1. 20-1 *See PowerPoint Image Slides for all figures and tables pre-inserted into PowerPoint without notes. CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 20 Lecture Outlines* Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 20-2 Chapter 20 Thermodynamics: Entropy, Free Energy and the Direction of Chemical Reactions

3. 20-3 Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 20.1 The Second Law of Thermodynamics: Predicting Spontaneous Change 20.2 Calculating the Change in Entropy of a Reaction 20.3 Entropy, Free Energy, and Work 20.4 Free Energy, Equilibrium, and Reaction Direction

4. 20-4 Limitations of the First Law of Thermodynamics E system = E k + E p  E system = heat + work  E system = q + w E universe = E system + E surroundings  E system = -  E surroundings The total energy-mass of the universe is constant. But, this does not tell us anything about the direction of change in the universe.

5. 20-5 Figure 20.1 A spontaneous endothermic chemical reaction water Ba(OH) 2 8H 2 O( s ) + 2NH 4 NO 3 ( s ) Ba 2+ ( aq ) + 2NO 3 - ( aq ) + 2NH 3 ( aq ) + 10H 2 O( l ).  H 0 rxn = + 62.3 kJ Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

6. 20-6 The Concept of Entropy (S) Entropy refers to the state of order. A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process. solid liquid gas more orderless order crystal + liquid ions in solution more orderless order more orderless order crystal + crystal gases + ions in solution

7. 20-7 Figure 20.2 The number of ways to arrange a deck of playing cards

8. 20-8 Figure 20.3 1 atmevacuated Spontaneous expansion of a gas stopcock closed stopcock opened 0.5 atm

9. 20-9 1877 Ludwig Boltzman S = k ln W S = entropy W = number of ways of arranging the components of a system k = Boltzman constant A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy. A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy. 2nd law of thermodynamics The universe is going to a higher state of entropy  S universe =  S system +  S surroundings > 0

10. 20-10 Figure 20.4 Random motion in a crystal 3 rd Law of thermodynamics A perfect crystal has zero entropy at a temperature of absolute zero. S system = 0 at 0 K

11. 20-11 Predicting Relative S 0 Values of a System 1. Temperature changes 2. Physical states and phase changes 3. Dissolution of a solid or liquid 5. Atomic size or molecular complexity 4. Dissolution of a gas S 0 increases as the temperature rises. S 0 increases as a more ordered phase changes to a less ordered phase. S 0 of a dissolved solid or liquid is usually greater than the S 0 of the pure solute. However, the extent depends upon the nature of the solute and solvent. A gas becomes more ordered when it dissolves in a liquid or solid. In similar substances, increases in mass relate directly to entropy. In allotropic substances (e.g. O 2 & O 3 ), increases in complexity (e.g. bond flexibility) relate directly to entropy.

12. 20-12 Figure 20.5 The increase in entropy from solid to liquid to gas

13. 20-13 Figure 20.6 The entropy change accompanying the dissolution of a salt pure solid pure liquid solution MIX

14. 20-14 Figure 20.7 Ethanol WaterSolution of ethanol and water The small increase in entropy when ethanol dissolves in water

15. 20-15 Figure 20.8 The large decrease in entropy when a gas dissolves in a liquid O 2 gas

16. 20-16 Figure 20.9 NO NO 2 N2O4N2O4 Entropy and vibrational motion

17. 20-17 Sample Problem 20.1: SOLUTION: Predicting Relative Entropy Values PROBLEM:Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]: (a) 1mol of SO 2 ( g ) or 1mol of SO 3 ( g ) (b) 1mol of CO 2 ( s ) or 1mol of CO 2 ( g ) (c) 3mol of oxygen gas (O 2 ) or 2mol of ozone gas (O 3 ) (d) 1mol of KBr( s ) or 1mol of KBr( aq ) (e) Seawater in midwinter at 2 0 C or in midsummer at 23 0 C (f) 1mol of CF 4 ( g ) or 1mol of CCl 4 ( g ) PLAN:In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature. (a) 1mol of SO 3 ( g ) - more atoms (b) 1mol of CO 2 ( g ) - gas > solid (c) 3mol of O 2 ( g ) - larger #mols (d) 1mol of KBr( aq ) - solution > solid (e) 23 0 C - higher temperature (f) CCl 4 - larger mass

18. 20-18 Sample Problem 20.4: SOLUTION: Calculating  G 0 from Enthalpy and Entropy Values ( 1of 2) PROBLEM:Potassium chlorate, one of the common oxidizing agents in explosives, fireworks, and matchheads, undergoes a solid- state redox reaction when heated. In this reaction, note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation): 4KClO 3 ( s ) 3KClO 4 ( s ) + KCl( s )  +7+5 Use  H 0 f and S 0 values to calculate  G 0 sys (  G 0 rxn ) at 25 0 C for this reaction. PLAN:Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve.  H 0 rxn =  m  H f 0 products -  n  H f 0 reactants  H 0 rxn = (3mol)(-432.8kJ/mol) + (1mol)(-436.7kJ/mol) - (4mol)(-397.7kJ/mol)  H 0 rxn = -144kJ

