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Chemical
Thermodynamics
First Law of Thermodynamics
• You will recall 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.
Chemical
Thermodynamics
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
Chemical
Thermodynamics
Spontaneous Processes
Processes that are
spontaneous in one
direction are
nonspontaneous in
the reverse
direction.
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
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.
Changes are
infinitesimally small in
a reversible process.
Chemical
Thermodynamics
Irreversible Processes
• Irreversible processes cannot be undone by
exactly reversing the change to the system.
• All Spontaneous processes are irreversible.
• All Real processes are irreversible.
Chemical
Thermodynamics
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
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
Entropy
• Like total energy, E, and enthalpy, H,
entropy is a state function.
• Therefore,
S = Sfinal  Sinitial
Chemical
Thermodynamics
Entropy
• For a process occurring at constant
temperature (an isothermal process):
qrev = the heat that is transferred when the
process is carried out reversibly at a constant
temperature.
T = temperature in Kelvin.
Chemical
Thermodynamics
Second Law of Thermodynamics
The second law of thermodynamics:
The entropy of the universe does not
change for reversible processes
and
increases for spontaneous processes.
Reversible (ideal):
Irreversible (real, spontaneous):
Chemical
Thermodynamics
Second Law of Thermodynamics
Reversible (ideal):
Irreversible (real, spontaneous):
“You can’t break even”
Chemical
Thermodynamics
Second Law of Thermodynamics
The entropy of the universe increases (real,
spontaneous processes).
But, entropy can decrease for individual systems.
Reversible (ideal):
Irreversible (real, spontaneous):
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
microstates, W, associated with it.
• Entropy is
S = k lnW
where k is the Boltzmann constant, 1.38  1023 J/K.
Chemical
Thermodynamics
Entropy on the Molecular Scale
Implications:
• more particles
-> more states -> more entropy
• higher T
-> more energy states -> more entropy
• less structure (gas vs solid)
-> more states -> more entropy
Chemical
Thermodynamics
Entropy on the Molecular Scale
• The number of microstates and,
therefore, the entropy tends to increase
with increases in
Temperature.
Volume (gases).
The number of independently moving
molecules.
Chemical
Thermodynamics
Entropy and Physical States
• Entropy increases with
the freedom of motion
of molecules.
• Therefore,
S(g) > S(l) > S(s)
Chemical
Thermodynamics
Solutions
Dissolution of a solid:
Ions have more entropy
(more states)
But,
Some water molecules
have less entropy
(they are grouped
around ions).
Usually, there is an overall increase in S.
(The exception is very highly charged ions that
make a lot of water molecules align around them.)
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
Third Law of Thermodynamics
The entropy of a pure crystalline
substance at absolute zero is 0.
Chemical
Thermodynamics
Third Law of Thermodynamics
The entropy of a pure crystalline
substance at absolute zero is 0.
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Chemical
Thermodynamics
Standard Entropies
• These are molar entropy
values of substances in
their standard states.
• Standard entropies tend
to increase with
increasing molar mass.
Chemical
Thermodynamics
Standard Entropies
Larger and more complex molecules have
greater entropies.
Chemical
Thermodynamics
Entropy Changes
Entropy changes for a reaction can be
calculated the same way we used for H:
S° for each component is found in a table.
Note for pure elements:
Chemical
Thermodynamics
Practical uses: surroundings & system
Entropy Changes in Surroundings
• Heat that flows into or out of the system
also changes the entropy of the
surroundings.
• For an isothermal process:
Chemical
Thermodynamics
Practical uses: surroundings & system
Entropy Changes in Surroundings
• Heat that flows into or out of the system also changes
the entropy of the surroundings.
• For an isothermal process:
• At constant pressure, qsys is simply
H for the system.
Chemical
Thermodynamics
Link S and H: Phase changes
A phase change is isothermal
(no change in T).
Entropy
system
For water:
Hfusion = 6 kJ/mol
Hvap = 41 kJ/mol
If we do this reversibly: Ssurr = –Ssys
Chemical
Thermodynamics
Standard Entropy
• Standard entropy: absolute entropy of a
substance at 1 atm (typically at 25C)
• Complete list in Appendix 2 of text
What do you notice about entropy values
for elements and compounds?
• Units: J/K·mol
Chemical
Thermodynamics
Entropies
Chemical
Thermodynamics
Trends in Entropy
• Entropy for gas phase is greater than that of
liquid or solid of same substance
I2 (g) has greater entropy than I2 (s)
• More complex structures have greater
entropy
C2H6 (g) has greater entropy than CH4 (g)
• Allotropes - more ordered forms have lower
entropy
Diamond has lower entropy than graphite
Chemical
Thermodynamics
Entropy Changes in a System
Qualitative
• Ssolid < Sliquid
Chemical
Thermodynamics
Entropy Changes in a System
Qualitative
• Sliquid < Svapor
Chemical
Thermodynamics
Entropy Changes in a System
Qualitative
• Spure < Saqueous
Chemical
Thermodynamics
Entropy Changes in a System
Qualitative
Slower temp < Shigher temp
Chemical
Thermodynamics
Entropy Changes in a System
Qualitative
• Sfewer moles < Smore moles
Chemical
Thermodynamics
Entropy Changes in a System
Qualitative
Determine the sign of S for the following
(qualitatively)
1. Liquid nitrogen evaporates
2. Two clear liquids are mixed and a
solid yellow precipitate forms
3. Liquid water is heated from 22.5 C
to 55.8 C
Chemical
Thermodynamics
Entropy Changes in the System
Entropy can be calculated from the table
of standard values just as enthalpy
change was calculated.
S
rxn = nS
products  mS
reactants
Chemical
Thermodynamics
Standard Entropy
• Calculate the standard entropy change
for the following using the table of
standard values. (first, predict the sign
for S qualitatively)
2NH3(g)  N2(g) + 3H2(g)
nS
products
Chemical
Thermodynamics
Standard Entropy
2NH3(g)  N2(g) + 3H2(g)
S
rxn = nS
products  mS
reactants
= [(1)(191.5 J/K · mol) + (3)(131.0 J/K · mol)]
- [(2)(193.0 J/K · mol)]
= 584.5 J/K · mol - 386 J/K · mol
S
rxn = 198.5 J/K · mol (Entropy increases)
(2 mol gas  4 mol gas)
Chemical
Thermodynamics
Your Turn!
• Calculate the standard entropy change
for the following using the table of
standard values. (first, predict the sign
for S qualitatively)
2H2(g) + O2(g)  2H2O (g)
Chemical
Thermodynamics
Entropy Changes in the
Surroundings
• Change in entropy of surroundings is
directly proportional to the enthalpy of
the system.
Ssurroundings   Hsystem
Notice: exothermic process corresponds
to positive entropy change in
surroundings
Chemical
Thermodynamics
Entropy Changes in the
Surroundings
• Change in entropy of surroundings is
inversely proportional to temperature
Ssurroundings  1 / T
Combining the two expressions:
sys
surr
H
S
T

