Chapter 19 Chemical Thermodynamics

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Chemical

Thermodynamics

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Chapter 19

Chemical
Thermodynamics

Chemistry, The Central Science
, 11th edition

Theodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

John D. Bookstaver

St. Charles Community College

Cottleville, MO

Chemical

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First Law of Thermodynamics


You will recall from Chapter 5 that
energy cannot be created nor
destroyed.


Therefore, the total energy of the
universe is a constant.


Energy can, however, be converted
from one form to another or transferred
from a system to the surroundings or
vice versa.

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Spontaneous Processes


Spontaneous processes
are those that can
proceed without any
outside intervention.


The gas in vessel
B

will
spontaneously effuse into
vessel
A
, but once the
gas is in both vessels, it
will
not

spontaneously
return to vessel
B
.

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Spontaneous Processes


Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.

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Spontaneous Processes


Processes that are spontaneous at one
temperature may be nonspontaneous at other
temperatures.


Above 0

C it is spontaneous for ice to melt.


Below 0

C the reverse process is spontaneous.

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Reversible Processes


In a reversible
process the system
changes in such a
way that the system
and surroundings
can be put back in
their original states
by exactly reversing
the process.

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Irreversible Processes


Irreversible processes cannot be undone by
exactly reversing the change to the system.


Spontaneous processes are irreversible.

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Entropy


Entropy

(
S
) is a term coined by Rudolph
Clausius in the 19th century.


Clausius was convinced of the
significance of the ratio of heat
delivered and the temperature at which
it is delivered, .

q

T

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Entropy


Entropy can be thought of as a measure
of the randomness of a system.


It is related to the various modes of
motion in molecules.

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Entropy


Like total energy,
E
, and enthalpy,
H
,
entropy is a state function.


Therefore,


S

=
S
final



S
initial

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Entropy


For a process occurring at constant
temperature (an isothermal process), the
change in entropy is equal to the heat that
would be transferred if the process were
reversible divided by the temperature:



S

=

q
rev

T

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Second Law of Thermodynamics


The second law of thermodynamics
states that the entropy of the universe
increases for spontaneous processes,
and the entropy of the universe does
not change for reversible processes.

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Second Law of Thermodynamics

In other words:

For reversible processes:


S
univ

=

S
system

+

S
surroundings

= 0

For irreversible processes:


S
univ

=

S
system

+

S
surroundings

> 0


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Second Law of Thermodynamics


These last truths mean that as a result
of all spontaneous processes the
entropy of the universe increases.

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Entropy on the Molecular Scale


Ludwig Boltzmann described the concept of
entropy on the molecular level.


Temperature is a measure of the average
kinetic energy of the molecules in a sample.

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Entropy on the Molecular Scale


Molecules exhibit several types of motion:


Translational: Movement of the entire molecule from
one place to another.


Vibrational: Periodic motion of atoms within a molecule.


Rotational: Rotation of the molecule on about an axis or
rotation about


bonds.

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Entropy on the Molecular Scale


Boltzmann envisioned the motions of a sample of
molecules at a particular instant in time.


This would be akin to taking a snapshot of all the
molecules.


He referred to this sampling as a
microstate

of the
thermodynamic system.

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Entropy on the Molecular Scale


Each thermodynamic state has a specific number of
microstates,
W
, associated with it.


Entropy is

S

=
k

ln
W


where
k

is the Boltzmann constant, 1.38


10

23

J/K.

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Entropy on the Molecular Scale


The number of microstates and,
therefore, the entropy tends to increase
with increases in


Temperature.


Volume.


The number of independently moving
molecules.

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Entropy and Physical States


Entropy increases with
the freedom of motion
of molecules.


Therefore,


S
(
g
)

>
S
(
l
)

>
S
(
s
)

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Solutions


Generally, when
a solid is
dissolved in a
solvent, entropy
increases.

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Entropy Changes


In general, entropy
increases when


Gases are formed from
liquids and solids;


Liquids or solutions are
formed from solids;


The number of gas
molecules increases;


The number of moles
increases.

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Third Law of Thermodynamics

The entropy of a pure crystalline
substance at absolute zero is 0.

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Sample Exercise 19.3
Predicting the Sign of
Δ
S

Solution

Analyze:
We are given four equations and asked to predict the sign of
Δ
S

for each chemical reaction.

