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Chemistry
Third Edition
Julia Burdge
Lecture PowerPoints
Chapter 18
Entropy, Free Energy,
and Equilibrium
Copyright © 2012, The McGraw-Hill Compaies, Inc. Permission required for reproduction or display.
Thermodynamics
Energy conversion
Conversion of the chemical energy of fuels into heat energy of
combustion, and this heat is in turn converted to mechanical or
electrical work.
Aim f the this chapter
- To understand laws of thermodynamics
- How to use thermodynamic quantities to
determine whether or a process is
spontaneous (or energy efficient) or not?
CHAPTER
18
Entropy, Free Energy, and Equilibrium
18.1
18.2
18.3
18.4
18.5
18.6
18.7
Spontaneous Processes
Entropy
Entropy Changes in a System
Entropy Changes in the Universe
Predicting Spontaneity
Free Energy and Chemical Equilibrium
Thermodynamics in Living Systems (Omitted)
4
18.1
Spontaneous Processes
Topics
Spontaneous Processes
5
18.1
Spontaneous Processes
Spontaneous Processes
A process that does occur under a specific set of conditions is
called a spontaneous process.
One that does not occur under a specific set of conditions is
called nonspontaneous.
6
18.1
Spontaneous Processes
Spontaneous Processes
7
18.1
Spontaneous Processes
Spontaneous Processes
These exothermic reactions are spontaneous at room
temperature:
8
• Spontaneous Processes
H2(g) + 1/2O2(g)
H2O(l)
= -58.5 kJ/mole
This exothermic reaction is not spontaneous at temperatures above 0°C:
- Enthalpy change
- Entropy change
18.1
Spontaneous Processes
Spontaneous Processes
So, while exothermicity favors spontaneity, it cannot be the
sole factor that determines spontaneity.
The other factor which tells about spontaneity is Entropy (S).
10
What is Entropy?
• Entropy (S): is a
thermodynamic quantity
that measures the disorder
of a system.
In general, a greater
disorder means a greater
entropy, and a greater order
means a smaller entropy.
Melting of ice is explained in
terms of entropy change
Ordered
Disordered
H2O(S)
H2O(l)
= 6.01 kJ/mole
18.3
Entropy Changes in a System
Topics
Qualitatively Predicting the Sign of Ssys
Determination of Standard Entropy, S°
13
Qualitatively Predicting the Sign of Ssys
(Sometimes it’s useful just to know the sign
of S°rxn).
We must ask few questions while predicting the
sign of Entropy…
If the volume is increasing and the system is
becoming more disordered?
Is the system more spreading?
Is the system has more probability?
If the answer is yes, the entropy of system is increasing.
An increase in Entropy favors the reaction to occur!
18.3
Entropy Changes in a System
Qualitatively Predicting the Sign of Ssys
Several processes that lead to an increase in entropy are
• Melting
• Vaporization or sublimation
• Temperature increase
• Reaction resulting in a greater number of gas molecules
16
Volume change
Increase in volume increases disorder, and hence entropy
Entropy Change in a System
• One can predict qualitatively whether the entropy of a system
increases (ΔS > 0) or decreases (ΔS < 0).
(a) Melting:
ΔSsys > 0
solid
liquid
Particles are confined to fixed
positions. They are highly
ordered.
Became less ordered compared to the
solid state.
Ssolid < Sliquid
Entropy Change in a System
(b) Vaporization:
ΔSsys > 0
liquid
gas
Sliquid << Sgas
Particles in the gaseous phase
are highly disordered as
compared to solid and liquid
phases.
In general…
S(s)
<
S(l)
<<
S(g)
Entropy Change in a System
(c) Dissolving:
ΔSsys > 0
pure substance
When the highly ordered particles of the
pure substance and the ordered solvent
molecules are mixed, the order in both
are disrupted, and hence entropy
increases.
solution
18.3
Entropy Change in a System
Translational motion
(d) Heating:
Heating always increases Vibrational motion
both the energy and the
Rotational motion
entropy of a system.
ΔSsys > 0
Slower temp. <
Shigher temp.
Entropy Change in a System
(e) Chemical reactions:
A reaction that results in an increase in the number of
moles of gas always increases the entropy of the system.
