Entropy and Free Energy

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Entropy and Free Energy

• 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

2
Entropy and Free Energy

• Some processes are spontaneous
• others never occur. WHY ?

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

THERMODYNAMICS

• 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

4
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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Gas expansion - spontaneity from greater
probability
Consider distribution of 4 molecules in 2 flasks
P1 lt P2
P1 gt P2
P1 P2
With more molecules (gt1020) P1P2 is most
probable by far
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Directionality of Reactions

• How probable is it that reactant molecules will
  react?
• PROBABILITY suggests that a product-favored
  reaction will result in the dispersal of energy
  or dispersal of matter or both.

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Spontaneous Processes

• A process that is spontaneous in one direction is
  not spontaneous in the opposite direction.
• The direction of a spontaneous process can depend
  on temperature Ice turning to water is
  spontaneous at T gt 0?C, Water turning to ice is
  spontaneous at T lt 0?C.

9
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-1 mol-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-1 mol-1
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Entropy and Phase
So (J/Kmol) H2O (g) 188.8 H2O (l) 69.9 H2O
(s) 47.9

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

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Entropy and Temperature

• The entropy of a substance increases with
  temperature.

• Higher T means
• more randomness
• larger S

Molecular motions different temps.
12
Entropy and complexity
Increase in molecular complexity generally leads
to increase in S.
13
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-1 mol-1
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Entropy Changes for Phase Changes

• For a phase change, DS q/T
• where q heat transferred in phase change
• For H2O (liq) ---gt H2O(g)
• DH q 40,700 J/mol

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The Molecular Interpretation of Entropy
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Calculating ?S for a Reaction
DSo S So (products) - S So (reactants)

• Consider 2 H2(g) O2(g) ? 2 H2O(l)
• 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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The Molecular Interpretation of Entropy

• Energy is required to get a molecule to
  translate, vibrate or rotate.
• The more energy stored in translation, vibration
  and rotation, the greater the degrees of freedom
  and the higher the entropy.
• In a perfect crystal at 0 K there is no
  translation, rotation or vibration of molecules.
  Therefore, this is a state of perfect order.
• Third Law of Thermodynamics the entropy of a
  perfect crystal at 0 K is zero.
• Entropy changes dramatically at a phase change.

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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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2nd Law of Thermodynamics

• A reaction is spontaneous (product-favored) if DS
  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 DS for a Reaction
DSo ? So (products) - ? So (reactants)

• Consider 2 H2(g) O2(g) ---gt 2 H2O(l)
• 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.

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• 2 H2(g) O2(g) ---gt 2 H2O(l)
• DSosystem -326.9 J/K

22

• 2 H2(g) O2(g) ---gt 2 H2O(liq)
• DSosystem -326.9 J/K

23

• 2 H2(g) O2(g) ---gt 2 H2O(liq)
• DSosystem -326.9 J/K
• Can calc. that DHorxn DHosystem -571.7 kJ

24

• 2 H2(g) O2(g) ---gt 2 H2O(liq)
• DSosystem -326.9 J/K
• Can calc. that DHorxn DHosystem -571.7 kJ

25

• 2 H2(g) O2(g) ---gt 2 H2O(liq)
• DSosystem -326.9 J/K
• Can calc. that DHorxn DHosystem -571.7 kJ
• DSosurroundings 1917 J/K

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• 2 H2(g) O2(g) ---gt 2 H2O(l)
• DSosystem -326.9 J/K
• DSosurroundings 1917 J/K
• DSouniverse 1590. J/K
• The entropy of the universe is increasing,
• so the reaction is product-favored.

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• 2 H2(g) O2(g) ---gt 2 H2O(liq)
• DSosystem -326.9 J/K
• DSosurroundings 1917 J/K
• DSouniverse 1590. J/K
• The entropy of the universe is increasing,
• so the reaction is product-favored.

