Transcript Chapter

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 17
Free Energy
and
Thermodynamics
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
First Law of Thermodynamics
• you can’t win!
• First Law of Thermodynamics: Energy
cannot be Created or Destroyed
the total energy of the universe cannot change
though you can transfer it from one place to another
 DEuniverse = 0 = DEsystem + DEsurroundings
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First Law of Thermodynamics
• Conservation of Energy
• For an exothermic reaction, “lost” heat from the system
•
•
•
goes into the surroundings
two ways energy “lost” from a system,
converted to heat, q
used to do work, w
Energy conservation requires that the energy change in
the system equal the heat released + work done
 DE = q + w
 DE = DH + PDV
DE is a state function
internal energy change independent of how done
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Energy Tax
• you can’t break even!
• to recharge a battery with 100 kJ of
•
useful energy will require more than
100 kJ
every energy transition results in a
“loss” of energy
 conversion of energy to heat which is
“lost” by heating up the surroundings
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Heat Tax
fewer steps
generally results
in a lower total
heat tax
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Thermodynamics and Spontaneity
• thermodynamics predicts whether a process will
proceed under the given conditions
spontaneous process
nonspontaneous processes require energy input to go
• spontaneity is determined by comparing the free
energy of the system before the reaction with the free
energy of the system after reaction.
if the system after reaction has less free energy
than before the reaction, the reaction is
thermodynamically favorable.
• spontaneity ≠ fast or slow
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Comparing Potential Energy
The direction of
spontaneity can
be determined by
comparing the
potential energy
of the system at
the start and the
end.
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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
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Thermodynamics vs. Kinetics
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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).
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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.
 DH
• The entropy factors relates to the
randomness/orderliness of a system
 DS
• The enthalpy factor is generally more important
than the entropy factor
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•
•
•
•
•
•
•
Enthalpy
related to the internal energy
DH generally kJ/mol
stronger bonds = more stable molecules
if products more stable than reactants, energy released
exothermic
 DH = negative
if reactants more stable than products, energy absorbed
endothermic
 DH = positive
The enthalpy is favorable for exothermic reactions and
unfavorable for endothermic reactions.
Hess’ Law DH°rxn = S(DH°prod) - S(DH°react)
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Substance
DH°
kJ/mol
Substance
DH°
kJ/mol
Al(s)
Br2(l)
C(diamond)
CO(g)
Ca(s)
Cu(s)
Fe(s)
H2(g)
H2O(g)
HF(g)
HBr(g)
I2(s)
N2(g)
NO(g)
Na(s)
S(s)
0
0
+1.88
-110.5
0
0
0
0
-241.82
-268.61
-36.23
0
0
+90.37
0
0
Al2O3
Br2(g)
C(graphite)
CO2(g)
CaO(s)
CuO(s)
Fe2O3(s)
H2O2(l)
H2O(l)
HCl(g)
HI(g)
I2(g)
NH3(g)
NO2(g)
O2(g)
SO2(g)
-1669.8
+30.71
0
-393.5
-635.5
-156.1
-822.16
-187.8
-285.83
-92.30
+25.94
+62.25
-46.19
+33.84
0
-296.9
Entropy
• entropy is a thermodynamic function that increases as the
number of energetically equivalent ways of arranging the
components increases, S
 S generally J/mol
• S = k ln W
 k = Boltzmann Constant = 1.38 x 10-23 J/K
 W is the number of energetically equivalent ways, unitless