 
Chemical
Thermodynamics
Entropy Changes
If the entropy change for a system is
known to be 187.5 J/Kmol and the
enthalpy change for a system is known
to be 35.8 kJ/mol, is the reaction
spontaneous?
Spontaneous if: Suniv= Ssys + Ssurr > 0
sys
surr
H
S
T

 
Chemical
Thermodynamics
Entropy Changes
Is the reaction spontaneous?
Suniv= -187.5 + 120.0 < 0 so the reaction
is non-spontaneous
sys
35,800 J/mol
120.0J/K mol
298 K
S
   
Chemical
Thermodynamics
Entropy Change in the Universe
• The universe is composed of the system and
the surroundings.
Therefore,
Suniverse = Ssystem + Ssurroundings
• For spontaneous processes
Suniverse > 0
Practical uses: surroundings & system
Chemical
Thermodynamics
Practical uses: surroundings & system
= – Gibbs Free Energy
Chemical
Thermodynamics
Practical uses: surroundings & system
= – Gibbs Free Energy
Make this equation nicer:
Chemical
Thermodynamics
TSuniverse is defined as the Gibbs free
energy, G.
For spontaneous processes: Suniverse > 0
And therefore: G < 0
Practical uses: surroundings & system
…Gibbs Free Energy
G is easier to determine than Suniverse.
So:
Use G to decide if a process is spontaneous.
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
Standard Free Energy Changes
Standard free energies of formation, Gf
are analogous to standard enthalpies of
formation, Hf.
G can be looked up in tables,
or
calculated from S° and H.
Chemical
Thermodynamics
Free Energy Changes
Very key equation:
This equation shows how G changes with
temperature.
(We assume S° & H° are independent of T.)
Chemical
Thermodynamics
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 comes from the entropy term.
Chemical
Thermodynamics
Free Energy and Temperature
By knowing the sign (+ or -) of S and H,
we can get the sign of G and determine if a
reaction is spontaneous.
Chemical
Thermodynamics
Free Energy and Equilibrium
Remember from above:
If G is 0, the system is at equilibrium.
So G must be related to the equilibrium
constant, K (chapter 15). The standard free
energy, G°, is directly linked to Keq by:
Chemical
Thermodynamics
Free Energy and Equilibrium
Under non-standard conditions, we need to use
G instead of G°.
Q is the reaction quotiant from chapter 15.
Note: at equilibrium: G = 0.
away from equil, sign of G tells which way rxn goes
spontaneously.
Chemical
Thermodynamics
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.
Chemical
Thermodynamics
Copyright McGraw-Hill 2009
Example
For a reaction in which H = 125 kJ/mol
and S = 325 J/Kmol, determine the
temperature in Celsius above which the
reaction is spontaneous.
385 K  273 = 112C
125 kJ/mol
385 K
0.325 kJ/K×mol
H
T
S