19. 20-19 Sample Problem 20.4: Calculating  G 0 from Enthalpy and Entropy Values continued  S 0 rxn =  mS 0 products -  nS 0 reactants  S 0 rxn = (3mol)(151J/mol*K) + (1mol)(82.6J/mol*K) - (4mol)(143.1J/mol*K)  S 0 rxn = -36.8J/K  G 0 rxn =  H 0 rxn - T  S 0 rxn  G 0 rxn = -144kJ - (298K)(-36.8J/K)(kJ/10 3 J)  G 0 rxn = -133kJ

20. 20-20 Sample Problem 20.2: Calculating the Standard Entropy of Reaction,  S 0 rxn PROBLEM: Calculate  S 0 rxn for the combustion of 1mol of propane at 25 0 C. C 3 H 8 ( g ) + 5O 2 ( g ) 3CO 2 ( g ) + 4H 2 O( l ) PLAN:Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6mols of gas to 3mols of gas. SOLUTION:Find standard entropy values in the Appendix or other table.  S = [(3mol)(S 0 CO 2 ) + (4mol)(S 0 H 2 O)] - [(1mol)(S 0 C 3 H 8 ) + (5mol)(S 0 O 2 )]  S = [(3mol)(213.7J/mol*K) + (4mol)(69.9J/mol*K)] - [(1mol)(269.9J/mol*K) + (5mol)(205.0J/mol*K)]  S = - 374 J/K

21. 20-21 Figure B20.1 Lavoisier studying human respiration as a form of combustion

22. 20-22 Figure B20.2 A whole-body calorimeter

23. 20-23 Figure 20.10 Components of  S 0 universe for spontaneous reactions exothermic system becomes more disordered exothermic system becomes more ordered endothermic system becomes more disordered  S universe =  S system +  S surroundings

24. 20-24 Sample Problem 20.3: SOLUTION: Determining Reaction Spontaneity PROBLEM: At 298K, the formation of ammonia has a negative  S 0 sys ; Calculate  S 0 rxn, and state whether the reaction occurs spontaneously at this temperature. N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g )  S 0 sys = -197J/K PLAN:  S 0 universe must be > 0 in order for this reaction to be spontaneous, so  S 0 surroundings must be > 197J/K. To find  S 0 surr, first find  H sys ;  H sys =  H rxn which can be calculated using  H 0 f values from tables.  S 0 universe =  S 0 surr +  S 0 sys.  H 0 rx = [(2mol)(  H 0 f NH 3 )] - [(1mol)(  H 0 f N 2 ) + (3mol)(  H 0 f H 2 )]  H 0 rx = -91.8kJ  S 0 surr = -  H 0 sys /T = -(-91.8x10 3 J/298K)= 308J/K  S 0 universe =  S 0 surr +  S 0 sys = 308J/K + (-197J/K) = 111J/K  S 0 universe > 0 so the reaction is spontaneous.

25. 20-25 Different Ways to Calculate  G 0 rxn 0 = Std conds: 25 o C, 1 atm, 1 M solutions (Eqn 1)  G 0 rxn =  H 0 rxn - T  S 0 rxn (Eqn 2)  G 0 rxn =  m  G f 0 products -  n  G f 0 reactants (Eqn 3 )  G 0 = - RT lnK eq R = 8.314 J/mol. K When not at standard conditions (Eqn 4)  G rxn =  G 0 rxn + RT ln Q (Q = Reaction Quotient) Use these equations with equation #1  S 0 rxn =  mS 0 products -  nS 0 reactants  H 0 rxn =  m  H f 0 products -  n  H f 0 reactants Useful Thermodynamic Relationships

26. 20-26 Sample Problem 20.6: PROBLEM: PLAN: SOLUTION: Determining the Effect of Temperature on  G 0 (1 of 2) An important reaction 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 298K,  G 0 = -141.6kJ;  H 0 = -198.4kJ; and  S 0 = -187.9J/K (a) Use the data to decide if this reaction is spontaneous at 25 0 C, and predict how  G 0 will change with increasing T. (b) Assuming  H 0 and  S 0 are constant with increasing T, is the reaction spontaneous at 900. 0 C? The sign of  G 0 tells us whether the reaction is spontaneous and the signs of  H 0 and  S 0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b). (a) The reaction is spontaneous at 25 0 C because  G 0 is (-). Since  H 0 is (-) but  S 0 is also (-),  G 0 will become less spontaneous as the temperature increases.

