1. Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Chapters 5,8: Energy (E), Heat (q), Work (w), and Enthalpy (∆H)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapters 5,8: Energy (E), Heat (q), Work (w), and Enthalpy (∆H)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

3. Energy (E)
Enthalpy (H)
(kJ)
Entropy (S)
(J/K)
Free Energy (G) (kJ)

4. Energy (E) What is it? ability to do work OR transfer heat
Work (w): transfer of energy by applying a force over a distance.
Heat (q): transfer of energy by DT (high to low)
unit of energy: joule (J)
an older unit still in widespread use is… calorie (cal)
1 Cal = 1000 cal
1 cal = 4.18 J
2000 Cal ≈ 8,000,000 J ≈ 8 MJ!!!

5. System and Surroundings
molecules to be studied (reactants & products)
Surroundings:
everything else
(container, thermometer)

6. 1st Law of Thermodynamics
Energy is neither created nor destroyed.
total energy of an isolated system is constant
(universe)
(no transfer matter/energy)
(conserved)
Internal Energy (E):
E = KE PE
(motions)
(Thermal Energy)
(attractions)
(calculating E is too complex a problem)
E = Efinal − Einitial
released or absorbed

7. Changes in Internal Energy
Energy is transferred between the system and surroundings, as either heat (q) or work (w).
E = q + w
E = ?
E = (–) + (+)
Surroundings
E = +
System
q in (+)
q out (–)
E = q + w
w on (+)
w by (–)

8. Changes in Internal Energy
Efinal > Einitial
absorbed energy
(endergonic)
Efinal < Einitial
released energy
(exergonic)

9. Work (w)
The only work done by a gas at constant P is change in V by pushing on surroundings.
PV = −w ΔV
Zn + H+
Zn2+ + H2(g)

10. Enthalpy Enthalpy (H) is: H = E + PV H = E + PV E = q+w PV = −w
internal work done
energy
work done by system
heat/work energy in or out of system
H = E PV
E = q+w
PV = −w
By substituting at constant P :
H = E + PV
H = (q+w) + (−w)
H = q
(change in) enthalpy IS heat absorbed/released
H = heat

11. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Enthalpy of Reaction
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
enthalpy is…
…the heat transfer in/out of a system
(at constant P)
H = E + PV
H = q
H = Hproducts − Hreactants
exergonic
exothermic
endergonic
endothermic

12. Endothermic & Exothermic
Endothermic: H > 0 (+)
H(+) = Hfinal − Hinitial
products
reactants
Exothermic: H < 0 (–)
H(–) = Hfinal − Hinitial
reactants
products
H(–) is thermodynamically favorable

13. Enthalpy of Reaction
2 H2(g) + O2(g)  2 H2O(g)
Hrxn, is the enthalpy of reaction, or “heat” of reaction.
units: kJ
molrxn
kJ/molrxn
kJ∙molrxn
Demo
–242 kJ
per 1 mol O2
(OR)
H =
–242 kJ/molrxn
–242 kJ
per 2 mol H2
Demo – Zn in narrow mouth Erlenmeyer flask. Add 6 M H2SO4. Stretch out balloon well and place over mouth. Blow some (not much) O2 into balloon and tie.
(OR) –1
–121 kJ
per 1 mol H2

14. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Enthalpy
HW p. 207 #34,35,38,45
H depends on amount (moles, coefficients)
Hreverse rxn = –Hforward rxn
Hrxn depends
on the state (s, l, g) of
products & reactants
H1 = –802 kJ if 2 H2O(g)
because…
2 H2O(l)  2 H2O(g)
H = +88 kJ/molrxn
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)

15. Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies 2) Hess’s Law 3) Standard Heats of Formation (Hf ) 4) Calorimetry (lab) 

16. Overlap and Bonding When bonds/attractions form, energy is _________.
released – + + – + +
What repulsive forces?
What attractive forces?
Where is energy being released?
Where must energy be added?

