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Chapter 16
Spontaneity, entropy and free
energy
1st Law of Thermodynamics
When methane and oxygen react to form
carbon dioxide and water, the products have
lower potential energy than the reactants.
This change in potential energy results in
thermal energy flow (heat) to the
surroundings.
But the energy of the universe remains
“constant”
2
1st Law of Thermodynamics
Keeps track of the energy

How much energy is involved in
the change?

What form is the energy in?

Does the energy flow in or out of
the system?
***But does not tell us why?**
3
Spontaneous
A reaction that will occur without outside
intervention.
 We can’t determine how fast.
 We need both thermodynamics and
kinetics to describe a reaction
completely.
 Thermodynamics compares initial and
final states.
 Kinetics describes pathway between
(activation energy, [concentration], T,
catalysts.

The rate of a
reaction depends on
the pathway from
reactants to
products; this is the
domain of kinetics.
Thermodynamics tells
us whether a reaction
is spontaneous based
only on the properties
of the reactants and
products. The
predictions of
thermodynamics do
not require knowledge
of the pathway
between reactants
and products.
Thermodynamics
1st Law- the energy of the universe is
constant.
 Keeps track of thermodynamics doesn’t
correctly predict spontaneity.
 What makes a process spontaneous?
– A ball rolls down a hill, but never back up
– Steel rusts
– A gas fills a container uniformly
– Why????
– One common characteristic in all…

Entropy
The driving force for a spontaneous
process is an increase in the entropy of
the universe.
Entropy, S, can be viewed as a measure
of randomness, or disorder.
7
 Defined
Entropy
in terms of probability.
 Substances take the arrangement
that is most likely.
 The most likely is the most random.
 Calculate the number of
arrangements for a system.
Nature spontaneously proceeds toward
states that have the highest probability
of existing.
The expansion of an ideal gas into an
evacuated bulb.
Why is this
Spontaneous?
Exist in more
states, the
more spread
out the
molecules.
9
Physical evidence of entropy
4
possible
arrangements
(microstates)
 25% chance of
finding the left
empty
 1/22=1/4
 50 % chance of
them being
evenly
dispersed
2
4
atoms
 ½ = 1/16
 8% chance of
finding the left
empty
 50 % chance of
them being
evenly
dispersed
4
The greater the disorder the larger the entropy.
12
Entropy from physical evidence
Ssolid
 Gases
<Sliquid <<Sgas
have a huge number of positions
possible. (Positional probability)
 Pure solid dissolves in a solvent, the
entropy of the substance increases
– Except:Carbonates
 Entropy increases with increasing
molecular complexity (more e- moving)
 Increase # moles, increase entropy
Concept Check
Predict the sign of S for each of the
following, and explain:
+ a) The evaporation of alcohol
– b) The freezing of water
– c) Compressing an ideal gas at constant
temperature
+ d) Heating an ideal gas at constant
pressure
+ e) Dissolving NaCl in water
The Second Law of
Thermodynamics
. . . in any spontaneous process there is
always an increase in the entropy of
the universe.
Suniv > 0
for a spontaneous process.
Suniv= Ssys + Ssurr
15
Entropy
 Solutions
form because there are
many more possible arrangements of
dissolved pieces than if they stay
separate.
∆Suniv = ∆ Ssys + ∆ Ssurr
 If ∆ Suniv is positive the process is
spontaneous.
 If ∆ Suniv is negative the process is
spontaneous in the opposite direction.
Entropy Changes in Chemical
Reactions
N2(g) +3H2(g)  2NH3(g)
We go from 4 to 2 gas molecules.
What happens to the ∆S, (+,-)?
∆S –
The change in positional entropy is
dominated by the relative
number of molecules of gaseous
reactants and products.
(Although change is important we can assign absolute entropy values.)
Predict the sign of ΔS°
a.
CaCo3(s) →CaO(s) + CO2(g)
+
b.
2SO2(g) + O2(g) → 2SO3(g)
-

What if the # of molecules is
the same?
Generally, the more complex the
molecule, the higher the
standard entropy value.
18
Third Law of Thermo
 The
entropy of a pure crystal at 0 K is
0.
 Gives us a starting point for finding
values of S at other temperatures
which must be > 0.
 Remember Entropy is a state function.
Figure 16.5: (a) A perfect crystal of hydrogen
chloride at 0 K. (b) As the temperature rises
above 0 K, lattice vibrations allow some dipoles
to change their orientations, producing some
disorder and an increase in entropy.
Copyright©2000 by Houghton Mifflin Company. All rights reserved.
20
 S  reaction 
n
p
S  products   nr S  reac tan ts
Standard Entropies Sº ( at 298 K and 1 atm)
of substances are listed.
 ∆S is also an extensive property, so it is
dependent on the amount.

