Transcript Chapter - Suffolk County Community College

Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 17
Free Energy
and
Thermodynamics
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright  2011 Pearson Education, Inc.
First Law of Thermodynamics
• You can’t win!
• First Law of Thermodynamics: Energy
cannot be created or destroyed
the total energy of the universe cannot change
though you can transfer it from one place to another
 DEuniverse = 0 = DEsystem + DEsurroundings
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First Law of Thermodynamics
• Conservation of Energy
• For an exothermic reaction, “lost” heat from the
•
system goes into the surroundings
Two ways energy is “lost” from a system
converted to heat, q
used to do work, w
• Energy conservation requires that the energy
change in the system equal the heat released +
work done
 DE = q + w
 DE = DH + PDV
• DE is a state function
internal energy change independent of how done
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The Energy Tax
• You can’t break even!
• To recharge a battery with 100 kJ of
useful energy will require more than
100 kJ
 because of the Second Law of
Thermodynamics
• Every energy transition results in a
“loss” of energy
 an “Energy Tax” demanded by nature
 and conversion of energy to heat which
is “lost” by heating up the surroundings
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Heat Tax
fewer steps
generally results
in a lower total
heat tax
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Thermodynamics and Spontaneity
• Thermodynamics predicts whether a process will
occur under the given conditions
processes that will occur are called spontaneous
nonspontaneous processes require energy input to go
• Spontaneity is determined by comparing the
chemical potential energy of the system before
the reaction with the free energy of the system after
the reaction
if the system after reaction has less potential energy
than before the reaction, the reaction is
thermodynamically favorable.
• Spontaneity ≠ fast or slow
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Comparing Potential Energy
The direction of
spontaneity can
be determined
by comparing
the potential
energy of the
system at the
start and the
end
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Reversibility of Process
• Any spontaneous process is irreversible because
there is a net release of energy when it proceeds
in that direction
 it will proceed in only one direction
• A reversible process will proceed back and forth
between the two end conditions
 any reversible process is at equilibrium
 results in no change in free energy
• If a process is spontaneous in one direction, it
must be nonspontaneous in the opposite direction
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Thermodynamics vs. Kinetics
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Diamond → Graphite
Graphite is more stable than diamond, so the conversion
of diamond into graphite is spontaneous – but don’t worry,
it’s so slow that your ring won’t turn into pencil lead in
your lifetime (or through many of your generations)
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Spontaneous Processes
• Spontaneous processes occur because they
•
release energy from the system
Most spontaneous processes proceed from a
system of higher potential energy to a system at
lower potential energy
 exothermic
• But there are some spontaneous processes that
proceed from a system of lower potential energy to
a system at higher potential energy
 endothermic
• How can something absorb potential energy, yet
have a net release of energy?
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Melting Ice
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Melting is an
When
a solid process,
melts, the
Endothermic
particles
have more
yet ice will
freedom
of movement.
spontaneously
melt
above 0 °C.
More freedom of motion
increases the
randomness of the
system. When systems
become more random,
energy is released. We
call this energy,
entropy
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Factors Affecting Whether a
Reaction Is Spontaneous
• There are two factors that determine whether a
reaction is spontaneous. They are the
enthalpy change and the entropy change of
the system
• The enthalpy change, DH, is the difference in
the sum of the internal energy and PV work
energy of the reactants to the products
• The entropy change, DS, is the difference in
randomness of the reactants compared to the
products
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Enthalpy Change
 DH generally measured in kJ/mol
• Stronger bonds = more stable molecules
• A reaction is generally exothermic if the bonds in the
products are stronger than the bonds in the
reactants
 exothermic = energy released, DH is negative
• A reaction is generally endothermic if the bonds in
the products are weaker than the bonds in the
reactants
 endothermic = energy absorbed, DH is positive
• The enthalpy change is favorable for exothermic
reactions and unfavorable for endothermic reactions
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Entropy
• Entropy is a thermodynamic function
that increases as the number of
energetically equivalent ways of
arranging the components increases, S
 S generally J/mol
• S = k ln W
k = Boltzmann Constant = 1.38 x 10−23 J/K
W is the number of energetically equivalent
ways a system can exist
 unitless
• Random systems require less energy
than ordered systems
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W
These are energetically
equivalent states for the
expansion of a gas.
