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Chapter 17
Free Energy and
Thermodynamics
TMHsiung©2015
Chapter 17
Slide 1 of 66
Contents
1.
2.
3.
4.
5.
6.
7.
Nature’s Heat Tax: You Can’t Win and You Can’t
Break Even
Spontaneous and Nonspontaneous Processes
Entropy and the Second Law of Thermodynamics
Heat Transfer and Changes in the Entropy of the
Surroundings
Gibbs Free Energy
Entropy Changes in Chemical Reactions:
Calculating ΔSorxn
Free Energy Changes in Chemical Reactions:
Calculating ΔGorxn
TMHsiung©2015
Chapter 17
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Continued
8. Free Energy Changes for Nonstandard States: The
Relationship between ΔGorxn and ΔGrxn
9. Free Energy and Equilibrium: Relating ΔGorxn to
the Equilibrium Constant (K)
TMHsiung©2015
Chapter 17
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*****
1. Nature’s Heat Tax: You Can’t Win and You Can’t
Break Even
1) First Law of Thermodynamics
The total energy of the universe is constant that
energy cannot be created or destroyed, but you can
transfer it from one place to another:
ΔEuniverse = 0 = ΔEsystem + Δ Esurroundings
 There are two ways energy is “lost” from a system:
̶ Converted to heat, q
̶ Used to do work, w
 The energy change in the system:
ΔE = q + w or ΔE = ΔH + P ΔV
 Internal energy change (ΔE) is a state function, depends
only on the state of the system, not on how the system got
to that state.
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Chapter 17
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*****
2) Heat Tax
Conversion of energy to heat, which is “lost” by heating
up the surroundings.
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Chapter 17
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2. Spontaneous and Nonspontaneous Processes
1) Spontaneous and Nonspontaneous
 Spontaneous process: A process that occurs without
outside intervention.
 Nonspontaneous: A process that occurs must be
assisted by outside intervention.
*
If a process is spontaneous in one direction, it must
be nonspontaneous in the opposite direction.
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2)




Chemical Thermodynamics Information
Stability of particular substances
The Spontaneity of a chemical reaction
Equilibrium constant (Keq) of a chemical reaction
Predict the proportions of products and reactants at
equilibrium.
 The optimum temperature, pressure, and choice of
solvent etc. for a particular reaction.
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3) Kinetics versus Thermodynamics
 Thermodynamics tells us what processes are possible.
 Kinetics tells us whether the process is practical. (high
activation energy can effectively block a reaction
although that is thermodynamically favored)
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4) Exothermic reaction versus Endothermic reaction
Most spontaneous processes are exothermic, however,
some spontaneous processes are endothermic.
For example:
- the melting of ice above 0 oC
- the evaporation of liquid water to gaseous water
- the dissolution of sodium chloride in water
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Chapter 17
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5) Natural tendencies for spontaneous processes
 To achieve a lower energy state, i.e., decreasing
enthalpy (H)
- The enthalpy change, ΔH, is the difference in the
sum of the internal energy and PV work energy of
the reactants to the products.
 Toward a more disordered state, i.e., increasing
entropy (S)
- The entropy change, ΔS, is the difference in
randomness of the reactants compared to the
products.
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Chapter 17
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*****
3. Entropy and the Second Law of Thermodynamics
1) Entropy (S)
A thermodynamic function that is proportional to the
number of energetically equivalent ways to arrange
the components of a system to achieve a particular
state; a measure of the energy randomization or
energy dispersal in a system.
 S = k ln W
k = Boltzmann constant = 1.38 × 10−23 J/K
W: the number of energetically equivalent ways a
system can exist, unitless
S, unit generally in J/mol
 The greater spreading of the energy of the
microscopic particles, the greater the S of the system.
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Chapter 17
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Continued
 Each system has the same total energy (4 J), but System
A has only one possible microstate while System B
has two possible microstate.