• Random systems require less energy than ordered systems
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W
Energetically
Equivalent States for
the Expansion of a
Gas
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Macrostates → Microstates
These microstates
all have the same
macrostate
So there are 6
differentThis
particle
macrostate can be achieved through
arrangements
several different
that
arrangements of the particles
result in the same
macrostate
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Macrostates and Probability
There is only one possible
arrangement that gives State A
and one that gives State C
There are 6 possible
arrangements that give State B
Therefore State B has
higher entropy than either
State A or State B
The macrostate with the
highest entropy also has the
greatest dispersal of energy
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Changes in Entropy, DS
• entropy change is favorable when the result is a more
•
random system.
 DS is positive
Some changes that increase the entropy are:
reactions whose products are in a more disordered
state.
(solid > liquid > gas)
reactions which have larger numbers of product
molecules than reactant molecules.
increase in temperature
solids dissociating into ions upon dissolving
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Increases in Entropy
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The 2nd Law of Thermodynamics
• the total entropy change of the universe must be
positive for a process to be spontaneous
 for reversible process DSuniv = 0,
 for irreversible (spontaneous) process DSuniv > 0
• DSuniverse = DSsystem + DSsurroundings
• if the entropy of the system decreases, then the
entropy of the surroundings must increase by a
larger amount
 when DSsystem is negative, DSsurroundings is positive
• the increase in DSsurroundings often comes from the
heat released in an exothermic reaction
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Entropy Change in State Change
• when materials change state, the number of
macrostates it can have changes as well
for entropy: solid < liquid < gas
because the degrees of freedom of motion increases
solid → liquid → gas
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Entropy Change and State Change
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Heat Flow, Entropy, and the
nd
2
Law
Heat must flow from
water to ice in order for
the entropy of the
universe to increase
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Temperature Dependence of DSsurroundings
• when a system process is exothermic, it adds heat to
•
•
the surroundings, increasing the entropy of the
surroundings
when a system process is endothermic, it takes heat
from the surroundings, decreasing the entropy of the
surroundings
the amount the entropy of the surroundings changes
depends on the temperature it is at originally
 the higher the original temperature, the less effect addition or
removal of heat has
DSsurroundings 
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 DH system
T
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Gibbs Free Energy, DG
• maximum amount of energy from the system
available to do work on the surroundings
G = H – T∙S
DGsys = DHsys – TDSsys
DGsys = – TDSuniverse
DGreaction = S nDGprod – S nDGreact
• when DG < 0, there is a decrease in free
energy of the system that is released into the
surroundings; therefore a process will be
spontaneous when DG is negative
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Ex. 17.2a – The reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4
H2O(g) has DHrxn = -2044 kJ at 25°C.
Calculate the entropy change of the surroundings.
Given:
Find:
Concept Plan:
Relationships:
Solution:
DHsystem = -2044 kJ, T = 298 K
DSsurroundings, J/K
T, DH
DSsurr 
 DHsys
DS
T
  2044 kJ 
DSsurr 

