1. Fundamental relations: The thermodynamic functions The molecular partition function Using statistical thermodynamics Mean energies Heat capacities Equation of state Residual entropies Equilibrium constants Chapter 20 Statistical thermodynamics: the machinery

2. Exercises for Chapter 20 20.1(b), 20.3(a), 20.6(a), 20.10(b), 20.12(a), 20.15(b),20.17(a) 20.3, 20.6, 20.10, 20.16, 20.19

3. Fundamental relations The thermodynamic functions The molecular partition function

4. The thermodynamic functions: A and p The Helmholtz energy Independent molecules: (Distinguishable) (Indistinguishable) A= U - TS A(0)=U(0) The pressure

5. Thermodynamic Riddles (Ch.5) dU=TdS-pdV, dH=TdS+Vdp dG=Vdp-SdT, dA=-SdT-pdV Enjoy!! U=q+w, H=U+pV, G=H-TS, A=U-TS

6. Deriving an equation of state Derive the expression for the pressure of a gas of independent particles For a gas of independent particles: Where following relations have been used:

7. Classroom exercise Deriving the equation of state of a sample for which With where f depends on the volume

8. The thermodynamic function: H H=U+pV  For a gas of independent particles:

9. The thermodynamic functions: G G=H-TS=A+pV  For a gas of independent particles: G=A+pV  Or Define molar partition function

10. The molecular partition function Translational, Rotational, Vibrational, Electronic energies

11. The translational contribution At room temperature, O 2 in a vessel of 100 ml At room temperature, H 2

12. Independent states (  factorization of q) Three-dimensional box: Thermal wavelength (Translational partition function)

13. Typical Rotors

14. The Rotational Energy Levels Around a fixed-axis Around a fixed-point (Spherical Rotors)

15. Linear Rotors

16. The rotational contribution hcB/kT=0.05111 At room temperature, for H(1)Cl(35), B=10.591 1/cm kT/hc=207.22 1/cm (Linear rotors)

17. The rotational contribution: Approximation (Linear rotors) hcB/kT<<1

18. Symmetric Rotors

19. The rotational contribution: Approximation (Symmetric rotors) hcB/kT<<1

20. J K 0 1 2 3 4 J K 0 1 2 3 4 =

21. The rotational contribution: Approximation (Symmetric rotors)

22. The rotational contribution: Approximation (Asymmetric rotors) (If you have question here, just ignore it. The detailed derivation of this equation is beyond the Scope of this course.)

23. Rotational temperature The ``high temperature`` approximation means From Table 20.1, it is clear that this approximation is indeed valid unless the temperature is not too low (< ~10K).

25. Symmetry number How to avoid overestimating the rotational partition function? Symmetrical linear rotor After deducting the indistinguishable States, In general,

26. Symmetry number Nonlinear molecules: General cases: the number of rotational symmetry elements.

30. Symmetry Group and Symmetry Numbers C1C1 CICI CS:CS:1 D2D2 D2dD2d D2h:D2h:4 C : 1 C2C2 C2vC2v C2h:C2h:2 D3D3 D3dD3d D3h:D3h:6 D : 2 C3C3 C3vC3v C3h:C3h:3 D4D4 D4dD4d D4h:D4h:8 T, T h T d :12 C4C4 C4vC4v C4h:C4h:4 D5D5 D5dD5d D5h:D5h:10 O, O h :24 C5C5 C5vC5v C5h:C5h:5 D6D6 D6dD6d D6h:D6h:12 I, I h :60 C6C6 C6vC6v C6h:C6h:6 D7D7 D7dD7d D7h:D7h:14 S4:S4:2 C7C7 C7vC7v C7h:C7h:7 D8D8 D8dD8d D8h:D8h:16 S6:S6:3 C8C8 C8vC8v C8h:C8h:8 S8:S8:4

31. A=4.828 1/cm, B=1.0012 1/cm, C=0.8282 1/cm, T=298 K ABC=4.0033 1/cm

32. Classroom exercise N A=0.2014 1/cm, B=0.1936 1/cm, C=0.0987 1/cm, T=298 K ABC=0.004 1/cm

33. Quantum mechanical interpretation The wavefunction of fermions changes sign when exchanged whereas the wavefunction of bosons does not change sign when exchanged.

