1. Thermodynamics Beyond Simply Energy

2. 2 H 2 + O 2  2 H 2 O Given a chemical reaction: 1. Does it happen? 2. How fast does it happen? 3. Is it an equilibrium reaction? 4. How does it compare with a competing reaction? (If I mix H 2, O 2 and N 2, what do I get?)

3. Joe’s Rule of the Possible If it can happen, it will happen. But, that doesn’t tell you how much, how fast, how often, how easily… …Thermodynamics picks up where Joe’s Rule leaves off.

4. Thermodynamics Thermodynamics deals with energy, as the name implies, but not just energy. It includes the study of all the different possible states of a system and how the system moves between different states.

5. States I mix 1 molecule of O 2 and 1 molecule of H 2 in an evacuated 1 L flask. How many different states of this system are there? A nearly infinite number of them!

6. “States” of a system H2H2 H2H2 O2O2 O2O2

7. H2H2 O2O2 H2H2 O2O2

8. What the &^%* can we do? Thermodynamics deals with statistical analysis of ensembles of states. In our case, we are usually looking at a single representative state of the system that is the “most probable” state.

9. Putting the “thermo” in thermodynamics As the name implies, “thermo- dynamics” is about energy (thermo=heat). What does this mean for a reaction?

10. Reaction Energies The energy change associated with a chemical reaction is called the enthalpy of reaction and abbreviated H.  H = H final - H initial

11. Enthalpy of Reactions There are actually a number of different types of enthalpies because enthalpy depends on conditions. THEY ARE ALL JUST SPECIFIC TYPES OF A GENERAL CONCEPT CALLED “ENTHALPY”.  H = H final - H initial

12. Types of  H  H – generic version  H rxn – generic version  H º - enthalpy change under Standard Temperature and Pressure (298 K, 1 atm)  H f – enthalpy of formation, refers to a specific reaction type

13. General Reaction Scheme – “hot pack” Reaction Coordinate Energy Reactants Products EaEa ΔHΔH

14. Endothermic Reaction – “cold pack” Reaction Coordinate Energy Reactants Products EaEa ΔHΔH

15. Where does the Energy go? In the case of a chemical reaction, you need to keep the different types of energy separate in your mind: Bond energy – energy INSIDE the molecules Thermal energy (heat) – kinetic energy of the molecules Energy of the “bath” – kinetic energy of solvent or other molecules in the system

16. Energy changes  H represents the change in INTERNAL MOLECULAR ENERGY.  H = H final - H initial

17. Exothermic Reaction – “hot pack” Reaction Coordinate Energy Reactants Products EaEa ΔHΔH

18. Exothermic energy changes  H = H final – H initial < 0 H initial >H final This energy is internal to the molecule. The excess gets absorbed by the rest of the system as heat causing the molecules to move faster (more kinetic energy) and the temperature to increase.

19. Endothermic Reaction – “cold pack” Reaction Coordinate Energy Reactants Products EaEa ΔHΔH

20. Endothermic energy changes  H = H final – H initial > 0 H initial <H final This energy is internal to the molecule and must come from somewhere. The additional energy required by the system gets absorbed from the rest of the system as heat causing the molecules to move slower (less kinetic energy) and the temperature to decrease.

21. The hard part is getting over the hump. Reaction Coordinate Energy Reactants Products EaEa ΔHΔH

22. E a = Activation Energy The tale of a reaction is not limited strictly to the identity and energetics of the products and reactants, there is a path (reaction coordinate) that must get followed. The “hump” represents a hurdle that must be overcome to go from reactants to products.

23. How do you get over the hump? If you are at the top, it is easy to fall down into the valley (on either side), but how do you get to the top? Reaction Coordinate Energy Reactants Products EaEa ΔHΔH

24. How do you get over the hump? The molecules acquire or lose energy the same way: by colliding with each other! The energy comes from the “bath”, the rest of the system. Reaction Coordinate Energy Reactants Products EaEa ΔHΔH

25. Types of  H  H – generic version  H rxn – generic version  H º - enthalpy change under Standard Temperature and Pressure (298 K, 1 atm)  H f – enthalpy of formation, refers to a specific reaction type

26. Enthalpy is a “State Function” What’s a “state function”? A “state function” is a value that is a function only of the initial and final states of the system, not the path you take to get there!