T
298 K
3 J
DSsurr  6.86 kJ

6.86

10
K
K
 DH sys
Check: combustion is largely exothermic, so the entropy of
the surrounding should increase significantly
Free Energy Change and Spontaneity
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Gibbs Free Energy, DG
• process will be spontaneous when DG is negative
 DG will be negative when
•
•
 DH is negative and DS is positive
exothermic and more random
 DH is negative and large and DS is negative but
small
 DH is positive but small and DS is positive and
large
or high temperature
DG will be positive when DH is + and DS is −
never spontaneous at any temperature
when DG = 0 the reaction is at equilibrium
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DG, DH, and DS
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Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has
DH = +95.7 kJ and DS = +142.2 J/K at 25°C.
Calculate DG and determine if it is spontaneous.
Given:
Find:
Concept Plan:
DH = +95.7 kJ, DS = 142.2 J/K, T = 298 K
DG, kJ
T, DH, DS
DG
DG  DH  TDS
Relationships:
Solution: DG  DH  TDS



  95.7  103 J  298 K   142.2 KJ

 5.33  10 4 J
Answer: Since DG 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
DH = +95.7 kJ and DS = +142.2 J/K.
Calculate the minimum temperature it will be spontaneous.
Given:
Find:
Concept Plan:
DH = +95.7 kJ, DS = 142.2 J/K, DG < 0
T, K
DG, DH, DS
DG  DH  TDS
Relationships:
Solution:
T
3
J
K
 95.7 10 J   T
 142.2 
3
J
K
673 K  T
DG  DH  TDS  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
The 3rd Law of Thermodynamics
Absolute Entropy
• the absolute entropy of a substance
•
is the amount of energy it has due to
dispersion of energy through its
particles
the 3rd Law states that for a perfect
crystal at absolute zero, the absolute
entropy = 0 J/mol∙K
 therefore, every substance that is not a
perfect crystal at absolute zero has some
energy from entropy
 therefore, the absolute entropy of
substances is always +
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Standard Entropies
• S°
• extensive
• entropies for 1 mole at 298 K for a particular
state, a particular allotrope, particular molecular
complexity, a particular molar mass, and a
particular degree of dissolution
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Substance
S°
J/mol-K
Substance
S°
J/mol-K
Al(s)
Br2(l)
C(diamond)
CO(g)
Ca(s)
Cu(s)
Fe(s)
H2(g)
H2O(g)
HF(g)
HBr(g)
I2(s)
N2(g)
NO(g)
Na(s)
S(s)
28.3
152.3
2.43
197.9
41.4
33.30
27.15
130.58
188.83
173.51
198.49
116.73
191.50
210.62
51.45
31.88
Al2O3(s)
Br2(g)
C(graphite)
CO2(g)
CaO(s)
CuO(s)
Fe2O3(s)
H2O2(l)
H2O(l)
HCl(g)
HI(g)
I2(g)
NH3(g)
NO2(g)
O2(g)
SO2(g)
51.00
245.3
5.69
213.6
39.75
42.59
89.96
109.6
69.91
186.69
206.3
260.57
192.5
240.45
205.0
248.5
Relative Standard Entropies
States
• the gas state has a larger entropy than the liquid
state at a particular temperature
• the liquid state has a larger entropy than the
solid state at a particular temperature
Substance
S°,
(J/mol∙K)
H2O (g)
70.0
H2O (l)
188.8
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Relative Standard Entropies
Molar Mass
• the larger the molar mass,
the larger the entropy
• available energy states
more closely spaced,
allowing more dispersal
of energy through the
states
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Relative Standard Entropies
Allotropes
• the less
constrained the
structure of an
allotrope is, the
larger its entropy
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Relative Standard Entropies
Molecular Complexity
• larger, more complex
molecules generally
have larger entropy
• more available energy
states, allowing more
dispersal of energy
through the states
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Molar
S°,
Substance
Mass (J/mol∙K)
Ar (g)
39.948
154.8
NO (g)
30.006
210.8
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Relative Standard Entropies
Dissolution
• dissolved solids
generally have larger
entropy
• distributing particles
throughout the mixture
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Substance
S°,
(J/mol∙K)
KClO3(s)
143.1
KClO3(aq)
265.7
39
Substance
S, J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
Given: standard entropies from Appendix IIB
188.8
Ex. 17.4 –Calculate DS for the reaction
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
Find: DS, J/K
Concept Plan:
Relationships:
SNH3, SO2, SNO, SH2O,

 

DS
DS  Sn pS products  SnrS reactants

 

