Chapter 19
Principles of
Chemical Reactivity:
Entropy and Free
Energy
Jeffrey Mack
California State University,
Sacramento
Entropy & Free Energy
Chemical Thermodynamics
provides us with information
about equilibrium and whether
or not a reaction is
Spontaneous.
Chemical Kinetics provides us
with information about the rate
of reaction and how the
reaction proceeds.
Entropy & Free Energy
Chemical Thermodynamics provides us with information
about equilibrium and whether or not a reaction is
Spontaneous.
Spontaneous changes occur only in the direction
that leads to equilibrium.
Systems never change spontaneously in a direction
that takes them farther from equilibrium.
Example: Heat Transfer
Heat (thermal energy) always flows spontaneously from a
hot object to a cold object.
The process occurs until thermal equilibrium is achieved,
that is when both objects are at the same temperature.
Spontaneous Process
Additional Examples:
• Solvation of a Soluble salt: NH4NO3(s) dissolves
spontaneously even the process is endothermic (H
> 0)
• Expansion of a gas and Diffusion. Gasses mix to
form homogeneous mixtures spontaneously.
• Certain Phase Changes: Ice melts spontaneously
above 0oC. Water evaporates even though the
enthalpy of vaporization is endothermic.
• Certain Chemical Reactions: Na(s) reacts
vigorously when dropped in water. Iron rusts when
exposed to the atmosphere.
Spontaneous Process
But many spontaneous reactions or processes
are endothermic or even have ∆H  0.
NH4NO3(s) + heat  NH4NO3(aq)
∆H = 0
Thermodynamics
A Review of Concepts of Thermodynamics
• First law of thermodynamics: The law of conservation
of energy; energy cannot be created or destroyed.
• State Function: Quantity in which its determination is
path independent.
• U = q + w: The change in internal energy of a system
is a function of heat and work done on or by the system.
• H: Heat transferred at constant pressure.
• Exothermic Process: H < 0
• Endothermic Process: H > 0
Thermodynamics
Enthalpy alone does not predict spontaneity:
Some processes are energetically favored (rH < 0)
but not spontaneous.
Equilibrium alone cannot determine spontaneity:
Some processes are favored based on Equilibrium (K
>> 1) yet they are non-spontaneous.
There must be another factor that plays a role in
determination of spontaneity!
Thermodynamics & Kinetics
Diamond is
thermodynamically favored to
convert to graphite, but not
kinetically favored.
Paper burns once the
reaction is initiates. The
process is product-favored
& kinetically favored.
Kinetics
Thermodynamics & Kinetics
Reactants
Products
Thermodynamics
Thermodynamics & Kinetics
Factors that Affect Spontaneity
(Thermodynamic favorability):
1. Enthalpy: Comparison of bond energy (H)
2. Entropy: Randomness vs. Order of a system
(S)
In general, enthalpy is more important
than entropy.
Dispersal of Energy: Entropy
• In a spontaneous processes the
energy of the final state is more
dispersed.
• The system moves to a higher state
of disorder.
• The thermodynamic quantity
associated with disorder and
energy dispersal is called
ENTROPY, S.
• The 2nd law of thermodynamics
states that a spontaneous process
results in an increase in the entropy
of the universe. S > 0
Reaction of K
and water
Dispersal of Energy: Entropy
Observation of a spontaneous process shows
that it is associated with a dispersal of energy.
Energy Dispersal
Dispersal of Energy: Entropy
The change in entropy for a spontaneous process is
given by:
qrev
DS =
T
Where qrev is the heat gained or lost by the system
during the process and T is the absolute temperature.
A reversible process can be returned to its original
state. (chemical equilibrium), an irreversible process
cannot.
Example: The breaking of a coffee mug into many
pieces is an irreversible process.
Dispersal of Energy: Entropy
To begin, particle 1 has 2 units of energy and 2-4
have none.
Dispersal of Energy: Entropy
Particle 1 can transfer one unit of energy to particle 2,
then the other to 3 or 4.
Dispersal of Energy: Entropy
Particle 2 can initially have two units of energy.
Dispersal of Energy: Entropy
Particle 2 transfer one unit to particles 4 or 3. (2 to 1
has already been counted.)
Dispersal of Energy: Entropy
Particle 4 can initially have two units of energy.
Dispersal of Energy: Entropy
Particle 4 transfer one unit to particle 3. (4 to 1 & 4 to
2 have already been counted.)
Dispersal of Energy: Entropy
Particle 3 can start with two units of energy. Energy
transfers between particle 3 were previously counted.
Dispersal of Energy: Entropy
Each unique combination that results in a dispersion of energy
is called a microstate. There are 10 microstates in this
system. The greater the number of microstates, the greater the
entropy of the system.
Dispersal of Energy: Entropy
Dispersal of Energy: Entropy
As the size of the
container increases, the
number of microstates
accessible to the system
increases. Therefore the
entropy of the system
increases.
Dispersal of Energy: Entropy
• The entropy of liquid
water is greater than
the entropy of solid
water (ice) at 0˚ C.
• Energy is more
dispersed in liquid
water than in solid
water due to the lack
of an ordered network
as in the solid state.
Entropy & States of Matter
So (J/K•mol)
H2O(liq)
69.95
H2O(gas) 188.8
S (solids) < S (liquids) < S (gases)
Energy dispersal
Entropy & States of Matter
S˚(Br2 liq) < S˚(Br2 gas)
S˚(H2O sol) < S˚(H2O liq)
Dispersal of Energy: Entropy
Entropy and Microstates:
As the number of microstates increases, so
does the entropy of the system.
S = klnW
k = Boltzman’s constant (1.381 1023 J/K)
W = the number of microstates
Entropy, Entropy Change, & Energy
Dispersal: A Summary
