1. Chapter 16 Spontaneity, entropy and free energy

2. Spontaneous A reaction that will occur without outside intervention. A reaction that will occur without outside intervention. We can’t determine how fast. We can’t determine how fast. We need both thermodynamics and kinetics to describe a reaction completely. We need both thermodynamics and kinetics to describe a reaction completely. Thermodynamics compares initial and final states. Thermodynamics compares initial and final states. Kinetics describes pathway between. Kinetics describes pathway between.

3. Thermodynamics 1st Law- the energy of the universe is constant. 1st Law- the energy of the universe is constant. Keeps track of thermodynamics doesn’t correctly predict spontaneity. Keeps track of thermodynamics doesn’t correctly predict spontaneity. Entropy (S) is disorder or randomness Entropy (S) is disorder or randomness 2nd Law the entropy of the universe increases. 2nd Law the entropy of the universe increases.

4. Entropy Defined in terms of probability. Defined in terms of probability. Substances take the arrangement that is most likely. Substances take the arrangement that is most likely. The most likely is the most random. The most likely is the most random. Calculate the number of arrangements for a system. Calculate the number of arrangements for a system.

5. 2 possible arrangements 2 possible arrangements 50 % chance of finding the left empty 50 % chance of finding the left empty

6. l 4 possible arrangements l 25% chance of finding the left empty l 50 % chance of them being evenly dispersed

7. l 4 Molecules create 12 possible arrangements l 8% chance of finding the left empty l 50 % chance of them being evenly dispersed

8. Gases Gases completely fill their chamber because there are many more ways to do that than to leave half empty. Gases completely fill their chamber because there are many more ways to do that than to leave half empty. S solid <S liquid <<S gas S solid <S liquid <<S gas there are many more ways for the molecules to be arranged as a liquid than a solid. there are many more ways for the molecules to be arranged as a liquid than a solid. Gases have a huge number of positions possible. Gases have a huge number of positions possible.

9. Positional Entropy Probability Equation: Probability Equation: P =( 1) # of Particles P =( 1) # of Particles # of Positions # of Positions Example Problem. 5 Molecules of Carbon Dioxide are place in a 4 bulb container. What is the probability that all of the Molecules will be in one bulb? Example Problem. 5 Molecules of Carbon Dioxide are place in a 4 bulb container. What is the probability that all of the Molecules will be in one bulb?

10. Solution P = (1 ) # of Particles P = (1 ) # of Particles # of Positions # of Positions P = (1) 5 P = (1) 5 4 P=1/1024 P=1/1024

11. Positional Entropy Which would have the higher positional entropy, one Mole of Solid Carbon Dioxide and gaseous Carbon Dioxide? A mole of gaseous Carbon Dioxide has a greater volume then a mole of solid Carbon Dioxide, so by logic gaseous Carbon Dioxide would have a higher positional energy

12. Predicting Entropy Changes Solid sugar is added to water to form a solution, would the entropy be positive or negative? (greater or less) Sugar Molecules become randomly dispersed in the water, so they have a larger number of positions, so entropy would be greater (Positive)

13. Homework Book Exercises 16-21 to 16.24 Book Exercises 16-21 to 16.24

14. Entropy Solutions form because there are many more possible arrangements of dissolved pieces than if the solute and solvent stay separate. Solutions form because there are many more possible arrangements of dissolved pieces than if the solute and solvent stay separate. 2nd Law 2nd Law  S univ =  S sys +  S surr  S univ =  S sys +  S surr If  S univ is positive the process is spontaneous. If  S univ is positive the process is spontaneous. If  S univ is negative the process is spontaneous in the opposite direction. If  S univ is negative the process is spontaneous in the opposite direction.

15. For exothermic processes  S surr is positive. For exothermic processes  S surr is positive. For endothermic processes  S surr is negative. For endothermic processes  S surr is negative. Consider this process H 2 O(l)  H 2 O(g) Consider this process H 2 O(l)  H 2 O(g)  S sys is positive  S sys is positive  S surr is negative  S surr is negative  S univ depends on temperature.  S univ depends on temperature.

16. Example of the Second Law Is the organized state of life consistent with the second law? Suniv must be positive for this to be true. Ssys is negative in this case, but…. Ssurr is positive and much great then Ssys so…. Suniv is greater then zero, life is consistent with the second law.

