2. 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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3. 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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4. CONCEPT CHECK!
Describe the following as spontaneous/non-spontaneous/cannot tell, and explain.
A reaction that is:
Exothermic and becomes more positionally random
Exothermic and becomes less positionally random
Endothermic and becomes more positionally random
Endothermic and becomes less positionally random
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.

5. 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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6. ΔSsurr
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7. ΔSsurr
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8. ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH
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10. 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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11. Free Energy (G) ΔG = ΔH – TΔS (at constant T and P)
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12. Spontaneous Reactions
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13. Effect of ΔH and ΔS on Spontaneity
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14. 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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15. 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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16. 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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17. 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)
[2(50) + 131] – [2(51) + 2(70)] = –11 J/K
ΔS°= –11 J/K
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18. 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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19. 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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