It doesn’t matter, in terms
of potential energy,
whether the molecules
are all in one flask, or
evenly distributed
But one of these states is
more probable than the
other two
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Macrostates → Microstates
These
microstates all
have the same
macrostate
Thisare
macrostate
can be achieved through
So there
six
several
different arrangements of the particles
different
particle
arrangements
that result in the
same macrostate
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Macrostates and Probability
There is only one possible
arrangement that gives State
A and one that gives State B
There are six possible
arrangements that give State C
The
macrostate
with
the
Therefore
State
C has
highest
entropy
also than
has the
higher
entropy
greatest
dispersal
energy
either State
A orofState
B
There is six times the
probability of having the
State C macrostate than
either State A or State B
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Changes in Entropy, DS
 DS = Sfinal − Sinitial
• Entropy change is favorable when the result is a
more random system
 DS is positive
• Some changes that increase the entropy are
reactions whose products are in a more random
state
solid more ordered than liquid more ordered than gas
reactions that have larger numbers of product
molecules than reactant molecules
increase in temperature
solids dissociating into ions upon dissolving
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Increases in Entropy
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DS
• For a process where the final condition is more
random than the initial condition, DSsystem is
positive and the entropy change is favorable
for the process to be spontaneous
• For a process where the final condition is more
orderly than the initial condition, DSsystem is
negative and the entropy change is
unfavorable for the process to be
spontaneous
 DSsystem = DSreaction = Sn(S°products) − Sn(S°reactants)
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Entropy Change in State Change
• When materials change state, the number of
macrostates it can have changes as well
the more degrees of freedom the molecules have,
the more macrostates are possible
solids have fewer macrostates than liquids, which
have fewer macrostates than gases
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Entropy Change and State Change
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Practice – Predict whether DSsystem is + or −
for each of the following
• A hot beaker burning your fingers DS is +
• Water vapor condensing DS is −
• Separation of oil and vinegar salad dressing DS is −
DS is +
• Dissolving sugar in tea
• 2 PbO2(s)  2 PbO(s) + O2(g) DS is +
• 2 NH3(g)  N2(g) + 3 H2(g) DS is +
• Ag+(aq) + Cl−(aq)  AgCl(s) DS is −
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The 2nd Law of Thermodynamics
• The 2nd Law of Thermodynamics says that the
total entropy change of the universe must be
positive for a process to be spontaneous
for reversible process DSuniv = 0
for irreversible (spontaneous) process DSuniv > 0
• DSuniverse = DSsystem + DSsurroundings
• If the entropy of the system decreases, then the
entropy of the surroundings must increase by a
larger amount
when DSsystem is negative, DSsurroundings must be
positive and big for a spontaneous process
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Heat Flow, Entropy, and the 2nd Law
When ice is placed in
water, heat flows from
the water into the ice
According to the 2nd Law,
heat must flow from
water to ice because it
results in more dispersal
of heat. The entropy of
the universe increases.
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Heat Transfer and Changes in
Entropy of the Surroundings
• The 2nd Law demands that the entropy of the
•
•
•
universe increase for a spontaneous process
Yet processes like water vapor condensing are
spontaneous, even though the water vapor is more
random than the liquid water
If a process is spontaneous, yet the entropy change
of the process is unfavorable, there must have
been a large increase in the entropy of the
surroundings
The entropy increase must come from heat released
by the system – the process must be exothermic!