  

Chemical
Thermodynamics
Copyright McGraw-Hill 2009
Standard Free Energy Changes
Free energy can be calculated from the
table of standard values just as enthalpy
and entropy changes.
G
rxn = nG
products  mG
reactants
Chemical
Thermodynamics
Standard Free Energy Changes
Calculate the standard free-energy
change for the following reaction.
2KClO3(s)  2KCl(s) + 3O2(g)
G
rxn = nG
products  mG
reactants
= [2(408.3 kJ/mol) + 3(0)]  [2(289.9
kJ/mol)]
= 816.6  (579.8) = 236.8 kJ/mol
(spont)
Chemical
Thermodynamics
Free Energy and Chemical
Equilibrium
• Reactions are almost always in
something other than their standard
states.
• Free energy is needed to determine if a
reaction is spontaneous or not.
• How does free energy change with
changes in concentration?
Chemical
Thermodynamics
Free Energy and Equilibrium
G = G° + RT ln Q
• G = non-standard free energy
• G° = standard free energy (from
tables)
• R = 8.314 J/K·mole
• T = temp in K
• Q = reaction quotient
Chemical
Thermodynamics
Free Energy and Equilibrium
• Consider the reaction,
H2(g) + Cl2(g)  2 HCl(g)
How does the value of G change when
the pressures of the gases are altered
as follows at 25 C?
• H2 = 0.25 atm; Cl2 = 0.45 atm;
HCl = 0.30 atm
Chemical
Thermodynamics
Free Energy and Equilibrium
2 2
( ) (0.30)
HCl 0.80
( ) ( ) (0.25) (0.45)
H Cl
2 2
P
QP P P
  
First, calculate standard free energy:
H2(g) + Cl2(g)  2 HCl(g)
G° = [2(95.27 kJ/mol)]  [0 + 0] =
190.54 kJ/mol
Second, find Q:
Chemical
Thermodynamics
Free Energy and Equilibrium
Solve: G = G° + RT ln Q
G = 190,540 J/mol +
(8.314J/K·mol)(298 K) ln (0.80)
G =  191.09 kJ/mol (the reaction
becomes more spontaneous - free
energy is more negative)
Chemical
Thermodynamics
Your Turn!
• Consider the reaction,
O2(g) + 2CO(g)  2CO2(g)
How does the value of G change when
the pressures of the gases are altered
as follows at 25 C?
• O2 = 0.50 atm; CO = 0.30 atm;
CO2 = 0.45 atm
Chemical
Thermodynamics
Relationship Between G° and K
• At equilibrium, G = 0 and Q = K
• The equation becomes:
0 = G° + RT ln K
or
G° = – RT ln K
Chemical
Thermodynamics
Relationship Between G° and K
Chemical
Thermodynamics
Relationship Between G° and K
• Using the table of standard free
energies, calculate the equilibrium
constant, KP, for the following reaction
at 25 C.
2HCl(g) H2(g) + Cl2(g)
Chemical
Thermodynamics
Relationship Between G° and K
First, calculate the G°:
= [0 + 0]  [2(95.27 kJ/mol)]
= 190.54 kJ/mol (non-spontaneous)
Substitute into equation:
190.54 kJ/mol =  (8.314 x 103 kJ/K·mol)(298
K) ln KP
76.90 = ln KP = 3.98 x 1034
K < 1 reactants are favored
Chemical
Thermodynamics
Your Turn!
• Calculate the value of the equilibrium
constant, KP, for the following reaction
at 25 C.
2NO2(g) N2O4(g)
Chemical
Thermodynamics
18.6 Thermodynamics in Living
Systems
• Coupled reactions
• Thermodynamically favorable reactions
drives an unfavorable one
• Enzymes facilitate many
nonspontaneous reactions
Chemical
Thermodynamics
Thermodynamics in Living Systems
• Examples:
C6H12O6 oxidation: G° = 2880 kJ/mol
ADP  ATP G° = 31 kJ/mol
Synthesis of proteins: (first step)
alanine + glycine  alanylglycine
G° = 29 kJ/mol
Chemical
Thermodynamics
Thermodynamics in Living Systems
• Consider the coupling of two reactions:
ATP + H2O + alanine + glycine  ADP +
H3PO4 + alanylglycine
Overall free energy change:
G° = 31 kJ/mol + 29 kJ/mol = 2
kJ/mol
Protein synthesis is now favored.
Chemical
Thermodynamics
ATP-ADP Interconversions
Chemical
Thermodynamics
Key Points
• Spontaneous vs nonspontaneous
• Relate enthalpy, entropy and free
energy
• Entropy
Predict qualitatively
Calculate from table of standard values
Calculate entropy for universe
Chemical
Thermodynamics
Key Points
• Free energy
Calculate from table of standard values
Calculate from Gibbs equation
Calculate non-standard conditions
Calculate in relationship to equilibrium
constant

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