1. Chapter 19 Thermodynamics
Ch 14 Kinetics: How fast do reactions go?
Ch15 Equilibrium: How far do reactions proceed?
Ch 19 Chemical Thermodynamics: What makes reactions go?

2. Thermodynamics
The study of the relationships between heat, work, and the energy of a system.
You may recall that changes can be exothermic (-DH) or endothermic (+DH)

3. Exothermic: activation energy (A) is absorbed to break bonds, and energy is released as new bonds form (B). Heat is released to the surroundings, since B>A
-H
+H
Endothermic: activation energy (A) is absorbed to break bonds, and energy is released as new bonds form (B). Heat is absorbed from the surroundings, since B<A

4. Exothermic can usually continue once begun since they can recycle some of the energy released (B) as activation energy (A) to keep the reaction going. Changes like this are said to be spontaneous. ?
Endothermic do not release enough energy (B) to provide all the activation energy to keep reaction going, so outside energy must usually be added to drive the reaction. Changes like this are usually not spontaneous.

5. Actually the AP folks prefer you to use the word favorable in replace of spontaneous. Either way it means a reaction that will continue on its own once begun. ?
So exothermic reactions are favorable since they provide their own activation energy.

6. To clarify: Spontaneous in the English language implies “a process which begins on its own – without cause”.
Spontaneous in chemical vernacular means a process that continues on its own once it is caused to begin – by an outside force.
In order to avoid eliminate confusion the AP folks avoid using the word “spontaneous” even though the term is still commonly used in chemistry. Instead you will see the term “favorable”.

7. favorable
Exothermic: combustion
Spontaneous once begun
Endothermic: Cooking, Not spontaneous, cooking stops once heat is removed.
Not favorable

8. But, some endothermic reactions drive themselves! How?
Dissolving proceeds on its own without outside intervention.
favorable
Its
Entropy
in action
Entropy is “S”

10. Review of chapter 5 topics
Crash course chemistry #17: Energy & Chemistry
Crash course chemistry #18: Enthalpy
Chapter 19 topic: Gibbs Free energy
Crash course chemistry #20: Entropy

11. 19.1 Spontaneous Processes
(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 move to just one vessel.

12. Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
Spontaneous processes are irreversible.
favorable

13. Spontaneous Processes
favorable
Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
Above 0C it is spontaneous for ice to melt.
Below 0C the reverse process is spontaneous.
favorable

15. 19.2/3 Entropy
Entropy can be thought of as a measure of the randomness or disorder of a system.
Entropy increases as the number of modes of motion in molecules (microstates) increases
Vibrational
Rotational
Translational

16. The Entropy of a pure crystalline substance at absolute zero is zero.
why?
It has only one possible microstate.
As temperature increases the molecules begin to move and there become multiple possible microstates.

17. The first law was “Conservation of Energy”
( E system = E surroundings ): Chapter 5
The Second law (of thermodynamics): Entropy in the universe always increases
Entropy (unlike energy) is NOT conserved.
Spontaneous processes happen when randomness in the universe increases

18. If a change is spontaneous, the entropy of the universe is increasing.
Favorable
Un- Favorable
Because rusting is spontaneous, overall S must be positive. therefore the entropy of the surroundings must increase more than the entropy decrease of the system.
If a change is spontaneous, the entropy of the universe is increasing.

19. Molecules have more entropy (disorder) when:
Phase Changes from: S  L  G
Example: Sublimation
CO2(s)  CO2(g)
Dissolving occurs (solution forms):
Example:
NaCl(s)  Na+(aq) + Cl-(aq)
Matter spreads out
(more microstates)
Matter spreads out
(more microstates)

20. Molecules have more entropy (disorder) when:
3) Temperature increases
Example:
Fe(s) at 0 oC  Fe(s) at 25 oC
4) For Gases ONLY, when
Volume increases or Pressure decreases
Examples:
2 Liters He(g)  4 Liters He(g)
3 atm He(g)  1 atm He(g)
Energy spreads out
(more microstates)
matter spreads out (more microstates)