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Chapter 17
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Continued
The Spontaneous Mixing of Ideal Gases
 Intermolecular forces are negligible, no
significant enthalpy change, total energy of
the system remains unchanged
 However, the number of possibilities for the
distribution of that energy increases
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Chapter 17
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Continued
Spontaneous Formation of An Ideal Solution
 Entropy of the mixture is higher than the
entropies of the two substances separated.
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Chapter 17
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Continued
Vaporization of Water
 Evaporation is spontaneous because of the
increase in entropy
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Chapter 17
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2) Second Law of Thermodynamics
*****
All spontaneous process are accompanied by an increase
in entropy (S) of the universe.
*****
Universe
Spontaneous process:
Surrounding
ΔSuniv = ΔSsys + ΔSsurr > 0
System
TMHsiung©2015
*ΔS = Sfinal − Sinitial
Chapter 17
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3) Absolute Entropy of a Substance
 The particular S of a substance is calculated from the
amount of heat required to raise the temperature from 0
K.
 Standard molar entropy (So): The entropy of one mole
(or 1 M) of a substance in its standard state. (Appendix
IIB)
 In a particular phase, S slow increase with increasing T,
the temperature dependence of S:
S(T2) = S(T1) + Cp ln(T2 –T1)
Cp: Heat capacity of the substance
 S sharply change at phase change/transition:
S(phase 2) = S(phase 1) + ΔH/T
example: ΔSfusion = ΔHfusion/Tm)
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4) Entropy Change Associated with a Change in State
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Chapter 17
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5) Predicting the Trends of Entropy






*****
SSolid < SLiquid < SGas
S(small atom or molecule) < S(large atom or molecule)
S(simple molecule) < S(complex molecule)
S(low temperature) < S(high temperature)
S(solute) + S(solvent) < S(solution)
S(less amount of gases) < S(more amount of gases)
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*****
Example 17.1 Predicting the Sign of Entropy Change
Predict the sign of ΔS for each process:
a.
H2O(g) → H2O(l)
b.
Solid carbon dioxide sublimes.
c.
2 N2O(g) → 2 N2(g) + O2(g)
Solution
a. Since a gas has a greater entropy than a liquid, the entropy
decreases and ΔS is negative.
b. Since a solid has a lower entropy than a gas, the entropy
increases and ΔS is positive.
c. Since the number of moles of gas increases, the entropy
increases and ΔS is positive.
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Chapter 17
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4. Heat Transfer and Changes in the Entropy of the
Surroundings
1) Entropy of the Surroundings (ΔSsurr)
a) Observation:
• Water freezes at temperatures below 0 oC, the
entropy of the water decreases, the process is
spontaneous.
• Water vapor in air condenses into fog on a cold
night, the entropy of the water decreases, the
process is spontaneous.
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b) Explanation:
 When the entropy change
in a system is unfavorable
(negative), the entropy
change in the
surroundings must be
favorable (positive), and
large in order to allow the
process to be
spontaneous.
* An exothermic process increases the entropy of the
surroundings.
* An endothermic process decreases the entropy of the
surroundings.
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2) The Temperature Dependence of ΔSsurr
a) Observation:
Water freezes at temperatures below 0 oC
spontaneously, however, freezing of water becomes
nonspontaneous above 0 oC.
b) Explanation:
The greater the temperature, the smaller the
increase in entropy for a given amount of energy
dispersed into the surroundings.
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Chapter 17
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Continued
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Chapter 17
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*****
3) Quantifying Entropy Changes in the Surroundings
At Constant P and T
Ssurr   qsys
Ssurr =
Ssurr
1

T
 qsys
T
qsys  H sys
Ssurr =
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H sys
Chapter 17
T
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*****
Example 17.2 Calculating Entropy Changes in the Surroundings
Consider the combustion of propane gas:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔHrxn = –2044 kJ
a. Calculate the entropy change in the surroundings associated with this
reaction occurring at 25 oC.
b. Determine the sign of the entropy change for the system.
c. Determine the sign of the entropy change for the universe. Will the
reaction be spontaneous?
