1. 20-1 Due: CH 20 Connect - Take out Notes/POGIL Today: Solving Problems for Thermodynamics HW: Test on Wednesday Potluck Thursday

2. 20-2 Warm Up Which of the following reaction characteristics will always indicate a thermodynamically favored reaction? A) +∆H and + ∆SB) +∆H and - ∆S C) -∆H and + ∆SD) -∆H and - ∆S Which of the salts below will be considered spontaneous? Explain why. A) PbCl 2 B) Pb(NO 3 ) 2

3. 20-3 Three Laws of Thermodynamics First Law of Thermodynamics states that the total energy of the universe is constant. Energy is conserved, and is neither created nor destroyed. Energy is transferred in the form of heat and/or work. Cannot predict spontaneous change ∆E univ = ∆E system + ∆E surr =0 ∆E = q + w w = P∆V Second Law of Thermodynamics If we consider both the system and the surroundings, we find that all real processes occur spontaneously in the direction that increases the entropy of the universe. ∆S univ = ∆S system + ∆S surr > 0 ∆S system = q rev /T **reversible process Third Law of Thermodynamics A perfect crystal has zero entropy at absolute zero. ∆S system = 0 @ 0K

4. 20-4 Table 20.1 Reaction Spontaneity and the Signs of  H,  S, and  G HH SS-TS-TS GG Description –+––Spontaneous at all T +–++Nonspontaneous at all T ++–+ or –Spontaneous at higher T; nonspontaneous at lower T ––++ or –Spontaneous at lower T; nonspontaneous at higher T

5. 20-5

6. 20-6 Entropy Changes in the System The standard entropy of reaction,  S° rxn, is the entropy change that occurs when all reactants and products are in their standard states.  S° rxn =  mS° products −  nS° reactants where m and n are the amounts (mol) of products and reactants, given by the coefficients in the balanced equation.

7. 20-7 Sample Problem 20.2 Calculating the Standard Entropy of Reaction,  S° rxn PROBLEM: Predict the sign of  S° rxn and calculate its value for the combustion of 1 mol of propane at 25°C. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(l) PLAN:From the change in the number of moles of gas (6 mol yields 3 mol), the entropy should decrease (  S° rxn < 0). To find  S° rxn, we apply Equation 20.4, using the S° values from Appendix B. SOLUTION: ΔS° rxn = [(3 mol CO 2 )(S° of CO 2 ) + (4 mol H 2 O)(S° of H 2 O)] – [(1 mol C 3 H 8 )(S° of C 3 H 8 ) + (5 mol O 2 )(S° of O 2 )] = [(3mol)(213.7J/K·mol) + (4 mol)(69.9J/K·mol)] – [(1mol)(269.9 J/K·mol) + (5 mol)(205.0 J/K·mol)] = –374 J/K

8. 20-8 Entropy Changes in the Surroundings A decrease in the entropy of the system is outweighed by an increase in the entropy of the surroundings. The surroundings function as a heat source or heat sink. In an exothermic process, the surroundings absorbs the heat released by the system, and S surr increases. q sys 0 and  S surr > 0 In an endothermic process, the surroundings provides the heat absorbed by the system, and S surr decreases. q sys > 0; q surr < 0 and  S surr < 0

9. 20-9 Temperature at which Heat is Transferred Since entropy depends on temperature,  S° surr is also affected by the temperature at which heat is transferred. The impact on the surroundings is larger when the surroundings are at lower temperature, because there is a greater relative change in S surr. For any reaction, q sys = –q surr. The heat transferred is specific for the reaction and is the same regardless of the temperature of the surroundings.  S surr = – q sys T  S surr = –  H sys T for a process at constant P