(g)
(g)
(g)
ΔSsys > 0
Sfewer moles
<
Smore moles
Smaller
(f) Molar mass:
Ssmaller molar mass
<
Slarger molar mass
Bigger
18.3
Molecular complexity (F2 vs. O3)
Translational motion
Vibrational motion
Rotational motion
25
Predicting the Entropy Change of
Dissolution Process (ΔSsoln) of Ions
18.3
I. Solvation of nonionic compounds such as sugar.
The dissolving process causes dispersal of the molecules into a larger
volume - resulting in an increase in entropy
27
Entropy increases due to
the dispersion of ionic
species throughout the
solution
Entropy decreases due to
the less mobility of water
molecules
For example, when aluminum chloride dissolves in water,
each Al3+ ion becomes associated with six water molecules.
Usually, metals bearing more charges decrease the entropy (due to
decrease in entropy of water)
Important Trends
•
•
•
•
Look at the physical state (S of gas >> liquids >
solids).
If the physical state is same, look for number of
moles
If number of moles are same, look for molar mass
(higher molar mass – higher entropy)
If phase, moles & molar mass are same, look for
molecules with more complex structure
(complicated structure will be more disordered)
Exercise
For each process, determine the sign of S for the system:
(a) decomposition of CaCO3(s) to give CaO(s) and CO2(g) > 0
(b) heating bromine vapor from 45°C to 80°C > 0
(c) condensation of water vapor on a cold surface < 0
(d) reaction of NH3(g) and HCl(g) to give NH4Cl(s) < 0
(e) dissolution of sugar in water. > 0
31
Determine the sign of S for the system
C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O (g) > 0
O3(g)
O2(g) < 0
2O3(g)
3O2(g) > 0
C (s) + 2 Cl2 (g)
2F(g)
2 CCl4 (g) > 0
F2(g) < 0
What we learned in last class…
• Spontaneous process
• Entropy
• How to determine the sign of entropy
(physical states, number moles, molar mass,
structure complexity)
What we hope to learn today…
• Determination of Standard Entropy (S°)
• The second law of Thermodynamics
Determination of Standard Entropy, S°
18.3
Entropy Changes in a System
Standard Entropy, S°
It is possible to determine the absolute value of the
entropy of a substance, S; something we cannot do with
either energy or enthalpy.
Standard entropy is the absolute entropy of a
substance at 1 atm.
36
18.4
Entropy Changes in the Universe
Calculating Ssys
R = 8.314 J/K · mol, n = mole, V = Volume.
37
18.1
Determine the change in entropy for 1.0 mole of an ideal gas
originally confined to one-half of a 5.0-L container when the
gas is allowed to expand to fill the entire container at constant
temperature.
Setup
R = 8.314 J/K · mol, n = 1.0 mole, Vfinal = 5.0 L, and
Vinitial = 2.5 L.
38
18.1
Solution
39
18.3
Entropy Changes in a System
Standard Entropy, S°
Pay attention to the state of materials, e.g.,
Br(l) = 152.3
Br(g) = 245.13
40
SAMPLE PROBLEM
18.2
From the standard entropy values given below, calculate the
standard entropy changes for the following reactions at 25°C:
Setup
S°[CaCO3(s)] = 92.9 J/K · mol
S°[CO2(g)] = 213.6 J/K · mol
S°[H2(g)] = 131.0 J/K · mol
S°[Cl2(g)] = 223.0 J/K · mol
S°[CaO(s)] = 39.8 J/K · mol
S°[N2(g)] = 191.5 J/K · mol
S°[NH3(g)] = 193.0 J/K · mol
S°[HCl(g)] = 187.0 J/K · mol
41
SAMPLE PROBLEM
18.2
Solution
42
SAMPLE PROBLEM
18.2
Solution
43
SAMPLE PROBLEM
18.2
Solution
44
SAMPLE PROBLEM
- In reality, N2 and H2 react and form NH3 (exothermic &
spontaneous reaction, though slow)
- Entropy is not the sole factor
- We need to look beyond
18.4
Entropy Changes in the Universe
Topics
The Second Law of Thermodynamics
46
18.4
Entropy Changes in the Universe
The Second Law of Thermodynamics
Topics
The second law of thermodynamics says that for a process to
be spontaneous as written (in the forward direction), Suniv
must be positive.
ΔSuniv = ΔSsys + ΔSsurr > 0
“Spontaneous”
An equilibrium process is one that does not occur
spontaneously in either the net forward or net reverse direction
but can be made to occur by the addition or removal of energy
to a system at equilibrium.