Entropy and Microstates:
The change in entropy associated with a
process is a function of the number of final and
initial microstates of the system.
DS = Sfinal = Sinitial
DS = k × ln Wfinal - k × ln Wfinal
æ Wfinal ö
DS = k × ln ç
÷
W
è initial ø
If Wfinal > Winital, S > 0
If Wfinal < Winital, S < 0
Dispersal of Energy: Entropy
When a solute dissolves in a solvent the process is
spontaneous owing to the increase in entropy.
Matter (and energy) are more dispersed. The number
of microstates is increased.
Entropy Measurements & Values
The entropy of a substance increases with
temperature.
Molecular motions of
heptane, C7H16
Molecular motions of
heptane at different temps.
Entropy Measurements & Values
An Increase in molecular
complexity generally leads
to increase in S.
Entropy Measurements & Values
Entropies of ionic solids depend on
coulombic attractions.
S° (J/K•mol)
Mg2+ & O2-
Na+ & F-
MgO
26.9
NaF
51.5
Standard Molar Entropies
Defined by Ludwig Boltzmann, the third law states that a
perfect crystal at 0 K has zero entropy; that is, S =0.
The entropy of an element or compound under any other
temperature and pressure is the entropy gained by converting
the substance from 0 K to those conditions.
To determine the value of S, it is necessary to measure the
energy transferred as heat under reversible conditions for the
conversion from 0 K to the defined conditions and then to use
Equation 19.1
qrev
DS =
T
Because it is necessary to add energy as heat to raise the
temperature, all substances have positive entropy values at
temperatures above 0 K.
Standard Molar Entropies
The standard molar entropy, S°, of a substance is the entropy
gained by converting 1 mol of it from a perfect crystal at 0 K to
standard state conditions (1 bar, 1m for a solution) at the specified
temperature.
Entropy Changes for Phase
Changes
For a phase change,
qrev
DS =
T
where q = heat transferred in the
phase change
For H2O (liq)
H2O(g)
J
DH = q = + 40,700
mol
qrev
40,700 J / mol
J
DS =
=
= + 109
Tboil
373.15 K
mol × K
Entropy & Temperature
S increases
slightly with T
S increases a large
amount with phase
changes
Determining Entropy Changes in
Physical & Chemical Processes
Standard molar entropy values can be used to
calculate the change in entropy that occurs in various
processes under standard conditions.
The standard entropy change for a reaction (rS°)
can be found in the same manner as rH° were:
S   n  S(products)  m  S(reactants)
Where n & m are the stoichiometric balancing
coefficients.
This calculation is valid only under reversible
conditions.
Determining Entropy Changes in
Physical & Chemical Processes
Problem:
Consider the reaction of hydrogen and oxygen to form liquid
water.
What is the standard molar entropy for the reaction?
Determining Entropy Changes in
Physical & Chemical Processes
Problem:
Consider the reaction of hydrogen and oxygen to form liquid
water.
What is the standard molar entropy for the reaction?
r S   n  S(products)  m  S(reactants)
Determining Entropy Changes in
Physical & Chemical Processes
Problem:
Consider the reaction of hydrogen and oxygen to form liquid
water.
What is the standard molar entropy for the reaction?
r S   n  S(products)  m  S(reactants)
2H2 (g)  O2 (g)  2H2O(l)
J 
J
J
J
r S  2  69.95   2  130.7  205.1   326.6
K 
K
K
K
Determining Entropy Changes in
Physical & Chemical Processes
Problem:
Consider the reaction of hydrogen and oxygen to form liquid
water.
What is the standard molar entropy for the reaction?
2H2 (g) + O2 (g) ® 2H2O(l)
J æ
J
Jö
J
Dr S = 2 ´ 69.6 - ç 2 ´ 130.7 + 205.3 ÷ = -326.9
K è
K
Kø
K
o
The enthalpy change is negative (net decrease in dispersion)
due to the change in number of moles: 3 reduced to 2
nd
2
Law of Thermodynamics
The second law of thermodynamics states:..
S°universe = ∆S°system + ∆S°surroundings
Any change in entropy for the system plus the entropy
change for the surroundings must equal the overall
change in entropy for the universe.
A process is considered to be spontaneous under
standard conditions if S°(universe) is greater than
zero.
nd
2
Law of Thermodynamics
The solution process for
NH4NO3 (s) in water is an
entropy driven process.
∆S°universe =
∆S°system +
∆S°surroundings
nd
2
Law of Thermodynamics
Calculating S°Surroundings:
nd
2
Law of Thermodynamics
Calculating S°Surroundings:
Ssystem 
qsystem
T
Ssurroundings 
qsurroundings
T
nd
2
Law of Thermodynamics
Calculating S°Surroundings:
Ssystem 
qsystem
T
qsystem + qsurroundings = 0
Ssurroundings 
qsurroundings
T
qsystem = -qsurroundings
nd
2
Law of Thermodynamics
Calculating S°Surroundings:
Ssystem 
qsystem
T
qsystem + qsurroundings = 0
Ssurroundings 
qsurroundings
T
qsystem = -qsurroundings
at constant pressure, qsystem = rH°system
nd
2
Law of Thermodynamics
Calculating S°Surroundings:
Ssystem 
qsystem
Ssurroundings 
T
qsystem + qsurroundings = 0
qsurroundings
qsystem = -qsurroundings
at constant pressure, qsystem = rH°system
Therefore :
and
T
qsurroundings  rHsystem
Ssurroundings  
rHsystem
T
nd
2
Law of Thermodynamics
Problem:
Calculate the entropy change for the surroundings for
the reaction: 2H2(g) + O2(g)  2H2O(l)
nd
2
Law of Thermodynamics
Problem:
Calculate the entropy change for the surroundings for
the reaction: 2H2(g) + O2(g)  2H2O(l)
Ssurroundings  
rHsystem
T
The enthalpy change for the reaction is calculated
using standard molar enthalpies of formation:
nd
2
Law of Thermodynamics
Problem:
Calculate the entropy change for the surroundings for
the reaction: 2H2(g) + O2(g)  2H2O(l)
Ssurroundings  
rHsystem
T
The enthalpy change for the reaction is calculated
using standard molar enthalpies of formation:
Ssurroundings
103 K
571.7kJ 
1kJ  1917 J

