1. Spontaneity of Chemical and Physical Processes: The Second and Third Laws of Thermodynamics 1

2. Study of the energy changes that accompany chemical and physical processes. Based on a set of laws. A tool to predict the spontaneous directions of a chemical reaction. 2

3. Spontaneity refers to the ability of a process to occur on its own! Waterfalls “Though the course may change sometimes, rivers always reach the sea” Page/Plant ‘Ten Years Gone’. Ice melts at room temperature! 3

4. Spontaneous Process – the process occurs without outside work being done on the system. 4

5. Kelvin statement Impossible to construct an engine the sole purpose of which is to completely convert heat into work Clausius statement Impossible for heat to flow spontaneously from low temperature to high temperature 5

6. The First Law - conservation of energy changes.  U = q + w The First Law tells us nothing about the spontaneous direction of a process. 6

7. We will look at a new property (the entropy). Entropy is the reason why salts like NaCl (s), KCl (s), NH 4 NO 3 (s) spontaneously dissolve in water. 7

8. For the dissolution of KCl (s) in water KCl (s)  K + (aq) + Cl - (aq) 8 Low entropyHigh entropy The formation of a solution is always accompanied by an increase in the entropy of the system!

9. An imaginary engine 9

10. Isothermal Expansion (P 1, V 1, T h )  (P 2, V 2, T h ) Adiabatic Expansion (P 2, V 2, T h )  (P 3, V 3, T c ) Isothermal Compression (P 3, V 3, T c )  (P 4, V 4, T c ) Adiabatic Compre ssion (P 4, V 4, T c )  (P 1, V 1, T h ) 10

11. Cyclic Process  U = 0 q cycle = -w cycle q cycle = q 1 + q 3 w cycle = w 1 + w 2 + w 3 + w 4 11

12. The Carnot engine represents the maximum efficiency of a thermal process. 12

13. The thermal efficiency of the Carnot engine is a function of T h and T c 13

14. Run the Carnot engine in reverse as a heat pump. Extract heat from the cold temperature reservoir (surroundings) and deliver it to the high temperature reservoir. 14

15. The coefficient of performance of the Carnot heat pump quantity of heat delivered to the high temperature reservoir per amount of work required. 15

16. Two definitions 16

17. Use a Carnot cycle as a refrigerator. Extract heat from the cold temperature reservoir (inside) and deliver it to the high temperature reservoir (outside). 17

18. Again, two definitions 18

19. The entropy of the system is defined as follows 19

20. Changes in entropy are state functions  S = S f – S i S f = the entropy of the final state S i = the entropy of the initial state 20

21. Combine the first law of thermodynamics with the definition of entropy. 21

22. In general, we can write S as a function of T and V 22

23. Examine the first partial derivative 23

24. Under isochoric conditions, the entropy dependence on temperature is related to C V 24

25. For a system undergoing an isochoric temperature change 25 For a macroscopic system

26. Examine the second partial derivative 26

27. From the first law 27 For a reversible, isothermal process

28. For an isothermal process for an ideal gas,  U = 0 28

29. The entropy change is calculated as follows 29

30. For a general gas or a liquid or solid 30 We will revisit this equality later

31. In general, we can also write S as a function of T and P 31

32. The entropy of the system can also be rewritten 32

33. From the definition of enthalpy 33

34. From the mathematical consequences of H 34

35. Examine the first partial derivative 35

36. Under isobaric conditions, the entropy dependence on temperature is related to C P 36

37. For a system undergoing an isobaric temperature change 37 For a macroscopic system

38. Examine the second partial derivative 38

39. Under isothermal conditions 39

40. For an isothermal process for an ideal gas, (  H/  T) p = 0 40

41. The entropy change is calculated as follows 41

42. For a general gas or a liquid or solid 42

43. At the transition (phase-change) temperature only tr = transition type (melting, vapourization, etc.)  tr S =  tr H / T tr 43

44. The second law of thermodynamics concerns itself with the entropy of the universe (  univ S).  univ S unchanged in a reversible process  univ S always increases for an irreversible process 44

45.  univ S =  sys S +  surr S  sys S = the entropy change of the system.  surr S = the entropy change of the surroundings. 45

46. We need to obtain estimates for both the  sys S and the  surr S. Look at the following chemical reaction. C(s) + 2H 2 (g)  CH 4 (g) The entropy change for the systems is the reaction entropy change,  r S . How do we calculate  surr S? 46

47. Note that for an exothermic process, an amount of thermal energy is released to the surroundings! 47

48. A small part of the surroundings is warmed (kinetic energy increases). The entropy increases! 48

49. Note that for an endothermic process, thermal energy is absorbed from the surroundings! 49 Insulation

50. A small part of the surroundings is cooled (kinetic energy decreases). The entropy decreases! For a constant pressure process q p =  H  surr S   surr H  surr S  -  sys H 50

51. The entropy of the surroundings is calculated as follows.  surr S = -  sys H / T For a chemical reaction  sys H =  r H   surr S = -  r H  / T 51

52. For an adiabatic process, q = 0!! There is no exchange of thermal energy between the system and surroundings!  surr S = -q rev / T = 0 52

53. We can make the following generalizations for an adiabatic process  univ S is unchanged for an adiabatic, reversible process  univ S always increases for an adiabatic, irreversible process The entropy of the system can never decrease during an adiabatic process! 53

54. Mixing of two gases (and two liquids) is an common example of an irreversible process 54 X Valve Before VaVa VbVb

55. After the valve is opened! 55 Valve After V 2 = V a + V b

56. For gas 1 56 For gas 2

57. The total change in entropy for the two gases 57 Spontaneous mixing process -  mix S > 0

58. The Boltzmann probability 58 k B – the Boltzmann constant (R/N A )  - the thermodynamic probability

59. Allow a gas to expand from one small container to an extremely large container 59 Before expansion N cells

60. After 60 N ’ cells

61. Calculating the entropy change 61

62. Entropy is related to the dispersal of energy (degree of randomness) of a substance. Entropy is directly proportional to the absolute temperature. Cooling the system decreases the disorder. 62

63. At a very low temperature, the disorder decreases to 0 (i.e., 0 J/(K mole) value for S). The most ordered arrangement of any substance is a perfect crystal! 63

64. The Third Law - the entropy of any perfect crystal is 0 J /(K mole) at 0 K (absolute 0!) Due to the Third Law, we are able to calculate absolute entropy values. 64

65. For any system, we can write the following for the entropy change between two temperatures 0 and T 1. 65 Assuming a constant pressure heating

66. The Debye ‘T-cubed’ law 66 metals nonmetals This equation is valid to  15 K

67. Above 15 K, the heat capacity data are usually available 67

68. For a phase change between 0 – T 1, we add in the appropriate entropy change. 68

69. The entropy changes of all species in the thermodynamic tables are calculated in this manner 69

70. Burning ethane! C 2 H 6 (g) + 7/2 O 2 (g)  2 CO 2 (g) + 3 H 2 O (l) The entropy change is calculated in a similar fashion to that of the enthalpies 70

71. Units for entropy values  J / (K mole) Temperature and pressure for the tabulated values are 298.2 K and 1.00 atm. 71

72. For any gaseous reaction (or a reaction involving gases).  g > 0,  r S  > 0 J/(K mole).  g < 0,  r S  < 0 J/(K mole).  g = 0,  r S   0 J/(K mole). For reactions involving only solids and liquids – depends on the entropy values of the substances. 72

73. Note – entropy values are absolute! Note – the elements have NON-ZERO entropy values! e.g., for H 2 (g)  f H  = 0 kJ/mole (by def’n) S  = 130.58 J/(K mole) 73