Entropy and Free Energy
How to predict if a
reaction can occur,
given enough time?
THERMODYNAMICS
How to predict if a
reaction can occur at
a reasonable rate?
KINETICS
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Thermodynamics
• Is the state of a chemical system such that
a rearrangement of its atoms and
molecules would decrease the energy of
the system?
• If yes, system is favored to react — a
product-favored system.
• Most product-favored reactions are
exothermic.
• Often referred to as spontaneous
reactions.
• Spontaneous does not imply anything
about time for reaction to occur.
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Thermodynamics and Kinetics
Diamond is
thermodynamically
favored to convert to
graphite, but not
kinetically favored.
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Thermodynamics and Kinetics
Diamond is
thermodynamically
favored to convert to
graphite, but not
kinetically favored.
Paper burns — a
product-favored
reaction. Also
kinetically favored
once reaction is
begun.
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Product-Favored Reactions
In general, productfavored reactions are
exothermic.
Fe2O3(s) + 2 Al(s)
---> 2 Fe(s) + Al2O3(s)
DH = - 848 kJ
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Product-Favored Reactions
But many spontaneous reactions or
processes are endothermic or even
have DH = 0.
NH4NO3(s) + heat ---> NH4NO3(aq)
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Entropy, S
One property common to
product-favored processes
is that the final state is
more DISORDERED or
RANDOM than the original.
Spontaneity is related to
an increase in
randomness.
The thermodynamic property
related to randomness is
ENTROPY, S.
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Reaction of K
with water
The entropy of
liquid water is
greater than
the entropy of
solid water
(ice) at 0 C.
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Directionality of Reactions
How probable is it that reactant
molecules will react?
PROBABILITY suggests that a
product-favored reaction will
result in the dispersal of energy
or of matter or both.
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Directionality of Reactions
Probability suggests that a productfavored reaction will result in the
dispersal of energy or of matter or
both.
Matter Dispersal
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Directionality of Reactions
Probability suggests that a productfavored reaction will result in the
dispersal of energy or of matter or
both.
Matter Dispersal
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Directionality of Reactions
Probability suggests that a productfavored reaction will result in the
dispersal of energy or of matter or
both.
Energy Dispersal
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Directionality of Reactions
Probability suggests that a productfavored reaction will result in the
dispersal of energy or of matter or
both.
Energy Dispersal
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Directionality of
Reactions —
Energy Dispersal
Exothermic reactions involve a release of
stored chemical potential energy to the
surroundings.
The stored potential energy starts out in a few
molecules but is finally dispersed over a
great many molecules.
The final state—with energy dispersed—is
more probable and makes a reaction
product-favored.
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Entropy, S
S (gases) > S (liquids) > S (solids)
So (J/K•mol)
H2O(liq)
69.91
H2O(gas) 188.8
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Entropy, S
Entropy of a substance increases
with temperature.
Molecular motions
of heptane, C7H16
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Molecular motions of
heptane at different temps.
Entropy, S
Increase in molecular complexity
generally leads to increase in S.
So (J/K•mol)
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CH4
248.2
C2H6
336.1
C3H8
419.4
Entropy, S
Entropies of ionic solids depend on
coulombic attractions.
So (J/K•mol)
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MgO
26.9
NaF
51.5
Entropy, S
Entropy usually increases when a
pure liquid or solid dissolves in a
solvent.
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Entropy Changes for Phase Changes
For a phase change,
DS = q/T
where q = heat transferred in
phase change
For H2O (liq) ---> H2O(g)
DH = q = +40,700 J/mol
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Entropy Changes for Phase Changes
For a phase change,
DS = q/T
where q = heat transferred in
phase change
For H2O (liq) ---> H2O(g)
DH = q = +40,700 J/mol
q
40,700 J/mol
DS =
=
= + 109 J/K • mol
T
373.15 K
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Calculating DS for a Reaction
D So =
 So (products) -  So (reactants)
Consider 2 H2(g) + O2(g) ---> 2 H2O(liq)
DSo = 2 So (H2O) - [2 So (H2) + So (O2)]
DSo = 2 mol (69.9 J/K•mol) [2 mol (130.7 J/K•mol) +
1 mol (205.3 J/K•mol)]
DSo = -326.9 J/K
Note that there is a decrease in S because
3 mol of gas give 2 mol of liquid.