ΔSuniv = ΔSsys + ΔSsurr = 0
“Equilibrium”
47
18.4
Entropy Changes in the Universe
Topics
 The universe is made up of two parts:
the system and
the surroundings
System
Reactants
and
products
Surroundings
Reaction
container,
room, and
everything else
ΔSuniv = ΔSsys + ΔSsurr
48
Entropy Change in the
Surroundings (ΔSsurr)
• Change in entropy of the surroundings is directly proportional
to the enthalpy of the system. At constant temperature:
– an exothermic process corresponds to an increase
of the entropy of the surroundings. (ΔSsurr > 0)
– an endothermic process corresponds to a decrease
of the entropy of the surroundings. (ΔSsurr < 0)
ΔSsurr α ‒ ΔHsys
 Just think about it. Any system naturally
tends to lower its energy (lose energy). This
is in consistent with the second law of
thermodynamics which states that the
entropy of the universe is always positive.
①
18.4
Entropy Changes in the Universe
Calculating Ssurr
ΔSuniv = ΔSsys + ΔSsurr > 0
ΔSuniv = ΔSsys + ΔSsurr < 0
ΔSuniv = ΔSsys + ΔSsurr = 0
“Spontaneous”
“Non-spontaneous”
“Equilibrium”
50
What we learned in previous lectures…
• Spontaneous process
• Entropy
• How to determine the sign of entropy
(physical states, number moles, molar mass,
structure complexity)
• Determination of Standard Entropy (S°)
• The second law of Thermodynamics
What we hope to learn today…
•
•
•
•
The second law of Thermodynamics
Gibbs Free-Energy Change, G
Standard Free-Energy Changes, G°
Using G and G° to Solve Problems
(nH°(products) – nH°(reactant)
ΔSuniv = ΔSsys + ΔSsurr > 0
“Spontaneous”
ΔSuniv = ΔSsys + ΔSsurr < 0
“Non-spontaneous”
ΔSuniv = ΔSsys + ΔSsurr = 0
“Equilibrium”
SAMPLE PROBLEM
18.4
Determine if each of the following is a spontaneous process, a
nonspontaneous process, or an equilibrium process at the
specified temperature:
54
SAMPLE PROBLEM
18.4
Solution
ΔSuniv = ΔSsys + ΔSsurr > 0
55
SAMPLE PROBLEM
18.4
Solution
56
SAMPLE PROBLEM
18.4
Solution
57
SAMPLE PROBLEM
18.4
Solution
Suniv is zero; therefore, the reaction is an equilibrium process
at 98°C. In fact, this is the melting point of sodium.
58
18.5
Predicting Spontaneity
Topics
Gibbs Free-Energy Change, G
Standard Free-Energy Changes, G°
Using G and G° to Solve Problems
ΔSuniv = ΔSsys + ΔSsurr > 0
“Spontaneous”
59
18.5
Predicting Spontaneity
Gibbs Free-Energy Change, G
60
18.5
Predicting Spontaneity
Gibbs Free-Energy Change, G
In terms of system (reactants and
products)
- Can be used to predict spontaneity of reaction
- Can predict the temperature for the reaction to be
spontaneous
- Can be used to determine the change in Entropy
for a phase change
61
18.5
Predicting Spontaneity
Predict spontaneity of reaction
G < 0
The reaction is spontaneous in the forward
direction (and nonspontaneous in the reverse
direction).
G > 0
The reaction is nonspontaneous in the forward
direction (and spontaneous in the reverse
direction).
G = 0
The system is at equilibrium.
62
18.5
Predicting Spontaneity
Gibbs Free-Energy Change, G
63
Also: It can be used to determine the temperature at which
process would be spontaneous or non-spontaneous!
- First we determine Temp. at equilibrium
- Look at the sign of ΔH and ΔS
- Decide Temp.
SAMPLE PROBLEM
18.5
For a reaction in which H = 199.5 kJ/mol and
S = 476 J/K · mol, determine the temperature (in °C) above
which the reaction is spontaneous.
ΔG = ΔH - T ΔS
At equilibrium,
ΔG = 0
T = ΔH/ΔS
65
18.5
Predicting Spontaneity
SAMPLE PROBLEM
Predict temperature for spontaneity
66
18.5
Predicting Spontaneity
SAMPLE PROBLEM
Predict temperature for spontaneity
Think about: If ΔS
and ΔH were negative
At T > 835°C, G° becomes negative, indicating that the
reaction would then favor the formation of CaO and CO2.
67
SAMPLE PROBLEM
Also: It can be used to determine the change in Entropy
for a phase change!
SAMPLE PROBLEM
18.7
Calculate the change in entropy for a phase change
The molar heats of fusion and vaporization of benzene are
10.9 and 31.0 kJ/mol, respectively.
Calculate the entropy changes for the solid-to-liquid and
liquid-to-vapor transitions for benzene.