34. (for even J) (for odd J)

35. CO 2 Nuclear spin = 0 Only even J-states are admissible CO 2  Boson

36. Quantum mechanical interpretation There are 8 nuclear spin states:

37. Quantum mechanical interpretation

40. Generally, for a molecule with N R rotational elements (including the identity operation), the symmetry number Quantum mechanical interpretation

41. Symmetry Group and Symmetry Numbers C1C1 CICI CS:CS:1 D2D2 D2dD2d D2h:D2h:4 C : 1 C2C2 C2vC2v C2h:C2h:2 D3D3 D3dD3d D3h:D3h:6 D : 2 C3C3 C3vC3v C3h:C3h:3 D4D4 D4dD4d D4h:D4h:8 T, T h T d :12 C4C4 C4vC4v C4h:C4h:4 D5D5 D5dD5d D5h:D5h:10 O, O h :24 C5C5 C5vC5v C5h:C5h:5 D6D6 D6dD6d D6h:D6h:12 I, I h :60 C6C6 C6vC6v C6h:C6h:6 D7D7 D7dD7d D7h:D7h:14 S4:S4:2 C7C7 C7vC7v C7h:C7h:7 D8D8 D8dD8d D8h:D8h:16 S6:S6:3 C8C8 C8vC8v C8h:C8h:8 S8:S8:4

42. 10 points!! Derive from quantum mechanics the symmetry number of benzene:

43. The vibrational contribution v ＝ 0, 1, 2,

45. Normal modes 3N-6 vibrational degrees of freedom For a nonlinear molecule of N atoms, there are 3N degrees of freedom: 3 translatinal, 3 rotational and For a linear molecule of N atoms, there are 3N degrees of freedom: 3 translatinal, 2 rotational and 3N-5 vibrational degrees of freedom The total vibrational partition function is:

46. Exercise The wave number of the three vibrational modes of H 2 O 3656.7 1/cm, 1594.8 1/cm, and 3755.8 1/cm. Calculate vibrational partition function at 1500 K. The total vibrational partition function then is: 1.031x1.276x1.028=1.353 At 1500 K, most molecules are at their vibrational ground state!

47. Classroom exercise The three vibrational normal modes of CO 2 are 1388 1/cm, 667.4 1/cm (doubly degenerate), 2349 1/cm. Calculate the vibrational partition function at 1500K.

48. Low temperature approximation High temperature approximation Only the zero-point level is occupied.

49. The electronic contribution For most cases, the excited energy is much larger than kT and the electronic energy level of the ground is not degenerate:

50. Degenerate case: NO

52. The overall partition function elec vib rot (Linear rotor, Single vib mode) (Nonlinear rotor, multiple vib mode)

53. m M=N A *m

54. Exercise Calculate the value of molar Gibbs energy for H2O(g) at 1500 K given that A=27.8788 1/cm, B=14.5092 1/cm, and C=9.2869 1/cm and the information of normal modes given in last exercise.

55. Classroom Exercise Calculate the value of molar Gibbs energy for CO2(g) at 1500 K given that B=0.3902 1/cm. The three vibrational normal modes of CO2 are 1388 1/cm, 667.4 1/cm (doubly degenerate), 2349 1/cm

56. Classroom Exercise Calculate the value of molar Gibbs energy for CO2(g) at 1500 K given that B=0.3902 1/cm. The three vibrational normal modes of CO2 are 1388 1/cm, 667.4 1/cm (doubly degenerate), 2349 1/cm

57. II. Using statistical thermodynamics Mean energies Heat capacities Equation of state Residual entropies Equilibrium constants

58. Mean energies M=T,R,V,orE The mean translational energy 1D case: 3D case:

59. The mean rotational energy (Linear rotors)

61. The mean rotational energy At high temperature, the partition function of a linear rotor: Classroom exercise: the mean rotational energy of an asymmetric rotor at high temperature:

62. The mean vibrational energy

64. Heat capacities Translational contribution:

65. The rotational contribution: At high temperature: Common cases

66. At low temperature: The vibrational contribution: Common cases Rare cases: Cv,m  R

67. The overall heat capacity At fairly high temperature, the vibrational contribution Is cloase to zero: Common cases: θ R <<T<<θ V

68. The overall heat capacity (diatomic molecules) Common cases

69. Exercise Estimate the molar constant-volume heat capacity of water vapor at 100C. The wave number of the three vibrational modes of H 2 O: 3656.7 1/cm, 1594.8 1/cm, and 3755.8 1/cm. The rotational constants of H2O: A=27.9 1/cm, B=14.5 1/cm And C=9.3 1/cm. Common cases: θ R <<T<<θ V Experimental value: 26.1 J/mol/K

