Entropy and Free Energy for a Reaction

Contributed by:

Jonathan James Fri, Jan 14, 2022 11:31 PM UTC

The highlights are:
• Spontaneous vs. non-spontaneous
• Thermodynamics vs. kinetics
• Entropy = randomness (So)
• Gibbs free energy
• Thermodynamics of coupled reactions

1. 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
Entropy & Free Energy (Ch 20) -
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2. 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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3. Product-Favored Reactions
In general, product-
favored reactions
are exothermic.
E.g. thermite reaction
Fe2O3(s) + 2 Al(s) 
2 Fe(s) + Al2O3(s)
H = - 848 kJ
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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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5. 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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6. 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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7. Entropy and Phase
Vapour
So (J/K•mol)
H2O(gas) 188.8
Ice Water H2O(liq) 69.9
H2O (s) 47.9
S (gases) > S (liquids) > S (solids)
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8. Entropy and Temperature
The entropy of a substance increases
with temperature.
Molecular motions of
heptane at different
Higher T means :
• more randomness
• larger S 9_heptane.mov
20m04an2.mov
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9. Entropy and complexity
Increase in molecular complexity
generally leads to increase in S.
9_alkmot.mov
20m04an3.mov
So (J/K•mol)
CH4 248.2
C2H6 336.1
C3H8 419.4
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10. Entropy of Ionic Substances
• Ionic Solids : Entropy depends on extent
of motion of ions. This depends on the
strength of coulombic attraction.
ion pairs So (J/K•mol)
MgO Mg2+ / O2- 26.9
NaF Na+ / F- 51.5
• 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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11. Entropy Change in a Phase Changes
For a phase change,
S = q/T
where q = heat transferred in phase change
H2O (liq)  H2O(g)
For vaporization of water:
H = q = +40,700 J/mol
q 40, 700 J/mol
S = = = + 109 J/K • mol
T 373.15 K
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12. Calculating S for a Reaction
So =  So (products) -  So (reactants)
Consider 2 H2(g) + O2(g)  2 H2O(liq)
So = 2 So (H2O) - [2 So (H2) + So (O2)]
So = 2 mol (69.9 J/K•mol) -
[2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]
So = -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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13. 2nd Law of Thermodynamics
A reaction is spontaneous (product-favored) if
S for the universe is positive.
Suniverse = Ssystem + Ssurroundings
Suniverse > 0 for product-favored process
First calc. entropy created by matter dispersal
(Ssystem)
Next, calc. entropy created by energy dispersal
(Ssurround)
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14. Calculating S(universe)
2 H2(g) + O2(g)  2 H2O(liq)
Sosystem = -326.9 J/K
q surroundings - H o system
S o surroundings = =
T T
Can calculate that Horxn = Hosystem = -571.7 kJ
o - (-571.7 kJ)(1000 J/kJ)
S surroundings =
298.15 K
Sosurroundings = +1917 J/K
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15. Calculating S(universe) (2)
2 H2(g) + O2(g)  2 H2O(liq)
Sosystem = -326.9 J/K (less matter dispersal)
Sosurroundings = +1917 J/K (more energy dispersal)
Souniverse = +1590 J/K
The entropy of the universe increases
so the reaction is spontaneous.
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16. 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
in every spontaneous process.
 STOTAL =  Ssystem +  Ssurroundings > 0
3. The entropy (S) of a pure, perfectly crystalline
compound at T = 0 K is ZERO. (no disorder)
ST=0 = 0 (perfect xll)
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17. Gibbs Free Energy, G
Suniv = Ssurr + Ssys
 Hsys
Suniv = + Ssys
T
Multiply through by -T
-TSuniv = Hsys - TSsys
-TSuniv = change in Gibbs free energy
for the system = Gsystem
Under standard conditions —
The Gibbs
G = H - TS
o o o
Equation
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18. 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
• Gof 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 - Gof for elements in standard states = 0
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19. Sign of G for Spontaneous processes
2nd LAW requirement for SPONTANEITY is :
STOTAL = Ssystem + Ssurroundings > 0
Multiply by T TSsystem + TSsurroundings > 0
and Ssurroundings = Hosystem/T
Thus TSsystem - Hosystem > 0
Multiply by -1 (->reverse > to <), drop subscript “system”
Ho -TS < 0 and Go = Ho -TS
Go < 0 for all SPONTANEOUS processes
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20. Sign of Gibbs Free Energy, G
Go = Ho - TSo
• change in Gibbs free energy =
(total free energy change for system - free energy lost in
disordering the system)
• If reaction is exothermic (Ho is -ve) and
entropy increases (So is +ve), then
Go must be -ve and reaction CAN proceed.
• If reaction is endothermic (Ho is +ve), and
entropy decreases (So is -ve), then
Go must be +ve; reaction CANNOT proceed.
