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Modified 02/02/12 by Laura Peck to accompany
Silberberg: Principles of General Chemistry Chapter 20
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
1
Thermodynamics
• If the state of a chemical system is such
that a rearrangement of its atoms and
molecules would decrease the energy of
the system--• AND the K is greater than 1,
• 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
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• The sign of ΔH cannot predict spontaneous change.
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
ΔH°rxn = - 802kJ
• N2O5(s)  2NO2(g) + 1/2O2(g)
ΔH°rxn = +109.5 kJ K>1
K>1
• *****Both are product favored***
• Both product- and reactant-favored reactions can proceed to
equilibrium in a spontaneous process.
• AgCl(s)  Ag+(aq) + Cl–(aq)
K<1 Reactant favored
Reaction is not product-favored, but it moves
spontaneously toward equilibrium.
• Spontaneous does not imply anything about time for reaction
to occur.
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)
∆H = - 848 kJ
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Spontaneous Reactions
But many spontaneous reactions or
processes are endothermic or even
have ∆H = 0.
NH4NO3(s) + heat --->
NH4NO3(aq)
Discuss your labs using urea &
water or vinegar & baking soda!
Entropy, S
One property common to
spontaneous 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.
Reaction of K
with water
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The entropy of
liquid water is
greater than
the entropy of
solid water
(ice) at 0˚ C.
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Entropy and the Number of Microstates
• Ludwig Boltzmann defined the entropy (S) of a system in
terms of W:
• S= klnW
• K is the universal gas constant R divided by Avogadro’s
number.
It equals 1.38x10-23J/K
• A system with fewer microstates (smaller W) among
which to spread its energy has lower entropy (lower S)
• A system with more microstate (larger W) among which
to spread its energy has higher entropy (higher S)
• Smore microstates > Sless microstates
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Directionality of Reactions
How probable is it that reactant
molecules will react?
PROBABILITY suggests that a
spontaneous reaction will result in
the dispersal
* of energy
* or of matter
* or of energy & matter.
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Directionality of Reactions
Probability suggests that a spontaneous
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 spontaneous
reaction will result in the dispersal of
energy or of matter or both.
Energy Dispersal
14
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.
Entropy, S
(Δssys = Sfinal – Sinitial)
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So (J/K•mol)
H2O(liq)
69.95
H2O(gas) 188.8
S (gases) > S (liquids) > S (solids)
ΔSsys = SliquidH2O – SgasH2O <0
= 69.95J/K*mol – 188.8J/K*mol = -118.9 J/K*mol
Δssys = SgasH2O – SliquidH2O >0
= 188.8J/K*mol – 69.95J/K*mol = +118.9J/K*mol
(discuss in light of microstates)
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Entropy and States of Matter
S˚(Br2 liq) < S˚(Br2 gas)
S˚(H2O sol) < S˚(H2O liq)
Entropy, S
17
Entropy of a substance increases
with temperature.
Molecular motions
of heptane, C7H16
Molecular motions of
heptane at different temps.
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-
Na+ & F-
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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Practice problems
• Choose the member with the higher
entropy in each of the following
pairs, and justify your choice
(assume constant temperature
except E)
• A) 1 mol of SO2(g) or 1 mol of SO3(g)
• B) 1 mol of CO2(s) or 1 mol of CO2(g)
• C) 3 mol of O2(g) or 2 mol of O3(g)
• D) 1 mol of KBr(s) or 1 mol of KBr(aq)
• E) Seawater at 2°C or at 23°C
• F) 1 mol of CF4(g) or 1 mol of CCl4(g)
SO3(g) >atoms
CO2(g) s<l<g
3 mol O2(g) >#mols
KBr(aq) s<l<g
23°C
>T
CCl4(g) >molar mass
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Where does the entropy go?
• The second law of thermodynamics states that entropy of
the universe will increase.
• An exothermic reaction will release energy (E) to the
environment and thus, entropy within the system will
decrease. The system becomes more ordered.
• An endothermic reaction will absorb energy (E) from the
environment and thus, entropy within the system will
INCREASE. The system becomes more disordered.
• In both cases, the total S will be greater than zero.
• ΔSuniv = ΔSsys + ΔSsurr >0
• (discuss combustion & urea/water reactions)
Standard Molar Entropies
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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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Entropy and Temperature
S increases
slightly with T
S increases a
large amount
with phase
changes
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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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Practice Problem
• Calculate ΔS°rxn for the combustion of 1 mol
of propane (C3H8) at 25°C
Write a balanced reaction. Since its combustion we know The products
have to be CO2(g) and H2O(l), the other reactant Has to be O2(g)
C3H8(g) + 5O2(g)  3CO2(g) +
*hint: since there’s 6 mols of gas reactants
4H2O(l) And only 3 mols of gas products – what do
You think entropy should do?
