Entropy and Free Energy Kotz Ch 20 Lecture

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Entropy and Free Energy (Kotz Ch 20) - Lecture 2

• Spontaneous vs. non-spontaneous
• thermodynamics vs. kinetics
• entropy randomness (So)

• Gibbs free energy (?Go)
• ?Go for reactions - predicting spontaneous
  direction
• thermodynamics of coupled reactions
• ?Grxn versus ?Gorxn
• predicting equilibrium constants from ?Gorxn

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Entropy and Free Energy ( Kotz Ch 20 )

• some processes are spontaneous others never
  occur. WHY ?

• How can we predict if a reaction can occur, given
  enough time?

THERMODYNAMICS
9-paper.mov 20m02vd1.mov

• Note Thermodynamics DOES NOT say how quickly (or
  slowly) a reaction will occur.
• To predict if a reaction can occur at a
  reasonable rate, one needs to consider

KINETICS
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Product-Favored Reactions
In general, product-favored reactions are
exothermic.

• E.g. thermite reaction
• Fe2O3(s) 2 Al(s) ? 2 Fe(s) Al2O3(s)
• DH - 848 kJ

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Non-exothermic spontaneous reactions

• But many spontaneous reactions or processes are
  endothermic . . .

NH4NO3(s) heat ? NH4 (aq) NO3-
(aq) ?Hsol 25.7 kJ/mol
or have ?H 0 . . .
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PROBABILITY - predictor of most stable state
WHY DO PROCESSES with ? H 0 occur ? Consider
expansion of gases to equal pressure
This is spontaneous because the final state, with
equal molecules in each flask, is much more
probable than the initial state, with all
molecules in flask 1, none in flask 2
SYSTEM CHANGES to state of HIGHER PROBABILITY For
entropy-driven reactions - the more RANDOM state.
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Standard Entropies, So

• Every substance at a given temperature and in a
  specific phase has a well-defined Entropy
• At 298o the entropy of a substance is called
• So - with UNITS of J.K-1.mol-1
• The larger the value of So, the greater the
  degree of disorder or randomness
• e.g. So (in J.K-1mol-1) Br2 (liq)
  152.2
• Br2 (gas) 245.5
• For any process

?So ? So(final) - ? So(initial)
?So(vap., Br2) (245.5-152.2) 93.3 J.K-1mol-1
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Entropy and Phase
So (J/Kmol) H2O(gas) 188.8 H2O(liq) 69.9 H2O
(s) 47.9

• S (gases) gt S (liquids) gt S (solids)

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Entropy and Temperature

• The entropy of a substance increases with
  temperature.

Molecular motions of heptane at different temps.

• Higher T means
• more randomness
• larger S

9_heptane.mov 20m04an2.mov
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Entropy and complexity
Increase in molecular complexity generally leads
to increase in S.
9_alkmot.mov 20m04an3.mov
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Entropy of Ionic Substances

• Ionic Solids Entropy depends on extent of
  motion of ions. This depends on the strength of
  coulombic attraction.

• Entropy increases when a pure liquid or solid
  dissolves in a solvent.

NH4NO3(s) ? NH4 (aq) NO3- (aq) ?Ssol
So(aq. ions) - So(s) 259.8 - 151.1 108.7
J.K-1mol-1
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Entropy Change in a Phase Changes

• For a phase change,
• DS q/T
• where q heat transferred in phase change

H2O (liq) ? H2O(g)
For vaporization of water ?H q 40,700 J/mol
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Calculating ?S for a Reaction
DSo S So (products) - S So (reactants)

• Consider 2 H2(g) O2(g) ? 2 H2O(liq)
• DSo 2 So (H2O) - 2 So (H2) So (O2)
• DSo 2 mol (69.9 J/Kmol) -
• 2 mol (130.7 J/Kmol) 1 mol (205.3 J/Kmol)
• DSo -326.9 J/K
• Note that there is a decrease in S because 3 mol
  of gas give 2 mol of liquid.

If S DECREASES, why is this a SPONTANEOUS
REACTION??
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2nd Law of Thermodynamics
A reaction is spontaneous (product-favored) if
?S for the universe is positive.