DS  Sn pS products  SnrS reactants
Solution:
 [4(S NO(g ) )  6(S H 2 O(g ) )]  [4(S NH3 ( g ) )  5(SO 2 ( g ) )]
 [4(210.8 K )  6(188.8 K )]  [4(192.8 K )  5(205.2 K )]
J
J
J
J
 178.8 K
J
Check: DS is +, as you would expect for a reaction with
more gas product molecules than reactant molecules
Calculating DG
• at 25C:
DGoreaction = SnGof(products) - SnGof(reactants)
• at temperatures other than 25C:
assuming the change in DHoreaction and DSoreaction is
negligible
DGreaction = DHreaction – TDSreaction
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Substance
DG°f
kJ/mol
Substance
DG°f
kJ/mol
Al(s)
Br2(l)
C(diamond)
CO(g)
Ca(s)
Cu(s)
Fe(s)
H2(g)
H2O(g)
HF(g)
HBr(g)
I2(s)
N2(g)
NO(g)
Na(s)
S(s)
0
0
+2.84
-137.2
0
0
0
0
-228.57
-270.70
-53.22
0
0
+86.71
0
0
Al2O3
Br2(g)
C(graphite)
CO2(g)
CaO(s)
CuO(s)
Fe2O3(s)
H2O2(l)
H2O(l)
HCl(g)
HI(g)
I2(g)
NH3(g)
NO2(g)
O2(g)
SO2(g)
-1576.5
+3.14
0
-394.4
-604.17
-128.3
-740.98
-120.4
-237.13
-95.27
+1.30
+19.37
-16.66
+51.84
0
-300.4
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DGf, kJ/mol
-50.5
0.0
-394.4
-228.6
163.2
Substance
CH4(g)
O2(g)
CO2(g)
H2O(g)
O3(g)
Ex. 17.7 –Calculate DG at 25C for the
reaction
CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g)
Given: standard free energies of formation from Appendix IIB
Find: DG, kJ
Concept Plan:
Relationships:
Solution:

DGf of prod & react

 
DG

DG  Sn p DGf products  Snr DGf reactants
 

DG   Sn p DG f products  Snr DG f reactants
 [(DG f CO 2 )  2(DG f H 2 O)  (DG f O3 )]  [(DG f CH 4 )  8(DG f O 2 )]
 [(394.4 kJ )  2(228.6 kJ )  (163.2 kJ)]  [(50.5 kJ )  8(0.0 kJ )]
 148.3 kJ
Ex. 17.6 – The reaction SO2(g) + ½ O2(g)  SO3(g) has
DH = -98.9 kJ and DS = -94.0 J/K at 25°C.
Calculate DG at 125C and determine if it is spontaneous.
Given:
Find:
Concept Plan:
DH = -98.9 kJ, DS = -94.0 J/K, T = 398 K
DG, kJ
T, DH, DS
DG  DH  TDS
Relationships:
Solution:
DG
DG   DH  TDS



  98.9  103 J  398 K   94.0 K
J

 61.5  103 J  61.5 kJ
Answer: Since DG is -, the reaction is spontaneous at this
temperature, though less so than at 25C
DG Relationships
• if a reaction can be expressed as a series of reactions,
the sum of the DG values of the individual reaction is
the DG of the total reaction
 DG is a state function
• if a reaction is reversed, the sign of its DG value
•
reverses
if the amounts of materials is multiplied by a factor, the
value of the DG is multiplied by the same factor
 the value of DG of a reaction is extensive
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Free Energy and Reversible Reactions
• the change in free energy is a theoretical limit as
to the amount of work that can be done
• if the reaction achieves its theoretical limit, it is
a reversible reaction
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Real Reactions
• in a real reaction, some of the free energy is
“lost” as heat
if not most
• therefore, real reactions are irreversible
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DG under Nonstandard Conditions
 DG = DG only when the reactants and products
are in their standard states
 there normal state at that temperature
 partial pressure of gas = 1 atm
 concentration = 1 M
 under nonstandard conditions, DG = DG + RTlnQ
 Q is the reaction quotient
 at equilibrium DG = 0
 DG = ─RTlnK
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Example - DG
• Calculate DG 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
DH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
DG° = -92380 J - (700 K)(-198.2 J/K)
DG° = +46400 J
DG = DG° + RTlnQ
DG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
DG = -46300 J = -46 kJ
50
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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)
DH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J
DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K
DG° = -92380 J - (700 K)(-198.2 J/K)
DG° = +46400 J
DG° = -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
52
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
DHrxn  1  DSrxn
ln K  
 
R T
R
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