1. Chapter 17: Free Energy & Thermodynamics CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University

2. Overview Second Law of Thermodynamics Gibbs Free Energy Spontaneity & Free Energy

3. The 2 nd Law of Thermodynamics The 2 nd Law of Thermodynamics says that the total entropy change of the universe must be positive for a process to be spontaneous –for reversible process  S univ = 0 –for irreversible (spontaneous) process  S univ > 0  S universe =  S system +  S surroundings If the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount –when  S system is negative,  S surroundings must be positive and big for a spontaneous process 3Tro: Chemistry: A Molecular Approach, 2/e

4. Heat Flow, Entropy, and the 2 nd Law According to the 2 nd Law, heat must flow from water to ice because it results in more dispersal of heat. The entropy of the universe increases. 4 When ice is placed in water, heat flows from the water into the ice Tro: Chemistry: A Molecular Approach, 2/e

5. Heat Transfer and Changes in Entropy of the Surroundings The 2 nd Law demands that the entropy of the universe increase for a spontaneous process Yet processes like water vapor condensing are spontaneous, even though the water vapor is more random than the liquid water If a process is spontaneous, yet the entropy change of the process is unfavorable, there must have been a large increase in the entropy of the surroundings The entropy increase must come from heat released by the system – the process must be exothermic! 5Tro: Chemistry: A Molecular Approach, 2/e

6. Entropy Change in the System and Surroundings 6 When the entropy change in system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive), and large to allow the process to be spontaneous Tro: Chemistry: A Molecular Approach, 2/e

7. Heat Exchange and  S surroundings When a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings When a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings The amount the entropy of the surroundings changes depends on its original temperature –the higher the original temperature, the less effect addition or removal of heat has 7Tro: Chemistry: A Molecular Approach, 2/e

8. Temperature Dependence of  S surroundings When heat is added to surroundings that are cool it has more of an effect on the entropy than it would have if the surroundings were already hot Water freezes spontaneously below 0 °C because the heat released on freezing increases the entropy of the surroundings enough to make  S positive –above 0 °C the increase in entropy of the surroundings is insufficient to make  S positive 8Tro: Chemistry: A Molecular Approach, 2/e

9. Quantifying Entropy Changes in Surroundings The entropy change in the surroundings is proportional to the amount of heat gained or lost –q surroundings = −q system The entropy change in the surroundings is also inversely proportional to its temperature At constant pressure and temperature, the overall relationship is 9Tro: Chemistry: A Molecular Approach, 2/e

10. Example 17.2a: Calculate the entropy change of the surroundings at 25ºC for the reaction below C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)  H rxn = −2044 kJ 10 combustion is largely exothermic, so the entropy of the surroundings should increase significantly  H system = −2044 kJ, T = 25 ºC = 298 K  S surroundings, J/K Check: Solution: Conceptual Plan: Relationships: Given: Find: SST,  H Tro: Chemistry: A Molecular Approach, 2/e

11. Practice – The reaction below has  H rxn = +66.4 kJ at 25 °C. (a) Determine the  s surroundings, (b) the sign of  S system, and (c) whether the process is spontaneous 2 O 2 (g) + N 2 (g)  2 NO 2 (g) 11Tro: Chemistry: A Molecular Approach, 2/e

12. Practice – The reaction 2 O 2 (g) + N 2 (g)  2 NO 2 (g) has  H rxn = +66.4 kJ at 25 °C. Calculate the entropy change of the surroundings. Determine the sign of the entropy change in the system and whether the reaction is spontaneous.  H sys = +66.4 kJ, T = 25 ºC = 298 K  S surr, J/K,  S react + or −,  S universe + or − Solution: Conceptual Plan: Relationships: Given: Find: SST,  H the major difference is that there are fewer product molecules than reactant molecules, so the  S reaction is unfavorable and (−) because  S surroundings is (−) it is unfavorable both  S surroundings and  S reaction are (−),  S universe is (−) and the process is nonspontaneous 12Tro: Chemistry: A Molecular Approach, 2/e

13. Gibbs Free Energy and Spontaneity It can be shown that −T  S univ =  H sys −T  S sys The Gibbs Free Energy, G, is the maximum amount of work energy that can be released to the surroundings by a system –for a constant temperature and pressure system –the Gibbs Free Energy is often called the Chemical Potential because it is analogous to the storing of energy in a mechanical system   G sys =  H sys −T  S sys Because  S univ determines if a process is spontaneous,  G also determines spontaneity –  S univ is + when spontaneous, so  G is − 13Tro: Chemistry: A Molecular Approach, 2/e

14. Gibbs Free Energy,  G A process will be spontaneous when  G is negative   G will be negative when –  H is (-) and  S is (+) [Exothermic and more random] –  H is (-) and large while  S is (-) but small –  H is (+) but small and  S is (+) and large or high temperature  G will be (+) when  H is + and  S is − –never spontaneous at any temperature When  G = 0 the reaction is at equilibrium 14Tro: Chemistry: A Molecular Approach, 2/e

15.  G,  H, and  S 15Tro: Chemistry: A Molecular Approach, 2/e

16. Free Energy Change and Spontaneity 16Tro: Chemistry: A Molecular Approach, 2/e

17. Example 17.3a: The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = +95.7 kJ and  S = +142.2 J/K at 25 °C. Calculate  G and determine if it is spontaneous. 17 Because  G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.  H = +95.7 kJ,  S = 142.2 J/K, T = 298 K  G, kJ Answer: Solution: Conceptual Plan: Relationships: Given: Find: GGT,  H,  S Tro: Chemistry: A Molecular Approach, 2/e