CHEMICAL THERMODYNAMICS

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CHEMICAL THERMODYNAMICS
The Second Law of Thermodynamics The is an
inherent direction in which any system not at
equilibrium moves
Processes that are spontaneous in one direction
are not spontaneous in the reverse direction
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CHEMICAL THERMODYNAMICS
Spontaneity, Enthalpy, and Entropy

• Reactions that are exothermic are generally
  spontaneous, ?H lt 0

• Reactions that are not exothermic may also be
  spontaneous

?H 0
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CHEMICAL THERMODYNAMICS
?H gt 0
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CHEMICAL THERMODYNAMICS
Spontaneity, Enthalpy, and Entropy
Spontaneity must, therefore, be a function of the
degree of randomness in a system.
Entropy (S) is a state function that describes
the randomness in a system such that, ?S ?S
final - ?S initial
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CHEMICAL THERMODYNAMICS
The Second Law of Thermodynamics tells us that
In any spontaneous process, there is always an
increase in the entropy of the universe
?S universe ?S system ?S surroundings gt 0
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CHEMICAL THERMODYNAMICS
A Molecular Interpretation of Entropy
translational
vibrational
rotational
The Third Law of Thermodynamics S(0 K) 0
The entropy of the lattice increases with
temperature because the number of possible energy
states in which the molecules or atoms are
distributed is larger
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CHEMICAL THERMODYNAMICS
Calculation of Entropy Changes
The standard entropy (S?) is expressed in units
of J/mol-K
For any given reaction aA bB ? cC dD
...
?S? cS?(p) dS ?(Q) - aS?(A) bS?(B)

Sample exercise Calculate the ?S for the
synthesis of ammonia from N2 and H2 N2(g)
3H2(g)? 2NH3(g)
?S? 2S?(NH3)- S?(N2) S?(H2)
?S? 2(192 J/mol-K)- 191.5 J/mol-K 3(130.58
J/mol-K)
?S? -198.2 J/K
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• Ice melting
• water condensing
• mixing salt water
• diffusion
• butane burning
• CH4g 2O2g? CO2g 2H2Og

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CHEMICAL THERMODYNAMICS
Gibbs Free Energy
Whether a reaction occurs spontaneously is
determined by the changes in enthalpy and entropy
for that reaction such that
I get it!!
G H -TS
The change in free energy is therefore
?G ?H - T?S

• If ?G is negative, the reaction is spontaneous in
  the forward direction

• If ?G is positive, the reaction is not
  spontaneous in the forward direction
• Work must be supplied from the surroundings to
  make it occur.

• If ?G is zero, the reaction is at equilibrium,
  there is no driving force

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CHEMICAL THERMODYNAMICS
Calculation of ?G? using standard free energy
change of formation
The standard free energy change (? G?) for any
reaction
aA bB ? cC dD ...
?G? cG?f (p) dG?f (Q) - aG?f (A)
bG?f (B)
Hip Deep In Alligators ?
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CHEMICAL THERMODYNAMICS
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CHEMICAL THERMODYNAMICS
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Free Energy and the Equilibrium Constant
What happens when you cant describe ?G under
standard conditions ?
?G ?G RT ln Q (R 8.314 J/K-mol)
When ?G 0 then ?G RT ln K
?G is negative K gt 1
?G is zero K 0
?G is positive K gt 1
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Sample exercise Calculate ?G at 298 K for the
following reaction if the reaction mixture
consists of 1.0 atm N2, 3.0 atm H2 ,and 1.0 atm
NH3 N2(g) 3H2 ? 2NH3
Q (1.0)2/ (1.0)(3.0)3 3.7 x 10-2
?G ?G RT ln Q where R 8.314 J/K-mol
?G -33.32 kJ (8.314 J/K-mol)(298 K)ln(3.7 x
10-2)
?G -41.49 kJ
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CHEMICAL THERMODYNAMICS
1st Law of Thermodynamics
Chapter 5
Hesss Law
Ek ? mv2
?E E final - E initial
?E q w
if ? V 0 then, ?E q v
?H ? ?E
?H H products - H reactants
q n C?T where C is J/mol-?C
q m S?T where is C J/g-?C
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CHEMICAL THERMODYNAMICS
2nd Law of Thermodynamics
Chapter 19
3rd Law of Thermodynamics
?S ?S final - ?S initial
?S universe ?S system ?S surroundings gt 0
For any given reaction aA bB ? cC dD
...
?S? cS?(p) dS ?(Q) - aS?(A) bS?(B)
where S? J/mol-K

• If ?G spontaneous

?G ?H - T?S

• If ?G not spontaneous

• If ?G zero, K 0

for any reaction
aA bB ? cC dD ...
?G? cG?f (p) dG?f (Q) - aG?f (A)
bG?f (B)
?G ?G RT ln Q where R 8.314 J/K-mol
?G is negative K gt 1
?G is zero K 0
if ?G 0 then ?G RT ln K
?G is positive K gt 1