1. Chapter 18 Entropy, Free Energy and Equilibrium
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2. 18.1 Spontaneous Processes
Spontaneous: process that does occur under a specific set of conditions
Nonspontaneous: process that does not occur under a specific set of conditions
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3. Spontaneous Processes
Often spontaneous processes are exothermic, but not always….
Methane gas burns spontaneously and is exothermic
Ice melts spontaneously but this is an endothermic process…
There is another quantity!
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4. Spontaneous vs Nonspontaneous
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18.2 Entropy
Entropy (S): Can be thought of as a measure of the disorder of a system
In general, greater disorder means greater entropy
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Probability- what is the chance that the order of the cards can be restored by reshuffling?
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Entropy is a state function just as enthalpy
S = Sfinal  Sinitial
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Standard Entropy
Standard entropy: absolute entropy of a substance at 1 atm (typically at 25C)
Complete list in Appendix 2 of text
What do you notice about entropy values for elements and compounds?
Units: J/K·mol
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Entropies
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Trends in Entropy
Entropy for gas phase is greater than that of liquid or solid of same substance
I2 (g) has greater entropy than I2 (s)
More complex structures have greater entropy
C2H6 (g) has greater entropy than CH4 (g)
Allotropes - more ordered forms have lower entropy
Diamond has lower entropy than graphite
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11. Entropy Changes in a System Qualitative
Ssolid < Sliquid
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12. Entropy Changes in a System Qualitative
Sliquid < Svapor
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13. Entropy Changes in a System Qualitative
Spure < Saqueous
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14. Entropy Changes in a System Qualitative
Slower temp < Shigher temp
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15. Entropy Changes in a System Qualitative
Sfewer moles < Smore moles
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16. Entropy Changes in a System Qualitative
Determine the sign of S for the following (qualitatively)
1. Liquid nitrogen evaporates
2. Two clear liquids are mixed and a solid yellow precipitate forms
3. Liquid water is heated from 22.5 C to 55.8 C
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17. 18.3 The Second and Third Laws of Thermodynamics
System: the reaction
Surroundings: everything else
Both undergo changes in entropy during physical and chemical processes
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18. Second Law of Thermodynamics
Entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.
Equilibrium process: caused to occur by adding or removing energy from a system that is at equilibrium
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19. Second Law of Thermodynamics
Mathematically speaking:
Spontaneous process:
Suniverse = Ssystem + Ssurroundings > 0
Equilibrium process:
Suniverse = Ssystem + Ssurroundings = 0
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20. Entropy Changes in the System
Entropy can be calculated from the table of standard values just as enthalpy change was calculated.
Srxn = nS products  mS reactants
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Standard Entropy
Calculate the standard entropy change for the following using the table of standard values. (first, predict the sign for S qualitatively)
2NH3(g)  N2(g) + 3H2(g)
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Standard Entropy
2NH3(g)  N2(g) + 3H2(g)
Srxn = nS products  mS reactants
= [(1)(191.5 J/K · mol) + (3)(131.0 J/K · mol)]
- [(2)(193.0 J/K · mol)]
= J/K · mol J/K · mol
Srxn = J/K · mol (Entropy increases)
(2 mol gas  4 mol gas)
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Your Turn!
Calculate the standard entropy change for the following using the table of standard values. (first, predict the sign for S qualitatively)
2H2(g) + O2(g)  2H2O (g)
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24. Third Law of Thermodynamics
Entropy of a perfect crystalline substance is zero at absolute zero.
Importance of this law: it allows us to calculate absolute entropies for substances
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18.4 Gibbs Free Energy
G = H – T S
The Gibbs free energy, expressed in terms of enthalpy and entropy, refers only to the system, yet can be used to predict spontaneity.
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Gibbs Free Energy
If G < 0,negative, the forward reaction is spontaneous.
If G = 0, the reaction is at equilibrium.
If G > 0, positive, the forward reaction is nonspontaneous
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Predicting Sign of G
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28. Predicting Temperature from Gibbs Equation
Set G = 0 (equilibrium condition)
0 = H – T S
Rearrange equation to solve for T- watch for units!