Plan:
The sign of
Δ
S

will be positive if there is an increase in temperature, an increase in the volume in
which the molecules move, or an increase in the number of gas particles in the reaction. The question states
that the temperature is constant. Thus, we need to evaluate each equation with the other two factors in mind.

Solve:

(a)
The evaporation of a liquid is accompanied by a large increase in volume. One mole of water (18 g)
occupies about 18 mL as a liquid and if it could exist as a gas at STP it would occupy 22.4 L. Because the
molecules are distributed throughout a much larger volume in the gaseous state than in the liquid state, an
increase in motional freedom accompanies vaporization. Therefore,
Δ
S

is positive.

(b)
In this process the ions, which are free to move throughout the volume of the solution, form a solid in
which they are confined to a smaller volume and restricted to more highly constrained positions. Thus,
Δ
S

is
negative.

(c)
The particles of a solid are confined to specific locations and have fewer ways to move (fewer
microstates) than do the molecules of a gas. Because O
2

gas is converted into part of the solid product
Fe
2
O
3
,
Δ
S

is negative.

(d)
The number of moles of gases is the same on both sides of the equation, and so the entropy change will
be small. The sign of
Δ
S

is impossible to predict based on our discussions thus far, but we can predict that
Δ
S

will be close to zero.

Predict whether
Δ
S

is positive or negative for each of the following processes, assuming each occurs at
constant temperature:


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Sample Exercise 19.3
Predicting the Sign of
Δ
S

Indicate whether each of the following processes produces an increase or decrease in the entropy of the
system:







Answer:
(a)
increase,
(b)
decrease,
(c)
decrease,
(d)
decrease

Practice Exercise

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Standard Entropies


These are molar entropy
values of substances in
their standard states.


Standard entropies tend
to increase with
increasing molar mass.

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Standard Entropies

Larger and more complex molecules have
greater entropies.

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Entropy Changes


Entropy changes for a reaction can be
estimated in a manner analogous to that by
which

H

is estimated:



S


=

n

S

(products)




m

S

(reactants)




where
n

and
m

are the coefficients in the
balanced chemical equation.

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Sample Exercise 19.5
Calculating
Δ
S

from Tabulated Entropies

Calculate
Δ
S
º

for the synthesis of ammonia from N
2
(
g
) and H
2
(
g
) at 298 K:








N
2
(
g
) + 3 H
2
(
g
) → 2 NH
3
(
g
)

Using the standard entropies in Appendix C, calculate the standard entropy change,
Δ
S
°
, for the following
reaction at 298 K:




Al
2
O
3
(
s
) + 3 H
2
(
g
) → 2 Al(
s
) + 3 H
2
O(
g
)

Answers:
180.39 J/K

Practice Exercise

Solution

Analyze:
We are asked to calculate the entropy change for the synthesis of NH
3
(
g
) from its constituent
elements.

Plan:
We can make this calculation using Equation 19.8 and the standard molar entropy values for the
reactants and the products that are given in Table 19.2 and in Appendix C.

Solve:
Using Equation 19.8, we have


Substituting the appropriate
S
°

values from
Table 19.2 yields

Δ
S
°

= 2
S
°
(NH
3
)
-

[
S
°
(N
2
) + 3
S
°
(H
2
)]


Δ
S
°

= (2 mol)(192.5 J/mol
-
K)
-

[(1 mol)(191.5 J/mol
-
K)

+ (3 mol)(130.6 J/mol
-
K)] =
-
198.3 J/K

Check:
The value for
Δ
S
°

is negative, in agreement with our qualitative prediction based on the decrease in
the number of molecules of gas during the reaction.

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Thermodynamics

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Entropy Change in the Universe


The universe is composed of the system
and the surroundings.


Therefore,


S
universe

=

S
system

+

S
surroundings


For spontaneous processes



S
universe
> 0


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Gibbs Free Energy


Energy available to do work



T

S
universe

is defined as the Gibbs free
energy,

G
.


When

S
universe

is positive,

G

is
negative.


Therefore, when

G

is negative, a
process is spontaneous.

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Gibbs Free Energy

1.
If

G

is negative, the
forward reaction is
spontaneous.