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Gibbs Free Energy, G

• DSuniv DSsurr DSsys

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Gibbs Free Energy, G

• DSuniv DSsurr DSsys

30
Gibbs Free Energy, G

• DSuniv DSsurr DSsys
• Multiply through by -T

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Gibbs Free Energy, G

• D Suniv D Ssurr D Ssys
• Multiply through by -T
• -T D Suniv D Hsys - T D Ssys

H
-D
sys
S






S

D
D
univ
sys
T
32
Gibbs Free Energy, G

• D Suniv D Ssurr D Ssys
• Multiply through by -T
• -T D Suniv D Hsys - T D Ssys
• -T D Suniv change in Gibbs free energy
• for the universe
• D Gsystem

33
Gibbs Free Energy, G

• DSuniv DSsurr DSsys
• Multiply through by -T
• -TDSuniv DHsys - TDSsys
• -TDSuniv change in Gibbs free energy for the
  universe DGuniv DGsystem
• Under standard conditions
• DGo DHo - TDSo

H
-D
sys
S






S

D
D
univ
sys
T
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Gibbs Free Energy, G

• DGo DHo - T DSo
• Gibbs free energy change DGo
• total energy change for system - energy lost
  in disordering the system
• If reaction is exothermic (DHo lt 0) and entropy
  increases (DSo gt 0), then DGo lt 0, that is,
  negative and reaction product-favored.
• If reaction is endothermic (DHo gt 0), and entropy
  decreases (DSo lt 0), then DGo gt 0
• and reaction is reactant-favored.

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Gibbs Free Energy, G

• DGo DHo - TDSo
• DHo DSo DGo Reaction
• exo(-) increase() - Prod-favored
• endo() decrease(-) React-favored
• exo(-) decrease(-) ? T dependent
• endo() increase() ? T dependent

36
Methods of calculating ?G
DGo DHo - TDSo
Two methods of calculating DGo

• 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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Example

• At what T is the following reaction spontaneous?

Br2(l) Br2(g)
where DH 30.91 kJ/mol, DS 93.2 J/mol.K

• Ans

DG DH - TDS
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Br2(l) ? Br2(g) where DH 30.91 kJ/mol,
DS 93.2 J/mol.K

• Try 298 K just to see

DG DH - TDS
DG 30.91 kJ/mol - (298K)(93.2 J/mol.K)
DG (30.91 - 27.78) kJ/mol 3.13 kJ/mol
gt 0
Not spontaneous at 298 K
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Example (cont.)

• At what T then?

DG DH - TDS
0
T DH/DS
T (30.91 kJ/mol) /(93.2 J/mol.K)
T 331.65 K 58.5oC
40
Calculating DG

• In our previous example, we needed to determine
  DHrxn and DSrxn to determine DGrxn

• Now, DG is a state function therefore, we can
  use known DG to determine DGrxn using

41
Standard DG of Formation DGf

• Like DHf and S, DGf is defined as the change
  in free energy that accompanies the formation of
  1 mole of that substance for its constituent
  elements with all reactants and products in their
  standard state.

• Like DHf, DGf 0 for an element in its
  standard state

Example DGf (O2(g)) 0
42
Example

• Determine the DGrxn for the following

C2H4(g) H2O(l) C2H5OH(l)

• Tabulated DGf from tables like Appendix D
• DGf(C2H5OH(l)) -175 kJ/mol
• DGf(C2H4(g)) 68 kJ/mol
• DGf(H2O (l)) -237 kJ/mol

43
Example (cont.)

• Using these values

C2H4(g) H2O(l) C2H5OH(l)
DGrxn DGf(C2H5OH(l)) DGf(C2H4(g))
DGf(H2O (l))
DGrxn -175 kJ 68 kJ (-237 kJ)
DGrxn -6 kJ
lt 0 therefore, spontaneous
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More DG Calculations

• Similar to DH, one can use the DG for various
  reactions to determine DG for the reaction of
  interest (a Hess Law for DG)

• Example

C(s, diamond) O2(g) CO2(g) DG -397 kJ
C(s, graphite) O2(g) CO2(g) DG -394 kJ
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More DG Calculations (cont.)
C(s, diamond) O2(g) CO2(g) DG -397 kJ
C(s, graphite) O2(g) CO2(g) DG -394 kJ
CO2(g) C(s, graphite) O2(g) DG 394
kJ
C(s, diamond) C(s, graphite) DG -3
kJ
DGrxn lt 0..rxn is spontaneous
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DGrxn ? Reaction Rate

• Although DGrxn can be used to predict if a
  reaction will be spontaneous as written, it does
  not tell us how fast a reaction will proceed.