Solution
a.
TMHsiung©2015
Chapter 17
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Continued
b. An increase in the number of moles of gas implies a positive ΔSsys.
c.
ΔSuniv is positive and the reaction is spontaneous.
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Chapter 17
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5. Gibbs Free Energy
1) The Equation
*****
At Constant T and P:
Since ΔSuniv = ΔSsurr + ΔSsys , therefore,
ΔH sys
ΔSuniv = ΔSsys 
T
multiple by T,
TΔSuniv = TΔSsys + ΔHsys = –(ΔHsys – TΔSsys)
Multiple by –1
–TΔSuniv = ΔHsys – TΔSsys
Let introduce another thermodynamic state function, Gibbs
free energy, G, which is defined as:
ΔG = –TΔSuniv
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*****
Thus,
ΔG = ΔHsys – TΔSsys
Now, let’s memorize this Gibbs equation as:
ΔG = ΔH – TΔS
2) A Summary
ΔG < 0 : spontaneous process
ΔG = 0 : at equilibrium
ΔG > 0 : nonspontaneous process
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Chapter 17
*****
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Continued
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Chapter 17
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3) The Effect of ΔH, ΔS, and T on Spontaneity
*****
ΔG = ΔH − TΔS
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Chapter 17
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*****
Example 17.3 Computing Gibbs Free Energy Changes and
Predicting Spontaneity from ΔH and ΔS
Consider the reaction for the decomposition of carbon tetrachloride gas:
CCl4(g) → C(s, graphite) + 2 Cl2(g) ΔH = + 95.7 kJ; ΔS = +142.2 J/K
a. Calculate ΔG at 25 oC and determine whether the reaction is spontaneous.
b. If the reaction is not spontaneous at 25 oC, determine at what temperature
(if any) the reaction becomes spontaneous.
Solution
a.
The reaction is not spontaneous.
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*****
Continued
b. Since ΔS is positive, ΔG becomes more negative with increasing
temperature.
The reaction is spontaneous above this temperature.
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*****
6. Entropy Changes in Chemical Reactions:
Calculating ΔSrxn
1) Standard State
• For Gas: pure gas at exactly 1 atm pressure.
• For Solid or Liquid: 1 mole pure solid or liquid in
its most stable form at exactly 1 atm pressure and
temperature of interest. Usually 25 oC
• Solution: Substance in a solution with
concentration 1 M.
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2) Third Law of Thermodynamics
*****
 For a perfect crystal at absolute zero (0 K), the
absolute entropy = 0 J/mol ∙ K.
S = k ln W = k ln 1 = 0
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3) Factors that affect the standard entropy
• State of the substance
• Molar mass of the substance
• Particular allotrope
• Molecular complexity
• Dissolution
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 State of the
substance
 Molar mass of
the substance
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 Particular allotrope
* Allotrope: Those
elements can exist
in two or more
forms.
•
•
Diamond: three-dimensional crystal structure.
Graphite: the atoms bond together in sheets, but the
sheets have freedom to slide past each other.
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 Molecular complexity
*****
• Ar: only translational motion.
• NO, translational motion, rotational motion, and
vibrational motions of the molecules
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 Dissolution
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4) The Standard Entropy Change, ΔS
*****
 The standard entropy change is the difference in
absolute entropy between the reactants and
products under standard conditions.
ΔSºreaction = (∑npSºproducts) − (∑nrSºreactants)
 The standard enthalpy of formation, ΔHf°, of an
element is 0 kJ/mol.
 The absolute entropy (So) at 25 oC is always positive.
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 Standard Absolute Entropies
*****
* More complete list can be found in Appendix IIB
** Intensive property: A property such as density that is independent of the
amount of a given substance.
Extensive property: A property that depends on the amount of a
given substance, such as mass.