21. Molecules have more entropy (disorder) when:
5) Rx results in more molecules/moles of gas
Examples:
2NH3(g)  N2(g) + 3H2(g)
CaCO3(s)  CaO(s) + CO2(g)
N2O4(g)  2 NO2(g)
This one is difficult to predict:
N2(g) + O2(g)  2 NO(g)
Vs. 4 moles
2 moles
matter spreads out (more microstates)
(1 2) matter spreads out (more microstates)
(2  2)

22. Molecules have more entropy (disorder) when:
When there are more moles
Example:
1 mole H2O(g) Vs 2 moles H2O(g)
7) When there are more atoms per molecule
Examples:
1 mole Ar(g) vs 1 mole HCl(g)
1 mole NO2(g) vs 1 mole N2O4(g)
More matter is more spread out (more microstates)
More matter is more spread out (more microstates)

23. Molecules have more entropy (disorder) when:
8) When an atom has a bigger atomic number
1 mole He(g) vs. 1 mole Ne (g)
More matter (P,N,e) = more spread out
(more microstates)
In spite of all of these, entropy changes during phase changes is the most important.
(solid  liquid or Aqueous  gases)

24. Again notice the typical change in entropy reduces to changes of state

27. Go figure: Why does the plot show vertical jumps at the melting and boiling points?
Separation of molecules produce the greatest increase in entropy 
(motions and microstates)
Molecules spread out
as temp increases?

28. 19.4 Entropy changes / Standard Entropies
Standard Conditions:
298 K, 1 atm, 1 Molar
The values for Standard Entropies (So) are expressed in J/mol-K.
Notice: Increase with increasing molar mass.

29. Entropy Changes DSo = S n Soproducts - S m Soreactants
Sum of all entropy in the products
- Sum of all entropy in the reactants
DSo = S n Soproducts - S m Soreactants
Be careful: S°units are in J/mol-K
Note for pure elements:
Even though:
(Joules Not kJ’s)
(recall that we ignored elements in Hess’ law probs?)

30. Entropy Changes DSo = S n Soproducts - S m Soreactants
To clarify: S has an actual value for elements since its based on number of possible microstates, while H is a relative value based on loss or gain of energy (not energy actually stored in the molecule) so the H for the lowest energy state element is set at zero.

31. Appendix C values
Al2O3(s) = Jmol-K-
H2(g) =
Al(s) = 28.32
H2O(g) =

32. Reference Tables Heat exchanged during temp change (calorimetry)
Entropy change
Enthalpy change (Hess’ law)
Free energy (Gibbs) change
Up next!
Free energy and the equilibrium constant
(coming soon)

33. 19.5 Gibbs Free Energy
Enthalpy and entropy both contribute to help drive reactions
Enthalpy
Matter tends to spread out increasing entropy
Exothermic changes lose heat to the surroundings

34. 19.5 Gibbs Free Energy Reactions which release heat are favorable
Entropy increases as temperature increases
T is combined with S into entropy term
TS
Enthalpy
Reactions which release heat are favorable
Entropy tends to increases as disorder increases

35. 19.5 Gibbs Free Energy The two are Combined into Gibb’s “Free” energy
Enthalpy
The two are Combined into Gibb’s “Free” energy
(energy free to do work on the surroundings)

36. 19.5 Gibbs Free Energy H + -(TS) H - TS G =  H - TS
Enthalpy
Taking the negative of the entropy term allows us to combine the two terms into one
H (TS)
H TS
G =  H - TS

37. 19.5 Gibbs Free Energy
Enthalpy
You should note that Free energy released into the surroundings (-G)
translates into an increase in the entropy of the universe as a whole (+Suniv).

38. 19.5 Gibbs Free Energy
Enthalpy
Spontaneous changes occur when the universe becomes more disorderly.

39. 19.5 Gibbs Free Energy
Enthalpy
Since the universe is a difficult thing to view as a whole, we can focus just on the changes occurring in the system: G

40. DGo = DHo – TDSo 19.5 Gibbs Free Energy
Use G to decide if a process is spontaneous
favorable
G = negative value = spontaneous
Favorable – free energy released to the surroundings?