This equation will also be useful to calculate temperature of a phase change.
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Example
For a reaction in which H = 125 kJ/mol and S = 325 J/Kmol, determine the temperature in Celsius above which the reaction is spontaneous.
385 K  273 = 112C
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30. Standard Free Energy Changes
Free energy can be calculated from the table of standard values just as enthalpy and entropy changes.
Grxn = nG products  mG reactants
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31. Standard Free Energy Changes
Calculate the standard free-energy change for the following reaction.
2KClO3(s)  2KCl(s) + 3O2(g)
Grxn = nG products  mG reactants
= [2(408.3 kJ/mol) + 3(0)]  [2(289.9 kJ/mol)]
= 816.6  (579.8) = 236.8 kJ/mol (spont)
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32. 18.5 Free Energy and Chemical Equilibrium
Reactions are almost always in something other than their standard states.
Free energy is needed to determine if a reaction is spontaneous or not.
How does free energy change with changes in concentration?
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33. Free Energy and Equilibrium
G = G° + RT ln Q
G = non-standard free energy
G° = standard free energy (from tables)
R = J/K·mole
T = temp in K
Q = reaction quotient
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34. Free Energy and Equilibrium
Consider the reaction,
H2(g) + Cl2(g)  2 HCl(g)
How does the value of G change when the pressures of the gases are altered as follows at 25 C?
H2 = 0.25 atm; Cl2 = 0.45 atm;
HCl = 0.30 atm
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35. Free Energy and Equilibrium
First, calculate standard free energy:
H2(g) + Cl2(g)  2 HCl(g)
G° = [2(95.27 kJ/mol)]  [0 + 0] =
 kJ/mol
Second, find Q:
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36. Free Energy and Equilibrium
Solve: G = G° + RT ln Q
G = 190,540 J/mol + (8.314J/K·mol)(298 K) ln (0.80)
G =  kJ/mol (the reaction becomes more spontaneous - free energy is more negative)
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Your Turn!
Consider the reaction,
O2(g) + 2CO(g)  2CO2(g)
How does the value of G change when the pressures of the gases are altered as follows at 25 C?
O2 = 0.50 atm; CO = 0.30 atm;
CO2 = 0.45 atm
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38. Relationship Between G° and K
At equilibrium, G = 0 and Q = K
The equation becomes:
0 = G° + RT ln K or
G° = – RT ln K
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39. Relationship Between G° and K
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40. Relationship Between G° and K
Using the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C.
2HCl(g) H2(g) Cl2(g)
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41. Relationship Between G° and K
First, calculate the G°:
= [0 + 0]  [2(95.27 kJ/mol)]
= kJ/mol (non-spontaneous)
Substitute into equation:
kJ/mol =  (8.314 x 103 kJ/K·mol)(298 K) ln KP
76.90 = ln KP = 3.98 x 1034
K < 1 reactants are favored
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Your Turn!
Calculate the value of the equilibrium constant, KP, for the following reaction at 25 C.
2NO2(g) N2O4(g)
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43. 18.6 Thermodynamics in Living Systems
Coupled reactions
Thermodynamically favorable reactions drives an unfavorable one
Enzymes facilitate many nonspontaneous reactions
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44. Thermodynamics in Living Systems
Examples:
C6H12O6 oxidation: G° = 2880 kJ/mol
ADP  ATP G° = 31 kJ/mol
Synthesis of proteins: (first step)
alanine + glycine  alanylglycine
G° = 29 kJ/mol
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45. Thermodynamics in Living Systems
Consider the coupling of two reactions:
ATP + H2O + alanine + glycine  ADP + H3PO4 + alanylglycine
Overall free energy change:
G° = 31 kJ/mol + 29 kJ/mol = 2 kJ/mol
Protein synthesis is now favored.
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46. ATP-ADP Interconversions
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Key Points
Spontaneous vs nonspontaneous
Relate enthalpy, entropy and free energy
Entropy
Predict qualitatively
Calculate from table of standard values
Calculate entropy for universe
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Key Points
Free energy
Calculate from table of standard values
Calculate from Gibbs equation
Calculate non-standard conditions
Calculate in relationship to equilibrium constant
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