K
298.15K
nd
2
Law of Thermodynamics
Problem:
Calculate the entropy change for the surroundings for
the reaction: 2H2(g) + O2(g)  2H2O(l)
As calculated previously:
Ssurroundings  1917 J
K
S°universe = 327 J/K + 1917 J/K =1590. J/K
Ssystem  327 J
K
nd
2
Law of Thermodynamics
Problem:
Calculate the entropy change for the surroundings for
the reaction: 2H2(g) + O2(g)  2H2O(l)
S°universe = 327 J/K + 1917 J/K = 1590. J/K
The entropy of the universe increases ( > 0) therefore
the process is spontaneous at standard state
conditions.
The process is spontaneous due to the entropy
change in the surroundings, not the system.
Spontaneous or Not?
Gibbs Free Energy, G
The method used so far to determine whether a
process is spontaneous required evaluation of two
quantities, S°(system) and S°(surroundings).
J. Willard Gibbs asserted that that the
maximum non-PV work available to the
system must be a function of Enthalpy
and Entropy:
G = H – TS
J. Willard Gibbs
1839-1903
G = Gibbs Free energy of the system, H = system
enthalpy and S = the entropy of the system.
Gibbs Free Energy, G
Since it is impossible to measure individual values of
enthalpy, we often express free energy in terms of the
changes of thermodynamic quantities.
(G = H – TS)
At constant temperature:
G  H  TS
G° = Gibbs Free energy change of the system,
H° = enthalpy change and S° = the entropy
change at SS conditions.
Gibbs Free Energy, G
G  H  TS
If the reaction is exothermic (H < 0)
And the change in entropy is positive (H > 0)
at a given temperature, then S < 0.
We then can assert that if G < 0 that the
reaction is spontaneous as well as product
favored!
Spontaneity is a function of energy and
dispersion!
Gibbs Free Energy, G
G  H  TS
H
S
G
Reaction

