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Contents and Concepts
Learning Objectives
1. First Law of Thermodynamics
• First Law of Thermodynamics;
Enthalpy
Spontaneous Processes and Entropy
A spontaneous process is one that occurs by itself.
As we will see, the entropy of the system
increases in a spontaneous process.
2. Entropy and the Second Law of
Thermodynamics
3. Standard Entropies and the Third Law of
Thermodynamics
a. Define internal energy, state function, work,
and first law of thermodynamics.
b. Explain why the work done by the system
as a result of expansion or contraction
during a chemical reaction is -P∆V.
1
4
Free-Energy Concept
1. First Law of Thermodynamics;
Enthalpy (cont)
The quantity ∆H – T∆S can function as a criterion
for the spontaneity of a reaction at constant
temperature, T, and pressure, P. By defining a
quantity called the free energy, G = H – TS, we
find that ∆G equals the quantity ∆H – T∆S, so the
free energy gives us a thermodynamic criterion of
spontaneity.
•
•
•
c. Relate the change of internal energy,
∆U, and heat of reaction, q.
d. Define enthalpy, H.
e. Show how heat of reaction at constant
pressure, qp, equals the change of
enthalpy, ∆H.
4. Free Energy and Spontaneity
5. Interpretation of Free Energy
2
Free Energy and Equilibrium Constants
5
Spontaneous Processes and Entropy
2. Entropy and the Second Law of
Thermodynamics
The total free energy of the substances in a
reaction mixture decreases as the reaction
proceeds. As we discuss, the standard free-energy
change for a reaction is related to its equilibrium
constant.
•
•
•
6. Relating ∆G° to the Equilibrium Constant
7. Change of Free Energy with Temperature
•
3
a. Define spontaneous process.
b. Define entropy.
c. Relate entropy to disorder in a
molecular system (energy dispersal).
d. State the second law of
thermodynamics in terms of system plus
surroundings.
6
1
2. Entropy and the Second Law of
Thermodynamics (cont)
–
•
•
5. Interpretation of Free Energy
e. State the second law of
thermodynamics in terms of the system
only.
f. Calculate the entropy change for a
phase transition.
g. Describe how ∆H - T∆S functions as a
criterion of a spontaneous reaction.
–
–
a. Relate the free-energy change to
maximum useful work.
b. Describe how the free energy changes
during a chemical reaction.
7
3. Standard Entropies and the Third Law
of Thermodynamics
–
–
–
–
–
–
a. State the third law of thermodynamics.
b. Define standard entropy (absolute
entropy).
c. State the situations in which the
entropy usually increases.
d. Predict the sign of the entropy change
of a reaction.
e. Express the standard change of
entropy of a reaction in terms of standard
entropies of products and reactants.
f. Calculate ∆So for a reaction.
8
10
Free Energy and Equilibrium Constants
6. Relating ∆Go to the Equilibrium
Constant
–
–
–
a. Define the thermodynamic equilibrium
constant, K.
b. Write the expression for a
thermodynamic equilibrium constant.
c. Indicate how the free-energy change of
a reaction and the reaction quotient are
related.
11
Free-Energy Concept
4. Free Energy and Spontaneity
–
–
–
–
–
–
–
a. Define free energy, G.
b. Define the standard free-energy
change.
c. Calculate ∆Go from ∆Ho and ∆So.
d. Define the standard free energy of
formation, ∆Go.
e. Calculate ∆Go from standard free
energies of formation.
f. State the rules for using ∆Go as a
criterion for spontaneity
g. Interpret the sign of ∆Go.
6. Relating ∆Go to the Equilibrium
Constant (cont)
–
–
–
9
d. Relate the standard free-energy
change to the thermodynamic equilibrium
constant.
e. Calculate K from the standard freeenergy change (molecular equation).
f. Calculate K from the standard freeenergy change (net ionic equation).
12
2
Consider this Reaction
7. Change of Free Energy with
Temperature
–
–
–
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
a. Describe how ∆Go at a given
temperature (∆GoT) is approximately
related to ∆Ho and ∆So at that temperature.
b. Describe how the spontaneity or
nonspontaneity of a reaction is related to
each of the four possible combinations of
signs of ∆Ho and ∆So.
c. Calculate ∆Go and K at various
temperatures.
Concerning this reaction:
1. Does this reaction naturally occur as written?
2. Will the reaction mixture contain sufficient
amount of product at equilibrium?
