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Free Energy and Thermodynamics
Perpetual motion is forbidden by the Laws of Thermodynamics
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Chapter 16
Spontaneity, Entropy And Free Energy
1st Law of Thermodynamics
Energy can be neither created nor destroyed.
Energy of the universe is constant
However not all energy is usable
The 1st Law is an energy bookkeeping system
Helps answer questions:
How much energy in involved?
Does energy flow in or out of system?
What form does the energy assume?
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Spontaneous
• A reaction is spontaneous if it occurs without
outside intervention
Ex.
Fe + H2O + O2  Fe2O3 (Rust)
• Does not mean fast
• We will be using Thermochemistry (Ch.6), Kinetics
(Ch.12) & Thermodynamics (Ch. 16) to describe a
reaction completely.
• Thermodynamics tells the direction not speed
• Thermodynamics compares initial & final states
• Kinetics describes pathway between reactant &
product.
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Formula Review
E( Internal Energy)  q( heat )  w( work )
H ( Enthalpy )  E  PV
q  mcT
[ A]
n
Rate  
 k [ A]
[t ]
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Spontaneity
• A process that is spontaneous in one direction will
not be spontaneous in the other.
• Ex. If wood burns into CO2 and H2O, Why can we not put CO2
and H2O together and make wood?
• Processes will proceed without outside assistance
• If steel is exposed to air and moisture it rusts
• It’s independent of time
• The characteristic common in all spontaneous
processes is an increase in the property called
Entropy (S)
If S(universe) > 0, the reaction is spontaneous
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Entropy
• Entropy is a measure of molecular randomness
or disorder
• High Disorder = High Entropy
• Entropy is a thermodynamic function that
describes the number of arrangements
(positions) available to a system, it’s probability.
• Entropy is a measure of chaos in a system
• Entropy is a state function
• Driving force in all spontaneous processes is
increase in Entropy (S) of the universe
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Entropy
• Formal definition: Entropy (S) is a thermodynamic
function that increases with the number of
energetically equivalent ways to arrange the
components of a system to achieve a particular
state.
• First expressed by Ludwig Boltzmann in early 1900
S = k ln W
• k is Boltzmann constant = gas constant (8.31) /
Avogadro’s number = 1.38 X 10-23 J/K
• W = number of energetically equivalent ways
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Probability
Consider the number of arrangements for a
system in the following evacuated bulbs
1 MOLECULE OF GAS
• 2 possible arrangements
• 50 % chance of finding
the left empty
• Each configuration that
gives a particular
arrangement is called a
microstate
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Probability
2 MOLECULES OF GAS
4
possible arrangements
 25% chance of finding
the left empty
 50 % chance of them
being evenly dispersed
 25% chance of finding
the right empty
See Table 16.1 p.753
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Positional Entropy
• Positional Entropy (S) – depends upon the number
of microstates a set of molecules can have in space
1. Solids have the fewest possible number of microstates
in space, atoms are very ordered, crystal structures
2. Liquids have many more ways for molecules to be
arranged than a solid
3. Gases have a huge number of positions possible
S solid < S liquid << S gas
• A chemical system proceeds in a direction that
increases the entropy of the universe.
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Positional Entropy
• Choose the substance with the higher
positional entropy:
1. CO2(s) vs. CO2(g)
2. N2(g) at 1 atm vs. N2(g) at 1.0 X 10-2 atm
• As the change in entropy (∆S) becomes more
positive the system/universe becomes more
disorder.
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Entropy Values @ 298K J/mol K
Solids
Liquids
Gases
C(graph.)
5.7
H2O
69.9
H2
130.7
Ca
41.4
CH3OH
126.8
O2
205.1
MgO
NaCl
26.9
72.1
Br2
152.2
Cl2
223.0
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“Entropy Ain’t What It Used To Be”
• 2nd Law of Thermodynamics: Entropy of the
universe is always increasing
• If process is spontaneous Total entropy must increase
SUniverse  S system  S surroundings
• ∆S(universe) is always increasing
Entropy defined in terms of probability
– Describes the number of arrangements available to a
system existing in a given state
– The most likely of the particles is the most random
– Nature proceeds toward the system with the highest
number of possible states
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Predicting Entropy Changes
• Predict the sign of the entropy ∆S for each of the
following processes.
1. Solid sugar is added to water to form a solution.
2. Iodine vapor condenses on a cold surface to form a crystal
Remember: S solid < S liquid << S gas
Use this formula to help determine ∆S
∆S = S(Final) – S(Initial)
Which has a greater number of arrangement:
Final state or Initial state
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Entropy
• Solutions form because there are many more
possible arrangements of dissolved pieces than if
they stay separate.
Remember 2nd Law of Thermodynamics, Entropy of universe is always increase
Suniverse = Ssystem + Ssurroundings
If Suniv process is spontaneous in forward direction
If suniv process is spontaneous in opposite direction
If suniv = 0, process has no tendency to occur = equilibrium
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Effect of Temperature on Spontaneity
Exothermic processes (heat out) Ssurr is positive
Endothermic processes (heat ) Ssurr is negative
• Consider this process: H2O(l)H2O(g)
• Is it spontaneous?
•
i.e. Is suniv positive)?
How much volume will 1 mole of liquid and gas occupy? (Stoichiometry)
• Ssys is positive (liquid  gas,  entropy)
• Ssurr is negative (vaporization is an endothermic process)
ssys = sproducts - sreactants
suniv = ssys + ssurr
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Entropy Changes In The Surroundings
• Energy changes in the surroundings are primarily
determined by flow of heat. i.e the sign of ∆Ssurr
depends on direction of heat flow.
– An exothermic process is favored because by giving up
heat, the entropy of the surroundings increases.
– Summary
Exothermic Process Ssurr =+
Quantity of Heat J
H