17. Homework Book Problems 16.11 and 16.12 Book Problems 16.11 and 16.12

18. Temperature and Spontaneity Entropy changes in the surroundings are determined by the heat flow. Entropy changes in the surroundings are determined by the heat flow. An exothermic process is favored because by giving up heat the entropy of the surroundings increases. An exothermic process is favored because by giving up heat the entropy of the surroundings increases. The size of  S surr depends on temperature The size of  S surr depends on temperature  S surr = -  H/T  S surr = -  H/T  S surr = - (Change in enthalpy)/Temp in K  S surr = - (Change in enthalpy)/Temp in K

19.  S sys  S surr  S univ Spontaneous? +++ --- +-? +-? Yes No, Reverse At Low temp. At High temp.

20. Example Problem In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores: In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores: Sb 2 S 3 (s) + 3 Fe(s)  2Sb(s) + 3FeS(s) Sb 2 S 3 (s) + 3 Fe(s)  2Sb(s) + 3FeS(s)  H = -125 kJ  H = -125 kJ

21. Example Continued Carbon is used as the reducing agent for oxide ores: Sb4O6(s) + 6 C(s)  4 Sb(s) + 6 CO(g) H = 778 kJ Calculate Ssurr for each of these reactions at 25 C and 1 atm.

22. Solution  S surr = -  H/T  S surr = -  H/T T = 25 + 273 = 298 K T = 25 + 273 = 298 K Sulfide Ore Reaction Sulfide Ore Reaction  S surr = -(-125 kJ/298 K) = 0.419 kJ/K = 419 J/K  S surr = -(-125 kJ/298 K) = 0.419 kJ/K = 419 J/K Notice it is positive, so it is exotermic, heat is released into the surroundings, increasing randomness of the surroundings. Notice it is positive, so it is exotermic, heat is released into the surroundings, increasing randomness of the surroundings. Oxide Ore Reaction Oxide Ore Reaction  S surr = -(778 kJ/298 K) = -2.61 kJ/K = -2610 J/K  S surr = -(778 kJ/298 K) = -2.61 kJ/K = -2610 J/K Notice it is negative so heat flows from the surroundings to the system. Notice it is negative so heat flows from the surroundings to the system.

23. Homework Book Questions 16.25 and 16.26 Book Questions 16.25 and 16.26

24. Gibb's Free Energy G=H-TS G=H-TS Never used this way. Never used this way.  G=  H-T  S at constant temperature  G=  H-T  S at constant temperature Divide by –T Divide by –T  G/-T =  H/-T-T  S/-T  G/-T =  H/-T-T  S/-T -  G/T = -  H/T+  S -  G/T = -  H/T+  S -  G/T =  S surr +  S -  G/T =  S surr +  S -  G/T =  S univ -  G/T =  S univ If  G is negative at constant T and P, the Process is spontaneous. If  G is negative at constant T and P, the Process is spontaneous.

25. Let’s Check For the reaction H 2 O(s)  H 2 O(l) For the reaction H 2 O(s)  H 2 O(l)  Sº = 22.1 J/K mol  Hº =6030 J/mol  Sº = 22.1 J/K mol  Hº =6030 J/mol Calculate  G at 10ºC and -10ºC Calculate  G at 10ºC and -10ºC Look at the equation  G=  H-T  S Look at the equation  G=  H-T  S  G=  J/mol –(283K*22.1 J/K*mol)  G=  J/mol –(283K*22.1 J/K*mol)  G= 6030 J/mol – 6254 J/mol = -224  G= 6030 J/mol – 6254 J/mol = -224  G=  J/mol –(263K*22.1 J/K*mol)  G=  J/mol –(263K*22.1 J/K*mol)  G= 6030 J/mol – 5812 J/mol = 218  G= 6030 J/mol – 5812 J/mol = 218 Spontaneity can be predicted from the sign of  H and  S. Spontaneity can be predicted from the sign of  H and  S.

26.  G=  H-T  S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous

27. Sample Problem At what temperature is the following process spontaneous at 1 atm? At what temperature is the following process spontaneous at 1 atm? Br 2 (l)  Br 2 (g)  H o = 31.0 kJ/mol  S o = 93.0 J/K mol What is the normal boiling point? What is the normal boiling point?

