Chapter 17
Free Energy
and
Thermodynamics
Thermodynamics vs. Kinetics
2
CHEMICAL THERMODYNAMICS
The first law of thermodynamics:
Energy and matter can be neither created nor
destroyed; only transformed from one form to another.
The energy and matter of the universe is constant.
The second law of thermodynamics:
In any spontaneous process there is always an increase
in the entropy of the universe.
The entropy is increasing.
The third law of thermodynamics:
The entropy of a perfect crystal at 0 K is zero.
There is no molecular motion at absolute 0 K.
STATE FUNCTIONS
A property of a system which depends only on its
present state and not on its pathway.
H = Enthalpy = heat of reaction = qp
A measure of heat (energy) flow of a system relative
to its surroundings.
H° standard enthalpy
Hf° enthalpy of formation
H° = n Hf° (products) -  m Hf° (reactants)
H = U + PV
U represents the Internal energy of the particles, both
the kinetic and potential energy. U = q + w
HEAT
energy transfer as a
result of a temperature
difference
qp
VS
WORK
energy expanded to
move an object against
a force
w = F x d
endothermic (+q)
work on a system
(+w)
exothermic (-q)
work by the system
(-w)
qc = -qh
w = -PV
SPONTANEOUS PROCESSES
A spontaneous process occurs without outside intervention.
The rate may be fast or slow.
Entropy
A measure of randomness or disorder in a system.
Entropy is a state function with units of J/K and it can be
created during a spontaneous process.
Suniv = Ssys + Ssurr
The relationship between Ssys and Ssurr
Ssys
Ssurr
Suniv
Process spontaneous?
+
+
+
Yes
No (Rx will occur in
opposite direction)
+
?
Yes, if Ssys > Ssurr
+
?
Yes, if Ssurr > Ssys
Comparing Potential Energy
The direction of
spontaneity can
be determined
by comparing
the potential
energy of the
system at the
start and the
end.
7
Reversibility of Process
• any spontaneous process is irreversible
– it will proceed in only one direction
• a reversible process will proceed back and forth
between the two end conditions
– equilibrium
– results in no change in free energy
• if a process is spontaneous in one direction, it must be
nonspontaneous in the opposite direction
8
Diamond → Graphite
Graphite is more stable than diamond, so the
conversion of diamond into graphite is spontaneous –
but don’t worry, it’s so slow that your ring won’t turn
into pencil lead in your lifetime (or through many of
your generations).
9
Predicting Relative S0 Values of a System
1. Temperature changes
S0 increases as the temperature rises.
2. Physical states and phase changes
S0 increases as a more ordered phase changes to a less ordered
phase.
3. Dissolution of a solid or liquid
S0 of a dissolved solid or dissolved liquid is usually greater than the
S0 of the pure solute. However, the extent depends upon the nature
of the solute and solvent.
4. Dissolution of a gas
A gas becomes more ordered when it dissolves in a liquid or solid.
5. Atomic size or molecular complexity
In similar substances, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g. bond
flexibility) relate directly to entropy.
Factors Affecting Whether a
Reaction Is Spontaneous
• The two factors that determine the thermodynamic
favorability are the enthalpy and the entropy.
• The enthalpy is a comparison of the bond energy
of the reactants to the products.
– bond energy = amount needed to break a bond.
– H
• The entropy factors relates to the
randomness/orderliness of a system
– S
• The enthalpy factor is generally more important
than the entropy factor
15
Entropy Changes in the System
S0rxn - the entropy change that occurs when all reactants
and products are in their standard states.
S0rxn = S0products - S0reactants
The change in entropy of the surroundings is directly
related to an opposite change in the heat of the system
and inversely related to the temperature at which the
heat is transferred.
Ssurroundings = -
Hsystem
T
Increases in Entropy
19
Entropy
S = Sf - Si
S > q/T
S = H/T
For a reversible (at equilibrium) process
H - T  S < 0
For a spontaneous reaction at constant T & P
 H - T S
If the value for  H - T S is negative for a reaction
then the reaction is spontaneous in the direction of the
products.
If the value for  H - T S is positive for a reaction
then the reaction is spontaneous in the direction of the
reactants. (nonspontaneous for products)
A spontaneous endothermic chemical reaction.
water
Ba(OH)2.8H2O(s) + 2NH4NO3(s)
Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
H0rxn = +62.3 kJ
APPLICATION OF THE 3RD LAW OF THERMODYNAMICS
S° = standard entropy = absolute entropy
S is usually positive (+) for Substances, S can be negative
(-) for Ions because H3O+ is used as zero
Predicting the sign of S°
The sign is positive if:
1. Molecules are broken during the Rx
2. The number of moles of gas increases
3. solid  liquid liquid  gas solid  gas
an increase in order occurs
1.
2.
3.
4.
Ba(OH)2•8H2O (s) + 2NH4NO3(s) 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)
2SO2(g) + O2(g)  2SO3(g)
HCl(g) + NH3(g)  NH4Cl(s)
CaCO3(s)  CaO(s) + CO2(g)
Ex. 17.2a – The reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4
H2O(g) has Hrxn = -2044 kJ at 25°C.
Calculate the entropy change of the surroundings.
Given:
Find:
Concept Plan:
Hsystem = -2044 kJ, T = 298 K
Ssurroundings, J/K
T, H
Relationships:
Ssurr 
Solution:
 H sys
Ssurr 
Ssurr
T
 6.86 kJK
 H sys
S
T
  2044 kJ 