17. Potential Energy of Bonds
High PE
chemical bond
Low PE
(energy released when bonds form)
High PE
(energy absorbed when bonds break) + +

18. Bond Enthalpy (BE) p.330 BE: ∆H for the breaking of a bond (all +)
aka…
bond dissociation energy

19. Enthalpies of Reaction (∆H)
BE: ∆H for the breaking of a bond (all +)
To determine H for a reaction:
compare the BE of bonds broken (reactants) to the BE of bonds formed (products).
Hrxn = (BEreactants)  (BEproducts)
(bonds broken)
(bonds formed)
(released)
(stronger)
H(+) = BEreac − BEprod
(NOT on equation sheet)
H(–) = BEreac − BEprod
(stronger)

20. Enthalpies of Reaction (∆H)
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
Hrxn =
[4(C—H) + (Cl—Cl)] 
 [3(C—H) + (C—Cl) + (H—Cl)]
= [4(413) + (242)]  [3(413) + (328) + (431)] = (655)  (759) Hrxn = 104 kJ/molrxn
HW p. 339 #66, 68

21. (NOT on equation sheet)
Hess’s Law
p. 191
H = Hfinal − Hinitial
prod.
react.
Hrxn is independent of route taken
Hrxn = sum of H of all steps
(NOT on equation sheet)
Hoverall = H1 + H2 + H3 …

22. Calculation of H by Hess’s Law
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
∆Hcomb = ? 
Given: +
3 C(gr.) + 4 H2(g)  C3H8(g) ∆H1= –104 kJ
C(gr.) + O2(g)  CO2(g) ∆H2= –394 kJ
H2(g) + ½ O2(g)  H2O(l) ∆H3= –286 kJ
3( )
3( )
4( )
4( )
Used:
C3H8(g)  3 C(gr.) + 4 H2(g) ∆H1= +104 kJ
3 C(gr.) + 3 O2(g)  3 CO2(g) ∆H2= –1182 kJ
4 H2(g) + 2 O2(g)  4 H2O(l) ∆H3= –1144 kJ
∆Hcomb =
–2222 kJ
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)

23. Standard Enthalpy of Formation (Hf )
heat released or absorbed by the formation of a compound from its pure elements in their standard states. o
(25oC , 1 atm) o
Hf = 0 for all elements in standard state
3 C(gr.) + 4 H2(g)  C3H8(g) ∆Hf = –104 kJ o o o
Hf = 0
Hf = 0
Hf = –104 kJ
H = Hfinal − Hinitial
Recall…
…therefore --->
prod.
react.

24. Calculation of H by Hf’s
…we can use Hess’s law in this way:
H = nHf(products) – mHf(reactants)
where n and m are the stoichiometric coefficients.
“sum”   
(on equation sheet)

25. Calculation of H by Hf’s
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
∆Hcomb = ?  
H = nHf(products) – mHf(reactants)
Appendix C (p )    
H = [3(Hf CO2) + 4(Hf H2O)] – [1(Hf C3H8) + 5(Hf O2)] H = [(–1180.5) + (–1143.2)] – [(–103.85) + (0)] H = (–2323.7) – (–103.85) H = – kJ
HW p. 209 #60,63,66, 72,73

26. Calorimeter nearly isolated
Calorimetry
We can’t know the exact enthalpy of reactants and products, so we measure H by calorimetry,
the measurement of heat flow.
By reacting in solution in a calorimeter, we indirectly determine H of system by measuring ∆T & calculating q of the surroundings (calorimeter).
Calorimeter nearly isolated
(on equation sheet)
heat (J)
q = mcT
mass (g)
[of sol’n]
Tf – Ti (oC)
[of surroundings]

27. Specific Heat Capacity (c)
(or specific heat)
energy required to raise temp of 1 g by 1C.
(for water)
c = 4.18 J/goC
J of heat
Metals have much lower c’s b/c they transfer heat and change temp easily.