28. Solution If  G is negative, vaporization is spontaneous. If  G is negative, vaporization is spontaneous.  S o is positive, so it favors vaporization.  S o is positive, so it favors vaporization.  H o is positive, so it favors condensation.  H o is positive, so it favors condensation. Opposite tendencies exactly balance at boiling point. Opposite tendencies exactly balance at boiling point. We can determine the boiling point by setting  G o = 0. We can determine the boiling point by setting  G o = 0.

29. Equation  G o =  H o -T  S o  G o =  H o -T  S o  =  H o –T  S o  =  H o –T  S o  H o = T  S o  H o = T  S o  =  H o /  S  =  H o /  S (3.10 x 10 4 J/mol) / 93.0 J/K mol = 333 K (3.10 x 10 4 J/mol) / 93.0 J/K mol = 333 K > 333K vaporization is dominant > 333K vaporization is dominant < 333K condensation is dominant < 333K condensation is dominant = 333K both coexist, neither is dominant = 333K both coexist, neither is dominant

30. Homework 16.29 and 16.31 16.29 and 16.31

31. Third Law of Thermo The entropy of a pure crystal at 0 K is 0. The entropy of a pure crystal at 0 K is 0. Gives us a starting point. Gives us a starting point. All others must be>0. All others must be>0. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. Products - reactants to find  Sº (a state function). Products - reactants to find  Sº (a state function). More complex molecules higher Sº. More complex molecules higher Sº.

32. Example Predict the sign of  Sº for the following reaction: Predict the sign of  Sº for the following reaction: CaCO 3 (s)  CaO(s) + CO 2 (g) Since in this reaction a gas is produced from a solid reactant, the positional entropy increases, and  Sº is positive. Since in this reaction a gas is produced from a solid reactant, the positional entropy increases, and  Sº is positive.

33. Another Example In the oxidation of Sulfur Dioxide in air, is the  Sº positive or negative? In the oxidation of Sulfur Dioxide in air, is the  Sº positive or negative? 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) In this case, three molecules of reactants produce 2 molecules of Products. Since the number of gas molecules decreases, positional entropy decreases, and  Sº is negative In this case, three molecules of reactants produce 2 molecules of Products. Since the number of gas molecules decreases, positional entropy decreases, and  Sº is negative

34. More Homework 16.33 and 16.34 16.33 and 16.34

35. Free Energy in Reactions  Gº = standard free energy change.  Gº = standard free energy change. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Can’t be measured directly, can be calculated from other measurements. Can’t be measured directly, can be calculated from other measurements.  Gº=  Hº-T  Sº  Gº=  Hº-T  Sº Use Hess’s Law with known reactions. Use Hess’s Law with known reactions.

36. Free Energy in Reactions There are tables of  Gº f. There are tables of  Gº f. Products-reactants because it is a state function. Products-reactants because it is a state function. The standard free energy of formation for any element in its standard state is 0. The standard free energy of formation for any element in its standard state is 0. Remember- Spontaneity tells us nothing about rate. Remember- Spontaneity tells us nothing about rate.

37.  G=  H-T  S HH SS Spontaneous? +- At all Temperatures ++ At high temperatures, “entropy driven” -- At low temperatures, “enthalpy driven” +- Not at any temperature, Reverse is spontaneous

38. Third Law of Thermo The entropy of a pure crystal at 0 K is 0. The entropy of a pure crystal at 0 K is 0. Gives us a starting point. Gives us a starting point. All others must be>0. All others must be>0. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. Products - reactants to find  Sº (a state function) Products - reactants to find  Sº (a state function) More complex molecules higher Sº. More complex molecules higher Sº.

39. Yet Another Equation Entropy is a state function of the system Entropy is a state function of the system It is not pathway-dependent It is not pathway-dependent Entropy for a chemical reaction can be calculated by taking the difference between the values of the products and the reactants. Entropy for a chemical reaction can be calculated by taking the difference between the values of the products and the reactants.  S o reaction =  n p S o products –  n r S o reactants  S o reaction =  n p S o products –  n r S o reactants n p = moles of products n p = moles of products n r = moles of reactants n r = moles of reactants Entropy is an extensive property Entropy is an extensive property It depends on how much of the substance is present. It depends on how much of the substance is present.