41. DGo = DHo – TDSo 19.5 Gibbs Free Energy G = zero = at equilibrium
For reversible reactions

42. DGo = DHo – TDSo 19.5 Gibbs Free Energy
G = positive value = not spontaneous
Note: o under standard conditions
Equation can be used without also.
unfavorable

43. DGo = DHo – TDSo Gibbs Free Energy If DG is negative
DG = is the maximum amount of energy ‘free’ to do work by the reaction (energy out)
2. If G is positive
DG = is the minimum amount of work needed to make the reaction happen (energy in)
DGo = DHo – TDSo

44. DGo = DHo – TDSo Gibbs Free Energy In our tables, units are:
DGo = kJ/mol
DHo = kJ/mol
DSo = J/mol-K
DGo = DHo – TDSo

45. What causes a reaction to be spontaneous?
favorable
Think Humpty Dumpty
System tend to seek:
 Minimum Enthalpy
Exothermic Rx, DH = negative
 Maximum Entropy
More disorder, DS = positive
Because: DGo = DHo – TDSo
- = (-) (+)
Energy spreads out
(into the surroundings)
matter spreads out

47. Notice the unit conversion

51. Subtracting a positive entropy value makes G a larger negative

52. 19.6 Free Energy and Temperature
By knowing the sign (+ or -) of S and H,
we can get the sign of G and determine if a reaction is spontaneous.

53. Favorable under all conditions Continues once begun.
Ex1: Combustion - Exothermic with production of gases
CH(s) compound + O2(g)  CO2(g) + H2O(g)
- DHo / +DSo
DGo = DHo – TDSo
= (–) – (+)
DGo always negative
Favorable under all conditions
Continues once begun.
When both enthalpy and entropy changes are favorable, the overall reaction must be favorable under all conditions!

54. These reactions are said to be under “kinetic control”
Even though reactions may be thermodynamically favorable, they may still not proceed if their reaction rates are very slow.
Reactions with high activation energies may be too slow and still require a catalyst to speed things up!
Recall from chapter 14 that speed of a reaction depends on activation energy?
These reactions are said to be under “kinetic control”

56. “unfavorable under all conditions Stops without input of light
Ex2: Photosynthesis - endothermic with production of a solid
CO2(g) + H2O(g) CH(s) compound + O2(g)
+ DHo / -DSo
When enthalpy and entropy changes are both unfavorable the reaction is unfavorable not matter what the conditions. Continuous energy input is necessary to drive the reaction.
DGo = DHo – TDSo
= (+) – (–)
DGo always positive
“unfavorable under all conditions
Stops without input of light

57. How can unfavorable (non-spontaneous) reactions occur?
Drive them with an outside energy source:
EX: Electrolytic decomposition of water
2H2O + energy  2H2 + O2
Batteries
can be
recharged using outside energy

58. Coupling reactions
Another way to make unfavorable reaction become favorable is to couple reactions. Hess’s law tells us that the overall DG is the sum of the individual values
Ex: CaCO3  CaO + CO2
DG = +130 kJ
Calcium oxide is used to make concrete and mortar for building
But by heating with carbon:
C + O2  CO2
DG = -394
CaCO3 + C + O2  CaO + 2CO2 DG = -264 kJ
… and the reaction proceeds favorably

60. DGo = DHo – TDSo = (–) – (–)
A “mixed” problem: one favorable factor, the other unfavorable.
Ex3: Haber process - exothermic with decrease of molecule #
N2(g) + 3H2 (g)  2NH3(g)
- DHo / -DSo
The reaction is exothermic (favorable energy release)
but we are combining to form fewer molecules (entropy is decreasing – unfavorable)
DGo = DHo – TDSo
Or +
depends
= (–) – (–)
at lower temps (– DHo) > (– TDSo) so DGo is (-)
(larger -) – (smaller -)  (-) + (+) = (-)
Reaction will proceed spontaneously to the right
Products > reactants at equilibrium
At lower temps the unfavorable energy term is minimized

61. DGo = DHo – TDSo = (–) – (–)
Ex: Haber process - exothermic with decrease of molecule #
N2(g) + 3H2 (g)  2NH3(g)
- DHo / -DSo
DGo = DHo – TDSo
= (–) – (–)
at higher temps (– DHo) < (– TDSo) so DGo is (+)
(smaller -) – (larger -)  (-) + (+) = (+)
Reaction will not proceed spontaneously
Or equilibrium shifts so that reactants > products at equilibrium
At higher temps the unfavorable entropy term dominates .