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Chapter 17
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° )
Example 17.4 Calculating Standard Entropy Changes (ΔSrxn
*****
Calculate°ΔSrxn for the balanced chemical equation:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Solution
Appendix IIB.
°
* ΔSrxn is positive, the number of moles of gas increases. OK.
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Chapter 17
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7. Free Energy Changes in Chemical Reactions:
Calculating ΔGorxn
1) Standard free energies of formation (Gof)
 The change in free energy when 1 mol (or 1 M) of a
compound in its standard state forms from its
constituent elements in their standard states.
 The free energy of formation listed in Appendix IIB .
 Same as standard formation enthalpy (Hfo), Gof of the
pure elements in their standard states is zero.
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*****
2) Calculating Gorxn with Tabulated Values of Free
Energies of Formation (Gof)
ΔGorxn = ∑npΔGfo(products) − ∑nrΔGfo(reactants)
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*****
°
Example 17.7
Ozone in the lower atmosphere is a pollutant that can form by the following
reaction involving the oxidation of unburned hydrocarbons:
CH4(g) + 8 O2(g) → CO2(g) + 2 H2O(g) + 4 O3(g)
Use the standard free energies of formation°to determine ΔGrxn for this reaction at
25 oC.
Solution
Appendix IIB
* Gof of a pure element is zero.
°
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Chapter 17
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*****
3) Calculating Standard Free Energy Changes with
ΔGorxn = ΔHorxn − TΔSorxn
Example 17.5 At 25 oC
One of the possible initial steps in the formation of acid rain is the oxidation of
the pollutant SO2 to SO3 by the reaction:
Calculate ΔGrxn at 25 oC and determine whether the reaction is spontaneous.
Solution
Appendix IIB
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Chapter 17
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Continued
The reaction is spontaneous at this temperature because ΔGrxn is negative.
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Chapter 17
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Example 17.6 Other than 25
oC
*****
For the reaction in Example 17.5, estimate the value of ΔGorxn at 125 oC. Is the
reaction more or less spontaneous at °this elevated temperature; that is, is the
value of ΔGorxn more negative (more spontaneous) or more positive (less
spontaneous)?
Solution
ΔGorxn at 125 oC is less negative (or more positive) than the value of ΔGorxn at
25 oC (which is –70.9 kJ), the reaction is less spontaneous.
*
The different temperature resulted in relative small variations
of ΔSorxn and ΔHorxn (25oC).
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4) Calculating ΔGorxn for a Stepwise Reaction
° for a Stepwise Reaction
Example 17.8 Calculating ΔGrxn
°
Find ΔG
rxn for the reaction.
3 C(s) + 4 H2(g) → C3H8(g)
Use the following reactions with known ΔG’s:
Solution
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5) Coupled Reactions
 Coupled reactions: Use a thermodynamically favorable
reaction to drive an unfavorable one.
Example1:
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Chapter 17
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Example 2 (a biochemical cycle)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
ADP + H3PO4 → ATP + H2O(l)
C6H12O6(s) + 6O2(g) + ADP + H3PO4
→ 6CO2(g) + ATP + 7H2O(l)
Alnaine + Glycine → Alnanylglycine
ATP + H2O → ADP + H3PO4
ATP + H2O + Alnaine + Glycine
→ Ananylglycine + ADP + H3PO4
TMHsiung©2015
Chapter 17
ΔGo = –2280 kJ
ΔGo = +31 kJ
ΔGo = –2249 kJ
ΔGo = +29 kJ
ΔGo = –31 kJ
ΔGo = –2 kJ
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8. Free Energy Changes for Nonstandard States: The
Relationship between ΔGorxn and ΔGrxn
Q. Spilled water spontaneously
evaporates even though ΔGorxn for
the vaporization of water is
positive. Why?
A. Because ordinary conditions are
not standard conditions and
ΔGorxn applies only to standard
conditions.
Gorxn describes the free energy change at standard state.
