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Copyright 2011 Pearson Education, Inc.
Chapter 17
Free Energy
and
Thermodynamics
Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MACopyright 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
Endothermic process,
yet ice will
spontaneously melt
above 0 °C.
When a solid melts, the
particles have more
freedom of movement.
More freedom of motion
increases the
randomness of the
system. When systems
become more random,
energy is released. We
call this energy,
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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 systems15Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
W
These are energetically
equivalent states for the
expansion of a gas.
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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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This macrostate can be achieved through
several different arrangements of the particles
Macrostates → Microstates
17
These
microstates all
have the same
macrostate
So there are six
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
Therefore State C has
higher entropy than
either State A or State B
The macrostate with the
highest entropy also has the
greatest dispersal of energy
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
• Water vapor condensing
• Separation of oil and vinegar salad dressing
• Dissolving sugar in tea
• 2 PbO2(s) 2 PbO(s) + O2(g)
• 2 NH3(g) N2(g) + 3 H2(g)
• Ag+(aq) + Cl−(aq) AgCl(s)
DS is +
DS is −
DS is −
DS is +
DS is +
DS is +
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 process25Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Heat Flow, Entropy, and the 2nd Law
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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When ice is placed in
water, heat flows from
the water into the ice
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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!27Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Entropy Change in the System and
Surroundings
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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
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combustion is largely exothermic, so the entropy of
the surroundings should increase significantly
DHsystem = −2044 kJ, T = 25 ºC = 298 K
DSsurroundings, J/K
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DST, DH
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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.
DHsys = +66.4 kJ, T = 25 ºC = 298 K
DSsurr, J/K, DSreact + or −, DSuniverse + or −
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DST, DH
the major difference is that there are
fewer product molecules than
reactant molecules, so the DSreaction
is unfavorable and (−)
because DSsurroundings is
(−) it is unfavorable
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 equilibrium36Tro: Chemistry: A Molecular Approach, 2/e
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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.
39
Because DG is +, the reaction is not spontaneous
at this temperature. To make it spontaneous, we
need to increase the temperature.
DH = +95.7 kJ, DS = 142.2 J/K, T = 298 K
DG, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGT, DH, DS
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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.
Because DG is −, the reaction is spontaneous at
this temperature. To make it nonspontaneous, we
need to increase the temperature.
DH = −847.6 kJ and DS = −41.3 J/K, T = 298 K
DG, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGT, DH, DS
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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.
42
the temperature must be higher than 673K for
the reaction to be spontaneous
DH = +95.7 kJ, DS = +142.2 J/K, DG < 0
T, K
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
TDG, DH, DS
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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.
any temperature above 1775 C will make the
reaction nonspontaneous
DH = −847.6 kJ and DS = −41.3 J/K, DG > 0
T, C
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
TDG, DH, DS
2048 − 273 = 1775 C
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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 interestusually 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
SubstanceS°,
(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
SubstanceMolar
Mass
S°,
(J/mol∙K)
Ar (g) 39.948 154.8
NO (g) 30.006 210.8
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SubstanceMolar
Mass
S°,
(J/mol∙K)
CO (g) 28.01 197.7
C2H4 (g) 28.05 219.3
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Relative Standard Entropies
Dissolution
• Dissolved solids
generally have larger
entropy
• Distributing particles
throughout the mixture
SubstanceS°,
(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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Example 17.4: Calculate DS for the
reaction
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
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DS is +, as you would expect for a reaction with more gas
product molecules than reactant molecules
standard entropies from Appendix IIB
DS, J/K
Check:
Sol’n:
Conceptual
Plan:
Relationships:
Given:
Find:
DSSNH3, SO2, SNO, SH2O
Substance S, J/molK
NH3(g) 192.8
O2(g) 205.2
NO(g) 210.8
H2O(g) 188.8
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Practice – Calculate the DS for the reaction
2 H2(g) + O2(g) 2 H2O(g)
Substance S, J/molK
H2(g) 130.7
O2(g) 205.2
H2O(g) 188.8
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Example 17.4: Calculate DS for the
reaction
2 H2(g) + O2(g) 2 H2O(g)
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DS is −, as you would expect for a reaction with more
gas reactant molecules than product molecules
standard entropies from Appendix IIB
DS, J/K
Check:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DSSH2, SO2, SH2O
Substance S, J/molK
H2(g) 130.6
O2(g) 205.2
H2O(g) 188.8
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Calculating DG
• At 25 C
DGoreaction = SnDGo
f(products) - SnDGof(reactants)
• At temperatures other than 25 C
assuming the change in DHoreaction and DSo
reaction is negligible
DGreaction = DHreaction – TDSreaction
• or
DGtotal = DGreaction 1 + DGreaction 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.
