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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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
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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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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)
DH = kJ
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

4. Non-exothermic spontaneous reactions
But many spontaneous reactions or processes are endothermic . . .
NH4NO3(s) + heat  NH4+ (aq) + NO3- (aq)
Hsol = kJ/mol
or have H =
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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.
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

6. At 298o the entropy of a substance is called
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) = ( ) = 93.3 J.K-1mol-1
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

7. Entropy & Free Energy (Ch 20) - lect. 2
Entropy and Phase
Ice
Water
Vapour
So (J/K•mol)
H2O(gas) 188.8
H2O(liq)
H2O (s)
S (gases) > S (liquids) > S (solids)
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

8. 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
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10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

9. Entropy and complexity
Increase in molecular complexity generally leads to increase in S.
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So (J/K•mol) CH
C2H
C3H
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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+ / O
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) =
= J.K-1mol-1
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

11. 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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

12. 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/K•mol) -
[2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]
DSo = 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??
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

13. 2nd Law of Thermodynamics
A reaction is spontaneous (product-favored) if
S for the universe is positive.
DSuniverse = DSsystem + DSsurroundings
DSuniverse > 0 for product-favored process
First calc. entropy created by matter dispersal (DSsystem)
Next, calc. entropy created by energy dispersal (DSsurround)
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

14. Calculating S(universe)
2 H2(g) + O2(g)  2 H2O(liq)
DSosystem = J/K
Can calculate that DHorxn = DHosystem = kJ
DSosurroundings = J/K
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

15. Calculating S(universe) (2)
2 H2(g) + O2(g)  2 H2O(liq)
DSosystem = J/K (less matter dispersal)
DSosurroundings = J/K (more energy dispersal)
DSouniverse = J/K
The entropy of the universe increases
so the reaction is spontaneous.
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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 INCREASE
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)
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

17. Entropy & Free Energy (Ch 20) - lect. 2
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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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
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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

19. Sign of DG 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 < and Go = Ho -TS
DGo < 0 for all SPONTANEOUS processes
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

20. 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.
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

21. Gibbs Free Energy changes for reactions
DGo = DHo - TDSo
DHo DSo DGo 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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

22. 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)
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

23. Entropy & Free Energy (Ch 20) - lect. 2
Calculating DGorxn
EXAMPLE: Combustion of acetylene
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g)
From standard enthalpies of formation: DHorxn = kJ
From standard molar entropies: DSorxn = kJ/K
Calculate DGorxn from DGo = Ho - TSo
DGorxn = kJ - (298 K)( kJ/K)
= kJ
Reaction is product-favored in spite of negative DSorxn. Reaction is “enthalpy driven”
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

24. 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?
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10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

25. DGorxn for NH4NO3(s)  NH4NO3(aq)
From tables of thermodynamic data we find
DHorxn = kJ
DSorxn = J/K or kJ/K
DGorxn = kJ - (298 K)( kJ/K)
= kJ
Reaction is product-favored
. . . in spite of positive DHorxn.
Reaction is “entropy driven”
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

26. Entropy & Free Energy (Ch 20) - lect. 2
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 = kJ - [ ]
Note that free energy of formation of an element in its standard state is 0.
DGorxn = kJ
Reaction is product-favored as expected.
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

27. Free Energy and Temperature
2 Fe2O3(s) C(s)  4 Fe(s) CO2(g)
DHorxn = kJ DSorxn = J/K
DGorxn = kJ - (298K)(0.560kJ/K) = 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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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) DGorxn = kJ
with a thermodynamically allowed reaction:
3/2 C(s) + 3/2 O2 (g)  3/2 CO2(g) DGorxn = kJ
Overall :
Fe2O3(s) + 3/2 C(s)  2 Fe(s) + 3/2 CO2(g)
DGorxn= +301 kJ @ 25oC
BUT DGorxn < 0 kJ for T > 563oC
See Kotz, pp for analysis of the thermite reaction
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

29. Other examples of coupled reactions:
Copper smelting
Cu2S (s)  2 Cu (s) + S (s) DGorxn= kJ
(FORBIDDEN)
Couple this with:
S (s) + O2 (g)  SO2 (s) DGorxn= kJ
Overall: Cu2S (s) + O2 (g)  2 Cu (s) + SO2 (s)
DGorxn= kJ kJ = 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 = oC
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

30. Thermodynamics and Keq
Keq is related to reaction favorability.
If DGorxn < 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 ?
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

31. Entropy & Free Energy (Ch 20) - lect. 2
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
DGorxn = - RT lnK
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

32. Thermodynamics and Keq (2)
2 NO2  N2O4
DGorxn = -4.8 kJ
pure NO2 has DGrxn < 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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10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

33. Thermodynamics and Keq (3)
N2O4 2 NO DGorxn = +4.8 kJ
pure N2O4 has DGrxn < 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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10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

34. 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/K•mol
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

35. Thermodynamics and Keq (5)
DGorxn = - RT lnK
Calculate K for the reaction
N2O4  2 NO2 DGorxn = +4.8 kJ
DGorxn = J = - (8.31 J/K)(298 K) ln K
K = 0.14
When Gorxn > 0, then K < 1 - reactant favoured
When Gorxn < 0, then K >1 - product favoured
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2

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
10 Nov 97
Entropy & Free Energy (Ch 20) - lect. 2