• Example
• C(s, diamond) O2(g) ? CO2(g)

DGrxn -397 kJ
ltlt0
But diamonds are forever.
DGrxn ? rate
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Calculating DG

• Combustion of acetylene
• C2H2(g) 5/2 O2(g) --gt 2 CO2(g) H2O(g)
• Use enthalpies of formation to calculate
• DHorxn -1238 kJ lt 0
• Use standard molar entropies to calculate
• DSorxn -97.4 J/K -0.0974 kJ/K lt 0
• DGorxn -1238 kJ - (298 K)(-0.0974 J/K)
• -1209 kJ lt 0
• Reaction is product-favored in spite of negative
  DSorxn.
• Reaction is enthalpy driven

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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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The Concentration Dependence of Spontaneity

• As with ?H and S, ?G values have been
  tabulated for the standard state
• very few reactions occur at standard conditions
• noting that ?H and S change very little with
  changing T, we can make the assumption

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The Concentration Dependence of Spontaneity

• This works for changes in T, but ?G will change
  significantly with changing concentration and/or
  pressure
• For any reaction
• aA bB ? cC dD

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The Concentration Dependence of Spontaneity

• One possible source of acid rain is the reaction
  between NO2, a pollutant from automobile
  exhausts, and water.
• 3 NO2(g) H2O(l) ? 2 HNO3(g) NO(g)
• Determine whether this is thermodynamically
  feasible
• (a) under standard conditions and
• (b) at 298 K, with each product gas
  present at P 1.00?10-6 atm.
• Given
• ?Gf(NO2(g)) 51.3 kJ/mol
• ?Gf(H2O(l)) -237.1 kJ/mol
• ?Gf(HNO3(g)) -73.5 kJ/mol
• ?Gf(NO(g)) 87.6 kJ/mol

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The Temperature Dependence of Spontaneity
?G ?H - T ?S
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Bioenergetics

• The basic processes of life can be thought of as
  making order out of disorder
• This seems to go against the second law
• To create order, systems must release heat to the
  surroundings
• Living systems use large amounts of energy to
  survive
• most common energy sources are carbohydrates and
  fats

57
Bioenergetics

• The reaction of glucose with oxygen
• is highly
  spontaneous
• C6H12O6 6 O2 ? 6 CO2 6 H2O ?G -2870
  kJ/mol
• as is the oxidation of palmitic acid
• C15H31COOH 23 O2 ? 16 CO2 16 H2O ?G
  -9790 kJ/mol

58
Bioenergetics

• These reactions release too much energy for a
  cell to handle
• some of the energy must be stored
• stored by converting ADP to ATP
• ADP H3PO4 ? ATP H2O ?G 30.6 kJ/mol

59
Bioenergetics

• Cells use energy stored in ATP to drive
  nonspontaneous reactions
• ATP H2O ? ADP H3PO4 ?G -30.6 kJ/mol
• ATP conversion can be coupled to other reactions
• transfers energy from one reaction to another
• Amino Acid synthesis

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Bioenergetics

• glutamic acid NH3 ? glutamine H2O ?G 14
  kJ/mol
• ATP H2O ? ADP H3PO4 ?G -30.6 kJ/mol
• glutamic acid NH3 ATP ? glutamine ADP
  H3PO4
• ?G -16.6 kJ/mol

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Bioenergetics

• Cells are not 100 efficient in storing energy as
  ATP
• 1 glucose molecule produces 36 ATP molecules
• C6H12O6 6 O2 ? 6 CO2 6 H2O ?G -2870
  kJ/mol
• 36 ADP 36 H3PO4 ? 36 ATP 36 H2O ?G1102
  kJ/mol
• C6H12O6 6 O2 36 ADP 36 H3PO4 ? 6 CO2 6
  H2O 36 ATP 36 H2O
• ?G -1768 kJ/mol
• 38 ? ATP, 62 as HEAT

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Extra slides
63
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 ?

64
?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.

66
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.

67
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
68
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
69
Entropy and Free Energy

• 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