59
because DG is −, the reaction is spontaneous at
this temperature, though less so than at 25 C
DH = −98.9 kJ, DS = −94.0 J/K, T = 398 K
DG, kJ
Answer:
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGT, DH, DS
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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 DH, kJ/mol S, J/mol
H2O2(g) −136.3 232.7
O2(g) 0 205.2
H2O(g) −241.8 188.8
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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)
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standard energies from Appendix IIB, T = 298 K
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDHf° Sº of prod & react DH, DSº
DH = 211.0 kJ, DS = −117.4 J/K, T = 298 K
DG, kJ
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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 reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)
63
standard free energies of formation from Appendix IIB
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDGf° of prod & react
Substance DGf°, kJ/mol
CH4(g) −50.5
O2(g) 0.0
CO2(g) −394.4
H2O(g) −228.6
O3(g) +163.2
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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)
Substance DG, kJ/mol
H2O2(g) −105.6
O2(g) 0
H2O(g) −228.6
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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)
65
standard free energies of formation from Appendix IIB
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDGf° of prod & react
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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:
Find:
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
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 x [C(s) + O2(g) CO2(g) ] DGº = 3(−394.4 kJ)
2 x [2 H2(g) + O2(g) 2 H2O(g)] DGº = 2(−457.1 kJ)
Given:
Find:
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
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 C(s) + 3 O2(g) 3 CO2(g) DGº = −1183.2 kJ
4 H2(g) + 2 O2(g) 4 H2O(g) DGº = −914.2 kJ
3 C(s) + 4 H2(g) C3H8(g) DGº = −23 kJ
67Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – Determine the free energy change in the
following reaction at 298 K
2 H2O(g) + O2(g) 2 H2O2(g)
Substance DG, kJ/mol
H2 (g) + O2(g) H2O2(g) −105.6
2 H2 (g) + O2(g) 2 H2O(g) −457.2
68Tro: Chemistry: A Molecular Approach, 2/e
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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:
H2(g) + O2(g) H2O2(g) DGº = −105.6 kJ
2 H2(g) + O2(g) 2 H2O(g) DGº = −457.2 kJ
DGº of 2 H2O(g) + O2(g) 2 H2O2(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: 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
69Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
70Tro: Chemistry: A Molecular Approach, 2/e
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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
71
DG° is less than DH°
because some of the
released heat energy
is lost to increase the
entropy of the
surroundings
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
72Tro: Chemistry: A Molecular Approach, 2/e
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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
73Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
74Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.75Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Example 17.9: Calculate DG at 298 K for
the reaction under the given conditions2 NO(g) + O2(g) 2 NO2(g) DGº = −71.2 kJ
76
non-standard conditions, DGº, T
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDG, PNO, PO2, PNO2
Substance P, atm
NO(g) 0.100
O2(g) 0.100
NO2(g) 2.00
Tro: Chemistry: A Molecular Approach, 2/e
Copyright 2011 Pearson Education, Inc.
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) 33
H2(g) 99
NH3(g) 2.0
77Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
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
78
non-standard conditions, DGº, T
DG, kJ
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DGDG, PNO, PO2, PNO2
Substance P, atm
N2(g) 33
H2(g) 99
NH3(g) 2.0
Tro: Chemistry: A Molecular Approach, 2/e
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Copyright 2011 Pearson Education, Inc.
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
79Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.80Tro: Chemistry: A Molecular Approach, 2/e
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Example 17.10: Calculate K at 298 K
for the reactionN2O4(g) 2 NO2(g)
81
standard free energies of formation from Appendix IIB
K
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
Substance DGf°, kJ/mol
N2O4(g) +99.8
NO2(g) +51.3
DGDGf of prod & react K
DGº = −RTln(K)
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
Practice – Estimate the equilibrium constant
for the given reaction at 700 K
N2(g) + 3 H2(g) 2 NH3(g) DGº = +46.4 kJ
82Tro: Chemistry: A Molecular Approach, 2/e
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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
83
DGº
K
Solution:
Conceptual
Plan:
Relationships:
Given:
Find:
DG K
DGº = −RTln(K)
Tro: Chemistry: A Molecular Approach, 2/e Copyright 2011 Pearson Education, Inc.
• 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
Why Is the Equilibrium Constant
Temperature Dependent?
84Tro: Chemistry: A Molecular Approach, 2/e
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