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2nd Law of Thermodynamics
A reaction is spontaneous (productfavored) if ²S for the universe is positive.
DSuniverse
=
DSsystem
+ DSsurroundings
DSuniverse
> 0 for product-favored
process
First calc. entropy created by matter
dispersal (DSsystem)
Next, calc. entropy created by energy
dispersal (DSsurround)
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2nd Law of Thermodynamics
Dissolving NH4NO3
in water—an
entropy driven
process.
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundi ngs
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qsurroundi ngs
-DH system
=
=
T
T
2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundi ngs
qsurroundi ngs
-DH system
=
=
T
T
Can calc. that DHorxn = DHosystem = -571.7 kJ
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundi ngs
qsurroundi ngs
-DH system
=
=
T
T
Can calc. that DHorxn = DHosystem = -571.7 kJ
DS
o
surroundi ngs
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- (-571.7 kJ)(1000 J/kJ)
=
298.15 K
2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DS
o
surroundi ngs
qsurroundi ngs
-DH system
=
=
T
T
Can calc. that DHorxn = DHosystem = -571.7 kJ
DS
o
surroundi ngs
- (-571.7 kJ)(1000 J/kJ)
=
298.15 K
DSosurroundings = +1917 J/K
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DSosurroundings = +1917 J/K
DSouniverse = +1590. J/K
The entropy of the universe
is increasing, so the
reaction is productfavored.
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2nd Law of Thermodynamics
2 H2(g) + O2(g) ---> 2 H2O(liq)
DSosystem = -326.9 J/K
DSosurroundings = +1917 J/K
DSouniverse = +1590. J/K
The entropy of the universe
is increasing, so the
reaction is productfavored.
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Gibbs Free Energy, G
²Suniv = ²Ssurr + ²Ssys
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Gibbs Free Energy, G
²Suniv = ²Ssurr + ²Ssys
DHsys
DSuniv =
+ DSsys
T
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Gibbs Free Energy, G
²Suniv = ²Ssurr + ²Ssys
DHsys
DSuniv =
+ DSsys
T
Multiply through by -T
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Gibbs Free Energy, G
D Suniv = D Ssurr + D Ssys
D Hsys
+ D Ssys
DSuniv =
T
Multiply through by -T
-T D Suniv = D Hsys - T D Ssys
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Gibbs Free Energy, G
D Suniv = D Ssurr + D Ssys
DHsys
DSuniv =
+ DSsys
T
Multiply through by -T
-T D Suniv = D Hsys - T D Ssys
-T D Suniv = change in Gibbs free energy
for the system = D Gsystem
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Gibbs Free Energy, G
DSuniv = DSsurr + DSsys
D Hsys
+ D Ssys
DSuniv =
T
Multiply through by -T
-TDSuniv = DHsys - TDSsys
-TDSuniv = change in Gibbs free energy
for the system = DGsystem
Under standard conditions —
DGo
= DHo - TDSo
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Gibbs Free Energy, G
o
DG
=
o
DH
- T
o
DS
Gibbs free energy change =
total energy change for system
- energy lost in disordering the system
If reaction is exothermic (DHo negative) and
entropy increases (DSo is +), then DGo must be
negative and reaction product-favored.
If reaction is endothermic (DHo is +), and
entropy decreases (DSo is -), then DGo must
be + and reaction is reactant-favored.
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Gibbs Free Energy, G
o
DG
=
o
DH
o
TDS
-
DHo
DSo
DGo Reaction
exo(-)
increase(+)
-
Prod-favored
endo(+)
decrease(-)
+
React-favored
exo(-)
decrease(-)
?
T dependent
endo(+)
increase(+)
?
T dependent
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Gibbs Free Energy, G
o
DG
=
o
DH
-
o
TDS
Two methods of calculating DGo
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Gibbs Free Energy, G
o
DG
=
o
DH
-
o
TDS
Two methods of calculating DGo
a)
Determine DHorxn and DSorxn and use
GIbbs equation.
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Gibbs Free Energy, G
o
DG
=
o
DH
-
o
TDS
Two methods of calculating DGo
a)
Determine DHorxn and DSorxn and use
GIbbs equation.
b)
Use tabulated values of free energies of
formation, DGfo.