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Entropy Change in the System and
Surroundings
When the entropy
change in system is
unfavorable (negative),
the entropy change in
the surroundings must
be favorable (positive),
and large to allow the
process to be
spontaneous
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Heat Exchange and DSsurroundings
• When a system process is exothermic, it adds heat to
•
•
the surroundings, increasing the entropy of the
surroundings
When a system process is endothermic, it takes heat
from the surroundings, decreasing the entropy of the
surroundings
The amount the entropy of the surroundings
changes depends on its
original temperature
 the higher the original
temperature, the less
effect addition or
removal of heat has
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Temperature Dependence of DSsurroundings
• When heat is added to surroundings that are
•
cool it has more of an effect on the entropy than
it would have if the surroundings were already
hot
Water freezes spontaneously below 0 °C
because the heat released on freezing
increases the entropy of the surroundings
enough to make DS positive
above 0 °C the increase in entropy of the
surroundings is insufficient to make DS positive
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Quantifying Entropy Changes in
Surroundings
• The entropy change in the surroundings is
proportional to the amount of heat gained or lost
qsurroundings = −qsystem
• The entropy change in the surroundings is also
•
inversely proportional to its temperature
At constant pressure and temperature, the
overall relationship is
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Example 17.2a: Calculate the entropy change of the
surroundings at 25ºC for the reaction below
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DHrxn = −2044 kJ
Given:
Find:
DHsystem = −2044 kJ, T = 25 ºC = 298 K
DSsurroundings, J/K
Conceptual
Plan:
T, DH
DS
Relationships:
Solution:
Check: combustion is largely exothermic, so the entropy of
the surroundings should increase significantly
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Practice – The reaction below has DHrxn = +66.4 kJ at
25 °C. (a) Determine the Dssurroundings, (b) the sign of
DSsystem, and (c) whether the process is spontaneous
2 O2(g) + N2(g)  2 NO2(g)
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Practice – The reaction 2 O2(g) + N2(g)  2 NO2(g) has DHrxn
= +66.4 kJ at 25 °C. Calculate the entropy change of the
surroundings. Determine the sign of the entropy change in
the system and whether the reaction is spontaneous.
Given: DHsys = +66.4 kJ, T = 25 ºC = 298 K
Find:
DSsurr, J/K, DSreact + or −, DSuniverse + or −
Conceptual
Plan:
Relationships:
Solution:
because DSsurroundings is
(−) it is unfavorable
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T, DH
DS
the major difference is that there are
fewer product molecules than
reactant molecules, so the DSreaction
is unfavorable and (−)
both DSsurroundings and DSreaction are (−),
DSuniverse is (−) and the process is
nonspontaneous
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Gibbs Free Energy and Spontaneity
• It can be shown that −TDSuniv = DHsys−TDSsys
• The Gibbs Free Energy, G, is the maximum
amount of work energy that can be released to the
surroundings by a system
 for a constant temperature and pressure system
 the Gibbs Free Energy is often called the Chemical
Potential because it is analogous to the storing of energy
in a mechanical system

•
DGsys = DHsys−TDSsys
Because DSuniv determines if a process is
spontaneous, DG also determines spontaneity
 DSuniv is + when spontaneous, so DG is −
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Gibbs Free Energy, DG
• A process will be spontaneous when DG is
negative
 DG will be negative when
 DH is negative and DS is positive
exothermic and more random
 DH is negative and large and DS is negative but small
 DH is positive but small and DS is positive and large
or high temperature
• DG will be positive when DH is + and DS is −
never spontaneous at any temperature
• When DG = 0 the reaction is at equilibrium
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DG, DH, and DS
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Free Energy Change and Spontaneity
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Example 17.3a: The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
DH = +95.7 kJ and DS = +142.2 J/K at 25 °C.
Calculate DG and determine if it is spontaneous.
Given:
Find:
DH = +95.7 kJ, DS = 142.2 J/K, T = 298 K
DG, kJ
Conceptual
Plan:
T, DH, DS
DG
Relationships:
Solution:
Answer: Because DG is +, the reaction is not spontaneous
at this temperature. To make it spontaneous, we
need to increase the temperature.
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Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.
Calculate DG and determine if it is spontaneous.
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Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.
Calculate DG and determine if it is spontaneous.
Given: DH = −847.6 kJ and DS = −41.3 J/K, T = 298 K
Find: DG, kJ
Conceptual
Plan:
T, DH, DS
DG
Relationships:
Solution:
Answer: Because DG is −, the reaction is spontaneous at
this temperature. To make it nonspontaneous, we
need to increase the temperature.
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Example 17.3b: The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
DH = +95.7 kJ and DS = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
Given: DH = +95.7 kJ, DS = +142.2 J/K, DG < 0
Find:
T, K
Conceptual
Plan:
DG, DH, DS
T
Relationships:
Solution:
Answer: the temperature must be higher than 673K for
the reaction to be spontaneous
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Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K.
Calculate the maximum temperature it will be spontaneous.
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Practice – The reaction Al(s) + Fe2O3(s)  Fe(s) + Al2O3(s) has
DH = −847.6 kJ and DS = −41.3 J/K at 25 °C.
Determine the maximum temperature it is spontaneous.