10. 20-10 Sample Problem 20.3 Determining Reaction Spontaneity PROBLEM: At 298 K, the formation of ammonia has a negative  S° sys ; N 2 (g) + 3H 2 (g) → 2NH 3 (g);  S° sys = –197 J/K Calculate  S° univ, and state whether the reaction occurs spontaneously at this temperature. PLAN: For the reaction to occur spontaneously,  S° univ > 0, so  S° surr must be greater than +197 J/K. To find  S° surr we need  H° sys, which is the same as  H° rxn. We use the standard enthalpy values from Appendix B to calculate this value.  H° rxn = [(2 mol)(  H° f of NH 3 )] – [(1 mol)(  H° f of N 2 ) + (3 mol)(  H° f of H 2 )] SOLUTION: = [(2 mol)(-45.9 kJ/mol)] – [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)] = –91.8 kJ

11. 20-11 Sample Problem 20.3  S surr = -  H sys T = – –91.8 kJ x 1000 J 1 kJ 298 K = 308 J/K  S° univ =  S° sys +  S° surr = –197 J/K + 308 J/K = 111 J/K Since  S° univ > 0, the reaction occurs spontaneously at 298 K. Although the entropy of the system decreases, this is outweighed by the increase in the entropy of the surroundings.

12. 20-12 Gibbs Free Energy The Gibbs free energy (G) combines the enthalpy and entropy of a system; G = H – TS. The free energy change (  G) is a measure of the spontaneity of a process and of the useful energy available from it.  G sys =  H sys – T  S sys  G < 0 for a spontaneous process  G > 0 for a nonspontaneous process  G = 0 for a process at equilibrium

13. 20-13 Calculating  G°  G° sys =  H° sys – T  S° sys  G° rxn can be calculated using the Gibbs equation:  G° rxn can also be calculated using values for the standard free energy of formation of the components.  G° rxn =  m  G° products –  n  G° reactants

14. 20-14 Sample Problem 20.4 Calculating  G° rxn from Enthalpy and Entropy Values PROBLEM:Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated. 4KClO 3 (s) 3KClO 4 (s) + KCl(s) +7–1–1+5 Δ PLAN: To solve for  G°, we need values from Appendix B. We use  H f ° to calculate  H° rxn, use S° values to calculate  S° rxn, and then apply the Gibbs equation. SOLUTION:  H° rxn =  m  H f ° (products) -  n  H f ° (reactants) = [(3 mol KClO 4 )(  H f ° of KClO 4 ) + (1 mol KCl)(  H f ° of KCl)] ‒ [(4 mol KClO 3 )(  H f ° of KClO 3 )] = [(3 mol)(-432.8 kJ/mol) + (1 mol)(–436.7 kJ/mol) – [(4 mol)(-397.7 kJ/mol)] = –144 kJ

15. 20-15 Sample Problem 20.4  S° rxn =  mS° (products) –  S° (reactants) = [(3 mol of KClO 4 )(S° of KClO 4 ) + (1 mol KCl)(S° of KCl)] ‒ [(4 mol KClO 3 )(S° of KClO 3 )] = [(3 mol)(151.0 J/mol·K) + (1 mol)(82.6 J/mol·K) – [(4 mol)(143.1 J/mol·K)] = –36.8 J/K  G° sys =  H° sys – T  S° sys = –144 kJ – (298 K)(–36.8 J/K) 1 kJ 1000 J = –133 kJ

16. 20-16 Sample Problem 20.5 Calculating  G° rxn from  G° f Values PLAN: We apply Equation 20.8 and use the values from Appendix B to calculate  G° rxn. SOLUTION: PROBLEM: Use  G° f values to calculate  G° rxn for the reaction in Sample Problem 20.4: 4KClO 3 (s) 3KClO 4 (s) + KCl(s) Δ  G° rxn =  m  G° products -  n  G° reactants = [(3 mol KClO 4 )(  G f ° of KClO 4 ) + (1 mol KCl)(  G f ° of KCl)] ‒ [(4 mol KClO 3 )(  G f ° of KClO 3 )] = [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] ‒ [(4 mol)(-296.3 kJ/mol)] = –134 kJ

17. 20-17  G° and Useful Work  G is the maximum useful work done by a system during a spontaneous process at constant T and P. In practice, the maximum work is never done. Free energy not used for work is lost to the surroundings as heat.  G is the minimum work that must be done to a system to make a nonspontaneous process occur at constant T and P. A reaction at equilibrium (  G sys = 0) can no longer do any work.