13
16
Chapter 13 Chemical Thermodynamics
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Spontaneous Chemical and Physical Processes
Entropy and Disorder
Entropy and the Second Law of Thermodynamics
Standard-State Entropies of Reaction
The Third Law of Thermodynamics
Calculating Entropy Changes for Chemical Reactions
Gibbs Free Energy
The Effect of Temperature on the Free Energy of a Reaction
Beware of Oversimplification
Stand-State Free Energies of Reaction
Equilibria Expressed in Partial Pressures
Interpreting Stand-State Free Energy of Reaction Data
Relationship Between Free Energy and Equilibrium Constants
Temperature Dependence of Equilibrium Constants
Gibbs Free Energies of Formation and Absolute Entropies 14
We can answer these questions with
heat measurements only!!!
1. We can predict the natural direction.
2. We can determine the composition of the
mixture at equilibrium.
How?
17
Consider the Laws of Thermodynamics
1st Law: The change in internal energy of a
system ∆U, equals q + w
∆U = q + w
You can’t win
Thermodynamics and Equilibrium
Thermodynamics: Define
The study of the relationship between
heat and other forms of energy
involved in a chemical or physical
process
2nd Law: The total entropy of a system and its
surroundings increases for a spontaneous
process.
You cant break even.
How do we use this knowledge in chemistry?
3rd Law: A substance that is perfectly crystalline
at 0 K has an entropy of zero.
You can’t quit the game
15
18
3
1st Law: The change in internal energy of a
system ∆U, equals q + w
∆U = q + w
Showing work is P∆V
w = -F xh
∆U = Internal energy = Sum of kinetic energy and
potential energy of the
particle making up the system.
= - F x ∆V/A
= -F/A x ∆V
Kinetic Energy = Energy of motion of electrons, protons
and molecules.
= -P∆V
Potential Energy = Energy from chemical bonding of
atoms and from attraction between
molecules.
19
22
∆U is a state function:
P∆V = -92 J
Most often we are interested in the change:
∆U = Uf - Ui
q = Energy that moves in and out of a system
q = 165 J
w = Force x distance = F x d = P∆V
20
∆U = q + w = (+165) + (-92) = +73 J
23
Heat of Reaction and Internal Energy
Zn (s) + HCl (l) → ZnCl2 (aq) + H2 (g)
Heat vs. Work
21
24
4
• Here the system expands and evolves
heat from A to B.
Diagram and explain the change in internal energy
for the following reaction.
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)
q = -890.2 kJ
= P∆V = +(1.01 x 105 pa)(24.5 l)(2)
= +(1.01 x 105 pa)(2.45 x 10-3m3)(2)
= +4.95 kJ
w
Zn2+(aq) + 2Cl-(aq) + H2(g)
∆V is positive, so work is negative.
q=-
∆U
= -890.2 kJ + (+4.95 kJ) = 885.2 kJ
25
28
Enthalpy and Enthalpy Change
w =-
Enthalpy is defined as qp
Enthalpy = H = U + PV
All are state functions.
∆H = Hf - Hi
∆H = (Uf + PVf) – (Ui + PVi) = (Uf –Ui) + P(Vf –Vi)
∆U = qp - P∆V
∆H = (qp - P∆V) + P∆V = qp
∆Hof = Σn∆Hof (products) - Σm∆Hof (reactants)
26
w
=
-P∆V
=
-(1.01 x 105 pa)(24.5 l)
=
-1.01 x 105 pa)(24.5 x 10-3m3)
=
-2.47 x 103 J
=
-2.47 kJ
∆Hof = Standard enthalpy change (25oC)
29
∆Hof = Σn∆Hof (products) - Σm∆Hof (reactants)
Calculate ∆Hof for the reaction in slide 15
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
∆Hof =
=
=
=
for NH3 (g)
for CO2 (g)
for NH2CONH2
for H2O (l)
=
=
=
=
-45.9 kJ
-393.5 kJ
-319.2 kJ
-285.8 kJ
∆Ho = [(-319.2 – 285.8) – (-2 x 45.9 – 393.50)] kJ = -119.7
q
=
-152 kJ
∆U
=
-152 kJ + (-2.47 kJ) = -154.9 kJ
Since ∆Ho has a negative sign, heat is evolved
27
30
5
Spontaneity and Entropy
Definition of Spontaneous Process:
Physical or chemical process that occurs by itself.
