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Chapter 17
Spontaneity, Entropy,
and Free Energy
Section 17.1
Spontaneous Processes and Entropy
Thermodynamics vs.
Kinetics
 Domain of Kinetics
 Rate of a reaction depends
on the pathway from
reactants to products.
 Thermodynamics tells us
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
Spontaneous Processes and Entropy
 Thermodynamics lets us predict the direction in
which a process will occur but gives no information
about the speed of the process.
 A spontaneous process is one that occurs without
outside intervention.
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Section 17.1
Spontaneous Processes and Entropy
CONCEPT CHECK!
Consider 2.4 moles of a gas contained in a 4.0 L bulb
at a constant temperature of 32°C. This bulb is
connected by a valve to an evacuated 20.0 L bulb.
Assume the temperature is constant.
a) What should happen to the gas when you open
the valve?
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Section 17.1
Spontaneous Processes and Entropy
CONCEPT CHECK!
Consider 2.4 moles of a gas contained in a 4.0 L bulb
at a constant temperature of 32°C. This bulb is
connected by a valve to an evacuated 20.0 L bulb.
Assume the temperature is constant.
b) Calculate ΔH, ΔE, q, and w for the process you
described above.
All are equal to zero.
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Section 17.1
Spontaneous Processes and Entropy
CONCEPT CHECK!
Consider 2.4 moles of a gas contained in a 4.0 L bulb
at a constant temperature of 32°C. This bulb is
connected by a valve to an evacuated 20.0 L bulb.
Assume the temperature is constant.
c) Given your answer to part b, what is the
driving force for the process?
Entropy
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Section 17.1
Spontaneous Processes and Entropy
The Expansion of An Ideal Gas Into an Evacuated Bulb
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Section 17.1
Spontaneous Processes and Entropy
Entropy
 The driving force for a spontaneous process is an
increase in the entropy of the universe.
 A measure of molecular randomness or disorder.
 A measure of how many micro states can exist.
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Section 17.1
Spontaneous Processes and Entropy
Entropy
 Thermodynamic function that describes the number
of arrangements that are available to a system
existing in a given state.
 Nature spontaneously proceeds toward the states
that have the highest probabilities of existing.
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Section 17.1
Spontaneous Processes and Entropy
The Microstates
That Give a
Particular
Arrangement
(State)
Section 17.1
Spontaneous Processes and Entropy
Positional Entropy
 A gas expands into a vacuum to give a uniform
distribution because the expanded state has the
highest positional probability of states available to
the system.
 Therefore: Ssolid < Sliquid << Sgas
Section 17.1
Spontaneous Processes and Entropy
Interactive Example 17.1 - Positional Entropy
 For each of the following pairs, choose the
substance with the higher positional entropy (per
mole) at a given temperature
a. Solid CO2 and gaseous CO2
b. N2 gas at 1 atm and N2 gas at 1.0×10–2 atm
Section 17.1
Spontaneous Processes and Entropy
CONCEPT CHECK!
Predict the sign of ΔS 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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Section 17.2
Entropy and the Second Law of Thermodynamics
Second Law of Thermodynamics
 In any spontaneous process there is 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 17.2
Entropy and the Second Law of Thermodynamics
ΔSsurr
 ΔSsurr = +; entropy of the universe increases
 ΔSsurr = -; process is spontaneous in opposite direction
 ΔSsurr = 0; process has no tendency to occur
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Section 17.2
Entropy and the Second Law of Thermodynamics
Entropy Changes in the Surroundings (ΔSsurr)
 ΔSsurr is determined by flow of energy as heat
 Exothermic process increases ΔSsurr
 Important driving force for spontaneity
 Endothermic process decreases ΔSsurr
 Impact of transfer of energy as heat to or from
the surroundings is greater at lower temperatures
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Section 17.3
The Effect 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 of Temperature on Spontaneity
CONCEPT CHECK!
Describe the following as spontaneous/non-spontaneous/cannot tell,
and explain.
A reaction that is:
a) Exothermic and becomes more positionally random
Spontaneous
b) Exothermic and becomes less positionally random
Cannot tell
a) Endothermic and becomes more positionally random
Cannot tell
a) Endothermic and becomes less positionally random
Not spontaneous
Explain how temperature affects your answers.
Section 17.3
The Effect 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.
 $50 to a millionaire or a high school student
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Section 17.3
The Effect of Temperature on Spontaneity
ΔSsurr
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Section 17.3
The Effect of Temperature on Spontaneity
ΔSsurr
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Section 17.3
The Effect of Temperature on Spontaneity
ΔSsurr
Heat flow (constant P) = change in enthalpy = ΔH
Ssurr
H
= 
T
If the reaction is exothermic:
ΔH has a negative sign
ΔSsurr is positive since heat flows into the
surroundings
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Section 17.3
The Effect of Temperature on Spontaneity
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Section 17.3
The Effect of Temperature on Spontaneity
Interactive Example 17.4 - Determining ΔSsurr
 In the metallurgy of antimony, the pure metal is
recovered via different reactions, depending on
the composition of the ore
 For example, iron is used to reduce antimony in sulfide
ores
Sb2S3 (s) + 3Fe(s)  2Sb(s) + 3FeS(s)
H =  125kJ
Section 17.3
