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Chapter 17
Spontaneity, Entropy,
and Free Energy
Chapter 17
Table of Contents
17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8
17.9
Spontaneous Processes and Entropy
Entropy and the Second Law of Thermodynamics
The Effect of Temperature on Spontaneity
Free Energy
Entropy Changes in Chemical Reactions
Free Energy and Chemical Reactions
The Dependence of Free Energy on Pressure
Free Energy and Equilibrium
Free Energy and Work
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Section 17.1
Spontaneous Processes and Entropy
Why does a reaction occur in a particular direction?
•
Kinetics
– Gives info about amount of
time required for the process.
– Rate of a reaction depends on
the pathway from reactants to
products.
•
Thermodynamics
– Predicts whether a reaction is
spontaneous based only on
the properties of reactants
and products.
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Section 17.1
Spontaneous Processes and Entropy
Spontaneity & Entropy
• spontaneous process - occurs without outside
intervention
– Spontaneous does NOT mean fast.
– Ex: heat flows from hot object to cooler one, but not
the reverse
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Section 17.1
Spontaneous Processes and Entropy
•
•
Entropy, S - a measure of molecular randomness or
disorder.
• The driving force for a spontaneous process is an
increase in the entropy of the universe
• Order (lower entropy) to disorder (higher entropy)
• Thermodynamic function that describes the number of
arrangements (positions and/or energy levels)
available to a system in a given state.
Associated with probability.
• Nature spontaneously proceeds toward states that
have the highest probabilities of existing.
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Section 17.1
Spontaneous Processes and Entropy
The Expansion of An Ideal Gas Into an Evacuated Bulb
Why do the gas
molecules evenly
spread throughout
the container when
the valve is open??
Why is it
spontaneous?
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Section 17.1
Spontaneous Processes and Entropy
The Microstates That Give a Particular Arrangement (State)
• Consider a system with 4
gas molecules in a twobulbed container.
• Microstate – a particular
arrangement.
• How many states
(arrangements) are
possible?
• Which state is most likely
to occur?
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Section 17.1
Spontaneous Processes and Entropy
Positional Entropy
• A gas expands into a vacuum because the
expanded state has the highest positional
probability (probability based upon
configurations in space) of states available to
the system.
• Changes of state: Ssolid < Sliquid << Sgas
• Formation of solutions: increase in entropy
because of increased volume available for a
given particle.
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Section 17.1
Spontaneous Processes and Entropy
Concept Check
Predict the sign of S for each of the
following, and explain:
+ a) The evaporation of alcohol
– b) The freezing of water
– c) Compressing an ideal gas at constant
temperature
+ d) Heating an ideal gas at constant
pressure
+ e) Dissolving NaCl in water
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9
Section 17.2
Atomic Masses
Entropy
and the Second Law of Thermodynamics
Second Law of Thermodynamics
• spontaneous process -always an increase in
the entropy of the universe.
• The entropy of the universe is increasing.
• The total energy of the universe is constant, but
the entropy is increasing.
Suniverse = ΔSsystem + ΔSsurroundings
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Section 17.3
The Effect
Mole of Temperature on Spontaneity
Concept Check
For the process A(l)
A(s), which direction
involves an increase in energy randomness?
Positional randomness? Explain your answer.
As temperature increases/decreases (answer for
both), which takes precedence? Why?
At what temperature is there a balance between
energy randomness and positional randomness?
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Section 17.3
The Effect
Mole of Temperature on Spontaneity
ΔSsurr
•
The sign of ΔSsurr depends on the
direction of the heat flow.
• The magnitude of ΔSsurr depends
on the temperature.
– directly proportional to quantity
of heat transferred
– Inversely proportional to
temperature
Ssurr
H
= 
T
Heat flow (constant P) =
change in enthalpy
(system) = ΔH
* Exothermic reactions increase the entropy of
the surroundings
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Section 17.3
The Effect
Mole of Temperature on Spontaneity
ΔSsurr
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Section 17.3
The Effect
Mole of Temperature on Spontaneity
Example:
Sb2S3 (s) + 3Fe (s)  2Sb (s) + 3FeS (s)
ΔH = -125 kJ
Calculate Δsurr for the reaction above.
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Section 17.3
The Effect
Mole of Temperature on Spontaneity
Interplay of S
sys
and S
surr
in Determining the Sign of S
univ
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Section 17.4
Free Energy
Free Energy (G)
•
•
•
Another thermodynamic function used to predict spontaneity.
Is the maximum usable energy that can be utilized or trapped to do work.
Useful when considering temperature dependence of spontaneity.
• ΔG = ΔH – TΔS (at constant T and P)
• ΔG° = ΔH ° – TΔS °
– ΔH at constant P = to energy flow as heat
– T = temp in K
– ΔS (ΔSsys) = change in entropy in J/K
– °=> standard state
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Section 17.4
Free Energy
Free Energy (G)
Suniv
G
= 
(at constant T and P )
T
• A process (at constant T and P) is spontaneous
in the direction in which the free energy
decreases.
– Spontaneous : ΔG <0
•
- ΔG means + ΔSuniv.
– At equilibrium: ΔG = 0
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Section 17.4
Free Energy
Concept Check
A liquid is vaporized at its boiling point. Predict
the signs of:
H
+
S
+
Ssurr
–
G
0
Explain your answers.
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Section 17.4
Free Energy
Will ∆G be negative (& process spontaneous)?
• Which dominates: ∆H or ∆S? Also can depend on T.
• *Always spontaneous: exothermic & increase in entropy; always
NONspontaneous: endothermic & decrease in entropy
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Section 17.4
Free Energy
Example:
• Predict whether each of the following sets of data
represent spontaneous or nonspontaneous processes.