27. 20-27 Sample Problem 20.6: continued (b)  G 0 rxn =  H 0 rxn - T  S 0 rxn  G 0 rxn = -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/10 3 J)  G 0 rxn = 22.0 kJ; the reaction will be nonspontaneous at 900. 0 C Determining the Effect of Temperature on  G 0 (2 of 2)

28. 20-28  G and the Work a System Can Do For a spontaneous process,  G is the maximum work obtainable from the system as the process takes place:  G = work max For a nonspontaneous process,  G is the maximum work that must be done to the system as the process takes place:  G = work max An example

29. 20-29 Table 20.1 Reaction Spontaneity and the Signs of  H 0,  S 0, and  G 0 H0H0 S0S0 -T  S 0 G0G0 Description -+-- +-++ ++-+ or - --+ Spontaneous at all T Nonspontaneous at all T Spontaneous at higher T; nonspontaneous at lower T Spontaneous at lower T; nonspontaneous at higher T

30. 20-30 Sample Problem 20.5: PROBLEM: Calculating  G 0 rxn from  G 0 f Values Use  G 0 f values to calculate  G rxn for the reaction in Sample Problem 20.4: 4KClO 3 ( s ) 3KClO 4 ( s ) + KCl( s )  PLAN: Use the  G summation equation. SOLUTION:  G 0 rxn =  m  G f 0 products -  n  G f 0 reactants  G 0 rxn = (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) - (4mol)(-296.3kJ/mol)  G 0 rxn = -134kJ

31. 20-31 Figure 20.11 The effect of temperature on reaction spontaneity

32. 20-32 The coupling of a nonspontaneous reaction to the hydrolysis of ATP. 1 st Rxn of Glycolysis catalyzed by Hexose Kinase: Glucose + HPO 4 2-  Glucose-6-phosphate + H 2 O  G 0 = +13.8 kJ ATP 4- + H 2 O  ADP 3- + HPO 4 2-  G 0 = -30.5 kJ Glucose + HPO 4 2-  Glucose-6-phosphate + H 2 O  G 0 = +13.8 kJ Glucose + ATP  Glucose-6-phosphate + ADP 3-  G 0 = -16.7 kJ Enzyme Glucose Glucose-6-phosphate Enzyme: hexose kinase

33. 20-33 Figure B20.4 The cycling of metabolic free energy through ATP

34. 20-34 Figure B20.5 Why is ATP a high-energy molecule?

35. 20-35 Free Energy, Equilibrium and Reaction Direction If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (  G < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (  G > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (  G = 0)  G = RT ln Q/K = RT lnQ - RT lnK Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and lnQ = 0 so...  G 0 = - RT lnK

36. 20-36 FORWARD REACTION REVERSE REACTION Table 20.2 The Relationship Between  G 0 and K at 25 0 C  G 0 (kJ) KSignificance 200 100 50 10 1 0 -10 -50 -100 -200 9x10 -36 3x10 -18 2x10 -9 2x10 -2 7x10 -1 1 1.5 5x10 1 6x10 8 3x10 17 1x10 35 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent Forward reaction goes to completion; essentially no reverse reaction

37. 20-37 Sample Problem 20.7: PROBLEM: Calculating  G at Nonstandard Conditions The oxidation of SO 2, which we considered in Sample Problem 20.6 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298K and at 973K. (  G 0 298 = -141.6kJ/mol of reaction as written using  H 0 and  S 0 values at 973K.  G 0 973 = -12.12kJ/mol of reaction as written.) (b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO 2, 0.0100atm of O 2, and 0.100atm of SO 3 and kept at 25 0 C and at 700. 0 C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate  G for the system in part (b) at each temperature. PLAN:Use the equations and conditions found on slide.

38. 20-38 SOLUTION: Sample Problem 20.7: Calculating  G at Nonstandard Conditions continued (2 of 3) (a) Calculating K at the two temperatures:  G 0 = -RTlnK so At 298, the exponent is -  G 0 /RT = - (-141.6kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(298K) = 57.2 = e 57.2 = 7x10 24 At 973, the exponent is -  G 0 /RT (-12.12kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(973K) = 1.50 = e 1.50 = 4.5

39. 20-39 Sample Problem 20.7: Calculating  G at Nonstandard Conditions continued (3 of 3) (b) The value of Q = pSO 3 2 (pSO 2 ) 2 (pO 2 ) = (0.100) 2 (0.500) 2 (0.0100) = 4.00 Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight. (c) The nonstandard  G is calculated using  G =  G 0 + RTlnQ  G 298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(298K)(ln4.00)  G 298 = -138.2kJ/mol  G 973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(973K)(ln4.00)  G 298 = -0.9kJ/mol

40. 20-40 Figure 20.12 The relation between free energy and the extent of reaction  G 0 1  G 0 > 0 K <1

41. 20-41 Different Ways to Calculate  G rxn (Eqn 1)  G 0 rxn =  H 0 rxn - T  S 0 rxn (Eqn 2)  G 0 rxn =  m  G f 0 products -  n  G f 0 reactants (Eqn 3)  G 0 = - RT lnK eq R = 8.314 J/mol. K (Eqn 4)  G =  G 0 + RT lnQ Q = Rxn Quotient Use these with equation #1  S 0 rxn =  mS 0 products -  nS 0 reactants  H 0 rxn =  m  H f 0 products -  n  H f 0 reactants Useful Thermodynamic Relationships