40. Example Calculate  S o at 25 C for the reaction: Calculate  S o at 25 C for the reaction: 2 NiS (s) + 3 O 2 (g)  2 SO 2 (g) + 2 NiO (s) Given the following standard entropy values: Given the following standard entropy values: Substance S o (J/K mol) SO 2 (g) 248 NiO (s) 38 O 2 (g) 205 NiS (s) 53

41. Solution  S o reaction =  n p S o products –  n r S o reactants  S o reaction =  n p S o products –  n r S o reactants  S o reaction = (2S o SO2 (g) + 2S o NiS (s) ) – (2S o NiS (s) + 3S o O2(g) )  S o reaction = (2S o SO2 (g) + 2S o NiS (s) ) – (2S o NiS (s) + 3S o O2(g) )  S o reaction = [2 mol(248 J/K mol)+2 mol(38 J/K mol)] – [2mol(53 J/K mol) + 3 mol(205 J/K mol)]  S o reaction = [2 mol(248 J/K mol)+2 mol(38 J/K mol)] – [2mol(53 J/K mol) + 3 mol(205 J/K mol)]  S o reaction = (496 J/K + 76 J/K) – (106 J/K + 615 J/K)  S o reaction = (496 J/K + 76 J/K) – (106 J/K + 615 J/K)  S o reaction = - 149 J/K  S o reaction = - 149 J/K We would expect a negative  S o since the number of gaseous molecules decreases. We would expect a negative  S o since the number of gaseous molecules decreases.

42. Another example Calculate  S o for the reduction of aluminum oxide by hydrogen gas: Calculate  S o for the reduction of aluminum oxide by hydrogen gas: Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) Given the following standard entropy values: Given the following standard entropy values: Substance S o (J/K mol) Al 2 O 3 (s) 51 H 2 (g) 131 Al(s)28 H 2 O(g) 189

43. Solution  S o reaction =  n p S o products –  n r S o reactants  S o reaction =  n p S o products –  n r S o reactants  S o reaction = (2S o (Al(s) + 3S o H2O(g) ) – (3S o H2(g) + S o Al2O3(s) )  S o reaction = (2S o (Al(s) + 3S o H2O(g) ) – (3S o H2(g) + S o Al2O3(s) )  S o reaction = [2 mol(28 J/K mol)+3 mol(189 J/K mol)] – [3mol(131 J/K mol) + 1 mol(51 J/K mol)]  S o reaction = [2 mol(28 J/K mol)+3 mol(189 J/K mol)] – [3mol(131 J/K mol) + 1 mol(51 J/K mol)]  S o reaction = [56 J/K + 567 J/K] – [393 J/K + 51 J/K]  S o reaction = [56 J/K + 567 J/K] – [393 J/K + 51 J/K]  S o reaction = 179 J/K  S o reaction = 179 J/K We would expect a positive number since the moles of solid went up. We would expect a positive number since the moles of solid went up.

44. Homework Problems 16.37, 16.39 and 16.40 Problems 16.37, 16.39 and 16.40

45. Free Energy in Reactions  Gº = standard free energy change.  Gº = standard free energy change. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Can’t be measured directly, can be calculated from other measurements. Can’t be measured directly, can be calculated from other measurements.  Gº=  Hº-T  Sº  Gº=  Hº-T  Sº Use Hess’s Law with known reactions. Use Hess’s Law with known reactions.

46. Free Energy in Reactions There are tables of  Gº f There are tables of  Gº f Products-reactants because it is a state function. Products-reactants because it is a state function. The standard free energy of formation for any element in its standard state is 0. The standard free energy of formation for any element in its standard state is 0. Remember- Spontaneity tells us nothing about rate. Remember- Spontaneity tells us nothing about rate.

