Chapter18

Chapter 18
Thermodynamics
and Equilibrium

Contents and Concepts
1. First Law of Thermodynamics
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
Copyright © Cengage Learning. All rights reserved.

18 | 2

Free−Energy Concept
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.
4. Free Energy and Spontaneity
5. Interpretation of Free Energy

18 | 3

Free Energy and Equilibrium Constants
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


18 | 4

Learning Objectives

1. First Law of Thermodynamics; Enthalpy
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.


18 | 5

1. First Law of Thermodynamics; Enthalpy
(cont.)
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.


18 | 6

Spontaneous Processes and Entropy
Thermodynamics
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.


18 | 7

Thermodynamics (cont.)
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.


18 | 8

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.

18 | 9

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.

18 | 10

5. Interpretation of Free Energy
a. Relate the free−energy change to
maximum useful work.
b. Describe how the free energy changes
during a chemical reaction.


18 | 11

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.


18 | 12

6. Relating ∆Go to the Equilibrium Constant
(cont.)
d. Relate the standard free−energy change to
the thermodynamic equilibrium constant.
e. Calculate K from the standard free−energy
change (molecular equation).
f. Calculate K from the standard free−energy
change (net ionic equation).


18 | 13

7. Change of Free Energy with Temperature
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.

18 | 14

Thermodynamics is the study of heat and other
forms of energy involved in chemical or physical
processes.


18 | 15

First Law of Thermodynamics
The first law of thermodynamics is essentially the
law of conservation of energy applied to a
thermodynamic system.
Recall that the internal energy, U, is the sum of
the kinetic and potential energies of the particles
making up the system:
∆U = Uf – Ui


18 | 16

Exchanges of energy
between the system and its
surroundings are of two
types: heat, q, and work, w.
Putting this in an equation
gives us the first law of
thermodynamics.
∆U = q + w


18 | 17

Sign Convention for q
When heat is evolved by the system, q is negative.
This decreases the internal energy of the system.
When heat is absorbed by the system, q is
positive. This increases the internal energy of the
system.


18 | 18

Sign Convention for w
Recall from Chapter 6 that w = –P∆V.
When the system expands, ∆V is positive, so w is
negative. The system does work on the
surroundings, which decreases the internal energy
of the system.
When the system contracts, ∆V is negative, so w is
positive. The surroundings do work on the system,
which increases the internal energy of the system.

18 | 19

Here the system expands and evolves heat from A
to B.

Zn2+(aq) + 2Cl−(aq) + H2(g)
∆V is positive, so work is negative.

18 | 20

At constant pressure: qP = ∆H
The first law of thermodynamics can now be
expressed as follows:
∆U = ∆H – P∆V


18 | 21

To understand why a chemical reaction goes in a
particular direction, we need to study spontaneous
processes.
A spontaneous process is a physical or chemical
change that occurs by itself. It does not require
any outside force, and it continues until equilibrium
is reached.


18 | 22


18 | 23

The first law of thermodynamics cannot help us to
determine whether a reaction is spontaneous as
written.
We need a new quantity—entropy.
Entropy, S, is a thermodynamic quantity that is a
measure of how dispersed the energy of a system
is among the different possible ways that system
can contain energy.


18 | 24

Examining some spontaneous processes will
clarify this definition.
First, heat energy from a cup of hot coffee
spontaneously flows to its surroundings—the table
top, the air around the cup, or your hand holding
the cup. The entropy of the system (the cup of hot
coffee) and its surroundings has increased.


18 | 25

The rock rolling down the hill is a bit more
complicated. As it rolls down, the rock’s potential
energy is converted to kinetic energy. As it collides
with other rocks on the way down, it transfers
energy to them. The entropy of the system (the
rock) and its surroundings has increased.


18 | 26

Now consider a gas in a flask connected to an
equal−sized flask that is evacuated. When the
stopcock is open, the gas will flow into the
evacuated flask. The kinetic energy has spread
out, and the entropy of the system has increased.


