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
CO2(s)  CO2(g)
2) Dissolving occurs (solution forms):
Example:
NaCl(s)  Na+(aq) + Cl-(aq)
Entropy
3) Temperature increases
Example:
Fe(s) at 0 oC  Fe(s) at 25 oC
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:
2NH3(g)  N2(g) + 3H2(g)
CaCO3(s)  CaO(s) + CO2(g)
N2O4(g)  2 NO2(g)
This one is difficult to predict:
N2(g) + O2(g)  2 NO(g)
Entropy
6) When there are more moles
Example:
1 mole H2O(g)  2 moles H2O(g)
7) When there are more atoms per molecule
Examples:
1 mole Ar(g)  1 mole HCl(g)
1 mole NO2(g)  1 mole N2O4(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 (So) are
expressed in J/mol-K.
• Note: Increase with
increasing molar mass.
Standard Entropies
Larger and more complex molecules have
greater entropies.
Entropy Changes
DSo = S n Soproducts - S m Soreactants
Be careful: S°units are in J/mol-K
Note for pure elements:
Gibbs Free Energy
DGo = DHo – TDSo
Use DG to decide if a process is spontaneous
DG = negative value = spontaneous
DG = zero = at equilibrium
DG = positive value = not spontaneous
Note: equation can be used w/o the o too.
Gibbs Free Energy
1. If DG is negative
DG = maximum amt of energy ‘free’ to do
work by the reaction
2. If DG is positive
DG = minimum amt of work needed
to make the reaction happen
DGo = DHo – TDSo
Gibbs Free Energy
DGo = DHo – TDSo
In our tables, units are:
DGo = kJ/mol
DHo = kJ/mol
DSo = J/mol-K
Free Energy and Temperature
• There are two parts to the free energy
equation:
 DH— the enthalpy term
 TDS — 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, DH = negative
 Maximum Entropy
More disorder, DS = positive
• Because: DGo = DHo – TDSo
- = (-) - (+)
Free Energy and Temperature
By knowing the sign (+ or -) of DS and DH,
we can get the sign of DG and determine if a
reaction is spontaneous.
At Equilibrium
DGo = DHo – TDSo
DGo = zero
Or:
Therefore: DHo = TDSo
DSo = DHo
T
Use this equation when asked to calculate enthalpy of
vaporization or enthalpy of fusion.
Standard Free Energy Changes
Be careful: Values for DGf are in kJ/mol
DGo = S n Gof
products
- S m Go f
reactants
DG can be looked up in tables or calculated
from S° and DH.
DGo = DHo – TDSo
Free Energy and Equilibrium
Remember from above:
If DG is 0, the system is at equilibrium.
So DG must be related to the equilibrium
constant, K (chapter 15). The standard free
energy, DG°, is directly linked to Keq by:
o
DG
= - RT ln K
Where R = 8.314 J/mol-K
Relationships
o
DG
= - 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.
Free Energy and Equilibrium
Under non-standard conditions, we need to use
DG instead of DG°.
Q is the reaction quotient from chapter 15.