28. (in J) of calorimeter or surroundings
Calorimetry
HW p. 208 #49, 52, 54
(in J) of calorimeter or surroundings
q = mcT
– q = Hrxn
(in kJ/mol) of system
When 4.50 g NaOH(s) is dissolved 200. g of water in a calorimeter, the temp. changes from 22.4oC to 28.3oC. Calculate the molar heat of solution, ∆Hsoln (in kJ/mol NaOH).
q = ( )(4.18)(28.3–22.4)
q = 5040 J
H = –5.04 kJ
mol
= –44.8 kJ
mol
H = –5.04 kJ
4.50 g NaOH x 1 mol = mol
40.00 g

29. Chp. 5,8: Calculate ∆H (4 Ways)
1) Bond Energies
Hrxn = (BEreactants)  (BEproducts)
2) Hess’s Law
Hoverall = Hrxn1 + Hrxn2 + Hrxn3 …
3) Standard Heats of Formation (Hf )
H = nHf(products) – mHf(reactants)
4) Calorimetry (lab)
q = mc∆T (surroundings or thermometer)
–q = ∆H ∆H/mol = kJ/mol (molar enthalpy)
(NOT)
(+ broken)
(– formed)
(NOT)    
(given)
(given)

30. Chapter 19: Thermodynamics (∆H, ∆S, ∆G, K)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 19: Thermodynamics (∆H, ∆S, ∆G, K)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

31. Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ)
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
ΔE = q + w
PΔV = –w
(at constant P)
ΔH = q
(heat)
ΔH = ?
ΔS = ?
ΔG = ?

32. Big Idea #5: Thermodynamics
Bonds break and form to lower free energy (∆G).
Chemical and physical processes are driven by:
a decrease in enthalpy (–∆H), or
an increase in entropy (+∆S), or
both.

33. 1st Law of Thermodynamics
Energy cannot be created nor destroyed (is conserved)
Therefore, the total energy of the universe is constant.
Hsystem = –Hsurroundings OR
Huniv = Hsystem + Hsurroundings = 0
if (+) then (–) = 0
if (–) then (+) = 0

34. Thermodynamically Favorable
Thermodynamically Favorable (spontaneous) processes are those that can proceed without any outside intervention.
The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously

35. Thermodynamically Favorable
Processes that are thermodynamically favorable (spontaneous) in one direction are NOT in the reverse direction.

36. Thermodynamically Favorable
Processes that are favorable (spontaneous) at one temperature…
…may not be at other temperatures.
HW p. 837 #7, 11
melting
freezing

37. Entropy (S) S = Sfinal  Sinitial (okay but oversimplified)
disorder/randomness
(more correct)
dispersal of matter & energy among various motions of particles in space at a temperature in J/K.
“The energy of the universe is constant.”
“The entropy of the universe tends toward a maximum.”
(ratio of heat to temp)
S = ∆H T
S = Sfinal  Sinitial
S = + therm fav
S = – not therm fav
(more dispersal)
(less dispersal)
(structure/organization)

38. Entropy (S)
height
AND
weight
(a part)
S = ∆H T
(the rest)
change in entropy (S) depends on heat transferred (∆H) AND temperature (T)
same ∆H
diff. ∆S
System A
(100 K)
50 J
System B
(25 K)
50 J
Surroundings (100 K)
Visualize a 1 ft tall dog. Now, that dog is a 100 lb. dachshund!
Surroundings (25 K)
∆S = ____J/K
+0.5
∆S = ____J/K
+2.0
(loud restaurant, less disturbed)
(quiet library, more disturbed)
Cough!

39. Entropy Example: melting 1 mol of ice at 0oC. Hhand = –6000 J
Hice = J

40. Entropy S = ∆H T +Hice = –Hhand Huniv = 0 Suniv = +
+Sice > –Shand
The melting of 1 mol of ice at 0oC.
Hfusion T
(1 mol)(6000 J/mol)
273 K
Sice = = =
+22.0 J/K
(gained by ice)
Assume the ice melted in your hand at 37oC.
Hfusion T
(1 mol)(–6000 J/mol)
310 K
Shand = = =
–19.4 J/K
(lost by hand)
Suniv = Ssystem + Ssurroundings
Suniv = (22.0 J/K) + (–19.4 J/K) =
(ice)
(hand)
Suniv
+2.6 J/K

