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11111
Chemistry 132 NT
I wish I would have a real tragic love affair and get so bummed out that I’d just quit my job and become a bum for a few years, because I was thinking about doing that anyway.
Jack Handey
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Thermodynamics and
EquilibriumChapter 18
Module 2
Sections 18.4, 18.5, 18.6 and 18.7
Photomicrograph of urea crystals under polarized light.
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Review
Calculating the entropy change for a phase transition
Predicting the sign of the entropy change of a reaction
Calculating So for a reaction
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Free Energy Concept
The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation G=H-TS.
This quantity gives a direct criterion for spontaneity of reaction.
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Free Energy and Spontaneity
Changes in H an S during a reaction result in a change in free energy, G , given by the equation
Thus, if you can show that G is negative at a given temperature and pressure, you can predict that the reaction will be spontaneous.
STHG
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Standard Free-Energy Change
The standard free energy change, Go, is the free energy change that occurs when reactants and products are in their standard states.
The next example illustrates the calculation of the standard free energy change, Go, from Ho and So.
ooo STHG
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A Problem To Consider
What is the standard free energy change, Go , for the following reaction at 25 oC? Use values of Hf
o and So, from Tables 6.2 and 18.1.
Place below each formula the values of Hfo
and So multiplied by stoichiometric coefficients.
)g(NH2)g(H3)g(N 322
So: 3 x 130.6191.5 2 x 193 J/K
Hfo: 00 2 x (-45.9) kJ
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88888
A Problem To Consider
What is the standard free energy change, Go , for the following reaction at 25 oC? Use values of Hf
o and So, from Tables 6.2 and 18.1.
You can calculate Ho and So using their respective summation laws.
)g(NH2)g(H3)g(N 322
)reactants(Hm)products(HnH of
of
o kJ 8.91kJ ]0)9.45(2[
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99999
A Problem To Consider
What is the standard free energy change, Go , for the following reaction at 25 oC? Use values of Hf
o and So, from Tables 6.2 and 18.1.
You can calculate Ho and So using their respective summation laws.
)g(NH2)g(H3)g(N 322
)reactants(Sm)products(SnS ooo J/K -197J/K )]6.13035.191(1932[
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1010101010
A Problem To Consider
What is the standard free energy change, Go , for the following reaction at 25 oC? Use values of Hf
o and So, from Tables 6.2 and 18.1.
Now substitute into our equation for Go. Note that So is converted to kJ/K.
)g(NH2)g(H3)g(N 322
kJ/K) 0.197K)( (298kJ 91.8
ooo STHG
kJ 33.1(see Exercise 18.6 and Problem 18.37)
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Standard Free Energies of Formation
The standard free energy of formation, Gfo,
of a substance is the free energy change that occurs when 1 mol of a substance is formed from its elements in their most stable states at 1 atm pressure and 25 oC.
By tabulating Gfo for substances, as in Table
18.2, you can calculate the Go for a reaction by using a summation law.
)reactants(Gm)products(GnG of
of
o
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1212121212
A Problem To Consider
Calculate Go for the combustion of 1 mol of ethanol, C2H5OH, at 25 oC. Use the standard free energies of formation given in Table 18.2.
)g(OH3)g(CO2)g(O3)l(OHHC 22252 Gf
o: -174.8 0 2(-394.4) 3(-228.6)kJ
Place below each formula the values of Gfo
multiplied by stoichiometric coefficients.
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1313131313
A Problem To Consider
Calculate Go for the combustion of 1 mol of ethanol, C2H5OH, at 25 oC. Use the standard free energies of formation given in Table 18.2.
)g(OH3)g(CO2)g(O3)l(OHHC 22252
You can calculate Go using the summation law.
)reactants(Gm)products(GnG of
of
o kJ )]8.174()6.228(3)4.394(2[Go
Gfo: -174.8 0 2(-394.4) 3(-228.6)kJ
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1414141414
A Problem To Consider
Calculate Go for the combustion of 1 mol of ethanol, C2H5OH, at 25 oC. Use the standard free energies of formation given in Table 18.2.