Concept Check:
You have a sample of solid iodine
at room temperature. Later you notice
that the iodine has sublimed. What
can you say about the entropy change
of the iodine?
Why??
Give several spontaneous processes:
Still cannot predict spontaneity…..
31
Entropy and the 2nd Law of Thermodynamics
2nd
34
2nd Law: The total entropy of a system and its
surroundings increases for a spontaneous
process.
Law: The total entropy of a system and its
surroundings increases for a spontaneous
process.
Entropy = S = A thermodynamic quantity that is a
measure of how dispersed the energy
is among the different possible ways
that a system can contain energy.
Consider:
1.
A hot cup of coffee on the table
2.
Rock rolling down the side of a hill.
3.
Gas expanding.
32
4.
Stretching a rubber band.
Process occurs naturally as a result of energy dispersal
In the system.
∆S = entropy created + q/T
∆S > q/T
For a spontaneous process at a given temperature, the
change in entropy of the system is greater than the heat
35
divided by the absolute temperature.
Entropy and Molecular Disorder
Flask connected to an evacuated flask by a
valve or stopcock
∆S = Sf - Si
H2O (s) → H2O (l)
∆S = (63 – 41) J/K = 22 J/K
33
36
6
Figure 13.2
Entropy Change for a Phase Transition
∆S > q/T
(at equilibrium)
What processes can occur under phase
change at equilibrium?
Solid to liquid
Liquid to gas
Solid to gas
37
Was thought that in order to be spontaneous
A reaction had to be exothermic.
∆H < 0
Criterion for a Spontaneous Reaction
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
For water undergoing fusion.
∆S = ∆Hfus/T = 6.0 x 103 J/273 K
∆S = 22J/K
Is this reaction spontaneous?
That is, does it go left to right as written?
The heat of vaporization, ∆Hvap, of carbon tetrachloride,
CCl4, at 25 oC is 39.4 kJ/mol.
CCl4 (l) → CCl4 (g)
40
∆S > q/T
qp = ∆H at constant pressure
If we know ∆S and ∆H we can make prediction.
∆Hvap = 39.4 kJ/mol
If 1 mol of liquid carbon tetrachloride has an entropy of
216 J/K at 25 oC, what is the entropy of 1 mol of vapor
38
at equilibrium with the liquid at this temperature?
Solution:
∆S > qp/T = ∆H/T
∆H - T∆S < 0
∆H/T - ∆S < 0
(spontaneous rxn, constant T and P)
41
Standard Entropies and the Third Law of Thermodynamics
∆S = ∆Hvap/T = (39.4 x 103 J/mol)/ 298 K
= 132 j/(mol·K)
To determine the entropy of a substance you first
measure the heat absorbed by the substance by
warming it at various temperatures (heat capacity
at different temperatures)
Entropy of Vapor = (216 + 132) J/(mol·K)
= 348 J/(mol·K)
Determination of entropy is based on the 3rd Law.
3rd Law: A substance that is perfectly crystalline
at 0 K has an entropy of zero.
See page 742 – Ebbing and Gammon Houghton Mifflin,
9th Edition
39
42
7
Predict The Entropy sign for the following reactions:
a. C6H12O11 (s) → 2CO2 (g)+ C2H5OH (l)
b. 2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
c. CO (g) + H2O (g) → CO2 (g) + H2 (g)
d. Stretching a rubber band
Exercise 13.2 p 578
Standard entropy of methyl chloride, CHCl3, at various
Temperatures (approximate schematic graph)
43
46
Calculating ∆So for a reaction
Entropy Change for a Reaction
∆So may be positive for reactions with the following:
∆So = Σn∆So (products) - Σm∆So (reactants)
Calculate the entropy change for the following reaction
At 25 oC.
1. The reaction is one in which a molecule is
broken into two or more smaller molecules.
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
2. The reaction is one in which there is an increase
in moles of gas.
So 2 x 193
214
174
70
∆So = Σn∆So (products) - Σm∆So (reactants)
3. The process is one in which a solid changes to
a liquid or a liquid changes to a gas.
∆So= [(174 + 70) - (2x 193 + 214)] J/K= -356 J/K
44
See exercise 13.3 p 584 and problems 8-12 and 23-26
47
Free Energy Concept
Standard Entropy (Absolute Entropy): Entropy
value for the standard state of a species.