27. Climbing Mt. Everest Suppose you start at Himalayan Base Camp #1, climb to the summit of Everest over the course of 3 weeks, then return to Himalayan Base Camp #1.

28. Climbing Mt. Everest Back at base camp, I figure out my altitude change. What is it? ZERO – I’m back where I started

29. Climbing Mt. Everest I did a lot of work along the way, but all that matters is I’m back where I’m started. The net change in altitude is NADA, ZERO, ZILCH!

30. Enthalpy as a State Function Enthalpy is like that. It doesn’t care how you got where you are going, it simply looks at the difference from where you started.

31. Path doesn’t matter! Reactants Products  H H Actual path

32. Energy Considerations Energy is an important consideration in any physical or chemical process. You need to “climb the hill”!

33. Have you ever seen a ball roll uphill? The universe is a lazy place! Everything seeks its lowest energy state. Given the chance, systems always seek the lowest energy possible.

34. Am I lying? How did the ball get to the top of the hill in the first place?

35. Little nit-picking definitions: There are always two (at least) regions to consider: 1. The system – the object under study 2. The surroundings – the rest of the universe

36. You can go up…but someone goes down. You can add energy to the system, but you must take it from the surroundings. The issue with my lazy man’s definition of the universe is a thing called “spontaneity”.

37. Spontaneity is… Spontaneity means that the observed change happens without a push. It naturally occurs without being forced. The ball spontaneously rolls down the hill. We can force it back up the hill, but we have to put in energy.

38. Still seems like I’m lying… If what I say is true, then ALL observed changes in the world around us that happen without being forced would have to be downhill in energy. Is that true?

39. If energy were the whole story… Why would water evaporate? It is an endothermic process with an activation barrier, so it requires energy to be put into the system. Yet, water spontaneously evaporates even at near freezing temperatures. (And actually sublimes when frozen!)

40. BUT… ENERGY CHANGES AREN’T THE WHOLE STORY!

41. The rest of the story… The energy of the molecules and their motions are one part of the story – the “thermo part”. There is also the distribution of atoms within the allowed states. It not only matters what the average energy of the system is, but which molecules have what energies and what positions!

42. The rest of the story… …is entropy (S) - is a measure of the distribution of states. Entropy is sometimes defined as “disorder” or “randomness”. It is really more complicated than that and represents the total number of different micro-states available to the system.

43. “States” of a system O2O2 O2O2 O2O2 O2O2

44. O2O2 O2O2 O2O2 O2O2

45. O2O2 O2O2 O2O2 O2O2

46. Imagine 3 molecules! O2O2 O2O2 O2O2 O2O2 O2O2 O2O2

47. O2O2 O2O2 O2O2 O2O2 O2O2 O2O2

48. Imagine a MOLE of MOLECULES!!! O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 BIG EFFING MESS!!!! THANK GOD FOR STATISTICS

49. Entropy is… …a state function. Entropy gets handled much the same as enthalpy. There are tables of entropy values, and it is usually the change ( S) that matters more than the absolute amount.

50. Some examples What has more entropy: 1 mole of water or 1 mole of steam? Why? 1 mole of steam – the molecules in steam are not associated with each other and are, therefore, free to explore more positions and energy states!

51. Some examples What has more entropy: 1 mole of water or ½ mole of water mixed with ½ mole of methanol? Why? The mixture – there are the same number of molecules in both systems, but the mixture allows for more possible distributions of the molecules!

52. Clicker question In the following process what is the sign of S : 2 H 2 (g) + O 2 (g)  2 H 2 O (g) A. Positive B. Negative C. Zero D. I don’t know, I’m just glad to be here.

53. Clicker If delta S is negative, and that is the only consideration, should the reaction: A. Happen B. Not Happen C. I don’t know D. You are a beautiful animal E. Your mother

54. Suppose I want an exact number? Appendix II

56. The Laws of Thermodynamics 1 st Law – Conservation of Energy 2 nd Law – The Entropy of the universe is always increasing for spontaneous changes. 3 rd Law – A perfect crystal at 0 K has no entropy.

57. 3 rd Law The third law is interesting (and important). Unlike enthalpy, we have an absolute zero for entropy. That’s why Appendix II shows S values not S values.