Grxn describes the free energy change at any specified state.
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Chapter 17
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 The Free Energy Change of a Reaction under
Nonstandard Conditions
*****
Grxn = Gorxn + RTlnQ
Q:
R:
T:
TMHsiung©2015
reaction quotient
8.314 Jmol–1K–1
K
Chapter 17
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Example
Grxn = Gorxn + RTlnQ
(a) Standard Conditions, PH2O = 1 atm, i.e., Q = 1
Grxn = Gorxn + RTlnQ = +8.59 kJ/mol + RT ln(1)
= +8.59 kJ/mol
(spontaneous in the reverse direction)
(b) At pressure of 0.0313 atm, i.e., Q = Kp = 0.0313
Grxn = Gorxn + RTlnQ = +8.59 kJ/mol + RT ln (0.0313)
= +8.59 kJ/mol - 8.59 kJ/mol = 0
(c) At pressure of 5.00x10-3 atm, i.e., Q = Kp = 5.00x10-3 atm
Grxn = Gorxn + RTlnQ = +8.59 kJ/mol + RT ln (5.00x10-3)
= +8.59 kJ/mol – 13.1 kJ/mol = – 4.5 kJ/mol
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Chapter 17
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Continued
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Chapter 17
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Example 17.9 Calculating ΔGrxn under Nonstandard Conditions
Consider the reaction at 298 K:
*****
Calculate ΔGrxn under these conditions:
PNO = 0.100 atm; PO2 = 0.100 atm; PNO2 = 2.00 atm
Is the reaction more or less spontaneous under these conditions than under
standard conditions?
Solution
°
The reaction is spontaneous under these conditions, but less spontaneous than
it would be under standard conditions.
°
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°
Chapter 17
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9. Free Energy and Equilibrium: Relating ΔGorxn *****
to
the Equilibrium Constant ( K )Free En ergy
1) Equation and Explanation
Grxn = Gorxn + RTlnQ
At equilibrium, Grxn = 0 and Q = K
0 = Gorxn + RTlnQ = Gorxn + RTlnK
Gorxn = －RTlnK = －2.303RTlogKeq
lnK = －Go/RT
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Gorxn = －RTlnK
a) When K < 1
 Gorxn is positive.
 Under standard
conditions (when Q =
1 ), the reaction is
spontaneous in the
reverse direction.
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Gorxn = －RTlnK
b) When K > 1
 Gorxn is negative.
 Under standard
conditions (when Q =
1 ), the reaction is
spontaneous in the
forward direction.
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Gorxn = －RTlnK
c) When K = 1
 Gorxn is zero.
 Under standard
conditions (when Q =
1 ), the reaction is
equilibrium.
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*****
Example 17.10 The Equilibrium Constant and ΔGrxn
Use tabulated free energies of formation to calculate the equilibrium constant for
the reaction at 298 K:
Solution
Appendix IIB
°
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2) The Temperature Dependence of the Equilibrium
Constant
 For a particular reaction, H and S do not change
much for different temperature.
 At standard state:
Gorxn = – RTlnK
(1)
Gorxn = Horxn – TSorxn (2)
Combing (1) and (2)
– RTlnK = Horxn – TSorxn = Gorxn
o
o
H rxn

S
1
 
rxn
ln K 

 
R T  R
y = mx + b
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Chapter 17
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Example:
For the reaction: CO(g) + H2O(g) → CO2(g) + H2(g)
lnKeq vs. 1/T plot
Slope  4810 = –Ho/R (R=8.3145 J mol–1K–1)
Ho  – 40 kJ (tabulated data is – 41.2 kJ).
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 Calculating K at different temperature
H  1  S
H  1 
ln K 

 
   Constant
R T  R
R T 
o
rxn
o
rxn
o
rxn
van't Hoff equation:
K 2 H
ln

K1
R
o
rxn
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Chapter 17
1 1
(  )
T1 T2
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End of Chapter 17
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