41. Universe (isolated system)
1st Law: J = J
Huniv = 0
Suniv = +
2nd Law: +22 J/K > –19 J/K
5290 J
(usable E)
(usable E)
6000 J +
(dispersed E)
710 J
(in hand)
(in ice)
+2.6 J/K x 273 K =
(∆Suniv) (T)
Universe (isolated system)
Initial Energy
“dispersed” energy
(unusable)
Free energy
(useful for work)
Final Energy

42. 2nd Law of Thermodynamics
Suniv = Ssystem + Ssurroundings
For thermodynamically favorable (spontaneous) processes…
… +∆S gained always greater than –∆S lost,
so…
Suniv = Ssystem + Ssurroundings > 0
2nd Law of Thermodynamics (formally stated):
All favorable processes
increase the entropy of the universe
(Suniv > 0)
HW p. 837 #20, 21

43. Entropy (Molecular Scale)
Ludwig Boltzmann described entropy with molecular motion.
Motion: Translational , Vibrational, Rotational 
He envisioned the molecular motions of a sample of matter at a single instant in time (like a snapshot) called a microstate.

44. Entropy (Molecular Scale)
S = k lnW
Boltzmann constant
1.38  1023 J/K
microstates (max number possible)
Entropy increases (+∆S) with the number of microstates in the system.
<
<
<

45. Entropy (Molecular Scale)
S : dispersal of matter & energy at T
The number of microstates and, therefore, the entropy tends to increase with…
↑Temperature (motion as KEavg)
↑Volume (motion in space)
↑Number of particles (motion as KEtotal)
↑Size of particles (motion of bond vibrations)
↑Types of particles (mixing)

46. Entropy (Molecular Scale)
S : dispersal of matter & energy at T
Maxwell-Boltzmann distribution curve:
∆S > 0 by adding heat as…
…distribution of KEavg increases

47. Entropy (Molecular Scale)
S : dispersal of matter & energy at T
Entropy increases with the freedom of motion.
S(s) < S(l) < S(g)
S(s) < S(l) < S(aq) < S(g)
gas
solid T
more microstates
(s) + (l)  (aq) V
H2O(g)
H2O(g)

48. Standard Entropy (So)
Standard entropies tend to increase with increasing molecular size.
larger molecules have more microstates

49. Entropy Changes (S) In general, entropy increases when
gases form from liquids and solids
liquids or solutions form from solids
moles of gas molecules increase
total moles increase
Predict the sign of S in these reactions:
1. Pb(s) + 2 HI(aq)  PbI2(s) + H2(g)
2. NH3(g) + H2O(l)  NH4OH(aq)
S = +
S = –

50. Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0. (not possible)
S = k lnW
S = k ln(1)
S = 0
increase
Temp.
only 1 microstate
0 K
S = 0
> 0 K
S > 0

51. Standard Entropy Changes (∆So)
HW p. 838
#29, 31, 40, 42, 48
Standard entropies, S.
(Appendix C)
So = nSo(products) – mSo(reactants)
(on equation sheet)
where n and m are the coefficients in the balanced chemical equation.

52. (dispersal of matter & energy at T)
Energy (E)
Enthalpy (H)
(kJ)
Entropy (S)
(J/K)
Free Energy (G) (kJ)
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
(disorder)
(microstates)
(dispersal of matter & energy at T)
ΔE = q + w
PΔV = –w
(at constant P)
ΔH = q
(heat)
ΔS = ΔH T
ΔS = ?
ΔG = ?

53. Big Idea #5: Thermodynamics
Bonds break and form to lower free energy (∆G).
Chemical and physical processes are driven by:
a decrease in enthalpy (–∆H), or
an increase in entropy (+∆S), or
both.