See Table B-13 p. B-17 and Table B-16 p. B-28
∆Ho – T∆So Can Serve as a criteria for Spontaneity
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
Entropy values of substances must be positive.
So must be >0 but Ho can be plus or minus
(Why?)
∆Ho = -119.7 kJ
∆So = -365 J/K = -0.365 kJ/K
∆Ho – T∆So = (-119.7 kJ) – (298 K) x (-0.365kJ/K
= -13.6 kJ
How about ionic species?
So for H3O+ is set at zero.
45
∆Ho – T∆So is a negativity quantity, from which we can
conclude that the reaction is spontaneous
under standard conditions.
48
8
∆Go = ∆Ho - T∆So
Free Energy and Spontaneity
∆Ho = Σn∆Ho (products) - Σm∆Ho (reactants)
Free Energy: Thermodynamic quantity defined by
the equation G = H - TS
= [2 x (-45.9) – 0]
= -91.8 kJ
∆So = ΣnHo (products) - ΣmHo (reactants)
∆G = ∆H - T∆S
= [2 x 192.7 – (191.6 + 3 x 130.60] J/K = -198.0 J/K
If you can show that ∆G for a reaction at a given
temperature and pressure is negative, you can
predict that the reaction will be spontaneous…
∆Go = ∆Ho - T∆So
∆Go = −91.8 kJ - (298 K)(-0.1980 kJ/K) = -32.8 kJ
See exercise 13.4-6 and problems 30-36
49
52
Standard Free Energies of Formation
Standard Free Energy Changes
∆Gof = The free energy change that occurs when
1 mol of substance is formed from its
elements in their most stable states at 1 atm
and at a specific temperature (normally 25 oC )
Standard Conditions:
1 atm pressure
1 atm partial pressure
1 M concentration
Temperature of 25 oC or 298 K
∆Go = Σn∆Gfo (products) - Σm∆Gfo (reactants)
Standard free energy is free-energy change that takes
place when reactants in their standard states are
converted to products in their standard states.
½ N2 (g) + 3/2 H2 (g) → NH3 (g)
∆Go for 2 mols of NH3 = -32.8 kJ
∆Go
=
∆Ho
-
T∆So
50
Calculate ∆Go from ∆Ho and ∆So
Calculate ∆Go for the following reaction:
C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)
∆Gfo = -174.9
0
2(-394.4)
3(-228.6)
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Use values of ∆Hfo and ∆So from tables 6.1 and 18.1
∆Go = Σn∆Gfo (products) - Σm∆Gfo (reactants)
N2 (g) + 3 H2 (g) → 2 NH3 (g)
So :
0
191.6
0
2 x (-45.9) kJ
3 x 130.6
2 x 192.7 J/K
for 1 mol of NH3 = -32.8 kJ/2 mol = -16.4 kJ 53
Calculation of ∆Go from Standard Free Energies of Formation
What is the standard free energy change ∆Go, for the
following reaction at 25 oC?
∆Hfo :
∆Go
∆Go = [2(-394.4) + 3(-228.6) – (- 174.9)] = -1299.7kJ
Calculate ∆Go for the following reaction:
CaCO3 (s) → CaO (s) + CO2 (g)
51
Do Exercise 13.7-8
∆Go = 130.9 kJ
See problems 45-46
54
9
∆Go as a Criterion for Spontaneity
∆Go = ∆Ho - T∆So
∆Go
1. When
is a large negative number (more
negative than about – 10 kJ), the reaction is spontaneous
as written, and reactants transform almost entirely into
products when equilibrium is reached.
2. When ∆Go is a large positive number (more
positive than about + 10 kJ), the reaction is not
spontaneous as written, and reactants transform almost
entirely into products when equilibrium is reached.
3. When ∆Go has a small positive or negative value(less
than about 10 kJ), the reaction mixture gives an
equilibrium mixture with significant amounts of both
55
reactants and products..
58
Interpretation of Free Energy
Interpreting the Sign of ∆Go
Theoretically, spontaneous reactions can be used to
obtain useful work.