62. H
Enthalpy S
Entropy
I think of the enthalpy/entropy interaction as scales. In a mixed problem you need to figure out which has a greater influence. Recognize that the only factor that can be changed though is S, by increasing or decreasing temperature.

63. Standard Free Energy Changes
Be careful: Values for Gf are in kJ/mol
DGo = S n Gof products - S m Go f reactants
G can be looked up in tables
or calculated from S° and H.
DGo = DHo – TDSo

65. DGo = DHo – TDSo [19.12] ?DGo = (-DHo) – (-TDSo)
-TS becomes larger at higher temps leading to a lower negative G

67. Q = (PNH3)2
(PN2)(PH2)3
Q= (1 atm) = 1
(1 atm)(1atm)3
Kp = 1.07 at 25oC Q < Kp reaction shifts right

68. Q = (PNH3)2
(PN2)(PH2)3
Q= (1 atm) = 1
(1 atm)(1atm)3
Kp = at 800K Q > Kp reaction shifts left

69. At Equilibrium (Haber process)
DGo = DHo – TDSo
DG = zero and DHo = TDSo
If DGo = negative then DHo >TDSo
If DGo = Positive then DHo <TDSo

70. What else does G tell us?
N2 + 3H2   2NH3
Since G = kJ/mol Rxn at 250C is negative
– in a mixture off all three gases with partial pressures of 1.0 atm, the forward reaction is spontaneous, and the reaction shifts right to reach equilibrium.

71. And….
N2 + 3H2   2NH3
Since G = +61 kJ/mol Rxn at 5000C is positive
– in a mixture of all three gases with partial pressures of 1.0 atm, the forward reaction is not spontaneous, and the reaction shifts left to reach equilibrium.

72. Ho  So
SO2 = kJ/mol J/molK
O2 = J/molK
SO3 = kJ/mol J/molK
Remember: assume that H and S at 298K same as at 400K
so G at 400K only depends on change to new temperature T (in TS)

73. 19.7 Free Energy and Equilibrium
Remember from above:
If DG is 0, the system is at equilibrium.
So DG must be related to the equilibrium constant, K (chapter 15). The standard free energy, DG°, is directly linked to Keq by:
DGo = - RT ln K
Where R = J/mol-K

74. (-) - RT(lnLarge K) DGo = - RT ln K Relationships .
( kJmol- )
-(8.314J/mol K)(298K)ln 100
If the free energy change is a negative value,
the reaction is spontaneous, ln K must be a positive value, and K will be a large number meaning the equilibrium mixture is mainly products

75. (+) - RT(lnsmall K) DGo = - RT ln K Relationships
( kJmol- )
-(8.314J/mol-K-)(298K)ln 0.01
If the free energy change is a positive value,
the reaction is not spontaneous, ln K must be a negative value, and K will be a small number meaning the equilibrium mixture is mainly reactants

80. Ho  So
CC4(l) = kJ/mol J/molK
CC4(g) = kJ/mol J/molK

82. Ho  So
Br2(l) = kJ/mol J/molK
Br2(g) = kJ/mol J/molK

83. DGo = - RT ln K
-33.3 kJ = - (8.314 J/molK)(298K) (1kJ/1000J) ln K
13.4 = ln K
6.6 x 10 5 = K

84. DGo = - RT ln K

89. Or ΔGo = -RT ln K
ΔGo = ln K
-RT
-9.1 kJ mol = ln K
-(8.314 J mol- K)(298 K)
= ln K
= K

90. No surprise, as temp increases, salt solubility Ksp increases.