Presentation Transcript

1. Chemistry 231

   Spontaneity of Chemical and Physical Processes: The Second and Third Laws of Thermodynamics
2. What Is Thermodynamics? 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.
3. What Is Spontaneity? 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!
4. Thermodynamic Definition Spontaneous Process – the process occurs without outside work being done on the system.
5. Physical Statements of the Second Law 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
6. The First Law of Thermodynamics The First Law - conservation of energy changes. U = q + w The First Law tells us nothing about the spontaneous direction of a process.
7. Entropy We will look at a new property (the entropy). Entropy is the reason why salts like NaCl (s), KCl (s), NH4NO3(s) spontaneously dissolve in water.
8. The Solution Process Low entropy High entropy The formation of a solution is always accompanied by an increase in the entropy of the system! For the dissolution of KCl (s) in water KCl (s)  K+(aq) + Cl-(aq)
9. The Carnot Engine An imaginary engine
10. The Four Steps of the Carnot Engine Isothermal Expansion (P1, V1, Th)  (P2, V2, Th) Adiabatic Expansion (P2, V2, Th)  (P3, V3, Tc) Isothermal Compression (P3, V3, Tc)  (P4, V4, Tc) Adiabatic Compression (P4, V4, Tc)  (P1, V1, Th)
11. First Law for the Carnot Engine Cyclic Process U = 0 qcycle = -wcycle qcycle = q1 + q3 wcycle = w1 + w2 + w3 + w4
12. The Efficiency of Any Thermal Process The Carnot engine represents the maximum efficiency of a thermal process.
13. The Efficiency of the Carnot Engine The thermal efficiency of the Carnot engine is a function of Th and Tc
14. The Carnot Heat Pump 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.
15. The Coefficient of Performance The coefficient of performance of the Carnot heat pump quantity of heat delivered to the high temperature reservoir per amount of work required.
16. The Coefficient of Performance Two definitions
17. The Carnot Refrigerator Use a Carnot cycle as a refrigerator. Extract heat from the cold temperature reservoir (inside) and deliver it to the high temperature reservoir (outside).
18. The Coefficient of Performance Again, two definitions
19. Mathematical Definition of Entropy The entropy of the system is defined as follows
20. Entropy Is a State Variable Changes in entropy are state functions S = Sf – Si Sf = the entropy of the final state Si = the entropy of the initial state
21. The Fundamental Equation of Thermodynamics Combine the first law of thermodynamics with the definition of entropy.
22. The Properties of S In general, we can write S as a function of T and V
23. Isochoric Changes in S Examine the first partial derivative
24. The Temperature dependence of the Entropy Under isochoric conditions, the entropy dependence on temperature is related to CV
25. Entropy changes Under Constant Volume Conditions For a macroscopic system For a system undergoing an isochoric temperature change
26. Isothermal Changes in S Examine the second partial derivative
27. Isothermal Changes in S For a reversible, isothermal process From the first law
28. The Ideal Gas Case For an isothermal process for an ideal gas, U = 0
29. The Ideal Gas Case (Finally) The entropy change is calculated as follows
30. The General Case We will revisit this equality later For a general gas or a liquid or solid
31. Properties of S(T,P) In general, we can also write S as a function of T and P
32. Rewriting the Entropy The entropy of the system can also be rewritten
33. Rewriting (cont’d) From the definition of enthalpy
34. Rewriting (cont’d) From the mathematical consequences of H
35. Isobaric Changes in S Examine the first partial derivative
36. The Temperature dependence of the Entropy Under isobaric conditions, the entropy dependence on temperature is related to CP
37. Entropy changes Under Constant Pressure Conditions For a macroscopic system For a system undergoing an isobaric temperature change
38. Isothermal Changes in S(T,P) Examine the second partial derivative
39. Isothermal Changes in S(T,P) Under isothermal conditions
40. The Ideal Gas Case For an isothermal process for an ideal gas, (H/ T)p = 0
41. The Ideal Gas Case (Finally) The entropy change is calculated as follows
42. The General Case For a general gas or a liquid or solid
43. Phase Equilibria At the transition (phase-change) temperature only tr = transition type (melting, vapourization, etc.) trS = trH / Ttr
44. The Second Law of Thermodynamics The second law of thermodynamics concerns itself with the entropy of the universe (univS). univS unchanged in a reversible process univS always increases for an irreversible process
45. What is univS? univS = sysS + surrS sysS = the entropy change of the system. surrS = the entropy change of the surroundings.
46. How Do We Obtain univS? We need to obtain estimates for both the sysS and the surrS. Look at the following chemical reaction. C(s) + 2H2 (g)  CH4(g) The entropy change for the systems is the reaction entropy change, rS. How do we calculate surrS?
47. Calculating surrS Note that for an exothermic process, an amount of thermal energy is released to the surroundings!
48. A small part of the surroundings is warmed (kinetic energy increases). The entropy increases!
49. Calculating surrS Insulation Note that for an endothermic process, thermal energy is absorbed from the surroundings!
50. A small part of the surroundings is cooled (kinetic energy decreases). The entropy decreases! For a constant pressure process qp = H surrS surrH surrS -sysH
51. The entropy of the surroundings is calculated as follows. surrS = -sysH / T For a chemical reaction sysH = rH surrS = -rH/ T
52. Adiabatic Processes For an adiabatic process, q = 0!! There is no exchange of thermal energy between the system and surroundings! surrS = -qrev / T = 0
53. Adiabatic Processes and univS We can make the following generalizations for an adiabatic process univS is unchanged for an adiabatic, reversible process univS always increases for an adiabatic, irreversible process The entropy of the system can never decrease during an adiabatic process!
54. Entropy of mixing Before X Vb Valve Va Mixing of two gases (and two liquids) is an common example of an irreversible process
55. After Valve V2= Va + Vb After the valve is opened!
56. The Entropy Changes For gas 1 For gas 2
57. The Mixing Entropy Spontaneous mixing process - mixS > 0 The total change in entropy for the two gases
58. Statistical Entropy kB – the Boltzmann constant (R/NA)  - the thermodynamic probability The Boltzmann probability
59. The Lattice Gas Model N cells Before expansion Allow a gas to expand from one small container to an extremely large container
60. N’ cells After
61. The Expansion From N to N’ Cells Calculating the entropy change
62. The Third Law of Thermodynamics 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.
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!
64. The Third Law of Thermodynamics 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.
65. Third Law Entropy Calculations Assuming a constant pressure heating For any system, we can write the following for the entropy change between two temperatures 0 and T1.
66. Heat Capacities at Low Temperatures metals nonmetals This equation is valid to  15 K The Debye ‘T-cubed’ law
67. Higher Temperatures Above 15 K, the heat capacity data are usually available
68. Adding in Phase Changes For a phase change between 0 – T1, we add in the appropriate entropy change.
69. The entropy changes of all species in the thermodynamic tables are calculated in this manner
70. The Entropy Change in a Chemical Reaction Burning ethane! C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (l) The entropy change is calculated in a similar fashion to that of the enthalpies
71. Finding S Values Units for entropy values  J / (K mole) Temperature and pressure for the tabulated values are 298.2 K and 1.00 atm.
72. Some Generalizations For any gaseous reaction (or a reaction involving gases). g> 0, rS > 0 J/(K mole). g < 0, rS < 0 J/(K mole). g = 0, rS  0 J/(K mole). For reactions involving only solids and liquids – depends on the entropy values of the substances.
73. Finding S Values Note – entropy values are absolute! Note – the elements have NON-ZERO entropy values! e.g., for H2 (g) fH = 0 kJ/mole (by def’n) S = 130.58 J/(K mole)