298 K
 6.86  103 KJ
Check: combustion is largely exothermic, so the entropy of
the surrounding should increase significantly
 S°=  n S°(product)- m S°(reactant)
1.
Acetone, CH3COCH3, is a volitale liquid
solvent. The standard enthalpy of formation of
the liquid at 25°C is -247.6 kJ/mol; the same
quantity for the vapor is -216.6 kJ/mol. What
is  S when 1.00 mol liquid acetone vaporizes?
2. Calculate  S° at 25° for:
a. 2 NaHCO3
(s)
b. 2 AgBr(s) +
 CO2 (g) + Na2CO3 (s) + H2O
I2(s) 
Br2(l) + 2 AgI(s)
Table of Sf
(g)
Gibbs Free Energy (G)
G, the change in the free energy of a system, is a
measure of the spontaneity of the process and of the
useful energy available from it.
G0system = H0system - TS0system
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process at equilibrium
G0rxn =  mG0products -  nG0reactants
STANDARD FREE ENERGY OF FORMATION
G°f
The free energy change that occurs when 1 mol
of substance is formed from the elements in
their standard state.
Calculate G° for:
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
Table of Gf
INTERPRETING G° FOR SPONTANEITY
1. When G° is very small (less than -10 KJ) the
reaction is spontaneous as written. Products dominate.
G° < 0
G°(R) > G°(P)
2. When G° is very large (greater than 10 KJ) the
reaction is non spontaneous as written. Reactants
dominate.
G° > 0
G°(R) < G°(P)
3. When G° is small (+ or -) at equilibrium then both
reactants and products are present.
G° = 0
Ba(OH)•8 H2O(s) + 2 NH4NO3(s)  2 NH3(g)+10 H2O(l) + Ba(NO3)3(aq)
Free Energy Change and Spontaneity
32
Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
H = +95.7 kJ and S = +142.2 J/K at 25°C.
Calculate G and determine if it is spontaneous.
Given:
Find:
Concept Plan:
H = +95.7 kJ, S = 142.2 J/K, T = 298 K
G, kJ
T, H, S
G
G  H  TS
Relationships:
Solution: G  H  TS