18 | 27

In each of the preceding examples, energy has
been dispersed (that is, spread out).


18 | 28

Entropy is a state function. It depends on
variables, such as temperature and pressure, that
determine the state of the substance.
Entropy is an extensive property. It depends on the
amount of substance present.


18 | 29

Entropy is measured in units of J/K.
Entropy change is calculated as follows:
∆S = Sf – Si


18 | 30

Concept Check 18.1
You have a sample of 1.0 mg of solid iodine at room
temperature. Later, you notice that the iodine has
sublimed (passed into the vapor state). What can
you say about the change of entropy of the iodine?

The iodine has spread out, so its entropy has
increased.


18 | 31

Second Law of Thermodynamics
The total entropy of a system and its surroundings
always increases for a spontaneous process.
Note: Entropy is a measure of energy dispersal,
not a measure of energy itself.


18 | 32

For a spontaneous process at a constant
temperature, we can state the second law of
thermodynamics in terms of only the system:
q
∆S = entropy created +
T

For a spontaneous process:
q
∆S > T


18 | 33

A

C

B

A pendulum is put in motion, with all of its molecules
moving in the same direction, as shown in Figures A and B.
As it moves, the pendulum collides with air molecules.
When it comes to rest in Figure C, the pendulum has
dispersed its energy. This is a spontaneous process.

18 | 34

D

E

F

Now consider the reverse process, which is not
spontaneous. While this process does not violate the first
law of thermodynamics, it does violate the second law
because the dispersed energy becomes more concentrated
as the molecules move together.

18 | 35

Entropy and Molecular Disorder
Entropy is essentially related to energy dispersal.
The entropy of a molecular system may be
concentrated in a few energy states and later
dispersed among many more energy states. The
energy of such a system increases.


18 | 36

In the case of the cup of hot coffee, as heat moves
from the hot coffee, molecular motion becomes
more disordered. In becoming more disordered,
the energy is more dispersed.


18 | 37

Likewise, when the gas moves from one container
into the evacuated container, molecules become
more disordered because they are dispersed over
a larger volume. The energy of the system is
dispersed over a larger volume.


18 | 38

When ice melts, the molecules become more
disordered, again dispersing energy more widely.
When one molecule decomposes to give two, as in
N2O4(g) → 2NO2(g)
more disorder is created because the two
molecules produced can move independently of
each other. Energy is more dispersed.


18 | 39

In each of these cases, molecular disorder
increases, as does entropy.
Note: This understanding of entropy as disorder
applies only to molecular situations in which
increasing disorder increases the dispersion of
energy. It cannot be applied to situations that are
not molecular—such as a desk.


18 | 40

Entropy Change for a Phase Transition
In a phase transition process that occurs very
close to equilibrium, heat is very slowly absorbed
or evolved. Under these conditions, no significant
new entropy is created.
q
∆S =
T

This concept can be applied to melting using ∆Hfus
for q and to vaporization using ∆Hvap for q.

18 | 41

?

Acetone, CH3COCH3, is a volatile liquid
solvent; it is used in nail polish, for
example. The standard enthalpy of
formation, ∆Hf°, of the liquid at 25°C is
–247.6 kJ/mol; the same quantity for
the vapor is –216.6 kJ/mol.
What is the entropy change when 1.00
mol liquid acetone vaporizes at 25°C?


18 | 42

CH3COCH3(l) → CH3COCH3(g)
∆Hf°
n
n∆Hf°

–247.6 kJ/mol

–216.6 kJ/mol

1 mol
–247.6 kJ

1 mol
–216.6 kJ

∆H° = 31.0 kJ

∆H
ΔS =
T

31.0 × 10 3 J
ΔS =
298 K
∆S = 104 J/K

18 | 43

Criterion for a Spontaneous Reaction
The criterion is that the entropy of the system and
its surroundings must increase.
q
ΔS >
T
ΔH
ΔS >
T

TΔS > ΔH

0 > Δ H - TΔ S or Δ H - TΔ S < 0


18 | 44

Third Law of Thermodynamics
A substance that is perfectly crystalline at zero
Kelvin (0 K) has an entropy of zero.