+

Product Favored
+

+
Reactant Favored
Temperature
dependant
Temperature
dependant


?
+
+
?
rGo & Equilibrium
Since rG° is related to S°universe, it follows
that:
• If rG° < 0: The process is spontaneous in the
direction written under standard conditions.
• If rG° = 0: The process is at equilibrium under
standard conditions and K=1
• If rG° > 0: The process is non-spontaneous in
the direction written under standard conditions.
• Conclusion: A reaction proceeds spontaneously
toward the minimum in free energy, which
corresponds to equilibrium.
∆G, ∆G°, Q, & K
Product Favored Reactions, ∆G° negative, K > 1
Q < K: Heading to
equilibrium G < 0
Q = K: At
equilibrium G = 0
Q > K: Heading
away from
equilibrium G > 0
∆G, ∆G°, Q, & K
•
•
•
•
Product-favored
2 NO2  N2O4
∆rG° = – 4.8 kJ
State with both reactants
and products present is
more stable than
complete conversion.
• K > 1, more products
than reactants.
∆G, ∆G°, Q, & K
Reactant Favored Reactions, ∆G° positive, K < 1
Q < K: Heading to
equilibrium G < 0
Q = K: At
equilibrium G = 0
Q > K: Heading
away from
equilibrium G > 0
∆G, ∆G°, Q, & K
• Reactant-favored
• N2O4  2 NO2
∆rG° = +4.8 kJ
• State with both
reactants and products
present is more stable
than complete
conversion.
• K < 1, more reactants
than products
∆G, ∆G°, Q, & K
rG° represents the free energy change for a
process at standard state conditions. (Equilibrium)
What if this is not the case?
Under nonstandard conditions:
r G  r G  RT ln(Q)
Where R is the
gas law constant
and Q =
C] [D]
[
Q=
a
b
[A ] [B]
c
for aA + bB
d
cC + dD
∆G, ∆G°, Q, & K
At equilibrium, we know that rG = 0 and Q = K
Therefore:
r G  r G  RT ln(Q)
0  r G  RT ln(K)
r G  RT ln(K)
So knowing one quantity yields the other.
∆G, ∆G°, Q, & K
Problem:
rG° for the formation of ammonia at SS conditions is 16.37
kJ/mol. What is the value of Kp at this temperature and
pressure? (1 mol, 20 °C and 1 atm)
∆G, ∆G°, Q, & K
Problem:
rG° for the formation of ammonia at SS conditions is 16.37
kJ/mol. What is the value of Kp at this temperature and
pressure? (1 mol, 20 °C and 1 atm)
3
1
H2 (g)  N2 (g)
2
2
NH3 (g)
r G  RT ln(K p )
∆G, ∆G°, Q, & K
Problem:
rG° for the formation of ammonia at SS conditions is 16.37
kJ/mol. What is the value of Kp at this temperature and
pressure? (1 mol, 20 °C and 1 atm)
r G  RT ln(K p )
r G
ln(K)  
RT

kJ 103 J 
16.37




G

 r 
mol
1
kJ
K p  exp  
 740.

  exp 
J
 RT 
 8.314
 298 K 
mol  K


∆G, ∆G°, Q, & K
Problem:
rG° for the formation of ammonia at SS conditions is 16.37
kJ/mol. What is the value of Kp at this temperature and
pressure? (1 mol, 20 °C and 1 atm)
r G  RT ln(K p )
K >> 1, product
favored as
predicted by G
r G
ln(K)  
RT

kJ 103 J 
16.37




G

 r 
mol
1
kJ
K p  exp  
 740.

  exp 
J
 RT 
 8.314
 298 K 
mol  K


Summary
The relation of ∆rG, ∆rG°, Q, K, reaction spontaneity,
and product- or reactant favorability.
Calculating & Using Free Energy
Standard Free Energy of Formation
C (graphite)
+ 2H2(g)
 CH4(g)
rH°
(kJ/mol)
0
0
 74.9
S° (J/K)
+56
+130.7
+ 186.3
Calculating & Using Free Energy
Standard Free Energy of Formation
C (graphite)
+ 2H2(g)
 CH4(g)
rH°
(kJ/mol)
0
0
 74.9
S° (J/K)
+56
+130.7
+ 186.3
rH   n   f H(products)  m   f H(reactants)
r S   n  S(products)  m  S(reactants)
Calculating & Using Free Energy
Standard Free Energy of Formation
C (graphite)
+ 2H2(g)
 CH4(g)
rHo
(kJ/mol)
0
0
 74.9
So (J/K)
+56
+130.7
+ 186.3
rH  74.9 kJ
J
r S  80.7
K
Calculating & Using Free Energy
Standard Free Energy of Formation
rH  74.9 kJ
J
 rS  80.7
K
r G  rH  T r S
J  1 kJ 