21
Calculate ΔS °
Al2O3(s) + 3H2(g) → 2Al(s) + 3H20(g)
 S  reaction 
Substance
Al2O3
H2
Al
H20
n
p
S  products   nr S  reac tan ts
S°(J/K·mol)
51
131
28
189
(56J/K + 567J/K) – (393J/K + 51J/K)
= 179J/K
The more complex the molecule, the higher
Is the standard entropy value.
22
Figure 16.6:
The H2O
molecule can
vibrate and
rotate in
several ways,
some of
which are
shown here.
Effect of Temperature
∆Suniv = ∆ Ssys + ∆ Ssurr
exothermic processes ∆H∆Ssurr is (+)
 endothermic processes ∆H+
∆ Ssurr is (-)
Consider this process:
H2O(l)  H2O(g)
These are in
∆ Ssys is positive
Opposition
Which controls
∆ Ssurr is negative
The situation?
∆ Suniv depends on temperature.

At 1atm water changes from liquid to gas:
Above 100°C it is spontaneous.
Below 100°C the opposite process is spontaneous

The lower the T, the greater the impact.
Δssys =
heat transferred
T at which the transfer occured
ΔSuni= ΔSsys+ ΔSsurr
ΔSsurr= + when E is added (exo)
= - when E is used (endo)
Express ΔSsurr in terms of enthalpy ΔH
Heat flow(q)(constant P) = change in enthalpy= ΔH
Into and out of the system
+ endothermic/- exothermic
At constant temperature and pressure:
 S surr
H
 
T
(-)Sign changes the point of
View from the sys. to the surr.
26
2H2 (g) + 2O2(g)  H2O(g)
Ignite & very fast! (spontaneous)
Ssys = -88.9 J/K
why???
Hsys = -483.6 kJ
H
 S surr  
Ssurr = ? 1620 J/K
T
Suniv = ? 1530J/K
Even though the entropy of the system
declines, the entropy of the surroundings
is so large that the overall change is
positive.
Bottom line:
A process is spontaneous in spite of a
negative entropy change as long as it
is extremely exothermic. Sufficient
exothermicity offsets system ordering.
Ssys
-H/T
Ssurr
Suniv
+
+
+
No, Reverse
+
-
?
At High temp.
-
+
?
Spontaneous?
Yes
At Low
temp.
Free Energy
G = H  TS
(from the standpoint of the system)
A process (at constant T, P) is
spontaneous in the direction in which free
energy decreases:
**S (refers to the system)**
G means +Suniv
30
G = H  TS
To see how this equation relates to
spontaneity, Divide by –T
G
H
H

 
 S
remember  S surr  
T
T
T
At constant T and P
G
H

 
  S   S surr   S   S uni
T
T
 S uni
G
 
T
31
This means;
A
process at constant T and P is
only spontaneous if ΔG is negative.
(decreasing)
ΔG > 0 entropy of universe decreases
ΔG < 0 entropy of universe increases
Lets use the free energy equation to
predict the spontaneity of melting of
ice:
H2O(s)→H2O(l)
32
H2O(s)→H2O(l)
Based on your daily experiences you would say melting ice is
spontaneous at T> 0,and not spontaneous at T< 0.
G = H  TS
ENTROPY
DECREASING
HEAT BEING
RELEASED
ENTROPY
INCREASING
33
Example
At what temperatures is the following
process spontaneous at 1 atm?
Br2(l)→Br2(g)
ΔH°=31.0 kJ/mol and ΔS°=93.0 J/K ·mol
What is the normal boiling point of liquid
Br2?
34
Br2(l)→Br2(g)
ΔH°=31.0 kJ/mol and ΔS°=93.0 J/K ·mol
1. Vaporization will occur at all T where ΔG is negative
2. ΔS° favors vaporization due to increase in positional
entropy.
3. ΔH° favors the opposite process (exothermic)
4. The opposite tendencies will balance at the boiling pt.
At this BP temp. liquid and gaseous Br2 are in equilibrium
ΔG °= 0.
ΔG °= ΔH °- TΔS °
0 = ΔH °- T ΔS °
ΔH °= T ΔS °
T= ΔH / ΔS °
3.1x104J/mol/93.0J/K·mol = 333K
35
 T>
333K, TΔS° has larger
magnitiude than ΔH °; ΔG ° is
negative. Above 333K , vaporization
is spontaneous; opposite occurs
spontaneously below this T.
 At 333K, liquid and gaseous Br2
coexist in equilibrium.
36
Summary
1.
2.
3.
T > 333K. ΔS° controls. Increase in
entropy when Br2(l) is vaporized.
T < 333K. The process is spontaneous
in the direction in which it is
exothermic. ΔH° controls.
T = 333K. ΔS° and ΔH° are balanced
(ΔG° = 0), liquid and gas coexist.
Normal BP.
37
G=H-TS
S
H
+
-
At all Temperatures
+
At high temperatures,
“entropy driven”
-
At low temperatures,
“enthalpy driven”
+
Not at any temperature,
Reverse is spontaneous
+
-
Spontaneous?
Free Energy in 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.
Why Free energy changes?




