1. CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics
Chapter 19
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2. Thermodynamics Thermodynamics
study of energy and its transformations
heat and energy flow
Helps determine natural direction of reactions
Allows us to predict if a chemical process will occur under a given set of conditions
Organized around three fundamental laws of nature
1st law of thermodynamics
2nd law of thermodynamics
3rd law of thermodynamics
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3. First Law of Thermodynamics
Energy can be neither created nor destroyed
Energy of the universe is constant
Important concepts from thermochemistry
Enthalpy
Hess’s law
Purpose of 1st Law
Energy bookkeeping
How much energy?
Exothermic or endothernic?
What type of energy?
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4. Spontaneous vs Nonspontaneous
Spontaneous process
Occurs without outside assistance in the form of energy (occurs naturally)
Product-favored at equilibrium
May be fast or slow
May be influenced by temperature
Nonspontaneous process
Does not occur without outside assistance
Reactant-favored at equilibrium
All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction
Spontaneous processes have a definite direction
Spontaneous processes are irreversible
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5. Spontaneous vs. Nonspontaneous
Spontaneous Processes
Gases expand into larger volumes
H2O(s) melts above 0C
H2O(l) freezes below 0C
NH4NO3 dissolves spontaneously in H2O
Steel (iron) rusts in presence of O2 and H2O
Wood burns to form CO2 and H2O
CH4 gas burns to form CO2 and H2O
Reverse processes are not spontaneous!
nonspontaneous
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6. Spontaneous vs. Nonspontaneous
Spontaneous Processes
H2O(s) melts above 0C (endothermic)
NH4NO3 dissolves spontaneously in H2O (endothermic)
Steel (iron) rusts in presence of O2 and H2O (exothermic)
Wood burns to form CO2 and H2O (exothermic)
CH4 gas burns to form CO2 and H2O (exothermic)
Heat change alone is not enough to predict spontaneity because energy is conserved
Many, but not all, spontaneous processes are exothermic
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7. Factors That Favor Spontaneity
Two thermodynamic properties of a system are considered when determining spontaneity:
Enthalpy, H
Many, but not all, spontaneous processes tend to be exothermic as already noted
Entropy, S (J/K)
Measure of the disorder of a system
Many, but not all, spontaneous processes tend to increase disorder of the system
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8. Entropy Entropy, S (J/K)
describes # of ways the particles in a system can be arranged in a given state
More arrangements = greater entropy
S = heat change/T = q/T
S = Sfinal - Sinitial
 S > 0 represents increased randomness or disorder
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9. Patterns of Entropy Change
For the same or similar substances: Ssolid < Sliquid < Sgas
solid
vapor
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liquid

10. Patterns of Entropy Change
Particles farther apart, occupy larger volume of space; even more positions available to particles
Rigidly held particles; few positions available to particles
Particles free to flow; more positions available for particles
solid
vapor
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liquid

11. Patterns of Entropy Change
least ordered
less ordered
most ordered
solid
vapor
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liquid

12. Patterns of Entropy Change
Solution formation usually leads to increased entropy
particles more disordered
solvent
solute
solution
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13. Patterns of Entropy Change
Chemical Reactions
If more gas molecules produced than consumed, S increases. (Srxn > 0)
If only solids, ions and/or liquids involved, S increases if total # particles increases.
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14. fewer arrangements possible so lower entropy
more arrangements possible so higher entropy
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15. Patterns of Entropy Change
Increasing temperature increases entropy
System at T1 (S1)
System at T2 (T2 > T1) (S2 > S1)
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16. Patterns of Entropy Change
more energetic molecular motions
less energetic molecular motions
System at T1 (S1)
System at T2 (T2 > T1) (S2 > S1)
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17. Third Law of Thermodynamics
If increasing T increases S, then the opposite should be true also.
Is it possible to decrease T to the point that S is zero?
At what T does S = 0?
If entropy is zero, what does that mean?
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18. Third Law of Thermodynamics
Entropy of a perfect crystalline substance at 0 K is zero
No entropy = highest order possible
Why?
Purpose of 3rd Law
Allows S to be measured for substances
S = 0 at 0 K
S = heat change/temperature = q/T
S = standard molar entropy
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19. Standard entropy, S° (J/K)50 40
Standard entropy, S° (J/K) 30 20 10 50
100
150
200
250
300
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Temperature (K)
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20. Gas Liquid Solid Standard entropy, S° (J/K) Temperature (K) 50 40 3020
Solid 10 50
100
150
200
250
300
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Temperature (K)
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21. Standard Molar Entropies of Selected Substances at 298 K
S (J/K)
H2O(l)
69.9
NaCl(s)
72.3
H2O(g)
188.8
NaCl(aq)
115.5
I2(s)
116.7
Na2CO3(s)
136.0
I2(g)
260.6
CH4(g)
186.3
Na(s)
51.5
C2H6(g)
229.5
K(s)
64.7
C3H8(g)
269.9
Cs(s)
85.2
C4H10(g)
310.0
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22. Entropy versus Probability
Systems tend to move spontaneously towards increased entropy. Why?
