Entropy and Free Energy
Chapter 17
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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Objectives
•
•
•
•
Spontaneity of Reactions
Define Entropy, S
Show how entropy affects molecules
Show how entropy affects phase
changes
• Second Law of Thermodynamics
• Enthalpy, Entropy and Free Energy
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Ch. 17.1 - Spontaneous
Processes and Entropy
• If the state of a chemical system is such
that a rearrangement of its atoms and
molecules would decrease the energy of
the system--• then this is a product-favored system.
• Most product-favored reactions are
exothermic
—but this is not the only criterion
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Thermodynamics
• Both product- and reactant-favored reactions can
proceed to equilibrium in a spontaneous
process.
AgCl(s)  Ag+(aq) + Cl–(aq)
Reaction is not product-favored, but it moves
spontaneously toward equilibrium.
• 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.
Paper burns — a
product-favored
reaction. Also
kinetically favored once
reaction is begun.
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Spontaneous Reactions
In general, spontaneous
reactions are
exothermic.
Fe2O3(s) + 2 Al(s) 
2 Fe(s) + Al2O3(s)
∆rH = - 848 kJ
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Spontaneous Reactions
But many spontaneous reactions or
processes are endothermic or even
have ∆H = 0.
∆H = 0
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NH4NO3(s) + heat  NH4NO3(aq)
Entropy, S
One property common to
spontaneous processes is that
the energy of the final state is
more dispersed.
In a spontaneous process energy
goes from being more
concentrated to being more
dispersed.
The thermodynamic property
related to energy dispersal is
ENTROPY, S.
2nd Law of Thermo — a
spontaneous process results in an
increase in the entropy of the
universe.
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Reaction of K
with water
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Directionality of Reactions
Probability suggests that a spontaneous
reaction will result in the dispersal of
energy.
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
spontaneous.
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Energy Dispersal
Each of the
ways to
distribute
energy is called
a microstate.
These
microstates give
a particular
arrangement
(State).
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Directionality of Reactions
Matter & energy dispersal
As the size of the
container increases,
the number of
microstates accessible
to the system
increases, and the
density of states
increases. Entropy
increases.
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• The entropy of
liquid water is
greater than the
entropy of solid
water (ice) at 0˚ C.
• Energy is more
dispersed in
liquid water than
in solid water.
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Entropy, S
So (J/K•mol)
H2O(liq)
69.95
H2O(gas) 188.8
S (solids) < S (liquids) < S (gases)
Energy dispersal
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Entropy and States of Matter
S˚(Br2 liq) < S˚(Br2 gas)
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S˚(H2O sol) < S˚(H2O liq)
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Ch. 17.2 - 2nd Law of
Thermodynamics
A reaction is spontaneous if ∆S for the universe is
positive.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Suniverse > 0 for spontaneous process
Calculate the entropy created by energy dispersal
in the system and surroundings.
If ΔSuniverse < 0, then the process is
spontaneous in the opposite
direction.
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Ch. 17.3 – Temperature and
Spontaneity
Discussion Questions!
For the process A(l)
A(s), which direction
involves an increase in energy dispersal? 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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- Cengage
Copyright
© Cengage Learning.
All rights reserved
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Entropy and Temperature
S increases
slightly with T
S increases a
large amount
with phase
changes
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ΔSsurr
• The sign of ΔSsurr depends on the
direction of the heat flow.
• The magnitude of ΔSsurr depends
on the temperature.
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ΔSsurr
H
Ssurr = 
T
Heat flow (constant P) = change in enthalpy = ΔH
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© 2009 Brooks/Cole
- Cengage
Copyright
© Cengage Learning.
All rights reserved
21
Entropy Changes for Phase Changes
For a phase change,
∆S = q/T
where q = heat transferred in
phase change
For H2O (liq)  H2O(g)
∆H = q = +40,700 J/mol
q
40,700 J/mol
S =
=
= +109 J/K•mol
T
373.15 K
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Definition of Entropy
Practice
Determine whether entropy increases or
decreases for the following reactions.
a) HCl(g) + NH3(g)  NH4Cl(s)
b) 2H2O2(l)  2H2O(l) + O2(g)
c) Cooling of nitrogen gas from -20⁰C to -50⁰C
d) NaCl(s)  Na+(aq) + Cl-(aq)
e) CaCO3(s)  CaO(s) + CO2(g)
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2nd Law of Thermodynamics
Dissolving NH4NO3
in water—an
entropy driven
process.