2.
If

G

is 0, the system
is at equilibrium.

3.
If

G

is positive, the
reaction is
spontaneous in the
reverse direction.

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Standard Free Energy Changes


Analogous to standard enthalpies of
formation are standard free energies of
formation,

G

.

f


G


=

n

G


(products)




m

G


(reactants)

f


f

where
n

and
m

are the stoichiometric
coefficients.

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Sample Exercise 19.7
Calculating Standard Free
-
Energy Change from


Free Energies of Formation

(a)
Use data from Appendix C to calculate the standard free
-
energy change for the following reaction at

298 K:




P
4
(
g
) + 6 Cl
2
(
g
) → 4 PCl
3
(
g
)


(b)
What is
Δ
G
°

for the reverse of the above reaction?


Solution

Analyze:
We are asked to calculate the free
-
energy change for the indicated reaction and then to determine
the free
-
energy change of its reverse.

Plan:
To accomplish our task, we look up the free
-
energy values for the products and reactants and use
Equation 19.14: We multiply the molar quantities by the coefficients in the balanced equation, and subtract
the total for the reactants from that for the products.

Solve:

(a)
Cl
2
(
g
) is in its standard state, so
Δ
G
°
f

is zero for this reactant. P
4
(
g
), however, is not in its standard
state, so
Δ
G
°
f

is not zero for this reactant. From the balanced equation and using Appendix C, we have:


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Sample Exercise 19.7
Calculating Standard Free
-
Energy Change from


Free Energies of Formation

By using data from Appendix C, calculate
Δ
G
°

at 298 K for the combustion of methane:

CH
4
(
g
) + 2 O
2
(
g
) → CO
2
(
g
) + 2 H
2
O(
g
).

Answer:

800.7 kJ

Practice Exercise

Solution
(continued)

The fact that
Δ
G
°

is negative tells us that a mixture of P
4
(
g
), Cl
2
(
g
), and PCl
3
(
g
) at 25
°
C, each present at
a partial pressure of 1 atm, would react spontaneously in the forward direction to form more PCl
3
.
Remember, however, that the value of
Δ
G
°

tells us nothing about the rate at which the reaction occurs.

(b)
Remember that
Δ
G

=
G

(products)


G

(reactants). If we reverse the reaction, we reverse the roles of the
reactants and products. Thus, reversing the reaction changes the sign of
Δ
G
, just as reversing the reaction
changes the sign of
Δ
H.

(Section 5.4) Hence, using the result from part (a):






4 PCl
3
(
g
) → P
4
(
g
) + 6 Cl
2
(
g
)
Δ
G
°

= +1102.8 kJ

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Free Energy Changes


At temperatures other than 25
°
C,



G
°

=

H




T

S




How does

G


change with temperature?

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Sample Exercise 19.6
Calculating Free
-
Energy Change from
Δ
H
°
,
T
,
Δ
S
°

Calculate the standard free energy change for the formation of NO(
g)
from N
2
(
g
) and O
2
(
g
) at 298 K:








N
2
(
g
) + O
2
(
g
) → 2 NO(
g
)


given that
Δ
H
°

= 180.7 kJ and
Δ
S
°

= 24.7 J/K. Is the reaction spontaneous under these circumstances?

A particular reaction has
Δ
H
°

= 24.6 kJ and
Δ
S
°

= 132 J/K at 298 K. Calculate
Δ
G
°
.

Is the reaction spontaneous under these conditions?


Answers:
Δ
G
°

=

14.7 kJ; the reaction is spontaneous.

Practice Exercise

Solution

Analyze:
We are asked to calculate
Δ
G
°

for the indicated reaction (given
Δ
H
°
,
Δ
S
°

and
T
) and to
predict whether the reaction is spontaneous under standard conditions at 298 K.

Plan:
To calculate
Δ
G
°
, we use Equation 19.12,
Δ
G
°

=
Δ
H
°



T

Δ
S
°
. To determine whether the
reaction is spontaneous under standard conditions, we look at the sign of
Δ
G
°
.

Solve:








Because
Δ
G
°

is positive, the reaction is not spontaneous under standard conditions at 298 K.

Comment:
Notice that we had to convert the units of the
T

Δ
S
°

term to kJ so that they could be added to
the
Δ
H
°

term, whose units are kJ.