Temperature K
T
Endothermic Process Ssurr = -
Quantity of Heat J
H

Temperature K
T
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Entropy Changes In The Surroundings
• How much of an impact the transfer of heat is on the
surroundings depends on how high the temperature is.
• If the transfer occurs at low temperature, Ssurr will be
greater.
• Magnitude of Ssurr depends directly on quantity of heat
transferred
• Magnitude of Ssurr depends inversely on temperature
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Enthalpy H
• Enthalpy (H) is the internal energy (E) of a system
plus the pressure times volume (PV)
• H=E+PV
• H = a measure of the total change of internal
energy of a system.
• H heat flow into system (endothermic)
• H heat flow into system (exothermic)
• Entropy (S) and Enthalpy (H): two factors that
determine spontaneity
• Nature will always transfer thermodynamic energy
from a higher to a lower temperature
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Determining Ssurr
S surr
H

T
In the metallurgy of antimony, the pure metal is
recovered via different reactions, depending on the
composition of the ore @ 25oC
Iron is used to reduce antimony in sulfide ores.
Sb2S3(s) + 3Fe(s)  2Sb(s) + 3FeS(s)
H= -125kJ
Carbon is used as the reducing agent for oxide ores.
Sb4O6(s) + 6C(s)  4Sb(s) + 6CO(g)
H= 778kJ
Label RedOx Values & Calculate Ssurr for each reaction
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Spontaneous or Not?
Ssys
Ssurr
Suniv Spontaneous?
Yes
+
+
+
+
-
-
+
-
Yes in
Reverse
?
Yes If
Ssys>Ssurr
?
Yes If
Ssys<Ssurr
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•
Gibb's Free Energy
Gibb's Free Energy is defined as: G=H-TS
G=Free Energy
H=Enthalpy
T=Temp.(K)
G=H-TS
S=Entropy
(@ constant temperature)
(H & S without subscripts means system)
• Study the derivation of formulas on p. 760
SUNIV
G