18 | 45

The standard entropy of a substance—its absolute
entropy, S°—is the entropy value for the standard
state of the species. The standard state is
indicated with the superscript degree sign.
For a pure substance, its standard state is 1 atm
pressure. For a substance in solution, its standard
state is a 1 M solution.


18 | 46


18 | 47


18 | 48

Standard Entropy of Bromine, Br2, at Various Temperatures


18 | 49

Entropy Change for a Reaction
Entropy usually increases in three situations:
1. A reaction in which a molecule is broken into
two or more smaller molecules.
2. A reaction in which there is an increase in
the number of moles of a gas.
3. A process in which a solid changes to a
liquid or gas or a liquid changes to a gas.


18 | 50

The opening of Chapter 6, on
thermo−chemistry, describes the
endothermic reaction of solid barium
hydroxide octahydrate and solid
ammonium nitrate:
Ba(OH)2  8H2O(s) + 2NH4NO3(s) →
2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)
Predict the sign of ∆S° for this reaction.

?

3 moles of reactants produces 13 moles of
products. Solid reactants produce gaseous,
liquid, and aqueous products.
∆S° is positive.

18 | 51

To compute ∆S° where n = moles:
ΔS =


∑ nS



(products) −


∑ nS



(reactants)

18 | 52

When wine is exposed to air in the
presence of certain bacteria, the ethyl
alcohol is oxidized to acetic acid, giving
vinegar. Calculate the entropy change
at 25°C for the following similar
reaction:
CH3CH2OH(l) + O2(g) → CH3COOH(l) + H2O(l)

?

The standard entropies, S°, of the
substances in J/(K  mol) at 25°C are
CH3CH2OH(l),161; O2(g), 205;
CH3COOH(l), 160; H2O(l), 69.9.

18 | 53

S°
161
J/(mol  K)
n mol
1
nS° J/K
161
366

205

160

69.9

1
205

1
160

1
69.9

229.9

∆S = 229.9 J/K – 366 J/K
∆S = –136 J/K

18 | 54

Free Energy and Spontaneity
Physicist J. Willard Gibbs introduced the concept
of free energy, G. Free energy is a
thermodynamic quantity defined as follows:
G = H – TS


18 | 55

As a reaction proceeds, G changes:
∆G = ∆H – T∆S
Standard free energy change:
∆G° = ∆H° – T∆S°


18 | 56

?

Using standard enthalpies of formation,
∆Hf° and the value of ∆S° from the
previous problem, calculate ∆G° for the
oxidation of ethyl alcohol to acetic acid.



18 | 57

∆Hf° kJ/mol –277.6
n mol
n∆Hf° kJ

0

1
1
–277.6 0
–277.6 kJ

–487.0

–285.8

1
–487.0

1
–285.8
–772.8 kJ

∆H° = –495.2 kJ
∆S° = –136 J/K
T = 298 K

18 | 58

∆H° = –495.2 kJ
∆S° = –136 J/K = –0.136 kJ/K
T = 298 K
∆G° = ∆H° – T∆S°
∆G° = –495.2 kJ – (298 K)(–0.136 kJ/K)
∆G° = –495.2 kJ + 40.5 kJ
∆G° = –454.7 kJ
The reaction is spontaneous.


18 | 59

Standard Free Energies of Formation, ∆Gf°
The standard free energy of formation is the
free−energy change that occurs when 1 mol of
substance is formed from its elements in their
standard states at 1 atm and at a specified
temperature, usually 25°C.
The corresponding reaction for the standard free
energy of formation is the same as that for
standard enthalpy of formation, ∆Hf°.


18 | 60


18 | 61


18 | 62

To find the standard free energy change for a
reaction where n = moles:
ΔG =


∑ nΔG (products) − ∑ nΔG (reactants)





18 | 63

?

