Chapter 19
Chemical
Thermodynamics
John D. Bookstaver
St. Charles Community College
St. Peters, MO
2006, Prentice Hall, Inc
.
Modified by S.A. Green, 2006
Modified by D. Amuso 2011
Thermodynamics
The study of the relationships between
heat, work, and the energy of a
system.
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.
Second Law of Thermodynamics
The total
Entropy
of the
universe increases
in
any
spontaneous, irreversible
process.
Entropy
•
Entropy
can be thought of as a
measure of the
randomness or disorder
of a system.
•
It is related to the various modes of
motion in molecules (microstates).
Vibrational
Rotational
Translational
Entropy
Molecules have more entropy (disorder)
when:
1)
Phase Changes from: S
L
G
Example: Sublimation
CO
2
(s)
CO
2
(g)
2)
Dissolving occurs (solution forms):
Example:
NaCl
(s)
Na
+
(
aq
) +
Cl
-
(
aq
)
Entropy
3) Temperature increases
Example:
Fe(s) at 0
o
C
Fe(s) at 25
o
C
4) For Gases ONLY, when
Volume increases or Pressure decreases
Examples:
2 Liters He(g)
4 Liters He(g)
3 atm He(g)
1 atm He(g)
Entropy
5) Rx results in more molecules/moles of gas
Examples:
2NH
3
(g)
N
2
(g) + 3H
2
(g)
CaCO
3
(s)
CaO(s) + CO
2
(g)
N
2
O
4
(g)
2 NO
2
(g)
This one is difficult to predict:
N
2
(g) + O
2
(g)
2 NO(g)
Entropy
6)
When there are more moles
Example:
1 mole H
2
O(g)
2 moles H
2
O(g)
7) When there are more atoms per molecule
Examples:
1 mole Ar(g)
1 mole HCl(g)
1 mole NO
2
(g)
1 mole N
2
O
4
(g)
Entropy
8) When an atom has a bigger atomic
number
1 mole He(g)
1 mole Ne (g)
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
move to just one vessel.
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.
Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.
Spontaneous
processes are
irreversible
.
Second Law of Thermodynamics
A
reversible
process results in
no
overall change
in
Entropy
while an
irreversible, spontaneous
process
results in an
overall increase
in
Entropy
.
Reversible:
Irreversible (Spontaneous):
Third Law of Thermodynamics
The
Entropy
of a pure crystalline
substance at absolute zero is zero.
why?
Third Law of Thermodynamics
Standard Entropies
•
Standard Conditions:
298 K, 1 atm, 1 Molar
•
The values for Standard
Entropies (S
o
) are
expressed in J/mol
-
K.
•
Note: Increase with
increasing molar mass.
Standard Entropies
Larger and more complex molecules have
greater entropies.
Entropy Changes
D
S
o
=
S
n
S
o
products
-
S
m
S
o
reactants
Be careful: S
°
units are in J/mol
-
K
Note for pure elements:
Gibbs Free Energy
Use
D
G
to decide if a process is spontaneous
D
G = negative value = spontaneous
D
G = zero = at equilibrium
D
G = positive value = not spontaneous
Note: equation can be used w/o the
o
too.
D
G
o
=
D
H
o
–
T
D
S
o
Gibbs Free Energy
1.
If
D
G
is negative
D
G
= maximum amt of energy ‘free’ to do
work by the reaction
2. If
D
G
is positive
D
G
= minimum amt of work needed
to make the reaction happen
D
G
o
=
D
H
o
–
T
D
S
o
Gibbs Free Energy
In our tables, units are:
D
G
o
= kJ/mol
D
H
o
= kJ/mol
D
S
o
= J/mol
-
K
D
G
o
=
D
H
o
–
T
D
S
o
Free Energy and Temperature
•
There are two parts to the free energy
equation:
D
H
—
the enthalpy term
T
D
S
—
the entropy term
•
The temperature dependence of free
energy comes from the entropy term.
What causes a reaction to be
spontaneous?
•
Think Humpty Dumpty
•
System tend to seek:
Minimum Enthalpy
Exothermic Rx,
D
H = negative
Maximum Entropy
More disorder,
D
S = positive
•
Because:
D
G
o
=
D
H
o
–
T
D
S
o
-
= (
-
)
-
(+)
Free Energy and Temperature
By knowing the sign (+ or
-
) of
D
S and
D
H,
we can get the sign of
D
G and determine if a
reaction is spontaneous.
At Equilibrium
D
G
o
= zero Therefore:
D
H
o
=
T
D
S
o
Or
:
D
S
o
=
D
H
o
T
Use this equation when asked to calculate enthalpy of
vaporization or enthalpy of fusion.
D
G
o
=
D
H
o
–
T
D
S
o
D
G
o
=
S
n
G
o
f
products
-
S
m
G
o
f reactants
Standard Free Energy Changes
Be careful: Values for
D
G
f
are in kJ/mol
D
G
can be looked up in tables
or
calculated
from
S
°
and
D
H
.
D
G
o
=
D
H
o
–
T
D
S
o
Free Energy and Equilibrium
Remember from above:
If
D
G
is 0, the system is at equilibrium.
So
D
G
must be related to the equilibrium
constant, K (chapter 15). The
standard
free
energy,
D
G
°
, is directly linked to K
eq
by:
D
G
o
=
-
RT ln K
Where R = 8.314 J/mol
-
K
D
G
o
=
-
RT ln K
If the free energy change is a negative value,
the reaction is spontaneous, ln K must be a
positive value, and K will be a large number
meaning the equilibrium mixture is mainly products.
If the free energy change is zero,
ln K = zero and K = one.
Relationships
Free Energy and Equilibrium
Under non
-
standard conditions, we need to use
D
G
instead of
D
G
°
.
Q is the reaction quotient from chapter 15.
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