• DSuniverse DSsystem DSsurroundings
• DSuniverse gt 0 for product-favored process
• First calc. entropy created by matter dispersal
  (DSsystem)
• Next, calc. entropy created by energy dispersal
  (DSsurround)

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Calculating ?S(universe)
2 H2(g) O2(g) ? 2 H2O(liq) DSosystem -326.9
J/K
Can calculate that DHorxn DHosystem -571.7 kJ
DSosurroundings 1917 J/K
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Calculating ?S(universe) (2)
2 H2(g) O2(g) ? 2 H2O(liq) DSosystem
-326.9 J/K (less matter dispersal) DSosurroundings
1917 J/K (more energy dispersal)
DSouniverse 1590 J/K
The entropy of the universe increases so the
reaction is spontaneous.
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The Laws of Thermodynamics
0. Two bodies in thermal equilibrium are at same
T 1. Energy can never be created or destroyed.
? E q w
2. The total entropy of the UNIVERSE (
system plus surroundings) MUST INCREASE in
every spontaneous process.
? STOTAL ? Ssystem ? Ssurroundings gt 0
3. The entropy (S) of a pure, perfectly
crystalline compound at T 0 K is ZERO.
(no disorder)
ST0 0 (perfect xll)
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Gibbs Free Energy, G

• DSuniv DSsurr DSsys

Multiply through by -T -TDSuniv DHsys -
TDSsys -TDSuniv change in Gibbs free
energy for the system DGsystem Under
standard conditions
The Gibbs Equation
DGo DHo - TDSo
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Standard Gibbs Free Energies, ?Gof

• Every substance in a specific state has a
• Gibbs Free Energy, G H - TS
• recall only ?H can be measured. Therefore
• there is no absolute scale for G
• only ?G values can be determined
• DGof the Gibbs Free Energy of formation (from
  elements) is used as the standard value
• We set the scale of G to be consistent with that
  for H -

DGof for elements in standard states 0
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Sign of DG for Spontaneous processes
2nd LAW requirement for SPONTANEITY is
?STOTAL ?Ssystem ?Ssurroundings gt 0
Multiply by T T?Ssystem T?Ssurroundings gt
0
and ?Ssurroundings ?Hosystem/T
Thus T?Ssystem - ?Hosystem gt 0
Multiply by -1 (-gtreverse gt to lt), drop subscript
system
DGo lt 0 for all SPONTANEOUS processes
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Sign of Gibbs Free Energy, ?G

• DGo DHo - TDSo
• change in Gibbs free energy
• (total free energy change for system - free
  energy lost in disordering the system)
• If reaction is exothermic (DHo is -ve) and
• entropy increases (DSo is ve), then
• DGo must be -ve and reaction CAN proceed.
• If reaction is endothermic (DHo is ve), and
• entropy decreases (DSo is -ve), then
• DGo must be ve reaction CANNOT proceed.

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Gibbs Free Energy changes for reactions
DGo DHo - TDSo
DHo DSo DGo Reaction
exo (-) increase() - Product-favored endo() de
crease(-) Reactant-favored exo
(-) decrease(-) ? T dependent endo() increase()
? T dependent
Spontaneous in last 2 cases only if Temperature
is such that ?Go lt 0
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Methods of calculating ?G
DGo DHo - TDSo
Two methods of calculating DGo
a) Determine DHorxn and DSorxn and use Gibbs
equation. b) Use tabulated values of free
energies of formation, DGfo.
?Gorxn S ?Gfo (products) - S ? Gfo (reactants)
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Calculating DGorxn

• EXAMPLE Combustion of acetylene
• C2H2(g) 5/2 O2(g) ? 2 CO2(g) H2O(g)
• From standard enthalpies of formation DHorxn
  -1238 kJ
• From standard molar entropies DSorxn -
  0.0974 kJ/K
• Calculate DGorxn from DGo ?Ho - T?So
• DGorxn -1238 kJ - (298 K)(-0.0974 kJ/K)
• -1209 kJ
• Reaction is product-favored in spite of negative
  DSorxn. Reaction is enthalpy driven

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Calculating DGorxn for NH4NO3(s)
EXAMPLE 2 NH4NO3(s) ? NH4NO3(aq)

• Is the dissolution of ammonium nitrate
  product-favored?
• If so, is it enthalpy- or entropy-driven?

9_amnit.mov 20 m07vd1.mov
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DGorxn for NH4NO3(s) ? NH4NO3(aq)
From tables of thermodynamic data we find
DHorxn 25.7 kJ DSorxn 108.7 J/K or
0.1087 kJ/K DGorxn 25.7 kJ - (298
K)(0.1087 kJ/K) -6.7 kJ Reaction is
product-favored . . . in spite of
positive DHorxn. Reaction is entropy driven
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Calculating DGorxn
DGorxn S DGfo (products) - S DGfo (reactants)
EXAMPLE 3 Combustion of carbon C(graphite)
O2(g) ? CO2(g) DGorxn DGfo(CO2) -
DGfo(graph) DGfo(O2) DGorxn -394.4 kJ -
0 0 Note that free energy of formation of
an element in its standard state is 0. DGorxn
-394.4 kJ Reaction is product-favored as
expected.
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Free Energy and Temperature