  95.7  103 J  298 K   142.2 KJ

 5.33  104 J
Answer: Since G is +, the reaction is not spontaneous at
this temperature. To make it spontaneous, we need
to increase the temperature.
Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
H = +95.7 kJ and S = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
Given:
Find:
Concept Plan:
H = +95.7 kJ, S = 142.2 J/K, G < 0
T, K
G, H, S
G  H  TS
Relationships:
Solution:
T
3
J
K
 95.7 10 J   T
 142.2 
3
J
K
673 K  T
G  H  TS  0
3
 95.7 10 J   T  142.2   0
 95.7 10 J   T  142.2 
J
K
Answer: The temperature must be higher than 673K for the
reaction to be spontaneous
GIBBS FREE ENERGY : G
G = H - TS
describes the temperature dependence of
spontaneity
Standard conditions (1 atm, if soln=1M & 25°):
G° = H° - TS°
A process ( at constant P & T) is spontaneous
in the direction in which the free energy
decreases.
1. Calculate H°, S° & G° for
2 SO2(g) + O2(g)  2 SO3(g) at 25°C & 1 atm
Example - G
• Calculate G at 427°C for the reaction below if the
PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm
N2(g) + 3 H2(g)  2 NH3(g)
PNH32
(2.0 atm)2
Q = P 1 x P 3 = (33.0 atm)1 (99.0)3 = 1.2 x 10-7
N2
H2
H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
G° = -92380 J - (700 K)(-198.2 J/K)
G° = +46400 J
G = G° + RTlnQ
G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
36
G = -46300
J = -46 kJ
G AND EQUILIBRIUM
The equilibrium point occurs at the lowest free energy
available to the reaction system.
When a substance undergoes a chemical reaction, the
reaction proceeds to give the minimum free energy at
equilibrium.
G = G° + RT ln (Q)
at equilibrium: G = 0
G° = -RT ln (K)
G° = 0
then K = 1
G° < 0
then K > 1
G° > 0
then K < 1
Q: Corrosion of iron by oxygen is
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
calculate K for this Rx at 25°C.
38
Free Energy, Equilibrium and Reaction Direction
•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0)
•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0)
•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (G = 0)
G = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1
and ln Q = 0 so
G0 = - RT lnK
Table 1 The Relationship Between G0 and K at 250C
G0(kJ)
K
100
3x10-18
50
2x10-9
10
2x10-2
1
7x10-1
0
1
-1
1.5
-10
5x101
-50
6x108
-100
3x1017
-200
1x1035
Essentially no forward reaction;
reverse reaction goes to completion
Forward and reverse reactions
proceed to same extent
Forward reaction goes to
completion; essentially no reverse
reaction
REVERSE REACTION
9x10-36
FORWARD REACTION
200
Significance
Example - K
• Estimate the equilibrium constant and position of
equilibrium for the following reaction at 427°C
N2(g) + 3 H2(g)  2 NH3(g)
H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
G° = -92380 J - (700 K)(-198.2 J/K)
G° = +46400 J
G° = -RT lnK
+46400 J = -(8.314 J/K)(700 K) lnK
lnK = -7.97
K = e-7.97 = 3.45 x 10-4
since K is << 1, the position of equilibrium favors reactants
41
Practice Problems on Go & K
1.Calculate Gº at 25ºC
AgI (s)  Ag+ (aq) + I-
(aq)
Table
BaSO4 (s) Ba2+(aq) + SO42-(aq)
What is the value for Ksp at 25ºC?
2.Calculate K at 25ºC for
Zn(s) + 2H+(aq) Zn2+(aq) + H2 (g).
Temperature Dependence of K
• for an exothermic reaction, increasing the
temperature decreases the value of the
equilibrium constant
• for an endothermic reaction, increasing the
temperature increases the value of the
equilibrium constant
Hrxn
ln K  
R

1

S
 
rxn

 
R
T
43
Gº & Spontaneity is dependent on Temperature
Hº
Sº
Gº
-
+
-
Spontaneous at all T
+
-
+
Non spontaneous at all T
-
-
+/-
At Low T= Spontaneous
At High T= Nonspontaneous
+
+
+/-
At low T= Nonspontaneous
At High T= Spontaneous
Q. Predict the Spontaneity for H2O(s)  H2O(l)
at (a) -10ºC , (b) 0ºC & (c) 10ºC.
Practice Problems on Temperature Relationships
1. At what temperature is the following process
spontaneous at 1 atm?
Br2 (l)  Br2 (g)
What is the normal boiling point for Br2 (l)?
2. Calculate Gº & Kp at 35ºC
N2O4 (g)  2 NO2 (g)
3. Calculate Hº, Sº & Gº at 25ºC and 650ºC.
CS2 (g) + 4H2 (g) CH4 (g) + 2H2S(g)
Compare the two values and briefly discuss the
spontaneity of the Rx at both temperature.