Entropy is related to probability
Disordered states are more probable than ordered states
S = k(lnW)
k (Boltzman’s constant) = 1.38 x J/K
W = # possible arrangements in system
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23. Consider why gases tend to isothermally expand into larger volumes.
Isothermal Gas Expansion
Consider why gases tend to isothermally expand into larger volumes.
Gas Container = two bulbed flask
Ordered State
Gas Molecules
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24. Isothermal Gas Expansion
Gas Container
Ordered State
S = k(ln 1) = (1.38 x J/K)(0) = 0 J/K
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30. Disordered States
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31. Disordered States
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32. Disordered States
More probable that the gas molecules will disperse between two halves than remain on one side
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33. Disordered States
Driving force for expansion is entropy (probability); gas molecules have a tendency to spread out
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34. Disordered States
S = k(ln 7) = (1.38 x J/K)(1.95) = 2.7 x J/K
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35. Total Arrangements
Stotal = k(ln 23) = k(ln 8) = (1.38 x J/K)(1.79) = x J/K
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36. 2nd Law of Thermodynamics
The entropy of the universe increases in any spontaneous process (Suniv > 0)
Increased disorder in the universe is the driving force for spontaneity
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37. 2nd Law of Thermodynamics
Suniv = Ssysyem + Ssurroundings
Suniv > 0, process is spontaneous
Suniv = 0, process at equilibrium
Suniv < 0, process is nonspontaneous
Process is spontaneous in reverse direction
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38. 2nd Law of Thermodynamics
Spontaneous Processes
H2O(s) melts above 0C (endothermic)
Steel (iron) rusts in presence of O2 and H2O (exothermic)
4Fe(s) + 3O2(g)  2Fe2O3(s)
CH4 gas burns to form CO2 and H2O (exothermic)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Each process increases Suniverse.
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39. 2nd Law of Thermodynamics
To determine Suniv for a process, both Ssystem and Ssurroundings need to be known:
Ssysyem
related to matter dispersal in system
Ssurroundings
determined by heat exchange between system and surroundings and T at which it occurs
Sign of Ssurroundings depends on whether process in system is endothermic or exothermic
Why?
Magnitude of Ssurroundings depends on T
Ssurroundings = -Hsystem/T
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40. Entropy Changes in a System
For any reaction:
S°rxn = nS°(products) - mS°(reactants)
Where n and m are stoichiometric coefficients
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41. Example 1 (1 on Example Problem Handout)
Using standard molar entropies, calculate S°rxn for the following reaction at 25°C:
2SO2(g) + O2(g) --> 2SO3(g)
S° = (J · K-1mol-1)
(Ans.: J/K)
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42. Gibbs Free Energy, G 2nd law: Suniv = Ssys + Ssurr = Ssys - Hsys/T
Rearrange (multiply by -T)
-TSuniv = -TSsys + Hsys = Hsys - TSsys
-TSuniv = G (-TSuniv = G)
G = Gibbs free energy (J or kJ)
G = H - TS
and
G = Hsys - TSsys
G° = H°sys - TS°sy (if at standard state)
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43. Gibbs Free Energy Summary of Conditions for Spontaneity G < 0
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44. Gibbs Free Energy Summary of Conditions for Spontaneity G < 0
reaction is spontaneous in the forward direction
(Suniv > 0)
G > 0
G = 0
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45. Gibbs Free Energy Summary of Conditions for Spontaneity G < 0
reaction is spontaneous in the forward direction
(Suniv > 0)
G > 0
reaction is nonspontaneous in the forward direction
(Suniv < 0)
G = 0
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46. Gibbs Free Energy Summary of Conditions for Spontaneity G < 0
reaction is spontaneous in the forward direction
(Suniv > 0)
G > 0
reaction is nonspontaneous in the forward direction
(Suniv < 0)
G = 0
system is at equilibrium
(Suniv = 0)
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47. Example 2 (2 in Example Problem Handout)
For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0?