Tells us relative tendancy of a reaction
to occur.
The more - ΔG° , the more to the right
the reaction will proceed.
Use standard state because energy
varies w/ pressure or concentration
We must compare all reactions under
the same pressure or concentration
conditions.
40
ΔG° not measured directly
(calculated 3 ways)
1.
2.
3.
G ° = H °  TS ° (constant T)
Free energy is also a state function,
therefore, we can use a process similar to
finding H using Hess’s Law.
Standard free energy of formation (Gf)
SG = SnpGf(products)  SnrGf(reactants)
41
G ° = H °  TS ° (constant T)
Consider the reaction
2SO2(g) + O2(g) → 2SO3(g)
Carried out at 25C and 1 atm. Calculate ΔH°,
ΔS°& ΔG° using the following data:
Substance
SO2(g)
O2(g)
SO3(g)
ΔH°f (kJ/mol)
-297
-396
0
S°(J/K.mol)
248
257
205
Copyright©2000 by Houghton Mifflin Company. All rights reserved.
42
ΔH°? From enthalpies of formation.
ΔH°= 2mol(-396kJ/mol) – 2mol(-297kJ/mol) – 0
= -198 kJ
ΔS°? Standard entropy values
ΔS°= 2mol(257) – 2mol(248) – 1mol(205)
= -187J/K
& ΔG° using the following data:
43
ΔG°?
G ° = H °  TS °
= -198kJ – (298K)(-7J/K)(1kJ/1000J)
= -142kJ
44
Free energy is also a state function, therefore, we can
use a process similar to finding ΔH using Hess’s Law.
Use the following data (at 25C)
Cdiamond(s) + O2(g) → CO2(g)
Cgraphite(s) + O2(g) → CO2(g)
ΔG°= -397kJ
ΔG°= -394kJ
Calculate ΔG° for the reaction
Cdiamomd (s)→ Cgraphite(s)
45
Cdiamond(s) + O2(g) → CO2(g)
ΔG°=-397kJ
CO2(g) → Cgraphite(s) + O2(g)
ΔG°=394kJ
=-3kJ
The reaction is spontaneous but very slow at
25C and 1atm.
Reverse occurs at high T and P
46
SG = SnpGf(products)  nrGf(reactants)
 Change
in free energy that
accompanies the formation of 1 mole
of that substance from its constituent
elements with all reactants and
products.
 Standard
free energy of formation of
an element in its standard state is
zero.
47
Methanol is a high-octane fuel used in
high-performance racing engines.
Calculate ΔG° for the reaction,
2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)
Free energies of formation are:
Substance
ΔG°(kJ/mol)
CH3OH
-163
O2
0
CO2
-394
H 2O
-229
= 2mol(-394kJ/mol) + 4mol(-229kJ/mol) – 3(0)+ 2mol(-163kJ/mol)
= -1378 kJ
48
Free Energy in Reactions
are tables of Gºf .
 Products-reactants because it is a state
function.
 The standard free energy of formation
for any element in its standard state is 0.
 Remember- Spontaneity tells us nothing
about rate.
 There
Dependence of Free Energy on
Pressure
•
•
•
A system at constant T and P will
proceed spontaneously in the
direction to lower G.
The equilibrium position represents
the lowest G value available to a
reaction system.
G changes as a reaction proceeds
because G is dependent on P and
concentration.
50
G = H - TS
•
How does pressure affect the
thermodynamic of free energy?
• For an ideal gas H is not P
dependent
• S depends on P, because of the
dependence on V.
• Therefore G depends on P.
51
Consider 1 mol of ideal gas
at a given Temp
•
At V=10.0L there are more positions
available than at V=1.0L
Slarger V > Ssmall V
Slow P > Shigh P
(P and V are inversely proportional.)
52
G = G° + RT ln(P)
 G°=
P at 1atm
R= gas constant
 G = P at P atm
T=K
To demonstrate how the G is pressure
dependent:
N2(g) + 3H2(g) → 2NH3(g)
G = SnpGf(products)  SnrGf(reactants)
53
N2(g) + 3H2(g) → 2NH3(g)
G = SnpG(products)  SnrG(reactants)
G = S2GNH3  S1GN2 + 3GH2
GNH3 = G°NH3 + RT ln(PNH3)
GN2 = G°N2 + RT ln(PN2)
GH2 = G°H2 + RT ln(PH2)
54
G = 2[G°NH3 + RT ln(PNH3)]  [G°N2 + RT
ln(PN2)] -3[G°H2 + RT ln(PH2)]
G =(2G°NH3- G°N2 -3G°H2) + RT[2ln(PNH3)ln(PN2)- 3ln(PH2)]
G= G° + RT[2ln(PNH3)- ln(PN2)- 3ln(PH2)]
55
N2(g) + 3H2(g) → 2NH3(g)
G= G° + RT[2ln(PNH3)- ln(PN2)- 3ln(PH2)]
 PNH3 2 