18. 20-18 Effect of Temperature on Reaction Spontaneity Reaction is spontaneous at all temperatures If  H 0  G < 0 for all T Reaction is nonspontaneous at all temperatures If  H > 0 and  S < 0  G > 0 for all T  G sys =  H sys - T  S sys

19. 20-19 Effect of Temperature on Reaction Spontaneity Reaction becomes spontaneous as T increases If  H > 0 and  S > 0  G becomes more negative as T increases. Reaction becomes spontaneous as T decreases If  H < 0 and  S < 0  G becomes more negative as T decreases.

20. 20-20 Sample Problem 20.6 Using Molecular Scenes to Determine the Signs of  H,  S, and  G PROBLEM:The following scenes represent a familiar phase change for water (blue spheres). (a) What are the signs of  H and  S for this process? Explain. (b) Is the process spontaneous at all T, no T, low T, or high T? Explain. PLAN: (a) From the scenes, we determine any change in the amount of gas, which indicates the sign of  S, and any change in the freedom of motion of the particles, which indicates whether heat is absorbed or released. (b)To determine reaction spontaneity we need to consider the sign of  G at different temperatures.

21. 20-21 Sample Problem 20.6 SOLUTION: (a)The scene represents the condensation of water vapor, so the amount of gas decreases dramatically, and the separated molecules give up energy as they come closer together.  S < 0 and  H < 0 (b)Since  S is negative, the –T  S term is positive. In order for  G to be < 0, the temperature must be low. The process is spontaneous at low temperatures.

22. 20-22 Sample Problem 20.7 Determining the Effect of Temperature on ΔG PROBLEM:A key step in the production of sulfuric acid is the oxidation of SO 2 (g) to SO 3 (g): 2SO 2 (g) + O 2 (g) → 2SO 3 (g) At 298 K,  G = –141.6 kJ;  H = –198.4 kJ; and  S = –187.9 J/K (a) Use the data to decide if this reaction is spontaneous at 25°C, and predict how  G will change with increasing T. (b) Assuming  H and  S are constant with increasing T (no phase change occurs), is the reaction spontaneous at 900.° C? PLAN: We note the sign of  G to see if the reaction is spontaneous and the signs of  H and  S to see the effect of T. We can then calculate whether or not the reaction is spontaneous at the higher temperature.

23. 20-23 Sample Problem 20.7 SOLUTION: (a)  G < 0 at 209 K (= 25°C), so the reaction is spontaneous. With  S 0 and this term will become more positive at higher T.  G will become less negative, and the reaction less spontaneous, with increasing T. (b)  G =  H – T  S  G = –198.4 kJ – [1173 K)(–0.1879 kJ/K) = 22.0 kJ Convert T to K: 900. + 273.15 = 1173 K Convert S to kJ/K: –187.9 J/K = –0.1879 kJ/K  G > 0, so the reaction is nonspontaneous at 900.°C.

24. 20-24 Figure 20.14The effect of temperature on reaction spontaneity. HSHS T = The sign of  G switches at

25. 20-25 Sample Problem 20.8 Finding the Temperature at Which a Reaction Becomes Spontaneous PROBLEM:At 25°C (298 K), the reduction of copper(I) oxide is nonspontaneous (  G = 8.9 kJ). Calculate the temperature at which the reaction becomes spontaneous. PLAN: We need to calculate the temperature at which  G crosses over from a positive to a negative value. We set  G equal to zero, and solve for T, using the values for  H and  S from the text. SOLUTION: HSHS T = = 58.1 kJ 0.165 kJ/K = 352 K At any temperature above 352 K (= 79°C), the reaction becomes spontaneous.

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