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21. Gibbs Free Energy changes for reactions
Go = Ho - TSo
Ho So Go Reaction
exo (-) increase(+) - Product-favored
endo(+) decrease(-) + Reactant-favored
exo (-) decrease(-) ? T dependent
endo(+) increase(+) ? T dependent
Spontaneous in last 2 cases only if
Temperature is such that Go < 0
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22. Methods of calculating G
G = H - TS
o o o
Two methods of calculating Go
a) Determine Horxn and Sorxn and use Gibbs
b) Use tabulated values of free energies of
formation, Gfo.
Gorxn =  Gfo (products) -   Gfo (reactants)
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23. Calculating Gorxn
EXAMPLE: Combustion of acetylene
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g)
From standard enthalpies of formation: Horxn = -1238 kJ
From standard molar entropies: Sorxn = - 0.0974 kJ/K
Calculate Gorxnfrom Go = Ho - TSo
Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)
= -1209 kJ
Reaction is product-favored in spite of negative
Sorxn. Reaction is “enthalpy driven”
Entropy & Free Energy (Ch 20) -
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24. Calculating Gorxn for NH4NO3(s)
EXAMPLE 2:
NH4NO3(s)  NH4NO3(aq)
Is the dissolution of ammonium nitrate
product-favored? 9_amnit.mov
20 m07vd1.mov
If so, is it enthalpy- or entropy-driven?
Entropy & Free Energy (Ch 20) -
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25. Gorxn for NH4NO3(s)  NH4NO3(aq)
From tables of thermodynamic data we find
Horxn = +25.7 kJ
Sorxn = +108.7 J/K or +0.1087 kJ/K
Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
Reaction is product-favored
. . . in spite of positive Horxn.
Reaction is “entropy driven”
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26. Calculating Gorxn
Gorxn =  Gfo (products) -  Gfo (reactants)
EXAMPLE 3: Combustion of carbon
C(graphite) + O2(g)  CO2(g)
Gorxn = Gfo(CO2) - [Gfo(graph) + Gfo(O2)]
Gorxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an element
in its standard state is 0.
Gorxn = -394.4 kJ
Reaction is product-favored as expected.
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27. Free Energy and Temperature
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
Horxn = +467.9 kJ Sorxn = +560.3 J/K
Gorxn = 467.9 kJ - (298K)(0.560kJ/K) = +300.8 kJ
Reaction is reactant-favored at 298 K
At what T does Gorxn just change from (+) to (-)?
i.e. what is T for Gorxn = 0 = Horxn - TSorxn
If Gorxn = 0 then Horxn = TSorxn
so T = Ho/So ~ 468kJ/0.56kJ/K = 836 K or 563oC
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28. 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) Gorxn = +742 kJ
with a thermodynamically allowed reaction:
3/2 C(s) + 3/2 O2 (g)  3/2 CO2(g) Gorxn = -592 kJ
Overall : Fe2O3(s) + 3/2 C(s)  2 Fe(s) + 3/2 CO2(g)
Gorxn= +301 kJ @ 25oC
BUT Gorxn < 0 kJ for T > 563oC
See Kotz, pp933-935 for analysis of the thermite reaction
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29. Other examples of coupled reactions:
Copper smelting
Cu2S (s)  2 Cu (s) + S (s) Gorxn= +86.2 kJ
Couple this with: (FORBIDDEN)
S (s) + O2 (g)  SO2 (s) Gorxn= -300.1 kJ
Overall: Cu2S (s) + O2 (g)  2 Cu (s) + SO2 (s)
Gorxn= +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 Horxn = -20 kJ
Sorxn = +34 J/K
Gorxn = -30 kJ @ 37oC
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30. Thermodynamics and Keq
• Keq is related to reaction favorability.
• If Gorxn < 0, reaction is product-favored.
• Gorxn 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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31. 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 < 0 then reaction proceeds to right
If Grxn > 0 then reaction proceeds to left
At equilibrium, Grxn = 0. Also, Q = K. Thus
Gorxn = - RT lnK
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32. Thermodynamics and Keq (2)
2 NO2  N2O4
Gorxn = -4.8 kJ
• pure NO2 has Grxn < 0.
• Reaction proceeds until Grxn =
0 - the minimum in G(reaction) -
see graph.
• At this point, both N2O4 and
NO2 are present, with more
N2O4.
9_G_NO2.mov
20m09an1.mov
• This is a product-favored
reaction.
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33. Thermodynamics and Keq (3)
N2O4 2 NO2 Gorxn = +4.8
kJ
• pure N2O4 has Grxn < 0.
• Reaction proceeds until Grxn
= 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.
9_G_N2O4.mov
20m09an2.mov
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34. Thermodynamics and Keq (4)
Keq is related to reaction favorability and so
to Gorxn.
The larger the value of Gorxn the larger the
value of K.
G rxn = - RT lnK
o
where R = 8.31 J/K•mol
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35. Thermodynamics and Keq (5)
Gorxn = - RT lnK
Calculate K for the reaction
N2O4  2 NO2 Gorxn = +4.8 kJ
Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K
4800 J
lnK = - = - 1.94
(8.31 J/K)(298K)
K = 0.14
When Gorxn > 0, then K < 1 - reactant favoured
When Gorxn < 0, then K >1 - product favoured
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36. 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
Entropy & Free Energy (Ch 20) -
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lect. 2

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