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Thermodynamics
Chapter 18
st
1
Law of Thermodynamics
Energy is conserved.
E = q + w
SPONTANEOUS: occur without
any outside intervention
Example: drop an egg
The REVERSE is not spontaneous!!
REVERSIBLE PROCESS: change
can be restored to its’ original state
by exactly reversing the change.
Example: ice water at 0o C
IRREVERSIBLE PROCESS:
cannot simply be reversed to
original state.
Example: gas expanding
Processes in which the
disorder of the system
increases tend to occur
spontaneously.
Ex: gas expanding, ice melting, salt dissolving
ENTROPY: (S) the change in
disorder. (Change in randomness)
The more disorder, the larger the
entropy.
S = Sfinal - Sinitial
S = > 0 when the final state is
in more disorder
S = < 0 when the final state
is more ordered than original
H2O (l)  H2O
+
Ag (aq)
+
Cl
(aq)
(s)
 AgCl(s)
- a solid melts
- a liquid vaporizes
- a solid dissolves in water
- a gas liquefies
For a process at constant
temperature, the entropy
change is the value of qrev
divided by the absolute
temperature.
S = qrev/T
Example: Calculate the entropy
change when 1 mol of water is
converted into 1 mol of steam at
1 atm pressure. (Hvap = 40.67
kJ/mol)
(1 mole)(40.67 kJ/ mol)(1000 J/1 kJ)
373 K
S = 109 J/K
The normal freezing point of
o
mercury is -38.9 C, an its
molar enthalpy of fusion is
Hfus = 2.331 kJ/mol. What
is the entropy change when
50.0 g of Hg(l) freezes at the
normal freezing point?
-2.48 J/K
The answer is negative because the
process brings more order
The normal boiling point of
ethanol, is 78.3oC and its molar
enthalpy of vaporization is 38.56
kJ/mol. What is the change in
entropy when 25.8 g of
C2H5OH(g) condenses to liquid at
the normal boiling point?
-61.4 J/K
The element gallium, Ga,
o
freezes at 29.8 C, and its
enthalpy of fusion is 5.59
kJ/mol. Calculate the
value of S for the
freezing of 90.0 g of Ga(l).
S = -23.8 J/K
2nd Law of Thermodynamics: In any
reversible process, Suniverse = 0. In any
irreversible (spontaneous) process,
Suniverse > 0.
Suniverse = Ssystem + Ssurroundings
On a Molecular Level
TRANSLATIONAL MOTION: movement of
molecules
VIBRATIONAL MOTION: the movement of
atoms within the molecule.
ROTATIONAL MOTION: the molecules
spinning
Increasing Temperature Increases Entropy
3rd Law of Thermodynamics: the
entropy of a pure crystalline
substance at absolute zero is zero.
S(0K) = 0
In general, the entropy increases when:
Liquids or solutions are formed from solids
Gases are formed from either solids or
liquids
The number of molecules of gas increases
during a chemical reaction.
CaCO3(s)  CaO(s) + CO2(g)
N2(g) + 3H2(g)  2NH3(g)
Standard molar entropies: (So) absolute
entropies for substances in their standard
state. (J/mol-K)
1. Unlike enthalpies of formation, the standard molar
entropies of elements are not zero.
2. The So of gases are greater than those of liquids and
solids.
3. The So generally increases with increasing molar mass.
4. The So generally increase with the number of atoms in
the formula.
o
S
o
nS (products)
=
o
- mS (reactants)
Calculate So for the synthesis of ammonia:
N2(g) + 3H2(g)  2NH3(g)
o
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
Using Appendix C, calculate the
standard entropy change, for the
following reaction:
Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g)
180.4 J/K
C2H4(g) + H2(g)  C2H6(g)
NH3(g) + HCl(g)  NH4Cl(s)
-120.5 J/K
-284.6 J/K
GIBBS FREE ENERGY
The spontaneity of a reaction involves both
enthalpy and entropy. The relationship is
known as free energy.
G = H - TS
1. If G is negative, the reaction is
spontaneous in the forward direction.
2. If G is zero, the reaction is at
equilibrium.
3. If G is positive, the reaction in the
forward direction is nonspontaneous; work
must be supplied from the surroundings to
make it occur. However, the reverse reaction
will be spontaneous.
G
Go = nGfo(products) - mGfo(reactants)
N2(g) + 3H2(g)  2NH3(g)
-33.32 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
-800.7 kJ
o
H
Assuming no change for
o
o
and S , what happens to G
with an increase in temperature?
N2(g) + 3H2(g)  2NH3(g)
Calculate G at 298K for
a reaction mixture that
consists of 1.0 atm N2, 3.0
atm H2 and 0.50 atm NH3.
N2(g) + 3H2(g)  2NH3(g)
Use standard free energies
of formation to calculate
the equilibrium constant
o
K at 25 C for the reaction
involved in the Haber
process.