54. Thermodynamically Favorable
Chemical and physical processes are driven by:
decrease in enthalpy (–∆Hsys)
increase in entropy (+∆Ssys)
causes (+∆Ssurr)
(+)
(+)
Suniv = Ssystem + Ssurroundings > 0
Thermodynamically Favorable: (defined as)
increasing entropy of the universe (∆Suniv > 0)
∆Suniv > 0
(+Entropy Change of the Universe)

55. (∆Suniv) & (∆Gsys) Hsystem
For all thermodynamically favorable reactions:
Suniverse = Ssystem + Ssurroundings > 0
(Boltzmann)
Hsystem T
Suniverse = Ssystem +
(Clausius)
multiplying each term by T:
–TSuniverse = –TSsystem + Hsystem
rearrange terms:
–TSuniverse = Hsystem – TSsystem
Gsystem = Hsystem – TSsystem
(Gibbs free energy equation)

56. (∆Suniv) & (∆Gsys) –TSuniv = Hsys – TSsys Gsys = Hsys – TSsys
(Gibbs free energy equation)
Gibbs defined TDSuniv as the change in
free energy of a system (Gsys) or G.
Free Energy (G) is more useful than
Suniv b/c all terms focus on the system.
If –Gsys , then +Suniverse . Therefore…
–G is thermodynamically favorable.
“Bonds break & form to lower free energy (∆G).”

57. (not react to completion)
Gibbs Free Energy (∆G)
∆G : free energy transfer of system as work
–∆G : work done by system (–w) favorably
+∆G : work done on system (+w) to cause rxn
(not react to completion)
+DG
–DG

58. Q & ∆G (not ∆Go) [P] R  P Q = [R] Q < K Q > K +DG –DG –DG Q = K
–DG (release), therm. fav.
+G (absorb), not therm. fav.
DG = 0, system at equilibrium.
(Q = K)
(not react to completion)
can cause with electricity/light
Q > K
+DG
–DG
–DG
Q = K
Gmin  0
DG = 0
DGo (1 M, 1 atm, 25oC)
Q = 1 = K (rare)

59. Standard Free Energy (∆Go) and Temperature (T)
(on equation sheet)
(consists of 2 terms)
DG = DH – TS
free energy (kJ/mol)
enthalpy term (kJ/mol)
entropy term (J/mol∙K)
units convert to kJ!!!
max energy used for work
energy transferred as heat
energy dispersed as disorder
The temperature dependence of free energy comes from the entropy term (–TS).

60. Standard Free Energy (∆Go) and Temperature (T)
DG = DH  TS
Thermodynamic Favorability
∆Go =
(∆Ho)
∆So
– T( )
( ) –T( )
(high T) –
(low T) +
(fav. at high T)
(unfav. at low T) + + =
( ) – T ( ) + + –
(unfav. at ALL T) + =
( ) – T( ) +
(fav. at ALL T) – =
( ) – T( ) – +
(high T) +
(low T) –
( ) –T( )
(unfav. at high T)
(fav. at low T) – – =
( ) – T ( ) – –

61. Calculating ∆Go (4 ways)
Standard free energies of formation, Gf :
Gibbs Free Energy equation:
From K value (next few slides)
From voltage, Eo (next Unit)
DG = SnG(products) – SmG(reactants) f
(given equation)
HW p. 840 #52, 54, 56, 60
DG = DH – TS
(given equation)
(may need to calc. ∆Ho & ∆So first)
(given equation)
(given equation)

62. + = Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K)
Free Energy (G) (kJ)
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
(disorder)
(microstates)
(energy of ΔH and ΔS at a T)
(dispersal of matter & energy at T)
(max work done by favorable rxn)
ΔE = q + w
PΔV = –w
(at constant P)
(–∆Gsys means +∆Suniv & K>1)
ΔH = q
(heat)
ΔS = ΔH T
ΔG = ΔH – TΔS
ΔG = ?