59. Suppose I want an exact number? 2 H 2 (g) + O 2 (g)  2 H 2 O (g) S 0 = ΣS 0 products – ΣS 0 reactants S 0 = 2*S 0 (H 2 O(g)) – [2*S 0 (H 2 (g)) + S 0 (O 2 (g))] S 0 = 2mol*188.8 J/mol*K – [2mol*130.7 J/mol*K + 1 mol*205.2 J/mol*K] S 0 = - 89 J/K

60. From a thermodynamic standpoint Delta S = - 89 J/K is… A. Good B. Bad C. Indifferent D. You are still a beautiful animal.

61. 2 H 2 (g) + O 2 (g)  2 H 2 O (g) S 0 rxn = -89.0 J/K Again, this assumes stoichiometric quantities of everything. Does this number make sense? What does a NEGATIVE change in entropy mean?

62. 2 H 2 (g) + O 2 (g)  2 H 2 O (g) S 0 rxn = -89.0 J/K What does a NEGATIVE change in entropy mean? S 0 rxn = S (products) – S (reactants)<0 S(products) < S(reactants) # of product states < # of reactant states Does this make sense? ABSOLUTELY – there are fewer molecules so there are fewer states!

63. Does this number make sense? 2 H 2 (g) + O 2 (g)  2 H 2 O (g) S 0 = - 89 J/K S is negative meaning… The products have LESS entropy than the reactants. 3 moles of gas vs. 2 moles of gas 2 different gases vs. 1 gas

64. 2 particles vs. 3 particles! H2H2 O2O2 H2H2 H2OH2O H2OH2O

65. 2 H 2 (g) + O 2 (g)  2 H 2 O (g) So, why does it happen if S is negative? H is also negative (H 0 = -241.8 kJ/mol) – the lazy man wins this one!

66. Thermodynamics is… All about balancing H and S to determine “spontaneity”.

67. Spontaneous change A spontaneous change is one that happens “naturally”, without being forced by an outside agent. Spontaneous change: Water evaporating at room temperature. A rock rolling down hill. Non-spontaneous change: Freezing water at room temperature. Rolling a rock uphill.

68. Spontaneous change A spontaneous change is thermodynamically favorable.

69. Spontaneous change Thermodynamics is all about balancing enthalpy and entropy. Some processes are enthalpically and entropically favorable. Some process are enthalpically and entropically unfavorable. What about when one property is favorable and the other is unfavorable?

70. Balancing entropy and enthalpy Gibb’s Free Energy:  G =  H -T  S If  G >0 then reaction is NOT spontaneous. If  G <0 then reaction IS spontaneous If  G =0 then…the reaction is at equilibrium – more later!

71. Balancing entropy and enthalpy Gibb’s Free Energy:  G =  H -T  S 4 possibilities:  H is -,  S is -  H is -,  S is +  H is +,  S is +  H is +,  S is -

72. Balancing entropy and enthalpy  G =  H -T  S  H is -,  S is -  H is -,  S is + this is the best!!!  H is +,  S is +  H is +,  S is - this is the worst!!!

73. Balancing entropy and enthalpy  G =  H -T  S  H is -,  S is + this is the best!!! G will ALWAYS be negative:  = (-) – (+)(+)

74. Balancing entropy and enthalpy  G =  H -T  S  H is +,  S is - this is the worst!!! G will ALWAYS be positive:  = (+) – (+)(-) = (+)+(+)

75. Balancing entropy and enthalpy  G =  H -T  S  H is +,  S is + It’s gonna depend on the temperature.  G= (+ H) – (+T)(+ S) = (+ H)-(+T S) Is T S bigger than or smaller than H?

76. Balancing entropy and enthalpy  G =  H -T  S  H is -,  S is - It’s gonna depend on the temperature.  G= (- H) – (+T)(- S) = (- H)+(+T S) Is T S bigger than or smaller than H?

77. 2 H 2 (g) + O 2 (g)  2 H 2 O (g) S 0 = - 89 J/K H 0 = - 241.8 kJ This reaction is spontaneous at some temperatures, not all!