• 2 Fe2O3(s) 3 C(s) ? 4 Fe(s) 3 CO2(g)
• DHorxn 467.9 kJ DSorxn 560.3 J/K
• DGorxn 467.9 kJ - (298K)(0.560kJ/K) 300.8
  kJ
• Reaction is reactant-favored at 298 K
• At what T does DGorxn just change from () to
  (-)?
• i.e. what is T for DGorxn 0 DHorxn - TDSorxn

If DGorxn 0 then DHorxn TDSorxn so T
DHo/DSo 468kJ/0.56kJ/K 836 K or 563oC
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?Go for COUPLED CHEMICAL REACTIONS
Reduction of iron oxide by CO is an example
of using TWO reactions coupled to each other in
order to drive a thermodynamically forbidden
reaction
Fe2O3(s) ? 4 Fe(s) 3/2 O2(g) DGorxn
742 kJ
with a thermodynamically allowed reaction
3/2 C(s) 3/2 O2 (g) ? 3/2 CO2(g) DGorxn
-592 kJ
Overall
Fe2O3(s) 3/2 C(s) ? 2 Fe(s) 3/2 CO2(g)
DGorxn 301 kJ _at_ 25oC
BUT DGorxn lt 0 kJ for T gt 563oC
See Kotz, pp933-935 for analysis of the thermite
reaction
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Other examples of coupled reactions
Copper smelting
Cu2S (s) ? 2 Cu (s) S (s) DGorxn
86.2 kJ (FORBIDDEN)
Couple this with
S (s) O2 (g) ? SO2 (s)
DGorxn -300.1 kJ
Overall Cu2S (s) O2 (g) ? 2 Cu (s)
SO2 (s) DGorxn 86.2 kJ -300.1 kJ
-213.9 kJ (ALLOWED)
Coupled reactions VERY COMMON in Biochemistry
e.g. all bio-synthesis driven by ATP ? ADP
for which DHorxn -20 kJ DSorxn 34
J/K DGorxn -30 kJ _at_ 37oC
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Thermodynamics and Keq

• Keq is related to reaction favorability.
• If DGorxn lt 0, reaction is product-favored.
• DGorxn is the change in free energy as reactants
  convert completely to products.
• But systems often reach a state of equilibrium in
  which reactants have not converted completely to
  products.
• How to describe thermodynamically ?

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?Grxn versus ?Gorxn
Under any condition of a reacting system, we can
define ?Grxn in terms of the REACTION QUOTIENT, Q
?Grxn ?Gorxn RT ln Q
If ?Grxn lt 0 then reaction proceeds to right If
?Grxn gt 0 then reaction proceeds to left
At equilibrium, ?Grxn 0. Also, Q K. Thus
DGorxn - RT lnK
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Thermodynamics and Keq (2)

• 2 NO2 ? N2O4
• DGorxn -4.8 kJ
• pure NO2 has DGrxn lt 0.
• Reaction proceeds until DGrxn 0 - the minimum
  in G(reaction) - see graph.
• At this point, both N2O4 and NO2 are present,
  with more N2O4.
• This is a product-favored reaction.

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Thermodynamics and Keq (3)

• N2O4 ?2 NO2 DGorxn 4.8 kJ
• pure N2O4 has DGrxn lt 0.
• Reaction proceeds until DGrxn 0 - the minimum
  in G(reaction) - see graph.
• At this point, both N2O4 and NO2 are present,
  with more NO2.
• This is a reactant-favored reaction.

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Thermodynamics and Keq (4)
Keq is related to reaction favorability and so to
DGorxn. The larger the value of DGorxn the larger
the value of K.
DGorxn - RT lnK
where R 8.31 J/Kmol
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Thermodynamics and Keq (5)
DGorxn - RT lnK

• Calculate K for the reaction
• N2O4 ? 2 NO2 DGorxn 4.8 kJ
• DGorxn 4800 J - (8.31 J/K)(298 K) ln K

K 0.14
When ?Gorxn gt 0, then K lt 1 - reactant
favoured When ?Gorxn lt 0, then K gt1 - product
favoured
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Entropy and Free Energy (Kotz Ch 20)

• Spontaneous vs. non-spontaneous
• thermodynamics vs. kinetics
• entropy randomness (So)

• Gibbs free energy (?Go)
• ?Go for reactions - predicting spontaneous
  direction
• thermodynamics of coupled reactions
• ?Grxn versus ?Gorxn
• predicting equilibrium constants from ?Gorxn