Calculate ∆Ho and ∆Go for the following reaction
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
Combustion of gasoline
Battery generating electricity
Biochemical reaction in muscle tissue
Interpret the signs of ∆Ho and ∆Go
∆Hfo are as follows: KClO3 (s) = -397.7 kJ/mol
KCl (s) = -436.7 kJ/mol
O2 (g) = 0
∆Gfo
∆Gfo = 86.60 kJ/mol
Often reaction are not carried out in in a way that does
useful work.
are as follows: KClO3 (s) = -296.3 kJ/mol
KCl (s) = -408.8 kJ/mol
O2 (g) = 0
Reactants simply poured together..
56
2 KClO3 (s)
∆Hfo
2 x (-397.70)
∆Gfo 2 x (-296.3)
→ 2 KCl (s)
2 x (-436.7)
2 x (-408.8)
+
59
Interpretation of Free Energy
3 O2 (g)
In principle, if a reaction is carried out to obtain the
maximum useful work, no entropy is produced.
0 kJ
0 kJ
It can be shown that maximum useful work, wmax,
for a spontaneous reaction is ∆G,
Then:
∆Ho = [2 x (-436.7) – 2 x (-397.7)] kJ = -78 kJ
The free energy change is the maximum energy
available, or free to do useful work.
∆Go = [2 x (-408.8) – 2 x (-296.3)] kJ = -225 kJ
The reaction is exothermic, liberating 78 kJ of heat. The
large negative value of ∆Go indicates that the equilibrium
is mostly KCl and O2.
57
The sign of w (work) is defined so that a negative
value means work (energy is subtracted from the
system) obtained from the system). You can obtain
work from a reaction if its ∆G is negative.
60
10
Free energy change during reaction
Write Kp and Kc
• Consider again
N2(g) + 3H2(g) ⇄ 2 NH3(g)
Kc =
[NH3]2
[N2][H2]3
61
64
Free energy change during reaction
• In terms of partial pressures,
N2(g) + 3H2(g) ⇄ 2 NH3(g)
Kp =
[PNH3]2
[PN2][PH2]3
62
65
Free Energy and Equilibrium Constant
•
•
•
•
Very important relation is the relation between
free energy and the equilibrium constant.
Thermodynamic Equilibrium Constant- the equilibrium
constant in which the concentration of gases are
expressed in partial pressures in atmospheres,
whereas the concentration of solutes in liquid
are expressed in molarities.
Kc and Kp are related.
PV = nRT
P = [gas]RT
Kp=Kc × (RT)∆n
• ∆n = (number of moles of product gas) –
(number of moles of reactant gas)
• For this reaction, ∆n = -2.
N2(g) + 3H2(g) ⇄ 2 NH3(g)
K = Kc for reactions involving only liquid solutions
K = Kp for reactions involving only gases
63
66
11
Relating ∆Go to the Equilibrium Constant
• For this reaction, ∆G° = -33.0 kJ.
∆G = ∆Go + RT ln Q
N2(g) + 3H2(g) ⇄ 2 NH3(g)
∆G0 can be calculated from thermodynamic data
If you want ∆G for a none standard state use the
above equation.
• What does this value of ∆G° tell us
about the reaction?
At ∆G = 0 the reaction is at equilibrium
0 = ∆Go + RT ln K
Reaction should be spontaneous.
And Q = K the equilibrium constant
∆Go = -RT ln K
67
Problem: Find the equilibrium constant for the following
reaction:
Writing the Expression for a Thermodynamic
Equilibrium Constant
a. 2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
K=
[NH2CONH2]
P2NH3PCO2
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
∆Go for the reaction is -13.6 kJ
using ∆Go = ∆Ho - T∆So
0 = ∆Go + RT ln K
b. AgCl (s) → Ag+ (aq) + Cl- (aq)
o
-13.6 kJ
ln K = ∆G =
-8.31 J/(K·mol) x 298 K
-RT
K = e5.49 = 2.42 x 102
K = [Ag+][Cl-]
Exercises 13.9-10 and problems 58-62
70
68
Do Exercise 13.11-13 and problems 72-74
71
• What is the relationship between ∆G°
and K?
Table 13.2
∆G° = -RT lnK
• ∆G° = -RT lnK
– Using tabulated ∆G°ac values, ∆G°reaction
is calculated
– With ∆G°reaction (= ∆G°), K can be
calculated
K = e-∆G°/RT
69
72
12
• ∆G° = -RT lnK
• ∆G° =∆H° – T∆S°
• K must depend on temperature.
– We have already seen qualitatively how K
changes with temperature when we discussed
Le Châtelier’s principle in Chapter 10.