r G  74.9 kJ   298 K   80.7   3 
K  10 J 


r G  50.9 kJ
mol
rG° is negative at 298 K, so the reaction is
predicted to be spontaneous under standard
conditions at this temperature. It is also predicted to
be product-favored at equilibrium.
Gibbs Free Energy, G
G  H  TS
Under reversible conditions, both enthalpy and
entropy are state functions. It follows that the
Gibbs free energy must also be.
Therefore we can write that:
r G   n   f G(products)  m   f G(reactants)
Free Energies of Formation
Note that ∆fG° for an element = 0
Free Energies of Formation
Using the fG° found in appendix L of your text,
calculate rG° for the following reaction:
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
Free Energies of Formation
Using the fG° found in appendix L of your text,
calculate rG° for the following reaction:
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
r G   n   f G(products)  m   f G(reactants)
Free Energies of Formation
Using the fG° found in appendix L of your text,
calculate rG° for the following reaction:
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
r G   n   f G(products)  m   f G(reactants)
r G  4   f G NO2 (g)  6   f G H2O(g)   4   f G NO2 (g)   0
Free Energies of Formation
Using the fG° found in appendix L of your text,
calculate rG° for the following reaction:
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)
r G   n   f G(products)  m   f G(reactants)
r G  4   f G NO2 (g)   6   f G H2O(g)    4   f G NO2 (g)   0 
r G  4 mol  (51.23 kJ / mol)  6  mol ( 228.59 kJ / mol)
 4 mol  ( 16.37 kJ / mol)
rG° = –1101.14 kJ/mol (product-favored
Calculating ∆rG°
NH4NO3(s) + heat  NH4NO3(aq)
Is the dissolution of ammonium nitrate productfavored?
If so, is it enthalpy- or entropy-driven?
Free Energy & Temperature
• By definition:
G = H − TS
• Indicating that free energy is a function of
temperature.
• rG° will therefore change with temperature.
• A consequence of this temperature
dependence is that, in certain instances,
reactions can be product-favored at
equilibrium at one temperature and reactantfavored at another.
Free Energy & Temperature
When a reaction has
rH° < 0
&
rS° > 0
at all temperatures
rG° is negative.
(Product favored)
Free Energy & Temperature
When a reaction has
When a reaction has
rH° > 0
rH° < 0
&
&
rS° > 0
rS° < 0
at high temperatures
rG° is negative.
(Product favored)
at low temperatures
rG° is negative.
(Product favored)
Free Energy & Temperature
Free Energy & Temperature
At what temperature will a reaction that is non-spontaneous
turn over to a reaction that is? (rG changes from + to )
Free Energy & Temperature
At what temperature will a reaction that is non-spontaneous
turn over to a reaction that is? (rG changes from + to )
Example: The reaction,
2Fe2O3 (s)  3C(s)
4Fe(s)  3CO2 (g)
Has the following thermodynamic values.
∆rH° = +470.5 kJ
∆rS° = +560.3 J/K
∆rG° = +301.3 kJ
Reaction is reactant-favored at 298 K
Free Energy & Temperature
At what temperature will a reaction that is non-spontaneous
turn over to a reaction that is? (rG changes from + to )
Example: The reaction,
2Fe2O3 (s) + 3C(s)
4Fe(s) + 3CO2 (g)
When rG  0, the reaction begins to become spontaneous.
Free Energy & Temperature
At what temperature will a reaction that is non-spontaneous
turn over to a reaction that is? (rG changes from + to )
Example: The reaction,
2Fe2O3 (s) + 3C(s)
4Fe(s) + 3CO2 (g)
When rG  0, the reaction begins to become spontaneous.
r G  rH  T r S  0
T
rH
r S
Free Energy & Temperature
At what temperature will a reaction that is non-spontaneous
turn over to a reaction that is? (rG changes from + to )
Example: The reaction,
2Fe2O3 (s)  3C(s)
4Fe(s)  3CO2 (g)
∆rH° = +467.9 kJ
∆rS° = +560.3 J/K
103 J
467.9 kJ 
rH
1 kJ  839.7 K
T

J
r S
560.3
K