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Gibbs Free Energy, G
o
DG
=
o
DH
-
o
TDS
Two methods of calculating DGo
a)
Determine DHorxn and DSorxn and use
GIbbs equation.
b)
Use tabulated values of free energies of
formation, DGfo.
²Gorxn =  ²Gfo (products) -  ²Gfo (reactants)
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Calculating D
o
G rxn
Combustion of acetylene
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
DHorxn = -1238 kJ
Use standard molar entropies to calculate
DSorxn = -97.4 J/K or -0.0974 kJ/K
DGorxn = -1238 kJ - (298 K)(-0.0974 J/K)
= -1209 kJ
Reaction is product-favored in spite of negative
DSorxn.
Reaction is “enthalpy driven”
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Calculating D
o
G rxn
NH4NO3(s) + heat ---> NH4NO3(aq)
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy- or entropy-driven?
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Calculating D
o
G rxn
NH4NO3(s) + heat ---> NH4NO3(aq)
From tables of thermodynamic data we find
DHorxn = +25.7 kJ
DSorxn = +108.7 J/K or +0.1087 kJ/K
DGorxn = +25.7 kJ - (298 K)(+0.1087 J/K)
= -6.7 kJ
Reaction is product-favored in spite of negative
DHorxn.
Reaction is “entropy driven”
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Calculating D
DGorxn =  ²Gfo (products) -  DGfo (reactants)
o
G rxn
Combustion of carbon
C(graphite) + O2(g) --> CO2(g)
DGorxn = DGfo(CO2) - [DGfo(graph) + DGfo(O2)]
DGorxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an
element in its standard state is 0.
DGorxn = -394.4 kJ
Reaction is product-favored as expected.
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Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
DHorxn = +467.9 kJ
DSorxn = +560.3 J/K
DGorxn = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does DGorxn just change from being
(+) to being (-)?
When DGorxn = 0 = DHorxn - TDSorxn
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Free Energy and Temperature
2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g)
DHorxn = +467.9 kJ
DSorxn = +560.3 J/K
DGorxn = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does DGorxn just change from being
(+) to being (-)?
When DGorxn = 0 = DHorxn - TDSorxn
DHrxn
467.9 kJ
T =
=
= 835.1 K
DSrxn
0.5603 kJ/K
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Thermodynamics and Keq
Keq is related to reaction favorability.
When DGorxn < 0, reaction moves
energetically “downhill”
²Gorxn is the change in free energy as
reactants convert completely to products.
But systems often reach a state of
equilibrium in which reactants have not
converted completely to products.
In this case DGrxn is < DGorxn , so state with
both reactants and products present is
more stable than complete conversion.
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Thermodynamics and Keq
Product-favored
reaction.
2 NO2 ---> N2O4
DGorxn = -4.8 kJ
Here DGrxn is less than
DGorxn , so the state
with both reactants
and products
present is more
stable than complete
conversion.
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Thermodynamics and Keq
Reactant-favored
reaction.
N2O4 --->2 NO2
DGorxn = +4.8 kJ
Here DGorxn is greater
than DGrxn , so the
state with both
reactants and
products present is
more stable than
complete conversion.
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Thermodynamics and Keq
Keq is related to reaction favorability and so
to DGorxn.
The larger the value of DGorxn the larger the
value of K.
o
DG rxn
= - RT lnK
where R = 8.31 J/K•mol
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Thermodynamics and Keq
DGorxn
= - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
DGorxn = +4.8 kJ
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Thermodynamics and Keq
DGorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
DGorxn = +4.8 kJ
DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K
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Thermodynamics and Keq
DGorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
DGorxn = +4.8 kJ
DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
lnK = = - 1.94
(8.31 J/K)(298K)
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Thermodynamics and Keq
DGorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
DGorxn = +4.8 kJ
DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
lnK = = - 1.94
(8.31 J/K)(298K)
K = 0.14
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Thermodynamics and Keq
DGorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
DGorxn = +4.8 kJ
DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
lnK = = - 1.94
(8.31 J/K)(298K)
K = 0.14
When DGorxn > 0, then K < 1
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