78. 2 H 2 (g) + O 2 (g)  2 H 2 O (g) S 0 = - 89 J/K H 0 = - 241.8 kJ  G =  H -T  S  G = (-241.8 kJ) – T(-0.089 kJ/K) 0 = (-241.8 kJ) + T(0.089 kJ/K) 241.8 kJ = T(0.089 kJ/K) T = 2717 K

79. 2 H 2 (g) + O 2 (g)  2 H 2 O (g)  G =  H -T  S  G = (-241.8 kJ) – T(-0.089 kJ/K) T = 2717 K When T<2717, the reaction is spontaneous. When T>2717, the T  S term is now bigger than  H and the reaction is no longer spontaneous. This reaction is better at lower temperatures!

80. Clicker Question Who is the most powerful superhero of all? A. Superman B. Batman C. Santa Claus D. Joe E. Anyone but Joe

81. What Holiday are you celebrating? A. Hanukkah B. Christmas C. Kwanzaa D. Christmukkah E. None of the above

82. Question Is the following reaction spontaneous at 298 K? 8 H 2 (g) + S 8 (s)  8 H 2 S (g) If I care about “spontaneous”, I care about G!

83.  G 0 =  H 0 - 298  S 0 But if you look in Appendix II, you’ll find the magic 3 rd column:  G f 0

84. 8 H 2 (g) + S 8 (s)  8 H 2 S (g) From Appendix II H 2 (g) S 8 (s) H 2 S (g)  G f 0 0 kJ/mol 49.7 kJ/mol -33.4 kJ/mol  H f 0 0 kJ/mol 102.3 kJ/mol -20.6 kJ/mol S 0 130.7 J/mol K 430.9 J/mol K 205.8 J/mol K

85. 8 H 2 (g) + S 8 (s)  8 H 2 S (g) G rxn 0 = Σ  G f 0 (products) - Σ  G f 0 (reactants) = 8 mol * -33.4 kJ/mol –[8 mol* 0 kJ/mol+1 mol* 49.7 kJ/mol ] = -267.2 kJ – 49.7 kJ G rxn 0 = -316.9 kJ G rxn 0 <0, reaction is spontaneous.

86.  G 0 =  H 0 - 298  S 0 8 H 2 (g) + S 8 (s)  8 H 2 S (g) H rxn 0 = Σ  H f 0 (products) - Σ  H f 0 (reactants) H rxn 0 = 8*-20.6kJ/mol – [8*0+102.3kJ/mol] H rxn 0 =-267.1 kJ S rxn 0 = Σ S 0 (products) - Σ S 0 (reactants) S rxn 0 =8*205.8 J/molK-[8*130.7 J/molK+430.9 J/molK] S rxn 0 =169.9 J/K

87.  G 0 =  H 0 - 298  S 0 H rxn 0 =-267.1 kJ S rxn 0 =169.9 J/K S rxn 0 =0.1699 kJ/K  G 0 = -267.1 kJ - 298 (0.1699 kJ/K)  G 0 = -317.73 kJ Compare to  G 0 calc using  G 0 f = -316.9 kJ

88. Why the difference? Remember H 0 f and G 0 f are relative. S 0 is absolute. By definition G 0 f is 0 for an element as is H 0 f but not S 0

89. Question Is the following reaction spontaneous at 500 °C? 8 H 2 (g) + S 8 (s)  8 H 2 S (g) If I care about “spontaneous”, I care about G! But now there’s no naught!

90. 8 H 2 (g) + S 8 (s)  8 H 2 S (g) Can we take the short way? G rxn 0 = Σ  G f 0 (products) - Σ  G f 0 (reactants) Won’t work – no naught! Only  G 0 =  H 0 - T  S 0 works if T is not 298.

91.  G 0 =  H 0 - 298  S 0 H rxn 0 =-267.1 kJ S rxn 0 =169.9 J/K S rxn 0 =0.1699 kJ/K Same as before…I only have one Appendix II!!! Can this be right? Yes and no.

92. ASSUMING…  H 0 and  S 0 are temperature independent. Is this a good assumption? Not perfect. We already said that heating a material usually increases its entropy. But it is a better assumption for H and the S has a T term so…it’s the best we can do!

93.  G 0 =  H 0 - T  S 0 H rxn 0 =-267.1 kJ S rxn 0 =169.9 J/K S rxn 0 =0.1699 kJ/K Same as before…I only have one Appendix II!!!  G 0 = -267.1 kJ – (500C+273.15K) (0.1699 kJ/K)  G 0 = -398 kJ Even “more spontaneous” at 500 K.