Given:
Find:
DH = −847.6 kJ and DS = −41.3 J/K, DG > 0
T, C
Conceptual
Plan:
DG, DH, DS
T
Relationships:
Solution:
2048 − 273 = 1775 C
Answer: any temperature above 1775 C will make the
reaction nonspontaneous
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Standard Conditions
• The standard state is the state of a
material at a defined set of conditions
• Gas = pure gas at exactly 1 atm pressure
• Solid or Liquid = pure solid or liquid in its
most stable form at exactly 1 atm pressure
and temperature of interest
usually 25 °C
• Solution = substance in a solution with
concentration 1 M
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The 3rd Law of Thermodynamics:
Absolute Entropy
• The absolute entropy of a
•
substance is the amount of
energy it has due to dispersion of
energy through its particles
The 3rd Law states that for a
perfect crystal at absolute zero,
the absolute entropy = 0 J/mol∙K
 therefore, every substance that is not
a perfect crystal at absolute zero has
some energy from entropy
 therefore, the absolute entropy of
substances is always +
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Standard Absolute Entropies
• S°
• Extensive
• Entropies for 1 mole of a substance at 298 K
for a particular state, a particular allotrope,
particular molecular complexity, a particular
molar mass, and a particular degree of
dissolution
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Standard Absolute Entropies
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Relative Standard Entropies:
States
• The gas state has a larger entropy than the
•
liquid state at a particular temperature
The liquid state has a larger entropy than the
solid state at a particular temperature
Substance
S°,
(J/mol∙K)
H2O (l)
70.0
H2O (g)
188.8
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Relative Standard Entropies:
Molar Mass
• The larger the molar
•
mass, the larger the
entropy
Available energy states
more closely spaced,
allowing more dispersal
of energy through the
states
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Relative Standard Entropies:
Allotropes
• The less constrained
•
the structure of an
allotrope is, the
larger its entropy
The fact that the
layers in graphite
are not bonded
together makes it
less constrained
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Relative Standard Entropies:
Molecular Complexity
• Larger, more complex
•
molecules generally
have larger entropy
More available energy
states, allowing more
dispersal of energy
through the states
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Molar
S°,
Substance
Mass (J/mol∙K)
CO
Ar (g)
(g) 39.948
28.01
154.8
197.7
CNO
(g)
2H4(g)
30.006
28.05
210.8
219.3
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Relative Standard Entropies
Dissolution
• Dissolved solids
•
generally have larger
entropy
Distributing particles
throughout the mixture
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Substance
S°,
(J/mol∙K)
KClO3(s)
143.1
KClO3(aq)
265.7
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The Standard Entropy Change, DS
• The standard entropy change is the difference
in absolute entropy between the reactants and
products under standard conditions
DSºreaction = (∑npSºproducts) − (∑nrSºreactants)
remember: though the standard enthalpy of
formation, DHf°, of an element is 0 kJ/mol, the
absolute entropy at 25 °C, S°, is always positive
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Substance
S, J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
Given: standard entropies from Appendix IIB
188.8
Example 17.4: Calculate DS for the
reaction
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Find: DS, J/K
Conceptual
SNH3, SO2, SNO, SH2O
Plan:
Relationships:
DS
Sol’n:
Check:
DS is +, as you would expect for a reaction with more gas
product molecules than reactant molecules
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Practice – Calculate the DS for the reaction
2 H2(g) + O2(g)  2 H2O(g)
Substance
H2(g)
O2(g)
H2O(g)
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S, J/molK
130.7
205.2
188.8
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Example 17.4: Calculate DS for the
reaction
2 H2(g) + O2(g)  2 H2O(g)
Substance
S, J/molK
H2(g)
130.6
O2(g)
205.2
H2O(g)
Given: standard entropies from Appendix IIB
Find: DS, J/K
Conceptual
SH2, SO2, SH2O
Plan:
188.8
DS
Relationships:
Solution:
Check:
DS is −, as you would expect for a reaction with more
gas reactant molecules than product molecules
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Calculating DG
• At 25 C
DGoreaction = SnDGof(products) - SnDGof(reactants)
• At temperatures other than 25 C
 assuming the change in DHoreaction and DSoreaction is
negligible
• or
DGreaction = DHreaction – TDSreaction
DGtotal = DGreaction 1 + DGreaction 2 + ...
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Example 17.6: The reaction SO2(g) + ½ O2(g)  SO3(g) has
DH = −98.9 kJ and DS = −94.0 J/K at 25 °C.