70. Classroom Exercise Estimate the molar constant-volume heat capacity of gaseous I 2 at 25C. The rotational constants of I 2 : B=0.037 1/cm. A rare case: θ V ~ T

71. Classroom Exercise Estimate the molar constant-volume heat capacity of gaseous I 2 at 25C. The rotational constants of I 2 : B=0.037 1/cm. A rare case: θ V ~ T, C v,m  R Experimental value: 29.06 J/mol/K

72. Equations of state Equation of state of perfect gas: For real gases, The purpose: to find expressions for B and C in terms of the intermolecular interactions. For perfect gases, N! should be dropped for systems of distinguishable particles.

73. Deriving an equation of state Derive the expression for the pressure of a gas of independent particles For a gas of independent particles: Where following relations have been used:

74. Pair interactions Further approximation:

75. Mayer function

76. Second virial coefficient B

77. Second virial coefficient B (spherical potential) The interaction potential depends on distance only.

78. Second virial coefficient B (Hard sphere potential) when r ≦ σ, E p = ∞: when r ≧ σ,E p = 0: σ

79. Application of virial coefficient B For perfect gases, all virial coefficients except the first one are zero. The thermodynamic properties that depend on intermolecular interactions are determined by second and higher order virial coefficients. (Classroom Exercise)

80. Answer

81. Residual entropy The entropy (disorder) at temperature zero. AB BAAB BA S=0 S>0 There are 2^N microscopic states for N molecules with two equally possible orientations at T=0 

82. General cases ABC ACBABC BAC There are s equally possible orientations at T=0:

83. The residual entropy of ice HOH H2O is nonlinear Hydrogen bonds are oriented. The acceptable arrangement: Out of the four hydrogen atoms around an oxygen atom, two are close and two are far.

84. There are sixteen (2^4) equally possible configurations for ice, but only six of them are allowed.

85. Equilibrium constants (gas-phase reactions) For species J, aA+bB+ …  cC+dD+ …

86. Standard reaction Gibbs energy G(0)=U(0) 

87. X 2 (g)→2X(g) A dissociation equilibrium (dissociation energy of bond X-X)

88. Equilibrium constant: an example Evaluate the equilibrium constant for the dissociation Na 2 (g)  2Na(g) at 1000 K with data: B=0.1547 1/cm, v=159.2 1/cm, D 0 =70.4 kJ/mol. The Na atoms have doublet ground terms.

89. Classroom exercise Evaluate the equilibrium constant for the dissociation Na 2 (g)  2K(g) at 1500 K with data: B=0.1547 1/cm, v=159.2 1/cm, D 0 =70.4 kJ/mol. The Na atoms have doublet ground terms.

90. Contributions to the equilibrium constant: density of states Gas-phase reaction: The density of states== The number of states in a given range of energy:

91. Similar densities of states for products and reactants The equilibrium is dominated by the species with lower zero- point energy.

92. Very different densities of states The equilibrium is dominated by the species with larger density of states.

93. Contributions to the equilibrium constant The number ratio of product to reactant molecules: which leads to the equilibrium constant:

94. Proof

95. Energy separation and state density on equilibria Gibss energy, rather than enthalpy, controls position of equilibrium.

96. Helix structure of polypeptides (proteins)

97. Helix-coil transition h c

98. The statistical physics of helix-coil transition: a tetrapeptide hhhh hhch hchh chhh hhhc cchh chch hchc hcch chhc hhcc cchc chcc hccc ccch cccc (stability parameter)

99. Helix-coil transition of proteins with n amino acid residues hhhhhchhchhh … hhhhhchhchhhhhhhcchhcchhhhhhhchcchhhhhhhhchhchhh Zipper model:

100. Zipper model hhhhhchhhchhhhhhhchhcchhhhhchchhcchhhccchchhcchh The coils are necessarily in a contiguous region.

101. Zipper model hhhhhchhhhhhhhhhhchhchhh Nucleation with equilibrium constant σ<<1. The number of possibilities of placing a coil with i amino acids In a peptide of n amino acids: n-i+1 cchhhchhhhhhhcchhchhhhhhhhcchchhhhhh

103. Degree of conversion (Classroom exercise)

104. Zimm-Bragg model hhcchchhhhcc Separate coil regions are considered