hsys -TSsys Gibbs free
energy (G)
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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aA + bB cC + dD
Calculating Free-Energy Changes Using
G = Hsys -TSsys
∆𝐻𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑 ∙ ∆𝐻𝑓
𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∙ ∆𝐻𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
∆𝑆𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑 ∙ 𝑆𝑓
𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡 ∙ 𝑆𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
G = Hsys -TSsys
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)
(a)Before doing any calculations, predict the sign of ∆𝑆𝑟𝑥𝑛𝑜
(b)What is the actual value of ∆𝑆𝑟𝑥𝑛𝑜 ?
(c) What is the value of ∆𝐻𝑟𝑥𝑛𝑜 ?
(d)What is the value of ∆𝐺𝑟𝑥𝑛𝑜 ?
(e) Is the reaction spontaneous at 298 K and 1 atm?
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Standard free energy of
formation (G0) is the free-energy
change that occurs when 1 mole
of the compound is formed from its
elements in their standard states.
f
Calculating Free-Energy Changes Using
∆𝐺𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑∆𝐺𝑓
𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡∆𝐺𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
G0 of any element in its most
stable allotropic form is zero, e.g.
graphite and not diamond
f
Sample Exercise 12.5: Calculating ∆𝐺𝑟𝑥𝑛𝑜 Using
Appropriate ∆𝐺𝑓𝑜 Values
∆𝐺𝑟𝑥𝑛𝑜 = Σ𝑛𝑝𝑟𝑜𝑑∆𝐺𝑓
𝑜 𝑝𝑟𝑜𝑑 − Σ𝑛𝑟𝑒𝑎𝑐𝑡∆𝐺𝑓𝑜 𝑟𝑒𝑎𝑐𝑡
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)
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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
Chapter Outline
G = H –TS rearranging terms to match
the general formula for a
straight line -
G = –TS + H
G = –ST + H
y = m x + b
m = –S
b = H
At what temperature
does the reaction
spontaneity change?
Temperature and Spontaneity
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At what temperature does the reaction
spontaneity change?
G = H –TS
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 –TS and
H = TS and therefore
T =
Always
Spontaneous:
H < 0 and S > 0
G
T
0.0
b = H (negative)
m = - S (overall
negative)
G < 0 spontaneous at
all temperatures
b = H
m = - S
G = H - T S
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Never
Spontaneous:
H > 0 and S < 0
G
T
0.0
b = H (positive)
m = - S (overall positive)
G > 0 not spontaneous
at all temperatures
b = H
m = - S
G = H - T S
Enthalpy-driven:
spontaneous at low T
H < 0 and S < 0
G
T
0.0
b = H (negative)
m = - S (overall positive)
G < 0 spontaneous only
at low T
b = H
m = - S
T = H
S
G = H - T S
G < 0
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Entropy-driven:
spontaneous at high T
H > 0 and S > 0
G
T
0.0
b = H (positive)
m = - S (overall negative)
G < 0 spontaneous only
at high T b = H
m = - S
T = H
S
G = H - T S
G < 0
G = H - T S
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G = 0
T
G
Homework Problem #80
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)