Understanding ∆Go as a criterion for Spontaneity
∆Go = -RT ln K
When K is <1, ln is neg, ∆Go is positive
• Consider this equilibrium
– 2NO2(g) ⇄ N2O4(g)
– NO2 is brown, N2O4 is colorless.
• When this equilibrium is cooled, the system
becomes colorless.
• When this equilibrium is heated, the system turns
dark brown.
When K is >1, ln is pos, ∆Go is negative
73
76
Table 13.3
2NO2(g) ⇄ N2O4(g)
a. No change
b. Increases
74
• ∆G° can increase with temperature
(∆H°>0 and ∆S°<0) with K also
increasing with temperature.
77
Change of Free Energy with Temperature
– Normally ∆G° decreases when K
increases.
Measuring ∆Go and K at different temperatures is difficult
• This is illustrated with this equilibrium
2CH4(g) ⇄ C2H6(g) + H2(g).
∆GTo = ∆Ho – T∆So (approximation for ∆GTo)
Table 13.4
Spontaneity and Temperature Change
75
78
13
• Instead of using ∆G°ac to calculate ∆G°, you
can use ∆G°f, the Gibbs free energy of
formation.
• Gibbs free energy of formation values are
tabulated in Appendix B.16.
aA + bB ⇄ cC + dD
0
(
∆G = [c ∆G 0f
)
C
(
+ d ∆G 0f
)
D
(
] − [a ∆G 0f
)
A
(
)
+ b ∆G 0f B ]
• Entropies of reaction can be calculated from
tabulated values of absolute entropies. These are
found in Appendix B.16.
aA + bB ⇄ cC + dD
79
∆S 0 = [cSC0 + dS D0 ] − [aS A0 + bS B0 ]
82
Calculation of ∆Go at Various Temperatures
Consider the following reaction:
CaCO3 (s) → CaO (s) + CO2 (g)
At 25 oC ∆Go = +130.9 and Kp = 1.1 x 10-23 atm
What do these values tell you about CaCO3?
What happens when the reaction is carried out at a
higher temperature?
Increase in temperature – increase in NO2
80
83
Calculating ∆Go and K at Various Temperatures
Consider the decomposition of dinitrogen tetroxide,N2O4,
to nitrogen dioxide, NO2:
N2O4 (g) → 2 NO2 (g)
a. What is ∆Go at 1000oC for the calcium carbonate reaction?
CaCO3 (s) → CaO (s) + CO2 (g)
How would you expect the spontaneity of the reaction
to behave with temperature change?
Is this reaction spontaneous at 1000oC and 1 atm?
Increase in temperature – increase in NO2
b. What is the value of Kp at 1000oC for this reaction?
What is the partial pressure of CO2?
81
84
14
Strategy for solution…
Where does the reaction change from spontaneous
to non-spontaneous?
a. Calculate ∆Ho and ∆So at 25 oC using standard
enthalpies of formation and standard entropies.
Then substitute into the equation for ∆Gfo.
∆Go = 0 = ∆Ho - T∆So
Solve for T;
b. Use the ∆Gfo value to find K (=Kp ) as in
Exercise 13.16.
T =
∆Ho
∆So
=
178.3 kJ
0.1590 kJ/K
T = 1121 K = 848 oC
85
a. From Table B-16 you have the following:
CaCO3 (s)
→
∆Hfo: -1206.9
So:
92.9
CaO (s) +
CO2 (g)
-635.1
-393.5 kJ
38.2
88
That’s it for Chapter 13
Quiz 2.
Explain how you are able to determine whether
a reaction is spontaneous.
213.7 J/K
∆Ho = [(-635.1 – 393.5) – (-1206.9)] = 178.3 kJ
∆So = [(38.2 + 213.7) – (92.9)] = 159.0 J/K
∆GTo = ∆Ho –T∆So
= 178.3kJ – (1273 K)(0.1590kJ/K) = -24.1 kJ
∆Go is negative – reaction is spontaneous
86
89
b. Substitute the values of ∆Go at 1273 K, which
equals -24.1 x 103 J, into the equation relating
ln K and ∆Go.
ln K =
-24.1 x 103 = 2.278
∆Go
=
-RT
-8.31 x 1273
K = Kp = e2.278 = 9.76
Kp = PCO2 = 9.76 atm
87
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