The Effect of Temperature on Spontaneity
Interactive Example 17.4 - Determining ΔSsurr (Continued)
 Carbon is used as the reducing agent for oxide ores:
Sb4O6 (s) + 6C(s)  4Sb(s) + 6CO(g )
ΔH = 778 kJ
 Calculate ΔSsurr for each of these reactions at 25°C
and 1 atm
Section 17.3
The Effect of Temperature on Spontaneity
Interactive Example 17.4 - Solution
ΔH
ΔSsurr = 
, where T = 25 + 273 = 298 K
T
 For the sulfide ore reaction,
 125 kJ
ΔSsurr = 
= 0.419 kJ/K = 419 J/K
298 K
 ΔSsurr is positive since this reaction is exothermic, and
heat flow occurs to the surroundings, increasing the
randomness of the surroundings
Section 17.3
The Effect of Temperature on Spontaneity
Interactive Example 17.4 - Solution (Continued)
 For the oxide ore reaction,
 778 kJ
ΔSsurr = 
=  2.61 kJ/K =  2.61 × 103 J/K
298 K
 In this case ΔSsurr is negative because heat flow occurs
from the surroundings to the system
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.
 Negative ΔG means positive ΔSuniv.
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Section 17.4
Free Energy
Free Energy (G)
 ΔG = ΔH – TΔS (at constant T and P)
 All quantities refer to the system
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Section 17.4
Free Energy
CONCEPT CHECK!
A liquid is vaporized at its boiling point. Predict the signs of:
–
w
+
q
+
ΔH
+
ΔS
–
ΔSsurr
0
ΔG
Explain your answers.
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Section 17.4
Free Energy
EXERCISE!
The value of ΔHvaporization of substance X is 45.7 kJ/mol,
and its normal boiling point is 72.5°C.
Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one
mole of this substance at 72.5°C and 1 atm.
ΔS = 132 J/K·mol
ΔSsurr = -132 J/K·mol
ΔG = 0 kJ/mol
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Section 17.4
Free Energy
Spontaneous Reactions
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Section 17.4
Free Energy
Effect of ΔH and ΔS on Spontaneity
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Section 17.4
Free Energy
Interactive Example 17.5 - Free Energy and Spontaneity
 At what temperatures is the following process
spontaneous at 1 atm?
Br2 (l )  Br2 (g )
ΔH o = 31.0 kJ/mol and ΔS o = 93.0 J/K  mol
 What is the normal boiling point of liquid Br2?
Section 17.4
Free Energy
Interactive Example 17.5 - Solution
 The vaporization process will be spontaneous at
all temperatures where ΔG°is negative
 Note that ΔS°favors the vaporization process
because of the increase in positional entropy, and
ΔH°favors the opposite process, which is exothermic
 These opposite tendencies will exactly balance at the
boiling point of liquid Br2, since at this temperature
liquid and gaseous Br2 are in equilibrium (ΔG°= 0)
Section 17.4
Free Energy
Interactive Example 17.5 - Solution (Continued 1)
 We can find this temperature by setting ΔG°= 0
in the following equation:
ΔG= ΔH   TΔS 
0  ΔH   TΔS
ΔH  =TΔS 
ΔH  3.10 ×104 J/mol
T=
=
= 333K
ΔS 
93.0 J/K  mol
Section 17.4
Free Energy
Interactive Example 17.5 - Solution (Continued 2)
 At temperatures above 333 K, TΔS°has a larger
magnitude than ΔH°, and ΔG°is negative
 Above 333 K, the vaporization process is spontaneous
 The opposite process occurs spontaneously below this
temperature
 At 333 K, liquid and gaseous Br2 coexist in
equilibrium
Section 17.4
Free Energy
Interactive Example 17.5 - Solution (Continued 3)
 Summary of observations (the pressure is 1 atm in
each case)
 T > 333 K
 The term ΔS°controls, and the increase in entropy when
liquid Br2 is vaporized is dominant
 T < 333 K
 The process is spontaneous in the direction in which it is
exothermic, and the term ΔH°controls
Section 17.4
Free Energy
Interactive Example 17.5 - Solution (Continued 4)
 T = 333 K
 The opposing driving forces are just balanced (ΔH°= 0), and
the liquid and gaseous phases of bromine coexist
 This is the normal boiling point
Section 17.5
Entropy Changes in Chemical Reactions
Entropy Changes and Chemical Reactions
 Positional probability determines the changes
that occur in a chemical system
 Fewer the molecules, fewer the possible
configurations
Ammonia synthesis : N (g) + 3H (g)  2NH (g)
2
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40
3
Section 17.5
Entropy Changes in Chemical Reactions
Figure 17.6 - Entropy of Water
 H2O molecule can vibrate
and rotate in several
ways
 Freedom of motion leads
to a higher entropy for
water
Section 17.5
Entropy Changes in Chemical Reactions
Entropy Changes in Reactions That Involve Gaseous
Molecules
 Change in positional entropy is dominated by the
relative numbers of molecules of gaseous
reactants and products
 If the number of product molecules is greater than the
number of reactant molecules:
 Positional entropy increases
 ΔS is positive
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Section 17.5
Entropy Changes in Chemical Reactions
Interactive Example 17.6 - Predicting the Sign of ΔS°
 Predict the sign of ΔS°for each of the following
reactions
a. Thermal decomposition of solid calcium carbonate
CaCO3 (s)  CaO(s) + CO2 (g )
b. Oxidation of SO2 in air
2SO2 (g ) + O2 (g )  2SO3 (g )
Section 17.5
Entropy Changes in Chemical Reactions
CONCEPT CHECK!
Gas A2 reacts with gas B2 to form gas AB at constant
temperature and pressure. The bond energy of AB is much
greater than that of either reactant.
Predict the signs of:
ΔH
ΔSsurr
–
+
ΔS
ΔSuniv
0
+
Explain.
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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.
 The entropy of a substance increases with temperature.
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Section 17.5
Entropy Changes in Chemical Reactions
Standard Entropy Values (S°)
 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 ΔS° 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
ΔS°= –11 J/K
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Section 17.6
Free Energy and Chemical Reactions
AP Learning Objectives, Margin Notes and References
 Learning Objectives






LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored
by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or
estimation of Go when needed.
LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable
by calculating the change in standard Gibbs free energy.
LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable
with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become
favorable.
LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled
reactions that share a common intermediate drive formation of a product.
LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a
common intermediate, based on the equilibrium constant for the combined reaction.
LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large
amounts of product (based on consideration of both initial conditions and kinetic effects), or why a
thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial
conditions.
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.
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°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:
ΔH°
ΔS°
–
–
ΔG°
–
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)
ΔG° = −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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Section 17.6
Free Energy and Chemical Reactions
Interactive Example 17.11 - Calculating ΔG°
 Methanol is a high-octane fuel used in highperformance racing engines
 Calculate ΔG °for the following reaction:
2CH3OH(g ) + 3O2 (g )  2CO2 (g ) + 4H2O(g )
 The following energies of formation are provided:
Section 17.6
Free Energy and Chemical Reactions
Interactive Example 17.11 - Solution
 Use the following equation:
ΔG = npΔGf (products)   npΔGf (reactants)
= 2ΔGf CO2  g  + 4ΔGf  H2O g   3ΔGf O2  g   2ΔGf CH3OH g 
= 2 mol  394 kJ/mol  + 4 mol  229 kJ/mol 
 3  0   2 mol  163 kJ/mol 
G = 1378 kJ
Section 17.6
Free Energy and Chemical Reactions
Interactive Example 17.11 - Solution (Continued)
 The large magnitude and the negative sign of
ΔG°indicate that this reaction is very favourable
thermodynamically
Section 17.7
The Dependence of Free Energy on Pressure
AP Learning Objectives, Margin Notes and References
 Learning Objectives

LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large
amounts of product (based on consideration of both initial conditions and kinetic effects), or why a
thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial
conditions.
 Additional AP References

LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)
Section 17.7
The Dependence of Free Energy on Pressure
Free Energy and Pressure
 System under constant P and T proceeds
spontaneously in the direction that lowers its free
energy
 Free energy of a reaction system changes as the
reaction proceeds
 Dependent on the pressure of a gas or on the concentration
of species in solution
 Equilibrium - Point where free energy value is at
its lowest
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Section 17.7
The Dependence of Free Energy on Pressure
Free Energy and Pressure (Continued 1)
 For ideal gases:
 Enthalpy is not pressure-dependent
 Entropy depends on pressure due to its dependence
on volume
 At a given temperature for 1 mole of ideal gas:
 Slarge volume > Ssmall volume
Or,
 Slow pressure > Shigh pressure
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Section 17.7
The Dependence of Free Energy on Pressure
Free Energy and Pressure (Continued 2)
G = G + RTln  P 




G° - Free energy of a gas at 1 atm
G - Free energy of the gas at a pressure of P atm
R - Universal gas constant
T - Temperature in Kelvin
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Section 17.7
The Dependence of Free Energy on Pressure
Free Energy and Pressure (Continued 3)
G = G+ RTln  Q 





Q - Reaction quotient
T - Temperature in Kelvin
R - Universal gas constant (8.3145 J/K·mol)
ΔG°- Free energy change at 1 atm
ΔG - Free energy change at specified pressures
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Section 17.7
The Dependence of Free Energy on Pressure
Interactive Example 17.13 - Calculating ΔG°
 One method for synthesizing methanol (CH3OH)
involves reacting carbon monoxide and hydrogen
gases
CO  g  + 2H2  g   CH3OH l 
 Calculate ΔG at 25°C for this reaction where carbon
monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm
are converted to liquid methanol
Section 17.7
The Dependence of Free Energy on Pressure
Interactive Example 17.13 - Solution
 To calculate ΔG for this process, use the following
equation:
G = G+ RTln  Q 
ₒ
 First compute ΔG from standard free energies of
formation
ΔGfo CH3OHl   166 kJ
ΔGfo H2  g   0
ΔGfoCO g   137 kJ
ΔGo = 166 kJ   137 kJ   0 =  29 kJ =  2.9×104 J
Section 17.7
The Dependence of Free Energy on Pressure
Interactive Example 17.13 - Solution (Continued 1)
 One might call this the value of ΔG°for one round of
the reaction or for 1 mole of the reaction
 Thus, the ΔG°value might better be written as
–2.9×104 J/mol of reaction, or –2.9×104 J/mol rxn
 Use this value to calculate the value of ΔG
Section 17.7
The Dependence of Free Energy on Pressure
Interactive Example 17.13 - Solution (Continued 2)
 ΔG°= – 2.9×104 J/mol rxn
 R = 8.3145 J/K·mol
 T = 273 + 25 = 298 K
1
1
2
Q=
=
= 2.2×10
2
2
 PCO  PH2
 5.0 3.0