)g(OH3)g(CO2)g(O3)l(OHHC 22252
You can calculate Go using the summation law.
)reactants(Gm)products(GnG of
of
o kJ 8.1299Go
Gfo: -174.8 0 2(-394.4) 3(-228.6)kJ
(see Exercise 18.7 and Problem 18.41)
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Go as a Criteria for Spontaneity
The following rules are useful in judging the spontaneity of a reaction.
1. When Go is a large negative number (more negative than about –10 kJ), the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached.
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Go as a Criteria for Spontaneity
The following rules are useful in judging the spontaneity of a reaction.
2. When Go is a large positive number (more positive than about +10 kJ), the reaction is nonspontaneous as written, and reactants do not give significant amounts of product at equilibrium.
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Go as a Criteria for Spontaneity
The following rules are useful in judging the spontaneity of a reaction.
3. When Go is a small negative or positive value (less than about 10 kJ), the reaction gives an equilibrium mixture with significant amounts of both reactants and products.
(see Exercise 18.8 and Problem 18.45)
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Maximum Work
Often reactions are not carried out in a way that does useful work.
As a spontaneous precipitation reaction occurs, the free energy of the system decreases and entropy is produced, but no useful work is obtained.
In principle, if a reaction is carried out to obtain the maximum useful work, no entropy is produced.
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Maximum Work
Often reactions are not carried out in a way that does useful work.
It can be shown that the maximum useful work, wmax , for a spontaneous reaction is G.
The term free energy comes from this result.
Gmaxw
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2020202020
Free Energy Change During Reaction
As a system approaches equilibrium, the instantaneous change in free energy approaches zero.
Figure 18.9 illustrates the change in free energy during a spontaneous reaction.As the reaction proceeds, the free energy eventually reaches its minimum value.
At that point, G = 0, and the net reaction stops; it comes to equilibrium.
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Figure 18.9 Free-energy change during a spontaneous reaction.
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Free Energy Change During Reaction
As a system approaches equilibrium, the instantaneous change in free energy approaches zero.
Figure 18.10 illustrates the change in free energy during a nonspontaneous reaction.Note that there is a small decrease in free energy as the system goes to equilibrium.
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Figure 18.10 Free-energy change during a nonspontaneous reaction.
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Relating Go to the Equilibrium Constant
The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change, Go, by the following equation.
Here Q is the thermodynamic form of the reaction quotient.
QlnRTGG o
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2525252525
Relating Go to the Equilibrium Constant
The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change, Go, by the following equation.
G represents an instantaneous change in free energy at some point in the reaction approaching equilibrium.
QlnRTGG o
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2626262626
Relating Go to the Equilibrium Constant
The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change, Go, by the following equation.
At equilibrium, G=0 and the reaction quotient Q becomes the equilibrium constant K.
QlnRTGG o
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Relating Go to the Equilibrium Constant
The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change, Go, by the following equation.
At equilibrium, G=0 and the reaction quotient Q becomes the equilibrium constant K.
KlnRTG0 o
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Relating Go to the Equilibrium Constant
This result easily rearranges to give the basic equation relating the standard free-energy change to the equilibrium constant.
When K > 1 , the ln K is positive and Go is negative.
When K < 1 , the ln K is negative and Go is positive.
KlnRTGo
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A Problem To Consider
Find the value for the equilibrium constant, K, at 25 oC (298 K) for the following reaction. The standard free-energy change, Go, at 25 oC equals –13.6 kJ.
)l(OH (aq)CONHNH )g(CO)g(NH2 22223
Rearrange the equation Go=-RTlnK to give
RTG
Klno
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3030303030
A Problem To Consider
Find the value for the equilibrium constant, K, at 25 oC (298 K) for the following reaction. The standard free-energy change, Go, at 25 oC equals –13.6 kJ.