∆Suniverse =
∆Ssystem + ∆Ssurroundings
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2nd Law of Thermodynamics
2 H2(g) + O2(g)  2 H2O(liq)
∆Sosystem = -326.9 J/K
S
o
surroundings
=
qsurroundings
T
=
-Hsystem
T
Can calc. that ∆Hro = ∆Hosystem = -571.7 kJ
S
o
surroundings
- (-571.7 kJ)(1000 J/kJ)
=
298.15 K
∆Sosurroundings = +1917 J/K
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2nd Law of Thermodynamics
2 H2(g) + O2(g)  2 H2O(liq)
∆Sosystem = -326.9 J/K
∆Sosurroundings = +1917 J/K
∆Souniverse = +1590. J/K
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The entropy of the
universe is
increasing, so
the reaction is
product-favored.
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Ch. 17.4 - Free Energy, G
∆Suniv = ∆Ssurr + ∆Ssys
Suniv =
Hsys
+ Ssys
T
Multiply through by -T
J. Willard Gibbs
1839-1903
-T∆Suniv = ∆Hsys - T∆Ssys
-T∆Suniv = change in Gibbs free energy
for the system = ∆Gsystem
Under standard conditions —
o
∆G
sys
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=
o
∆H
sys
-
o
T∆S
sys
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o
∆G
=
o
∆H
-
o
T∆S
• This means the a process is spontaneous in
the direction in which the free energy
decreases.
• Remember that ΔSuniv = - ΔG/T
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Free Energy and Spontaneity
Example
H2O(s)  H2O(l)
• ΔH° = 6.03x103 J/mol and ΔS°=22.1 J/K
• At 10°C, the ΔG° = 6030 J – 283K(22.1 J/K)
= 6030 J – 6254 J
= -224 J
So, ice melts spontaneously at 10°C, even
though it’s an endothermic reaction.
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Free Energy and Spontaneity
Example
H2O(s)  H2O(l)
• ΔH° = 6.03x103 J/mol and ΔS°=22.1 J/K
• At -10°C, the ΔG° = 6030 J – 263K(22.1 J/K)
= 6030 J – 5800 J
= 230 J
So, ice does not melt spontaneously at -10°C.
However, water would freeze spontaneously at 10°C because opposite is more
thermodynamically favorable.
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Free Energy and Spontaneity
Example
H2O(s)  H2O(l)
• ΔH° = 6.03x103 J/mol and ΔS°=22.1 J/K
• At 0°C, the ΔG° = 6030 J – 273K(22.1 J/K)
= 6030 J – 6030 J
=0J
So, neither process is favorable
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o
∆G
Gibbs free
=
o
∆H
-
o
T∆S
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energy change =
total energy change for system
- energy lost in energy dispersal
If reaction is
• exothermic (negative ∆Ho)
• and entropy increases (positive ∆So)
• then ∆Go must be NEGATIVE
• reaction is spontaneous (and product-favored).
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o
∆G
=
o
∆H
-
o
T∆S
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Gibbs free energy change =
total energy change for system
- energy lost in energy dispersal
If reaction is
• endothermic (positive ∆Ho)
• and entropy decreases (negative ∆So)
• then ∆Go must be POSITIVE
• reaction is not spontaneous (and is reactant-
favored).
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Gibbs Free Energy, G
o
∆G
=
o
∆H
-
o
T∆S
∆Ho
∆So ∆Go Process
Exo,<0
>0
<0
Spontaneous at all T
End,>0
<0
>0
Never Spontaneous
Exo, <0
<0
?
T dependent, spontaneous
at lower T’s
End, >0
>0-
?
T dependent, spontaneous
at higher T’s
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Spontaneous or Not? Practice
35
Use the table on the previous slide to
classify the following reactions as one of
the 4 types.
1. CH4(g) + 2O2(g)  2H2O(l) + CO2(g)
ΔH = -891 kJ/mol
ΔS = -242 J/K
2. 2Fe2O3(s) + 3C(s, graphite)  4Fe(s) + 3CO2(g)
ΔH = +468 kJ/mol
ΔS = +561 J/K
3. C(s, graphite) + O2(g)  CO2(g)
ΔH = -394 kJ/mol
ΔS = +3 J/K
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Ch. 17.5 – Entropy Changes in
Reactions
• The 3rd Law of Thermodynamics:
• The entropy of a pure crystal at 0 K is
0 (S=0)
–Gives us a starting point.