63. Free Energy (∆G) & Equilibrium (K)
Under any conditions, standard or nonstandard, the free energy change can be found by:
G = G + RT lnQ
Q =
[P]
[R]
RT is “thermal energy”
RT = ( kJ)(298)
= 2.5 kJ at 25oC
At equilibrium:
Q = K
G = 0
therefore:
0 = G + RT lnK
rearrange:
G = –RT lnK

64. Free Energy (∆G) & Equilibrium (K)
G = –RT ln K
(on equation sheet)
If G in kJ,
then R in kJ………
R = J∙mol–1∙K–1
= kJ∙mol–1∙K–1
–∆Go RT
= ln K
–∆Go RT
Solved for K :
(NOT on equation sheet)
K = e^

65. Free Energy (∆G) & Equilibrium (K)
G = –RT ln K
∆Go = –RT(ln K) K
@ Equilibrium – + =
–RT ( )
> 1
product favored
(favorable forward) + – =
–RT ( )
< 1
reactant favored
(unfavorable forward)

66. + = Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K)
Free Energy (G) (kJ)
ΔH = ΔE + PΔV
internal work by
energy system
(KE + PE) (–w)
(disorder)
(microstates)
(energy of ΔH and ΔS at a T)
(dispersal of matter & energy at T)
(max work done by favorable rxn)
ΔE = q + w
PΔV = –w
(at constant P)
(–∆Gsys means +∆Suniv & K>1)
ΔH = q
(heat)
ΔS = ΔH T
ΔG = ΔH – TΔS

67. p. 837 #6 What does x quantify? ΔGo What is significant at this point?
G of Reactants
What does x quantify?
ΔGo
What is significant at this point?
G of Products
ΔG = 0 (at equilibrium)

68. p. 837 #4 In what T range is this favorable? What happens at 300 K?
ΔG = ΔH – TΔS
T > 300 K
ΔH = TΔS
so…
ΔG = ΔH – TΔS
ΔG = 0 (at equilibrium)
HW p. 840 #63, 72, 74, 76

69. ∆Go & Rxn Coupling Rxn Coupling:
Unfav. rxns (+∆Go) combine with Fav. rxns (–∆Go) to make a Fav. overall (–∆Gooverall ).
goes up if coupled
(zinc ore)  (zinc metal)
(NOT therm.fav.)
ZnS(s)  Zn(s) + S(s)
∆Go = +198 kJ/mol
S(s) + O2(g)  SO2(g)
∆Go = –300 kJ/mol
ZnS(s) + O2(g)  Zn(s) + SO2(g)
∆Go = –102 kJ/mol
(therm.fav.)

70. ∆Go & Biochemical Rxn Coupling
(weak bond broken, stronger bonds formed)
ATP
ADP
ATP + H2O  ADP + H3PO4
∆Go = –31 kJ/mol
Alanine + Glycine  Alanylglycine
∆Go = +29 kJ/mol
(amino acids) (peptide/proteins)
ATP + H2O + Ala + Gly  ADP + H3PO4 + Alanylglycine
∆Go = –2 kJ/mol

71. ∆Go & Biochemical Rxn Coupling
Overall Rxn:
Glu + Pi  Glu-6-P
ATP  ADP + Pi
Glu + ATP  Glu-6-P + ADP
+14 (not fav)
–31 (fav)
–17 (fav)
Overall Reaction:
1st of 10 steps of Glycolysis is phosphorylation of glucose at C-6 (ends with ATP production).
∆Govr
∆Govr = ∆G1 + ∆G2

72. ∆Go & Biochemical Rxn Coupling
Glucose
(C6H12O6)
ATP
Proteins
+ O2
(oxidation)
Free Energy (G)
–∆G
(fav)
+∆G
(not fav)
–∆G
(fav)
+∆G
(not fav)
ADP
CO2 + H2O
Amino Acids

73. Thermodynamic vs Kinetic Control
Kinetic Control: (path 2: A  C )
A thermodynamically favored process (–ΔGo) with no measurable product or rate while not at equilibrium, must have a very high Ea .
A  B ∆Go = +10
Ea = +20
(kinetic product)
(initially pure reactant A)
path 1
(low Ea , Temp , time) B A
A  C ∆Go = –50
Ea = +50
(thermodynamic product)
Free Energy (G) 
+10 kJ
path 2
–50 kJ C
(–∆Go, Temp, Q<<K, time)

74. Thermodynamic vs Kinetic Control
Thermodynamic Product: ___ E
Kinetic Product: ___ D
Rxn A  E will be under ______________ control at low temp and Q > K .
kinetic (high Ea)
Pain 