)l(OH (aq)CONHNH )g(CO)g(NH2 22223
Substituting numerical values into the equation,
49.5K 298K)J/(mol 31.8
J106.13Kln
3
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3131313131
A Problem To Consider
Find the value for the equilibrium constant, K, at 25 oC (298 K) for the following reaction. The standard free-energy change, Go, at 25 oC equals –13.6 kJ.
)l(OH (aq)CONHNH )g(CO)g(NH2 22223
Hence,249.5 1042.2eK
(see Exercises 18.10 and 18.11 and Problems 18.53, 18.55, and 18.57)
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Spontaneity and Temperature Change
All of the four possible choices of signs for Ho and So give different temperature behaviors for Go.
Ho So Go Description
– + – Spontaneous at all T
+ – + Nonspontaneous at all T
– – + or – Spontaneous at low T;
Nonspontaneous at high T
+ + + or – Nonspontaneous at low T;
Spontaneous at high T
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Calculation of Go at Various Temperatures
In this method you assume that Ho and So are essentially constant with respect to temperature.
You get the value of GTo at any temperature T
by substituting values of Ho and So at 25 oC into the following equation.
oooT STHG
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3434343434
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
Place below each formula the values of Hfo
and So multiplied by stoichiometric coefficients.
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3535353535
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
You can calculate Ho and So using their respective summation laws.
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3636363636
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
)reactants(Hm)products(HnH of
of
o kJ 3.178kJ)]9.1206()5.3931.635[(
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3737373737
)reactants(Sm)products(SnS ooo
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
K/J 0.159)]9.92()7.2132.38[(
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3838383838
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
Now you substitute Ho, So (=0.1590 kJ/K), and T (=298K) into the equation for Gf
o.
oooT STHG
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3939393939
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
Now you substitute Ho, So (=0.1590 kJ/K), and T (=298K) into the equation for Gf
o.
)K/kJ 1590.0)(K 298(kJ3.178GoT
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4040404040
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
Now you substitute Ho, So (=0.1590 kJ/K), and T (=298K) into the equation for Gf
o.
kJ 9.130GoT
So the reaction is nonspontaneous at 25 oC.
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4141414141
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
Now we’ll use 1000 oC (1273 K) along with our previous values for Ho and So.
)K/kJ 1590.0)(K 1273(kJ3.178GoT
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4242424242
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
Now we’ll use 1000 oC (1273 K) along with our previous values for Ho and So.
kJ 1.24GoT
So the reaction is spontaneous at 1000 oC.
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4343434343
A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
To determine the minimum temperature for spontaneity, we can set Gf
o =0 and solve for T.
o
o
S
HT
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A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
To determine the minimum temperature for spontaneity, we can set Gf
o =0 and solve for T.
)C 848( K 1121K/kJ 1590.0
kJ 3.178T o
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A Problem To Consider
Find the Go for the following reaction at 25 oC and 1000 oC. Relate this to reaction spontaneity.
)g(CO)s(CaO)s(CaCO 23
So: 38.292.9 213.7 J/K
Hfo: -635.1-1206.9 -393.5 kJ
Thus, CaCO3 should be thermally stable until its heated to approximately 848 oC.
(see Exercises 18.12 and 18.13 and Problems 18.59 and 18.61)
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Operational Skills
Calculating Go from Ho and So
Calculating Go from standard free energies of formation
Interpreting the sign of Go
Writing the expression for a thermodynamic equilibrium constantCalculating K from the standard free energy changeCalculating Go and K at various temperatures
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Key Equations
wqU VPw
Tq
S (For a spontaneous process)
Tq
S (For an equilibrium process)
)reactants(Sm)products(SnS ooo
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Key Equations
ooo STHG
)reactants(Gm)products(GnG of
of
o
QlnRTGG o
oooT STHG
KlnRTGo Klog RT 303.2Go or
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Homework
Chapter 18 Homework: collected at the first exam.
Review Questions: 7, 11, 16.Problems: 37, 41, 45, 49, 53, 57, 61, 83, 91.
Time for a few review questions
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