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Free Energy and Temperature


There are two parts to the free energy
equation:



H



the enthalpy term


T

S




the entropy term


The temperature dependence of free
energy, then comes from the entropy
term.

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Free Energy and Temperature

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Free Energy and Equilibrium


Under any conditions, standard or
nonstandard, the free energy change
can be found this way:



G

=

G


+
RT
ln
Q


(Under standard conditions, all concentrations are 1
M
,
so
Q

= 1 and ln
Q

= 0; the last term drops out.)

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Free Energy and Equilibrium


At equilibrium,
Q

=
K
, and

G

= 0.


The equation becomes

0 =

G


+
RT

ln
K


Rearranging, this becomes


G


=

RT

ln
K


or,






K

= e

-

G


RT

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Sample Exercise 19.10
Relating
Δ
G

to a Phase change at Equilibrium

As we saw in Section 11.5, the
normal boiling point
is the temperature at which a pure liquid is in equilibrium
with its vapor at a pressure of 1 atm.
(a)
Write the chemical equation that defines the normal boiling point of
liquid carbon tetrachloride, CCl
4
(
l
).
(b)
What is the value of
Δ
G
°

for the equilibrium in part (a)?
(c)
Use
thermodynamic data in Appendix C and Equation 19.12 to estimate the normal boiling point of CCl
4
.


Solution

Analyze: (a)
We must write a chemical equation that describes the physical equilibrium between liquid and
gaseous CCl
4

at the normal boiling point.
(b)
We must determine the value of
Δ
G
°

for CCl
4
, in equilibrium
with its vapor at the normal boiling point.
(c)
We must estimate the normal boiling point of CCl
4
, based on
available thermodynamic data.

Plan: (a)
The chemical equation will merely show the change of state of CCl
4

from liquid to solid.
(b)
We
need to analyze Equation 19.16 at equilibrium (
Δ
G

= 0).
(c)
We can use Equation 19.12 to calculate
T
when
Δ
G

= 0.

Solve: (a)
The normal boiling point of CCl
4

is the temperature at which pure liquid CCl
4

is in equilibrium with its vapor at a pressure
of 1 atm:

(b)
At equilibrium
Δ
G

= 0. In any normal boiling
-
point equilibrium both the liquid and the vapor are in their
standard states (Table 19.2). Consequently,
Q

= 1,

Q

= 0, and
Δ
G

=
Δ
G
°

for this process. Thus, we conclude
that
Δ
G

= 0 for the equilibrium involved in the normal boiling point of any liquid. We would also find that

Δ
G

= 0 for the equilibria relevant to normal melting points and normal sublimation points of solids.

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Sample Exercise 19.10
Relating
Δ
G

to a Phase change at Equilibrium

Solution
(continued)

(c)
Combining Equation 19.12 with the result
from part (b), we see that the equality at the
normal boiling point,
T
b
, of CCl
4
(
l
) or any
other pure liquid is


Solving the equation for
T
b
, we obtain


Strictly speaking, we would need the values
of
Δ
H
°

and
Δ
S
°

for the equilibrium
between CCl
4
(
l
) and CCl
4
(
g
) at the normal
boiling point to do this calculation. However,
we can
estimate
the boiling point by using
the values of
Δ
H
°

and
Δ
S
°

for CCl
4

at 298
K, which we can obtain from the data in
Appendix C and Equations 5.31 and 19.8:


Notice that, as expected, the process is
endothermic (
Δ
H

> 0)and produces a gas in
which energy can be more spread out

(
Δ
S

> 0). We can now use these values to
estimate
T
b

for CCl
4
(
l
):

Note also that we have used the conversion factor between J and kJ to make sure that the units of
Δ
H
°

and
Δ
S
°

match.

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Sample Exercise 19.10
Relating
Δ
G

to a Phase change at Equilibrium

Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br
2
(
l
). (The
experimental value is given in Table 11.3.)

Answers:
330 K

Practice Exercise

Solution

(continued)

Check:
The experimental normal boiling point of CCl
4
(
l
) is 76.5
°
C. The small deviation of our estimate
from the experimental value is due to the assumption that
Δ
H
°

and
Δ
S
°

do not change with temperature.