2. LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. (Sec 17.1)
LO 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions. (Sec 17.3)
LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes. (Sec , 17.5)
LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed. (Sec 17.4, 17.6)
LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy. (Sec 17.4, 17.6)

3. LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable. (Sec 17.6, 17.9)
LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. (Sec 17.6)
LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. (Sec 17.6)
LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions. (Sec 17.1, )

4. LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process. (Sec 17.8)

5. AP Learning Objectives, Margin Notes and References
LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects.
LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes.
LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.
Additional AP References
LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)

6. Thermodynamics vs. Kinetics
Domain of Kinetics
Rate of a reaction depends on the pathway from reactants to products.
Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products.
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7. Spontaneous Processes and Entropy
Thermodynamics lets us predict the direction in which a process will occur but gives no information about the speed of the process.
A spontaneous process is one that occurs without outside intervention.
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8. What should happen to the gas when you open the valve?
CONCEPT CHECK!
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.
What should happen to the gas when you open the valve?
The gas should spread evenly throughout the two bulbs.
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9. Calculate ΔH, ΔE, q, and w for the process you described above.
CONCEPT CHECK!
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.
Calculate ΔH, ΔE, q, and w for the process you described above.
All are equal to zero.
All are equal to zero. Since it is a constant temperature process, H = 0 and E = 0. The gas is working against zero pressure (evacuated bulb) so w = 0. E = q + w, so q = 0.
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10. CONCEPT CHECK!
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.
c) Given your answer to part b, what is the driving force for the process?
Entropy
Some students are probably aware of the concept of entropy. This is a good introduction to the concept.
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11. The Expansion of An Ideal Gas Into an Evacuated Bulb
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12. Entropy
The driving force for a spontaneous process is an increase in the entropy of the universe.
A measure of molecular randomness or disorder.
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13. Entropy
Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state.
Nature spontaneously proceeds toward the states that have the highest probabilities of existing.
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14. The Microstates That Give a Particular Arrangement (State)

15. Positional Entropy
A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability of states available to the system.
Therefore: Ssolid < Sliquid << Sgas

16. Predict the sign of ΔS for each of the following, and explain:
CONCEPT CHECK!
Predict the sign of ΔS for each of the following, and explain:
The evaporation of alcohol
The freezing of water
Compressing an ideal gas at constant temperature
Heating an ideal gas at constant pressure
Dissolving NaCl in water + –
a) + (a liquid is turning into a gas)
b) - (more order in a solid than a liquid)
c) - (the volume of the container is decreasing)
d) + (the volume of the container is increasing)
e) + (there is less order as the salt dissociates and spreads throughout the water)
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17. AP Learning Objectives, Margin Notes and References
LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes.
Additional AP References
LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)

18. Second Law of Thermodynamics
In any spontaneous process there is always an increase in the entropy of the universe.
The entropy of the universe is increasing.
The total energy of the universe is constant, but the entropy is increasing.
Suniverse = ΔSsystem + ΔSsurroundings
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19. ΔSsurr ΔSsurr = +; entropy of the universe increases
ΔSsurr = -; process is spontaneous in opposite direction
ΔSsurr = 0; process has no tendency to occur
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20. AP Learning Objectives, Margin Notes and References
LO 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions.
LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes.
Additional AP References
LO 5.3 (see Appendix 7.2, “Thermal Equilibrium, the Kinetic Molecular Theory, and the Process of Heat”)
LO 5.12 (see Appendix 7.11, “Non-Spontaneous Reactions”)

21. CONCEPT CHECK!
For the process A(l) A(s), which direction involves an increase in energy randomness? Positional randomness? Explain your answer. As temperature increases/decreases (answer for both), which takes precedence? Why? At what temperature is there a balance between energy randomness and positional randomness?
Since energy is required to melt a solid, the reaction as written is exothermic. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid).
As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored.
There is a balance at the melting point.
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22. CONCEPT CHECK!
Describe the following as spontaneous/non-spontaneous/cannot tell, and explain.
A reaction that is:
Exothermic and becomes more positionally random
Spontaneous
Exothermic and becomes less positionally random
Cannot tell
Endothermic and becomes more positionally random
Endothermic and becomes less positionally random
Not spontaneous
Explain how temperature affects your answers.
a) Spontaneous (both driving forces are favorable). An example is the combustion of a hydrocarbon.
b) Cannot tell (exothermic is favorable, positional randomness is not). An example is the freezing of water, which becomes spontaneous as the temperature of water is decreased.
c) Cannot tell (positional randomness is favorable, endothermic is not). An example is the vaporization of water, which becomes spontaneous as the temperature of water is increased..
d) Not spontaneous (both driving forces are unfavorable).
Questions "a" and "d" are not affected by temperature. Choices "b" and "c" are explained above.

23. ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow.
The magnitude of ΔSsurr depends on the temperature.
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24. ΔSsurr
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25. ΔSsurr
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26. ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH
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27. Copyright © Cengage Learning. All rights reserved

28. AP Learning Objectives, Margin Notes and References
LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed.
LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy.

29. Free Energy (G)
A process (at constant T and P) is spontaneous in the direction in which the free energy decreases.
Negative ΔG means positive ΔSuniv.
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30. Free Energy (G) ΔG = ΔH – TΔS (at constant T and P)
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31. CONCEPT CHECK!
A liquid is vaporized at its boiling point. Predict the signs of: w q ΔH ΔS ΔSsurr ΔG Explain your answers. – +
As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium.
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32. EXERCISE!
The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.
Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm.
ΔS = 132 J/K·mol
ΔSsurr = -132 J/K·mol
ΔG = 0 kJ/mol
S = 132 J/K·mol
Ssurr = -132 J/k·mol
G = 0 kJ/mol
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33. Spontaneous Reactions
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34. Effect of ΔH and ΔS on Spontaneity
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35. AP Learning Objectives, Margin Notes and References
LO 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes.

36. CONCEPT CHECK!
Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant.
Predict the signs of:
ΔH ΔSsurr ΔS ΔSuniv
Explain.
Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written.
Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive. –
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37. Third Law of Thermodynamics
The entropy of a perfect crystal at 0 K is zero.
The entropy of a substance increases with temperature.
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38. Standard Entropy Values (S°)
Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
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39. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information:
EXERCISE!
Calculate ΔS° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
S° (J/K·mol)
Na(s)
H2O(l)
NaOH(aq)
H2(g)
ΔS°= –11 J/K
[2(50) + 131] – [2(51) + 2(70)] = –11 J/K
ΔS°= –11 J/K
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40. AP Learning Objectives, Margin Notes and References
LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed.
LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy.
LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.
LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product.
LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction.
LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.

41. AP Learning Objectives, Margin Notes and References
Additional AP References
LO 5.15 (see Appendix 7.11, “Non-Spontaneous Reactions”)
LO 5.16 (see Appendix 7.11, “Non-Spontaneous Reactions”)
LO 5.17 (see Appendix 7.11, “Non-Spontaneous Reactions”)
LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)

42. Standard Free Energy Change (ΔG°)
The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
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43. CONCEPT CHECK!
A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS° ΔG° Explain.
The reaction is exothermic, more ordered, and spontaneous.
Thus, the sign of H is negative; S is negative; and G is negative.
– – –
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44. Consider the following system at equilibrium at 25°C.
CONCEPT CHECK!
Consider the following system at equilibrium at 25°C.
PCl3(g) + Cl2(g) PCl5(g)
ΔG° = −92.50 kJ
What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain.
The ratio will decrease.
S is negative (unfavorable) yet the reaction is spontaneous (G is negative). Thus, H must be negative (exothermic, favorable). Thus, as the temperature is increased, the reaction proceeds to the left, decreasing the ratio of partial pressure of PCl5 to the partial pressure of PCl3.
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45. AP Learning Objectives, Margin Notes and References
LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.
Additional AP References
LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)

46. Free Energy and Pressure
G = G° + RT ln(P) or ΔG = ΔG° + RT ln(Q)
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47. Sketch graphs of: G vs. P H vs. P
CONCEPT CHECK!
Sketch graphs of:
G vs. P
H vs. P
ln(K) vs. 1/T (for both endothermic and exothermic cases)
G vs. P: A natural log graph (levels off as P increases).
H vs. P: no relationship (slope of zero).
lnK vs 1/T: endothermic - straight line, negative slope
exothermic - straight line, positive slope
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48. The Meaning of ΔG for a Chemical Reaction
A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.
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49. AP Learning Objectives, Margin Notes and References
LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.
LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process.
Additional AP References
LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)
LO 6.25 (see Appendix 7.11, “Non-Spontaneous Reactions”)

50. The equilibrium point occurs at the lowest value of free energy available to the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
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51. Change in Free Energy to Reach Equilibrium
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52. Copyright © Cengage Learning. All rights reserved

53. AP Learning Objectives, Margin Notes and References
LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.
Additional AP References
LO 5.15 (see Appendix 7.11, “Non-Spontaneous Reactions”)

54. Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.
wmax = ΔG
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55. All real processes are irreversible.
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.
All real processes are irreversible.
First law: You can’t win, you can only break even.
Second law: You can’t break even.
As we use energy, we degrade its usefulness.
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