47. Example Problem Consider the reaction: Consider the reaction: 2SO 2 (g) + O 2 (g)  2SO 3 (g) Carried out at 25 o C and 1 atm. Calculate  H o,  S o, and  G o using the following data: Carried out at 25 o C and 1 atm. Calculate  H o,  S o, and  G o using the following data: Substance  H f (kJ/mol)  S f (J/K*mol) SO 2 (g) -297248 O 2 (g) -396257 SO 3 (g) 0205

48.  Solution  H o reaction =  n p  H f o products –  n r  H f o reactants  H o reaction =  n p  H f o products –  n r  H f o reactants  H o reaction = (2  H f o (SO3(g) ) – (2  H f o (SO2(g) +  H f o (O2(g) )  H o reaction = (2  H f o (SO3(g) ) – (2  H f o (SO2(g) +  H f o (O2(g) )  H o reaction = [2 mol (-396 kJ/mol)] – [2 mol(-297 kJ/mol) + (1 mol * 0 kJ/mol)]  H o reaction = [2 mol (-396 kJ/mol)] – [2 mol(-297 kJ/mol) + (1 mol * 0 kJ/mol)]  H o reaction = [-792 kJ] – [-594 kJ + 0 kJ]  H o reaction = [-792 kJ] – [-594 kJ + 0 kJ]  H o reaction = -792 kJ + 594 kJ  H o reaction = -792 kJ + 594 kJ  H o reaction = -198 kJ  H o reaction = -198 kJ

49.  S Solution  S o reaction =  n p  S o products –  n r  S o reactants  S o reaction =  n p  S o products –  n r  S o reactants  S o reaction =  S o SO3 (g) – (  S o SO2 (g) + S o O2 (g) )  S o reaction =  S o SO3 (g) – (  S o SO2 (g) + S o O2 (g) )  S o reaction = [  mol(257 J/K mol)] – [2 mol(248 J/K mol) + 1 mol(205 J/K mol)]  S o reaction = [  mol(257 J/K mol)] – [2 mol(248 J/K mol) + 1 mol(205 J/K mol)]  S o reaction = [  J/K] – [496 J/K + 205 J/K]  S o reaction = [  J/K] – [496 J/K + 205 J/K]  S o reaction = [  J/K] – [701 J/K]  S o reaction = [  J/K] – [701 J/K]  S o reaction = -187  J/K  S o reaction = -187  J/K

50.  G Solution  G o =  H o – T  S o  G o =  H o – T  S o  S o = (-187 J/K)(1kJ/1000 J)  S o = (-187 J/K)(1kJ/1000 J)  G o = -198 kJ – [(298 K) (-.187 kJ)]  G o = -198 kJ – [(298 K) (-.187 kJ)]  G o = -198 kJ + 55.7 kJ = -142 kJ  G o = -198 kJ + 55.7 kJ = -142 kJ

51. Homework Exercises 16.45 and 16.46 Exercises 16.45 and 16.46

52. A Second Method Another way to calculate Gibbs Free Energy (  G) is to take advantage of the fact that it is a state function. Another way to calculate Gibbs Free Energy (  G) is to take advantage of the fact that it is a state function. We can use the procedure we used with Hess’s Law (  H o reaction =  n p  H f o products –  n r  H f o reactants ) We can use the procedure we used with Hess’s Law (  H o reaction =  n p  H f o products –  n r  H f o reactants )  G o reaction =  n p  G f o products –  n r  G f o reactants  G o reaction =  n p  G f o products –  n r  G f o reactants

53. Example Find the free energy change for the following reaction: Find the free energy change for the following reaction: 2CO(g) + O 2 (g)  2CO 2 (g) We must use the following data: We must use the following data: 2CH 4 (g) + 3O 2 (g)  2CO(g) + 4H 2 O(g)  G o = 1088 kJ CH 4 (g) + 2O 2 (g)  2CO 2 (g) + 2H 2 O(g)  G o = -801 kJ

54. Example continued Compare the first and second equation. Compare the first and second equation. They both consume oxygen and produce water vapor. They both consume oxygen and produce water vapor. Reverse the first equation: Reverse the first equation: 2CO(g) + 4H 2 O(g)  2CH 4 (g) + 3O 2 (g)  G o = -1088 kJ  G o = -1088 kJ Compare it to the second: Compare it to the second: CH 4 (g) + 2O 2 (g)  2CO 2 (g) + 2H 2 O(g) CH 4 (g) + 2O 2 (g)  2CO 2 (g) + 2H 2 O(g) Now Oxygen and water vapor cancel out. Now Oxygen and water vapor cancel out.

