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. Copyright © Cengage Learning. All rights reserved

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. Copyright © Cengage Learning. All rights reserved

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.

ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow. The magnitude of ΔSsurr depends on the temperature. Copyright © Cengage Learning. All rights reserved

ΔSsurr Copyright © Cengage Learning. All rights reserved

ΔSsurr Copyright © Cengage Learning. All rights reserved

ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved

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. Copyright © Cengage Learning. All rights reserved

Free Energy (G) ΔG = ΔH – TΔS (at constant T and P) Copyright © Cengage Learning. All rights reserved

Spontaneous Reactions To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Effect of ΔH and ΔS on Spontaneity Copyright © Cengage Learning. All rights reserved

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. Copyright © Cengage Learning. All rights reserved

Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. The entropy of a substance increases with temperature. Copyright © Cengage Learning. All rights reserved

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 Copyright © Cengage Learning. All rights reserved

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) 51 H2O(l) 70 NaOH(aq) 50 H2(g) 131 [2(50) + 131] – [2(51) + 2(70)] = –11 J/K ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved

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 Copyright © Cengage Learning. All rights reserved

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. Copyright © Cengage Learning. All rights reserved