Calculate the free−energy change,
∆G°, for the oxidation of ethyl alcohol
to acetic acid using standard free
energies of formation.



18 | 64

∆Gf°, kJ/mol
237.2
n, mol
n∆Gf°, kJ
237.2

–174.8

0

1
1
–174.8
0
–174.8 kJ

–392.5

–

1
1
–392.5
–
–629.7 kJ

∆G° = –629.7 – (–174.8)
∆G° = –454.9 kJ

18 | 65

∆G° as a Criterion for Spontaneity
The spontaneity of a reaction can now be
determined by the sign of ∆G°.
∆G° < –10 kJ: spontaneous
∆G° > +10 kJ: nonspontaneous
∆G° = very small or zero (< +10 kJ and
> –10 kJ): at equilibrium

18 | 66

Concept Check 18.2
Consider the reaction of nitrogen, N2, and oxygen,
O2, to form nitrogen monoxide, NO:
N2(g) + O2(g) → 2NO(g)
From the standard free energy of formation of NO,
what can you say about thisreaction?
For the reaction as written, ∆G° = 173.20 kJ.
For 1 mol NO(g), ∆Gf° = 86.60 kJ/mol.
The reaction is nonspontaneous.


18 | 67


18 | 68


18 | 69

Interpreting Free Energy
Theoretically, spontaneous reactions can be used
to perform useful work. In fact, we use reactions
such as the combustion of gasoline to move a
vehicle.
We can also use spontaneous reactions to provide
the energy needed for a nonspontaneous reaction.
The maximum useful work is
wmax = ∆G

18 | 70

The thermodynamic equilibrium constant is the
equilibrium constant in which the concentrations of
gases are expressed as partial pressures in
atmospheres and the concentrations of solutes in
solutions are expressed in molarities.
If only gases are present, K = Kp.
If only solutes in liquid solution are present, K = Kc.


18 | 71

?

Write the expression for the
thermodynamic equilibrium constant for
these reactions:
a. N2O4(g)  2NO2(g)
b. Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
a.

b.


K=

K=

2
PNO2

PN2O 4
[Zn 2+ ] PH

2

[H + ] 2

18 | 72

Standard free energy change is related to the
thermodynamic equilibrium constant, K, at
equilibrium:
∆G = ∆G° + RT ln Q
At equilibrium:
∆G = 0 and Q = K
∆G° = –RT ln K


18 | 73

?

Calculate the value of the
thermodynamic equilibrium constant at
25°C for the reaction
N2O4(g)  2NO2(g)
The standard free energy of formation
at 25°C is 51.30 kJ/mol for NO2(g) and
97.82 kJ/mol for N2O4(g).

∆G° = 2 mol(51.30 kJ/mol) – 1 mol(97.82 kJ/mol)
∆G° = 102.60 kJ – 97.82 kJ
∆G° = 4.78 kJ

18 | 74

∆G = –RT ln K
− ∆G
ln K =
RT

J 

3
−  4.78 × 10
÷
mol 

ln K =
J
8.315
(298 K)
mol K
ln K = −1.929
K = 0.145

18 | 75

?

Sodium carbonate, Na2CO3, can be
prepared by heating sodium hydrogen
carbonate, NaHCO3:

2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)
Estimate the temperature at which the
reaction proceeds spontaneously at 1 atm.
See Appendix C for data.


18 | 76

2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO(g)
∆Hf°, kJ/mol –947.7 –1130.8 –241.8
–393.5
n, mol
2
1
1
1
n∆Hf°, kJ –1895.4 –1130.8 –241.8
–393.5
–1895.4 kJ
–1766.1 kJ
∆H° = 129.3 kJ
Sf°, J/mol  K 102
n, mol
2
nSf°, J/K
204
204 J/K

139
1
139

188.7
1
188.7

541.4 J/K
∆S° = 337.4 J/K


213.7
1
213.7

18 | 77

ΔH °
T =
ΔS °

129.3 × 10 3 J
T=
J
337.4
K

T = 383 K

T = 110°C

18 | 78

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