55. Further with the Example This leaves us with the original equation: This leaves us with the original equation: 2CO(g) + O 2 (g)  2CO 2 (g) We now must adjust the energy, since in the first equation only 1 mole was produced, we must double the  G = 2(-801 kJ) = -1602 kJ We now must adjust the energy, since in the first equation only 1 mole was produced, we must double the  G = 2(-801 kJ) = -1602 kJ We now have  G for the reactants and products. We now have  G for the reactants and products.  G o reaction = -1602 kJ – (-1088 kJ)  G o reaction = -1602 kJ – (-1088 kJ)  G o reaction = - 514 kJ  G o reaction = - 514 kJ

56. Free energy and Pressure  G =  Gº +RTln(Q) where Q is the reaction quotients (P of the products /P of the reactants).  G =  Gº +RTln(Q) where Q is the reaction quotients (P of the products /P of the reactants). CO(g) + 2H 2 (g)  CH 3 OH(l) CO(g) + 2H 2 (g)  CH 3 OH(l) Would the reaction be spontaneous at 25ºC with the H 2 pressure of 5.0 atm and the CO pressure of 3.0 atm? Would the reaction be spontaneous at 25ºC with the H 2 pressure of 5.0 atm and the CO pressure of 3.0 atm?  Gº f CH 3 OH(l) = -166 kJ (Appendix 4)  Gº f CH 3 OH(l) = -166 kJ (Appendix 4)  Gº f CO(g) = -137 kJ  Gº f H 2 (g) = 0 kJ  Gº f CO(g) = -137 kJ  Gº f H 2 (g) = 0 kJ

57. More on Example  Gº = -166 kJ – (-137 kJ + 0 kJ) = -29 kJ  Gº = -166 kJ – (-137 kJ + 0 kJ) = -29 kJ Since this is the value for 1 mole of CO, then this unit becomes kJ/mol rxn Since this is the value for 1 mole of CO, then this unit becomes kJ/mol rxn -29 kJ/mol = -2.9 x 10 4 J/mol rxn -29 kJ/mol = -2.9 x 10 4 J/mol rxn  G =  Gº +RTln(Q)  G =  Gº +RTln(Q)  Gº = -2.9 x 10 4 J/mol rxn  Gº = -2.9 x 10 4 J/mol rxn R = 8.3145 J/K mol (Since we are working in J we must use the R for kPa R = 8.3145 J/K mol (Since we are working in J we must use the R for kPa T = 298 K T = 298 K

58. Calculating Q for this Reaction Remember Q is calculated in the same fashion as Keq Remember Q is calculated in the same fashion as Keq Q = [Products] Q = [Products] [P CO ][P H2 2 ] Q = [1]= 0.022 Q = [1]= 0.022 [5][3 2 ] Note Pure Ethanol is a liquid therefore not included in calculations Note Pure Ethanol is a liquid therefore not included in calculations

59. Solution  G =  Gº + -RTln(Q)  G =  Gº + -RTln(Q)  G = (-2.9 x 10 4 J/mol rxn) + {-(8.3145 J/K mol rxn)(298K) ln(0.022)  G = (-2.9 x 10 4 J/mol rxn) + {-(8.3145 J/K mol rxn)(298K) ln(0.022)  G = (-2.9 x 10 4 J/mol rxn) – (9400 J/mol rxn)  G = (-2.9 x 10 4 J/mol rxn) – (9400 J/mol rxn)  G = -38000 J/mol rxn = -38 KJ/mol rxn  G = -38000 J/mol rxn = -38 KJ/mol rxn

60. Homework 16.55, 16.57, 16.59 16.55, 16.57, 16.59

61. How far?  G tells us spontaneity at current conditions. When will it stop?  G tells us spontaneity at current conditions. When will it stop? It will go to the lowest possible free energy which may be an equilibrium. It will go to the lowest possible free energy which may be an equilibrium. At equilibrium  G = 0, Q = K At equilibrium  G = 0, Q = K  Gº = -RTlnK  Gº = -RTlnK

62.  Gº K =0=1 0 0 >0 0<0

63. Temperature dependence of K  Gº= -RTlnK =  Hº - T  Sº  Gº= -RTlnK =  Hº - T  Sº ln(K) =  Hº/R(1/T)+  Sº/R ln(K) =  Hº/R(1/T)+  Sº/R A straight line of lnK vs 1/T A straight line of lnK vs 1/T

64. Free energy And Work Free energy is that energy free to do work. Free energy is that energy free to do work. The maximum amount of work possible at a given temperature and pressure. The maximum amount of work possible at a given temperature and pressure. Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work. Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.