T
• If G < 0 @ constant T and P, the Process is
spontaneous in which the free energy is decreasing
- G means +S
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Let’s Check
• For the reaction H2O(s)  H2O(l)
• So = 22.1 J/K mol
• Ho =6030 J/mol
• Calculate G at 10oC, 0oC, & -10oC (convert to K)
Superscripto means substance is in standard state
• Use equation G0=H0-TS0
• Spontaneity can be predicted from the sign of H
and S.
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At What Temperature Is The Following
Process Spontaneous?
Br2(l)  Br2(g)
∆H0 = 31.0 kJ/mol
∆S0 = 93.0 J/K
What is the normal boiling point for Br2(l)
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At What Temperature Is The Following
Process Spontaneous?
CH3OH(l)  CH3OH(g)
1. Get Ho and So values from Appendix 4
2. Caution: Make sure your units are compatible
3. Plug values into formula
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Summary G=H-TS
S
H
+
+
-
+
+
Spontaneous
?
At all temp.
At high temp.
“entropy driven”
At low temp.
“enthalpy driven”
Not at any temp.
Reverse at all temp.
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In general, when a reaction involves gaseous
molecules, the change in positional entropy is
dominated by the relative number of molecules
of reactants and products.
• Predict the sign of ∆S for each of the following
reactions.
1. Thermal decomposition of solid calcium carbonate
CaCO3(s)  CaO(s) + CO2(g)
2. The oxidation of SO2 in air
2SO2(g) + O2(g) → 2SO3(g)
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Third Law of Thermodynamics
•
•
•
•
•
Entropy (S) Of A Pure Crystal At 0K = 0
Walther Hermann Nernst (1864-1941)
Gives us a starting point
All others must be >0
Standard Entropies values So (@298 K & 1 atm) of
substances are listed Appendix 4
Entropy is a state function of a system and ∆S of a
given reaction can be calculated using the formula:
S reaction   n p S
o
o
prod .
 n S
o
r react .
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Calculate ∆So
• Predict sign of ∆So
• Calculate ∆So at 25oC
2NiS(s) + 3O2(g) → 2SO2(g) + 2 NiO(s)
Substance
So(J/Kmol)
SO2(g)
NiO(s)
O2(g)
NiS(s)
248
38
205
53
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Calculate ∆So
• Calculate ∆So for the reduction of aluminum
oxide by hydrogen gas
Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)
Substance
Al2O3(s)
H2(g)
Al(s)
H2O(g)
So(J/Kmol)
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131
28
189
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Free Energy and Chemical Reactions
• Go = standard free energy change.
• Free energy change that will occur if reactants in
their standard state turn to products in their
standard state.
• Go can’t be measured directly, can be calculated
from other measurements.
• Go=Ho-TSo
• Can calculate Ho using enthalpy of formation f equation p.246
H   n p H
o
o
f ( products )
  nr H
o
f (reactants)
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Calculating Ho, So, & Go
• Consider the reaction:
2SO2(g) + O2(g)  2SO3(g)
Carried out at 25oC & 1 atm, Use the following data:
Substance
SO2
SO3
O2
Hfo(kJ/mol)
-297
-396
0
So(J/Kmol)
248
257
205
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Hess’s Law
“More than one way to skin a cat”
Because ∆G is a state function you can obtain ∆G for
reactions using Hess’s Law
2CO(g) + O2(g)  2CO2(g)
Use a couple of related equation and manipulate them to get ∆Go
2CH4(g) + 3O2(g)  2CO(g) + 4H2O(g) ∆Go = -1088kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆Go = -801kJ
See Ch 6.3 p. 242 for review on Hess’s Law
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Calculate
o
G
• Using the following data at 25oC
Cdiamond(s) + O2(g)  CO2(g)
Cgraphite(s) + O2(g)  CO2(g)
Go = -397 kJ
Go = -394 kJ
Calculate Go for the reaction:
Cdiamond(s)  Cgraphite(s)
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Free Energy in Reactions
• Standard free energy of formation (Gof) is change
in free energy that accompanies the formation of 1
mol of that substance from reactants and products
in their standard state Gof
• The standard free energy of formation for any
element in its standard state is 0, (see table on p.A19)
• Use the following formula:
G  n p G
o
o
prod .
 nr G
o
react .
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Calculate
o
G
Methanol is a high-octane fuel used in highperformance racing engines.
Calculate Go for the reaction, given the Gfo below:
2CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g)
Substance
CH3OH(g)
O2(g)
CO2(g)
H2O(g)
Gfo
-163
0
-394
-229
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Free Energy is Dependent on Pressure
• Entropy depends on pressure because it effects
volume.
• Slarge volume > Ssmall volume
• Because volume and pressure are inversely
related
• Slow pressure> Shigh pressure
Study the Derivation of G = Go +RTln(Q)
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Calculate G
• One method for synthesizing methanol CH3OH(l)
involves reacting carbon monoxide and hydrogen
gas @25oC. Calculate G for this reaction where
carbon monoxide gas at 5.0 atm and hydrogen gas
at 3.0 atm are converted to methanol.
CO(g) + H2(g) → CH3OH(l)
1st calc. Go Get Gfo values p. A19 & formula p. 769
2nd calc. G
Using G = Go +RTln(Q)
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How far?
As a reaction proceeds toward Equilibrium, the G
changes until it reaches zero which is the
Equilibrium point.
G = Gprod. - Greact. = 0
At Equilibrium G = 0,
Q=K
For reactions run under “nonstandard conditions”
G = Go + RTlnQ
c
d
C  D

R = 8.31J/molK
Q
a
b
T = Temp in K
 A  B 
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Free Energy & Equilibrium
• At equilibrium, ∆G = 0 and k = Q so you can plug in
0 for ∆G and k for Q, rearranging the formula you
get new formula that lets you quantitatively relate
free energy to equilibrium.
G   RT ln k
o
Like wise, you can relate the pressure of reactants and
products to the equilibrium constant k with the following
equation:
k
( Pprod . )( Pprod . )
( Preact . )( Preact . )
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Quantitative Relationship Between Free Energy
And Equilibrium Constant
o
G
K
o
G
=0
K=1
o
G
<0
K>1
o
G
>0
K<1
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Free Energy & Equilibrium
• Consider the ammonia synthesis reaction

 2 NH3( g )
N 2( g )  3H 2( g ) 

• Where Go = -33.3 kJ per mole of N2 consumed at
25oC. For each of the following mixtures of
reactants and products at 25oC, predict the
direction in which the system will shift to reach
equilibrium.
1. PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00x10-2 atm
2. PNH3=1.00 atm, PN2=1.00 atm, PH2= 1.00 atm
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Temperature Dependence Of K
Go = -RTln(K) = Ho - TSo
• Rearrangement gives
H 1 S
ln  K   

R T
R
o
o
• A plot of ln(K) vs 1/T yields a straight line
(Y=mx+b)
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Free energy And Work
• Free energy is energy free to do work
• The maximum amount of work possible at a
given temperature and pressure is:
wmax = G
• Never really achieved because some of the free
energy is changed to heat during a change, so it
can’t be used to do work.
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Summary of Thermodynamic Laws
First Law: You can’t win, you can only break even
Second Law: You can’t break even
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