ΔS°rxn = [(3mol CO2)(S°CO2) + (4mol H2O)(S°H2O)] –
[(1mol C3H8)(S°C3H8) + (5mol O2)(S°O2)]
ΔS°rxn = [(3mol)(213.7J/mol*K) + (4mol)(69.9J/mol*K)] –
[(1mol)(269.9J/mol*K) + (5mol)(205.0J/mol*K)]
ΔS°rxn = -374 J/K
LE CHATELIER'S PRINCIPLE
Le Chatelier's Principle:If a dynamic equilibrium is disturbed by
changing the conditions, the position of equilibrium moves to
counteract the change.
• 1) Using Le Chatelier's Principle with a change of concentration
• Suppose you have an equilibrium established between four
substances A, B, C and D.
• Increase concentration of A:
Increase concentration of B:
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Using Le Chatelier's Principle with a
change of pressure
(This only applies to reactions involving gases):
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Using Le Chatelier's Principle with a change of
temperature
• For this, you need to know whether heat is given out or absorbed
during the reaction. Assume that our forward reaction is exothermic
(heat is evolved):
What would happen if you changed the
conditions by increasing the
temperature?
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• According to Le Chatelier, the position of equilibrium will move in such a way
as to counteract the change. That means that the position of equilibrium will
move so that the temperature is reduced again.
• Suppose the system is in equilibrium at 300°C, and you increase the
temperature to 500°C. How can the reaction counteract the change you have
made? How can it cool itself down again?
• To cool down, it needs to absorb the extra heat that you have just put in. In the
case we are looking at, the back reaction absorbs heat. The position of
equilibrium therefore moves to the left. The new equilibrium mixture contains
more A and B, and less C and D.
What would happen if you changed the
conditions by decreasing the
temperature?
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• The equilibrium will move in such a way that the
temperature increases again.
• Suppose the system is in equilibrium at 500°C and you
reduce the temperature to 400°C. The reaction will tend to
heat itself up again to return to the original temperature. It
can do that by favoring the exothermic reaction.
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Le Chateliers principle
2nd Law of Thermodynamics
A reaction is spontaneous if ∆S for the
universe is positive.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Suniverse > 0 for spontaneous
process
First calc. entropy created by matter
dispersal (∆Ssystem)
Next, calc. entropy created by energy
dispersal (∆Ssurround)
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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
surroundi ngs
qsurroundi ngs
-H system
=
=
T
T
Can calc. that ∆Horxn = ∆Hosystem = -571.7 kJ
S
o
surroundi ngs
- (-571.7 kJ)(1000 J/kJ)
=
298.15 K
∆Sosurroundings = +1917 J/K
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Practice problem:
Determining Reaction Spontaneity
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• At 298K, the formation of ammonia has a negative ΔS°sys.
Calculate ΔSuniv, and state whether the reaction occurs
spontaneously at this temperature.
•
N2(g) + 3H2(g)  2NH3(g) ΔS°sys = -197J/K
ΔH°sys = ΔH°rxn
= [(2molNH3)(-45.9kJ/mol)] – [(3molH2)(0 kJ/mol) + (1mol N2)(0 kJ/mol)]
= -91.8kJ
ΔSsurr = - ΔH°sys = - (-91.8kJ)(1000 J)
T
(298K)(1kJ)
= 308 J/K
ΔSuniv = ΔS°sys + ΔSsurr = -197 J/K + 308 J/K = 111 J/K
ΔSuniv > 0, so the rxn is spontaneous at 298K
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
• The entropy of
the universe is
increasing, so
the reaction is
product-favored.
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Spontaneous or Not?
Remember that –∆H˚sys is proportional to ∆S˚surr
An exothermic process has ∆S˚surr > 0.
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Entropy and Gibbs free energy,
ΔG = ΔH - TΔS
What happens when one of the potential driving forces
behind a chemical reaction is favorable and the other is
not? We can answer this question by defining a new
quantity known as the Gibbs free energy (G) of the
system, which reflects the balance between these forces.
Favorable
ΔH° < 0
ΔS° > 0
Unfavorable
ΔH° > 0
ΔS° < 0
ΔG°< 0
ΔG° > 0
The entropy term is
therefore subtracted
from the enthalpy term
when calculating ΔG°
for a reaction
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When to use Gibbs