(ans.: a) G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) K or 188C)
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48. Gibbs Free Energy and Temperature
G = H - TS
S > 0 H < 0
Spontaneous at all T
S < 0 H > 0
Not spontaneous at any T
S > 0 H > 0
Spontaneous at high T; nonspontaneous at low T
S < 0 H < 0
Spontaneous at low T; nonspontaneous at high T
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49. Gibbs Free Energy and Temperature
Spontaneous Processes
H2O(s) melts above 0C (endothermic)
H > 0, S > 0
Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic)
4Fe(s) + 3O2(g)  2Fe2O3(s)
H < 0, S < 0
G < 0 for each process
T determines sign of G: G = H -TS
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50. Calculating Free Energy Changes
For any reaction:
G°rxn = Gf°(products) - mGf°(reactants)
Where n and m are stoichiometric coefficients and Gf° = standard free energy of formation
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51. Example 3 (3 on Example Problem Handout)
Calculate the standard free energy change for the reaction at 25°C:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
G°f = (kJ/mol)
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52. Gibbs Free Energy What does Gibbs free energy represent?
For a spontaneous process:
Maximum amount of energy released by the system that can do useful work on the surroundings
Energy available from spontaneous process that can be used to drive nonspontaneous process
For a nonspontaneous process:
Minimum amount of work that must be done to force the process to occur
In actuality, Gspont > Gnonspont
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53. Gibbs Free Energy Conversion of rust to iron
2Fe2O3  4Fe + 3O2 G = 1487 kJ (NS)
To convert iron to rust, G must be provided from spontaneous rxn
6CO + 3O2  6CO2 G = kJ (S)
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54. Gibbs Free Energy Conversion of rust to iron
2Fe2O3  4Fe + 3O2 G = 1487 kJ (NS)
To convert iron to rust, G must be provided from spontaneous rxn
2Fe2O3  4Fe + 3O2 G = 1487 kJ (NS)
6CO + 3O2  6CO2 G = kJ (S)
2Fe2O3 + 6CO  4Fe + 6CO2  G = -56 kJ (S)
Reactions are “coupled”
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55. Gibbs Free Energy Many biological rxns essential for life are NS
Spontaeous rxns used to “drive” the NS biological rxns
Example: photosynthesis
6CO2 + 6H2O  C6H12O6 + 6O2  G > 0
What spontaneous rxns drive photosynthesis?
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56. proteins, cells etc., lower S amino acids, sugars, etc., higher S
sun
big ball of G
= G released
high free energy
DGsurr < 0
proteins, cells etc., lower S
C6H12O6, O2
ATP NS Sp NS Sp NS
CO2, H2O
amino acids, sugars, etc., higher S
ADP
photosynthesis
solar nuclear reactions
low free energy
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57. Free Energy and Equilibrium
G (Nonstandard State)
G = G° + RTlnQ
R = gas constant (8.314 J/K·mol)
T = temperature (K)
Q = reaction quotient
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58. Example 4 (4 on Example Problem Handout)
Calculate Grxn for the reaction below:
2A(aq) + B(aq)  C(aq) + D(g)
if G°rxn = 9.9 x 103 J/mol and (a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05 M, and PD = 0.05 atm, and (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5 M, and PD = 0.5 atm. Is the reaction spontaneous under these conditions?
(ans.: a) –2121 J, spontaneous; b) J, nonspontaneous)
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59. Free Energy and Equilibrium
At equilibrium:
Grxn = 0 and Q = K
0 = G°rxn + RTlnK
G°rxn = -RTlnK
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60. Example 5 (5 on Example Problem Handout)
Calculate G°rxn for the ionization of acetic acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is this reaction spontaneous under standard state conditions?
(ans.: 27 kJ)
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61. Example 6 (6 on Example Problem Handout)
Calculate G° for the neutralization of a strong acid with a strong base at 25°C. Is this process spontaneous under these conditions?
For the reaction below, K = 1.0 x 1014.
H+ + OH-  H2O
(ans.: -80 kJ)
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62. Example 7
Calculate Keq for a reaction at (a) 25°C and (b) 250°C if H°rxn = 42.0 kJ and S°rxn = 125 J/K. At which temperature is this process product favored?
(ans.: a) 0.15; b) 216; 250C)
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63. DGrxn < 0 Spontaneous Reaction equilibrium position Pure reactants
Pure products
DGrxn < 0
Gproducts
Greactants
extent of reaction
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64. Spontaneous Reaction
At equilibrium, any change requires an uphill climb in energy, DGrxn > 0
Gproducts
Greactants
Pure reactants
Pure products
extent of reaction
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65. Calculating G for Processes
G may be calculated in one of several ways depending on the information known about the process of interest
G = H - TS; G° = H° - TS°
G°rxn = Gf°(products) - mGf°(reactants)
G = G° + RTlnQ
G°rxn = -RTlnK
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66. Entropy and Life Processes
If the 2nd law is valid, how is the existence of highly-ordered, sophisticated life forms possible?
growth of a complex life form represents an increase in order (less randomness)
lower entropy
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67. Entropy and Life Processes
Organisms “pay” for their increased order by increasing Ssurr.
Over lifetime, Suniv > 0.
CO2
H2O
heat
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