 G   G RT ln
 ( PN 2 )( PH2 ) 
2lnPNH3= lnPNH32
3
-3ln(PH2)= ln(1/PH2 )
PNH3 2
( PN 2 )( PH2 )
-ln(PN2)= ln(1/PN2 )
Q
Reaction quotient
Q=K no shift
Q>K left
Q<K right
56
Free Energy and Pressure
G = G + RT ln(Q)
57
One method for synthesizing
methanol (CH3OH) involves reacting
carbon monoxide and hydrogen
gases:
CO(g) +2H2(g) → CH3OH(l)
Calculate ΔG at 25°C for this reaction
where carbon monoxide gas at 5.0
atm and hydrogen gas at 3.0 atm are
converted to liquid methanol.
58
CO(g) +2H2(g) → CH3OH(l)
G = G + RT ln(Q)
Gf(CH3OH)= -166kJ
Gf(H2)= 0
Gf(CO)= -137kJ
= -166 – (-137) – 0= -29kJ = -2.9x104J
ΔG°= -2.9x104J/mol rxn.
R= 8.3145J/K mol
T=298K
Q=1/(P(co))(PH22)= 1/(5.0)(3.02)= 2.2x10-2
ΔG= -2.9x104J/mol rxn + (8.3145J/K mol)(298K)ln(2.2x10-2)
= -38kJ/mol rxn
**G is more negative than G°
59
•We have seen that the formation of CH3OH is
spontaneous.
•But will it really form in a flask under these
conditions?
-no!
-it is true that 1mol of CH3OH has
lower G than CO or H2.
However, when they are mixed under these
conditions there is an even lower free energy
available to this system than 1.0 mol of pure
CH3OH.
**The system can achieve lowest possible free
energy by going to equilibrium, not going to
completion.
60
Figure 16.7: Schematic representations
of balls rolling down two types of hills.
Analogous to
Phase change
Chemical Change
Lowest possible
Free energy
The system can achieve lowest possible free energy
by going to equilibrium, not going to completion
61
Free Energy and Equilibrium
Equilibrium- when the forward and
reverse reaction rates are equal.
Thermodynamics view (equilibrium
point) occurs at the lowest value of
free energy available to the reaction
system.
62
A(g)↔ B(g)
1.0 mol of A placed in a reaction vessel at a
pressure of 2.0 atm.
Free energy of A = GA= G°A + RTln(PA)
Free energy of B = GB= G°B + RTln(PB)
Total free energy of system =G =GA + GB
63
Figure 16.8: (a) The
initial free energies
of A and B. (b) As
A(g) changes to
B(g), the free energy
of A decreases and
that of B increases.
(c) Eventually,
pressures of A and
B are achieved such
that GA = GB, the
equilibrium
position.
Suppose, minimum free energy is reached when
75% of A has been changed to B. At this point,
PA=.25 its original P.
(.25)(2.0atm)=0.50atm
PB= (0.75)(2.0atm)= 1.5atm
Since this is the equilibrium position, we use P
to find K.
PBe
15
. atm
K e 
 3.0
0.50atm
PA
65
Figure 16.9: (a) The change in free energy to reach equilibrium,
beginning with 1.0 mol A(g) at PA = 2.0 atm. (b) The change in free
energy to reach equilibrium, beginning with 1.0 mol B(g) at PB = 2.0
atm. (c) The free energy profile for A(g) → B(g) in a system
containing 1.0 mol (A plus B) at PTOTAL = 2.0 atm. Each point on the
curve corresponds to the total free energy of the system for a
given combination of A and B.
66
Summary
The reaction proceeds to the minimum
G (equilibrium), which corresponds
to the point where:
Gprod.= Greact.
ΔG = G prod – Greact. = 0
67
Free Energy and Equilibrium
G = G  RT ln(Q)
at equilibrium G =0 and Q = K
So G =0 =G +RT ln(K)
G = RT ln(K)
K = equilibrium constant
68
G = RT ln(K)
1.
G = 0.
The free energies of the reactants
and products are equal when all
components are in the standard
states (1 atm for gases).
The system is at equilibrium when
the Preact and prod.=1atm.
K=1
69
Case 2
G< 0
G = (Gprod- Greact) is negative
Gprod < Greact
the system will adjust to the right to reach equilibrium.
K>1
since the Pprod. at equilibrium is >1atm and
Preact.<1atm
70
Case 3
G>0
G = (Gprod- Greact) is positive
Greact < Gprod
the system will adjust to the left to reach
equilibrium.
K<1
since the Preact.>1atm and
Pprod <1atm at equilibrium
71
72
Consider the ammonia synthesis reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
Where ΔG°= -33.3kJ/mol of N2 consumed
at 25C. For each of the following
mixtures of reactants and products at
25C, predict the direction in which the
system will shift to reach equilibrium.
a.
b.
PNH3= 1.00atm, PN2=1.47atm, PH2=1.00x102atm
PNH3= 1.00atm, PN2=1.00atm, PH2=1.00atm
73
N2(g) + 3H2(g) ↔ 2NH3(g)
G = G  RT ln(Q)
PNH3= 1.00atm, PN2=1.47atm, PH2=1.00x102atm
Q=(1.00)2/(1.47)(1.00x10-2)3= 6.80x105
T=298K
R = 8.3145 J/K mol
ΔG° = -33.3kJ/mol
ΔG= -3.33x104J/mol + (8.3145 J/K mol )
(298K)(ln 6.80x105) = 0
At equilibrium no shift
74
N2(g) + 3H2(g) ↔ 2NH3(g)
G = G  RT ln(Q)
PNH3= 1.00atm, PN2=1.00atm, PH2=1.00atm
Q=(1.00)2/(1.00)(1.00)3= 1.00atm
T=298K
R = 8.3145 J/K mol
ΔG° = -33.3kJ/mol
ΔG= G  RT ln(1) = G  0
ΔG= G  33.3kJ/mol
Since it’s (-), shift right, K>1
75
The overall reaction for the corrosion (rusting)
of iron by oxygen is
+ 3O2(g)↔ 2Fe2O3(s)
Using the following data, calculate the
equilibrium constant for this reaction at 25C.
4Fe(s)
Substance
Fe2O3
ΔHf° (kJ/mol)
-826
S° (J/K . Mol)
90
Fe
0
27
O2
0
205
76
Temperature Dependence of K
 G   RT ln( K )   H  T S 
 H  S -  H
 S
ln( K )  