ΔH (kJ)
ΔS (J/K)
T (K)
40
300
130
-40
-300
130
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Section 17.4
Free Energy
Example
• For mercury, the enthalpy of vaporization is 58.51 kJ/mol and
the entropy of vaporization is 92.92 J/K∙mol. What is the
normal boiling point of mercury?
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Section 17.4
Free Energy
Exercise
The value of Hvaporization of substance X is 45.7
kJ/mol, and its normal boiling point is 72.5°C.
Calculate S, Ssurr, and G for the
vaporization of one mole of this substance at
72.5°C and 1 atm.
S = 132 J/K·mol
Ssurr = -132 J/K·mol
G = 0 kJ/mol
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Section 17.5
Entropy Changes in Chemical Reactions
ΔSsys is related to positional probabilities of the reactants
• Reaction involving only gas phase: increase in total #
of moles of gas means higher entropy.
• Reaction with different phases: production of gas
generally increases entropy more than an increase
in # moles of a liquid or solid.
Example: Predict the sign of ΔS°for each of the reactions:
1. (NH4)2Cr2O7 (s)  Cr2O3 (s) + 4H2O (l) + N2 (g)
2. Mg(OH) 2 (s)  MgO (s) + H2O (g)
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Section 17.5
Entropy Changes in Chemical Reactions
Third Law of Thermodynamics
• The entropy of a perfect crystal at 0 K is zero.
• Standard entropy values: Represent the
increase in entropy that occurs when a
substance is heated from 0 K to 298 K at 1 atm
pressure.
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
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Section 17.5
Entropy Changes in Chemical Reactions
Exercise
Calculate S° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
S° (J/K·mol)
Na(s)
51
H2O(l)
70
NaOH(aq)
50
H2(g)
131
S°= –11 J/K
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25
Section 17.6
Free Energy and Chemical Reactions
Standard Free Energy Change (ΔG°)
• The change in free energy that will occur if the
reactants in their standard states are converted
to the products in their standard states.
• Three methods of calculating ΔG°:
1) ΔG° = ΔH° – TΔS°
2) By manipulating known equations,
(similar to Hess’s law)
3) ΔG°reaction = ΣnpGf (products) – ΣnrGf (reactants)
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Section 17.6
Free Energy and Chemical Reactions
Concept Check
A stable diatomic molecule spontaneously
forms from its atoms.
Predict the signs of:
H°
S°
–
–
G°
–
Explain.
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Section 17.6
Free Energy and Chemical Reactions
Concept Check
Consider the following system at
equilibrium at 25°C.
PCl3(g) + Cl2(g)
PCl5(g)
G° = −92.50 kJ
What will happen to the ratio of partial
pressure of PCl5 to partial pressure of PCl3
if the temperature is raised? Explain.
The ratio will decrease.
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28
Section 17.7
The Dependence of Free Energy on Pressure
• System at constant T and P proceed spontaneously in the
direction that lowers its free energy.
• Reactions proceed until equilibrium is reached.
– Equilibrium position: lowest free energy value available
to a particular reaction system.
• Free energy at nonstandard pressures:
– For ideal gas: entropy is affected by pressure but
enthalpy is not.
• Slarge volume > Ssmall volume
• Slow pressure > Shigh pressure
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Section 17.7
The Dependence of Free Energy on Pressure
Free Energy at NON Standard pressures
G = G° + RT ln(P)
or
ΔG = ΔG° + RT ln(Q)
ΔG°= free energy change at 1 atm
ΔG = free energy change at specified pressure P
T = temp in K
R = 8.3145 J/K∙mol
Q = reaction quotient in terms of P
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Section 17.7
The Dependence of Free Energy on Pressure
Example
CO (g) + 2H2 (g)  CH3OH (l)
Calculate ∆G at 25°C for this reaction where CO at 5.0 atm and H2
gas at 3.0 atm are converted to liquid methanol.
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Section 17.7
The Dependence of Free Energy on Pressure
The Meaning of ΔG for a Chemical Reaction
•
•
A system can achieve the lowest possible free energy by going to
equilibrium, not by going to completion.
– ∆G<0 does NOT mean system proceeds to pure products
– ∆G>0 does NOT mean system remains at pure reactants
System spontaneously goes to equilibrium, the lowest possible
free energy available. SO what is this value of ∆G at
equilibrium???
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Section 17.8
Free Energy and Equilibrium
•
The equilibrium point occurs at the lowest value of
free energy available to the reaction system.
Gproducts = Greactants or ΔG = 0 so…
ΔG = 0 = ΔG° + RT ln(K) so…
ΔG° = –RT ln(K)
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Section 17.8
Free Energy and Equilibrium
Qualitative Relationship Between the Change in Standard Free
Energy and the Equilibrium Constant for a Given Reaction
• Given ∆G, which direction will the system shift to reach equilibrium?
– Look at the value of K!
Shift
RIGHT
RIGHT
LEFT
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Section 17.8
Free Energy and Equilibrium
Example
• The rusting of iron by oxygen is:
4Fe (s) + 3O2 (g)  2Fe2O3 (s)
• Using the following data, calculate the equilibrium constant for
this reaction at 25ºC.
Substance
ΔHf° (kJ/mol)
S° (J/K*mol)
Fe2O3 (s)
-826
90
Fe (s)
0
27
O2 (g)
0
205
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Section 17.7
The Dependence of Free Energy on Pressure
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Section 17.9
Free Energy and Work
• Maximum possible useful work obtainable from
a process at constant temperature and pressure
is equal to the change in free energy.
wmax = ΔG
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Section 17.9
Free Energy and Work
• Achieving the maximum work available from a
spontaneous process can occur only via a
hypothetical pathway. Any real pathway wastes
energy.
• All real processes are irreversible.
• First law: You can’t win, you can only break
even.
• Second law: You can’t break even.
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