Calculate DG at 125 C and determine if it is more or less
spontaneous than at 25 °C with DG° = −70.9 kJ/mol SO3.
Given:
Find:
DH = −98.9 kJ, DS = −94.0 J/K, T = 398 K
DG, kJ
Conceptual
Plan:
T, DH, DS
DG
Relationships:
Solution:
Answer: because DG is −, the reaction is spontaneous at
this temperature, though less so than at 25 C
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g)  2 H2O2(g)
Substance
H2O2(g)
O2(g)
H2O(g)
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DH, kJ/mol
−136.3
0
−241.8
60
S, J/mol
232.7
205.2
188.8
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g)  2 H2O2(g)
Given:
Find:
standard
energies
from
Appendix
= 298
DH = 211.0
kJ, DS
= −117.4
J/K,IIB,
T = T298
K K
DG,
kJ
DG, kJ
Conceptual DH ° Sº of prod & react
f
Plan:
Relationships:
DH, DSº
DG
Solution:
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Standard Free Energies of Formation
• The free energy of formation (DGf°) is the change
in free energy when 1 mol of a compound forms
from its constituent elements in their standard
states
• The free energy
of formation of
pure elements in
their standard
states is zero
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Example 17.7: Calculate DG at 25 C
for the reaction
CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g)
Substance
CH4(g)
O2(g)
CO2(g)
H2O(g)
O3(g)
DGf°, kJ/mol
−50.5
0.0
−394.4
−228.6
+163.2
Given: standard free energies of formation from Appendix IIB
Find: DG, kJ
Conceptual
Plan:
Relationships:
DGf° of prod & react
DG
Solution:
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g)  2 H2O2(g)
DG, kJ/mol
−105.6
0
−228.6
Substance
H2O2(g)
O2(g)
H2O(g)
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g)  2 H2O2(g)
Given: standard free energies of formation from Appendix IIB
Find: DG, kJ
Conceptual
Plan:
Relationships:
DGf° of prod & react
DG
Solution:
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DG Relationships
• If a reaction can be expressed as a series of
reactions, the sum of the DG values of the
individual reaction is the DG of the total reaction
 DG is a state function
• If a reaction is reversed, the sign of its DG value
•
reverses
If the amount of materials is multiplied by a factor,
the value of the DG is multiplied by the same
factor
 the value of DG of a reaction is extensive
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Example 17.8: Find DGºrxn for the following reaction
using the given equations
3 C(s) + 4 H2(g)  C3H8(g)
Given: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) DGº = −2074 kJ
C(s) + O2(g)  CO2(g)
DGº = −394.4 kJ
2 H2(g) + O2(g)  2 H2O(g)
DGº = −457.1 kJ
Find: DGº of 3 C(s) + 4 H2(g)  C3H8(g)
Rel: Manipulate the given reactions so they add up to the
reaction you wish to find. The sum of the DGº’s is the
DGº of the reaction you want to find.
Solution: 3 CO2(g) + 4 H2O(g)  C3H8(g) + 5 O2(g) DGº = +2074 kJ
3 xC(s)
[C(s)
+ 3+ O
O2(g)
 3CO
CO
DGº
DGº==3(−394.4
−1183.2 kJ
kJ)
2(g) 
2(g)
2(g)]
2H
4
x 2(g)
[2 H+2(g)
2O
+2(g)
O2(g)

4H
22H
O2(g)
O(g)]
DGº
DGº
==
2(−457.1
−914.2 kJ
kJ)
3 C(s) + 4 H2(g)  C3H8(g)
DGº = −23 kJ
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g)  2 H2O2(g)
Substance
H2 (g) + O2(g)  H2O2(g)
2 H2 (g) + O2(g)  2 H2O(g)
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DG, kJ/mol
−105.6
−457.2
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Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g)  2 H2O2(g)
Given: H2(g) + O2(g)  H2O2(g)
2 H2(g) + O2(g)  2 H2O(g)
Find: DGº of 2 H2O(g) + O2(g)  2 H2O2(g)
DGº = −105.6 kJ
DGº = −457.2 kJ
Rel: Manipulate the given reactions so they add up to the
reaction you wish to find. The sum of the DGº’s is the
DGº of the reaction you want to find.