At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
70
SAMPLE PROBLEM
18.7
Setup
The melting point of benzene is 5.5 + 273.15 = 278.7 K and the
boiling point is 80.1 + 273.15 = 353.3 K.
ΔG = ΔH - T ΔS
At equilibrium,
ΔG = 0
ΔS = ΔH/ T
71
SAMPLE PROBLEM
Standard Free-Energy Changes, G°
• The standard free energy (ΔGrxn) of a system is the change in free
energy when reactants in their standard states are converted to
products in their standard states.
• Standard states are:
– Gases
– Liquids
– Solids
– Elements
– Solutions
1 atm
[ O2(g) , CO2(g) , CH4(g) ]
Pure liquid [ H2O (l) , C2H5OH (l) ]
Pure solid [ Na(s) , Mg(s)]
The most stable form at 1 atm and 25C
1 M concentration
18.5
Predicting Spontaneity
SAMPLE PROBLEM
Standard Free-Energy Changes, G°
Gf ° is the standard free energy of formation of a
compound—that is, the free-energy change that occurs when
1 mole of the compound is synthesized from its constituent
elements, each in its standard state.
73
SAMPLE PROBLEM
18.6
Calculate the standard free-energy changes for the following
reactions at 25°C:
Setup
75
SAMPLE PROBLEM
18.6
Solution
76
SAMPLE PROBLEM
18.6
Solution
77
SAMPLE PROBLEM
SUMMARY (Gibbs Free Change)
Note:
will only tell - reactant or products are favored not spontaneity
18.5
More practical aspect of Gibb’s free energy
• The standard free energy (ΔGrxn) of a system is the change in free
energy when reactants in their standard states are converted to
products in their standard states.
• Standard states are:
– Gases
– Liquids
– Solids
– Elements
– Solutions
1 atm
[ O2(g) , CO2(g) , CH4(g) ]
Pure liquid [ H2O (l) , C2H5OH (l) ]
Pure solid [ Na(s) , Mg(s)]
The most stable form at 1 atm and 25C
1 M concentration
80
18.6
Free Energy and Chemical Equilibrium
Relationship Between G and G°
Relationship Between G° and K
81
18.6
Free Energy and Chemical Equilibrium
Relationship Between G and G°
• Exercise:
Consider the reaction: H2(g) + Cl2(g)
2HCl(g)
How does the value of ΔG change when the pressures of the H2,
Cl2 and HCl gases are changed to 0.25 atm, 0.45 atm and 0.30
atm, respectively, at 25C? (ΔGf for HCl(g) is ‒ 95.27 kJ/mol)
PRODUCT FAVORED
 Calculate ΔG.
ΔG° = [ 2 ( ‒ 95.27 kJ/mol)] ‒ [0 + 0] = ‒ 190.54 kJ/mol
 Calculate Q .
(PHCl )2
(0.30)2
QP 

 0.80
(PH2 )(PCl2 ) (0.25)(0.45)
ΔG = ΔG + RT lnQ
ΔG = ‒ 190540 J/mol + (8.314 J/K·mol)(298 K) ln(0.80)
ΔG = ‒ 191.09 kJ/mol
The reaction becomes even more spontaneous at the given
non standard states since ΔG is more negative than ΔG. Thus,
more HCl will be formed with the given pressures.
The reaction will continue producing more HCl and consuming
more H2 and Cl2 until QP = KP .
H2(g) + Cl2(g)  2HCl(g)
[HCl]2
[H2][Cl2]
18.6
Relationship between ΔG and K
At equilibrium:
Q = K and ΔG = 0
Thus, at equilibrium:
ΔG = ΔG + RT lnQ
0 = ΔG + RT lnK
ΔG = ‒ RT lnK
The larger the value of K,
the more negative ΔG
becomes, and the reaction
proceeds more towards
the product side.
K = Ka, Kb, Kp, Kc, Ksp, and so froth
SAMPLE PROBLEM
18.9
Calculate the equilibrium constant, KP , for the following
reaction at 25°C:
ΔG = 474.4 kJ/mol
Setup
86
SAMPLE PROBLEM
18.9
Solution
87
SAMPLE PROBLEM
18.10
The equilibrium constant, Ksp, for the dissolution of silver
chloride in water at 25°C,
is 1.6 × 10–10. Calculate G° for the process.
Setup
88
18.4
Entropy Changes in the Universe
SAMPLE PROBLEM
The Third Law of Thermodynamics
According to the third law of thermodynamics, the entropy
of a perfect crystalline substance is zero at absolute zero.
NaCl
89
18.4
Entropy Changes in the Universe
SAMPLE PROBLEM
The Third Law of Thermodynamics
90