 Note that the pure liquid methanol is not included in
the calculation of Q
Section 17.7
The Dependence of Free Energy on Pressure
Interactive Example 17.13 - Solution (Continued 3)
G = G+ RTln  Q 
=  2.9×104 J/mol rxn 
+  8.3145 J/K  mol rxn  298K  ln  2.2×102 
=  2.9×104 J/mol rxn    9.4×103 J/mol rxn 
 3.8×104 J/mol rxn =  38 kJ/mol rxn
Section 17.7
The Dependence of Free Energy on Pressure
Interactive Example 17.13 - Solution (Continued 4)
 Note that ΔG is significantly more negative than
ΔG°, implying that the reaction is more
spontaneous at reactant pressures greater than 1
atm
 This result can be expected from Le Châtelier’s
principle
Section 17.7
The Dependence of Free Energy on Pressure
CONCEPT CHECK!
Sketch graphs of:
1. G vs. P
2. H vs. P
3. ln(K) vs. 1/T (for both endothermic and
exothermic cases)
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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.
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Section 17.8
Free Energy and Equilibrium
AP Learning Objectives, Margin Notes and References
 Learning Objectives


LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large
amounts of product (based on consideration of both initial conditions and kinetic effects), or why a
thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial
conditions.
LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to
estimate the magnitude of K and, consequently, the thermodynamic favorability of the process.
 Additional AP References


LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)
LO 6.25 (see Appendix 7.11, “Non-Spontaneous Reactions”)
Section 17.8
Free Energy and Equilibrium
 The equilibrium point occurs at the lowest value of free
energy available to the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
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Section 17.8
Free Energy and Equilibrium
Change in Free Energy to Reach Equilibrium
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Section 17.8
Free Energy and Equilibrium
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Section 17.8
Free Energy and Equilibrium
Interactive Example 17.15 - Free Energy and Equilibrium
II

Section 17.8
Free Energy and Equilibrium
Interactive Example 17.15 - Solution
 Calculate ΔG° from ΔG° = ΔH°– TΔS°
ΔH = 2ΔH f(Fe2O3 (s ))  3ΔH f(O2 (g ))  4ΔH f(Fe(s ))
= 2 mol  826 kJ/mol   0  0
=  1652 kJ =  1.652×106 J
ΔS = 2S Fe2O3  3S O2  4S Fe
= 2 mol  90 J/K  mol   3 mol  205 J/K  mol 
 4 mol  27 J/K  mol   543 J/K
T = 273 + 25 = 298 K
Section 17.8
Free Energy and Equilibrium
Interactive Example 17.15 - Solution (Continued 1)
 Therefore,
ΔG° = ΔH°– TΔS°
= (– 1.652×106 J) – (298 K) (– 543 J/K)
= – 1.490 ×106 J
ΔG=  RT ln  K  = 1.490 ×106 J
=  8.3145 J/K  mol  298K  ln  K 
1.490×106
ln  K  =
= 601
3
2.48×10
Section 17.8
Free Energy and Equilibrium
Interactive Example 17.15 - Solution (Continued 2)
 Therefore,
K = e601
 This is a very large equilibrium constant
 The rusting of iron is clearly very favourable from a
thermodynamic point of view
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
Reversible and Irreversible Processes
 Reversible process: Universe is exactly the same
as it was before a cyclic process
 Irreversible process: Universe is different after a
cyclic process
 All real processes are irreversible
 Characteristics of a real cyclic process
 Work is changed to heat
 Entropy of the universe increases
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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.
 As we use energy, we degrade its usefulness.
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78