–At all other temperatures, entropy must
be > 0. (As you raise T, entropy
increases)
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Standard Molar Entropies
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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.
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Entropy, S
Increase in molecular
complexity generally
leads to increase in S.
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Entropy, S
Entropies of ionic solids depend on
coulombic attractions.
So (J/K•mol)
Mg2+ & O2-
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Na+ & F-
MgO
26.9
NaF
51.5
Entropy, S
Liquids or solids dissolve in a solvent in a
spontaneous process owing to the increase
in entropy. Matter (and energy) are more
dispersed.
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Calculating ∆S for a Reaction
∆So =  So (products) -  So (reactants)
Consider 2 H2(g) + O2(g)  2 H2O(liq)
∆So = 2 So (H2O) - [2 So (H2) + So (O2)]
∆So = 2 mol (69.9 J/K·mol) [2 mol (130.7 J/K·mol) +
1 mol (205.3 J/K·mol)]
∆So = -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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Entropy of Reaction Practice
Calculate ∆S˚ for the following reaction:
Al2O3(s) + 3H2(g)2Al(s)+3H2O(g)
S⁰
51
131
28
189
∆S⁰rxn = 2(28) + 3(189) – 51 – 3(131)
= 56 + 567 – 51 – 393
= 179 J/K
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Ch. 17.6 – Free Energy and
Chemical Reactions
• Really want to determine the STANDARD
FREE ENERGY CHANGE, ΔG° for a reaction
• This represents the change in free energy
that will occur if the reactants in their
standard states are converted to the products
in their standard states.
• Note: ΔG is not measured directly.
• Use it to compare relative tendency for
reactions to occur.
• The more negative the ΔG of reaction, the
greater tendency of product to form
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Gibbs Free Energy, G
∆Go = ∆Ho - T∆So
Two methods of calculating ∆Go
a) Determine ∆rHo and ∆rSo and use Gibbs equation.
b) Use tabulated values of
free energies of
formation, ∆Gfo.
∆Gro =  ∆Gfo (products) -  ∆Gfo (reactants)
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Free Energies of Formation
Note that ∆Gf˚ for an element = 0
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Calculating ∆Grxno
Combustion of acetylene
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
∆Hrxno = -1238 kJ
Use standard molar entropies to calculate
∆Srxno = -97.4 J/K or -0.0974 kJ/K
∆Grxno = -1238 kJ - (298 K)(-0.0974 kJ/K)
= -1209 kJ
Reaction is product-favored in spite of negative ∆Srxno.
Reaction is “enthalpy driven”
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Calculating ∆Grxno
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 ∆rGo
NH4NO3(s) + heat  NH4NO3(aq)
From tables of thermodynamic data we find
∆rHo = +26 kJ
∆rSo = +109 J/K or +0.109 kJ/K
∆rGo = +26 kJ - (298 K)(+0.109 J/K)
= -6 kJ
Reaction is product-favored in spite of positive
∆Hrxno.
Reaction is “entropy driven”
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Gibbs Free Energy, G
∆Go = ∆Ho - T∆So
Two methods of calculating ∆Go
a) Determine ∆rHo and ∆rSo and use Gibbs equation.
b) Use tabulated values of
free energies of
formation, ∆fGo.
∆rGo =  ∆fGo (products) -  ∆fGo (reactants)
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Calculating ∆Gorxn
∆Gro =  ∆Gfo (products) -  ∆Gfo (reactants)
Combustion of carbon
C(graphite) + O2(g)  CO2(g)
∆Gro = ∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)]
∆rGo = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an
element in its standard state is 0.
∆Gro = -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)
∆Hro = +468 kJ
∆Sro = +560 J/K
∆Gro = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does ∆Gro just change from being
(+) to being (-)?
When ∆Gro = 0 = ∆Hro - T∆Sro
T=
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r H
r S
=
467.9 kJ
= 835.1 K
0.5603 kJ/K
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Practice - Calculate the ΔG for this
Reaction
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2SO2(g) + O2(g) 2SO3(g) at 227 oC
Ho = -196.6 kJ; So = -189.6 J/K
Go = -196.6 – 500 (-0.1896) kJ
Go = -101.8 kJ
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Free Energy, Enthalpy and
Entropy Practice
•
•
•
•
•
•
2000 AP Exam #6a-c
2003 AP Exam #7b-d
2004 AP Exam #2d-e
2005 AP Exam #8a-c
2005B AP Exam #7c
2006 AP Exam #2a-d (Best one)
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