(1 / T ) 
RT R
R
R
y
= mx + b y=ln(K), m=-ΔH/R=slope,
xx=1/T, and b= ΔS/R=intercept ; will be
linear if plotted
(H and S  independent of
temperature over a small temperature
range) Copyright©2000 by Houghton Mifflin Company. All rights reserved.
77
Free Energy and Work
using a chemical process to do work
Maximum possible useful work
obtainable from a process at constant
T and P is equal to the change in free
energy.
wmax = ΔG
 ΔG for a spontaneous process: energy
that is free to do useful work
 ΔG for a non-spontaneous process:
minimum amount of work that must be
expended to make the process occur.
78
 Knowing
∆G tells us about efficiency.
 Amount of work actually obtained
from a spontaneous process is
always less than the max possible
amount.
 Energy
is always wasted.
79
Chemical change in a battery can do work by
sending current to a starter motor.
( Hypothetical Reversible process)
Force current in opposite direction
To recharge battery
Current flowing through a wire:
Energy is wasted as heat
80
Reversible v. Irreversible
Processes
Reversible:
The universe is exactly the
same as it was before the cyclic process.
Irreversible: The universe is different
after the cyclic process.
All real processes are irreversible -(some work is changed to heat).
Surroundings have less of an ability to do
the work. Entropy is increasing!!!!
81
At 1500°C for the reaction
CO(g) + 2H2(g) → CH3OH(g)
the equilibrium constant is
Kp = 1.4 x 10-7. Is H° at this
temperature:
A. positive
B. negative
C. zero
D. can not be determined
The standard free energy (Grxn0 for the
reaction
N2(g) + 3H2(g) → 2NH3(g)
is -32.9 kJ. Calculate the equilibrium
constant for this reaction at 25oC.
A. 13.3
B. 5.8 x 105
C. 2.5
D. 4.0 x 10-6
E. 9.1 x 108