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

2. Limitations of the First Law of Thermodynamics  E = q + w E universe = E system + E surroundings  E system = -  E surroundings The total energy-mass of the universe is constant. However, this does not tell us anything about the direction of change in the universe.

3. 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

4. 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

5. The number of ways to arrange a deck of playing cards

6. 1 atmevacuated Spontaneous expansion of a gas stopcock closed stopcock opened 0.5 atm

7. 1877 Ludwig Boltzman S = k ln W where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/N A (R = universal gas constant, N A = Avogadro’s number. 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.  S universe =  S system +  S surroundings > 0 This is the second law of thermodynamics.

8. Random motion in a crystal The third law of thermodynamics. A perfect crystal has zero entropy at a temperature of absolute zero. S system = 0 at 0 K Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

9. 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, increases in complexity (e.g. bond flexibility) relate directly to entropy.

10. The increase in entropy from solid to liquid to gas Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

11. The entropy change accompanying the dissolution of a salt pure solid pure liquid solution MIX Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

12. Ethanol WaterSolution of ethanol and water The small increase in entropy when ethanol dissolves in water Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

13. The large decrease in entropy when a gas dissolves in a liquid O 2 gas Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

14. NO NO 2 N2O4N2O4 Entropy and vibrational motion

15. Sample 1Predicting 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 )

16. Sample 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 )

17. Components of  S 0 universe for spontaneous reactions exothermic system becomes more disordered exothermic system becomes more ordered endothermic system becomes more disordered

18. Sample 3Determining 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.  H 0 f NH 3 = -45.9 kJ/mol N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g )  S 0 sys = -197J/K

19.  G 0 system =  H 0 system - T  S 0 system  G 0 rxn =  m  G 0 products -  n  G 0 reactants

20. Sample 4 Calculating  G 0 from Enthalpy and Entropy Values 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 at 25 0 C for this reaction.

21. Sample 5 PROBLEM: Calculating  G 0 rxn from  G 0 f Values Use  G 0 f values to calculate  G rxn for the reaction: 4KClO 3 ( s ) 3KClO 4 ( s ) + KCl( s ) 

22. Sample 6 PROBLEM: Determining the Effect of Temperature on  G 0 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?

23. Sample 7 PROBLEM: Determining the Effect of Temperature on  G 0 A reaction is nonspontaneous at room temperature but is spontaneous at -40 0 C. What can you say about the signs and relative magnitudes of  H 0  S 0 and -T  S 0

24. 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

25. The effect of temperature on reaction spontaneity

26.  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

27. The coupling of a nonspontaneous reaction to the hydrolysis of ATP.

28. The cycling of metabolic free enery through ATP

29. 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

30. FORWARD REACTION REVERSE REACTION 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

31. Sample Problem 6 PROBLEM: Calculating  G at Nonstandard Conditions The oxidation of SO 2, which we considered in Sample Problem 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.

32. Sample Problem 7 PROBLEM: Calculating  G at Nonstandard Conditions At 298 K hypobromous acid (HBrO) dissociates in water with a K a of 2.3 x 10 -9. (a) Calculate  G 0 298 (b) Calculate  G if [H 3 O+] = 6.0 x 10 -4 M, [BrO-] = 0.1 M and [HBrO]=0.20 M

33. The relation between free energy and the extent of reaction  G 0 1  G 0 > 0 K <1