Solution: 2 H2(s) + 2 O2(g)  2 H2O2(g)
DGº = −211.2 kJ
2 H2O(g)  2 H2(g) + O2(g)
DGº = +457.2 kJ
2 H2O(g) + O2(g)  2 H2O2(g)
DGº = +246.0 kJ
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What’s “Free” About Free Energy?
• The free energy is the maximum amount of
•
•
energy released from a system that is available
to do work on the surroundings
For many exothermic reactions, some of the
heat released due to the enthalpy change goes
into increasing the entropy of the surroundings,
so it is not available to do work
And even some of this free energy is generally
lost to heating up the surroundings
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Free Energy of an Exothermic Reaction
•
•
•
•
C(s, graphite) + 2 H2(g) → CH4(g)
DH°rxn = −74.6 kJ = exothermic
DS°rxn = −80.8 J/K = unfavorable
DG°rxn = −50.5 kJ = spontaneous
DG° is less than DH°
because some of the
released heat energy
is lost to increase the
entropy of the
surroundings
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Free Energy and Reversible Reactions
• The change in free energy is a theoretical limit
as to the amount of work that can be done
• If the reaction achieves its theoretical limit, it is a
reversible reaction
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Real Reactions
• In a real reaction, some of the free energy is
“lost” as heat
if not most
• Therefore, real reactions are irreversible
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DG under Nonstandard Conditions
 DG = DG only when the reactants and products


are in their standard states
 their normal state at that temperature
 partial pressure of gas = 1 atm
 concentration = 1 M
Under nonstandard conditions, DG = DG + RTlnQ
 Q is the reaction quotient
At equilibrium DG = 0
 DG = −RTlnK
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Example 17.9: Calculate DG at 298 K for
the reaction under the given conditions
2 NO(g) + O2(g)  2 NO2(g) DGº = −71.2 kJ
Substance
P, atm
NO(g)
O2(g)
NO2(g)
0.100
0.100
2.00
Given: non-standard conditions, DGº, T
Find: DG, kJ
Conceptual
Plan:
Relationships:
DG, PNO, PO2, PNO2
DG
Solution:
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Practice – Calculate DGrxn for the given
reaction at 700 K under the given conditions
N2(g) + 3 H2(g)  2 NH3(g) DGº = +46.4 kJ
Substance
N2(g)
H2(g)
P, atm
33
99
NH3(g)
2.0
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Practice – Calculate DGrxn for the given
reaction at 700 K under the given
conditions
N2(g) + 3 H2(g)  2 NH3(g) DGº = +46.4 kJ
Substance
P, atm
N2(g)
H2(g)
33
99
NH3(g)
2.0
Given: non-standard conditions, DGº, T
Find: DG, kJ
Conceptual
Plan:
Relationships:
DG, PNO, PO2, PNO2
DG
Solution:
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DGº and K
• Because DGrxn = 0 at equilibrium, then
•
DGº = −RTln(K)
When K < 1, DGº is + and the reaction is
spontaneous in the reverse direction under
standard conditions
 nothing will happen if there are no products yet!
• When K > 1, DGº is − and the reaction is
•
spontaneous in the forward direction under
standard conditions
When K = 1, DGº is 0 and the reaction is at
equilibrium under standard conditions
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Example 17.10: Calculate K at 298 K
for the reaction
N2O4(g)  2 NO2(g)
Substance DGf°, kJ/mol
N2O4(g)
+99.8
NO2(g)
+51.3
Given: standard free energies of formation from Appendix IIB
Find: K
Conceptual
Plan:
Relationships:
DGf of prod & react
DG
K
DGº = −RTln(K)
Solution:
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Practice – Estimate the equilibrium constant
for the given reaction at 700 K
N2(g) + 3 H2(g)  2 NH3(g) DGº = +46.4 kJ
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Practice – Estimate the Equilibrium Constant
for the given reaction at 700 K
N2(g) + 3 H2(g)  2 NH3(g) DGº = +46.4 kJ
Given: DGº
Find: K
Conceptual
Plan:
Relationships:
DG
K
DGº = −RTln(K)
Solution:
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Why Is the Equilibrium Constant
Temperature Dependent?
• Combining these two equations
DG° = DH° − TDS°
DG° = −RTln(K)
• It can be shown that
• This equation is in the form y = mx + b
• The graph of ln(K) versus inverse T is a
straight line with slope
and y-intercept
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