Statistical Thermodynamics:
from Molecule to Ensemble
1
http://www.bodybuilding.com/fun/thermodynamics_training.htm
Introduction to thermodynamic state
functions
• Partitional function (q, [q]=?):
Encodes how the probabilities are partitioned
among the different microstates, based on their
individual energies (’sum over states’)
• (Internal) Energy (E, [E]=kJ/mol):
ΔE = w + q
• Enthalpy (H, [H]=kJ/mol):
H ≡ E + pV = E + RT
• Entropy (S, [S]=J/molK):
a measure of the number of specific ways in
which a thermodynamic system may be arranged
(’measure of disorder’)
• Gibbs Free energy (G, [G]=kJ/mol):
G ≡ H - TS
maximum energy can be attained only in a
completely reversible process (’chemical
potential’ ’available energy’)
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State Function?
Independently:
fi
3
Get the feeling of
Energy conversion
Sedentary
Glucose
C6H12O6 + 6 O2 = 6 CO2 + 6 H2O
0
∆𝑐𝑜𝑚𝑏 𝐻298.15𝐾 = −2805 kJ/mol
0
∆𝑐𝑜𝑚𝑏 𝐻298.15𝐾 = −15.6 kJ/g
Estimated Energy
Requirements (kJ per day)
Year
Female
Male
2-3
4602
4602
4-5
5021
5230
6-7
5439
5858
8-9
5858
6276
10-11
6276
7113
12-13
7113
7950
14-16
7322
9623
17-18
7322
10251
19-30
7950
10460
31-50
7531
9832
51-70
6904
8996
71+
6485
83684
Statistical
thermodynamics
• 1 Molecule
𝐸(𝑇) = 𝑁𝑘𝐵 𝑇 2
𝜕𝑙𝑛𝑞
𝜕𝑇
𝑞(𝑉, 𝑇)
𝜕𝑙𝑛𝑞
𝑆 𝑇 = 𝑁𝑘𝐵 + 𝑁𝑘𝐵 𝑙𝑛
+𝑁𝑘𝐵 𝑇
𝑁
𝜕𝑇
𝑉
𝑉
• 1 Mol
5
Statistical
thermodynamics
• 1 Molecule
𝐸(𝑇) = 𝑁𝑘𝐵 𝑇 2
𝜕𝑙𝑛𝑞
𝜕𝑇
𝑞(𝑉, 𝑇)
𝜕𝑙𝑛𝑞
𝑆 𝑇 = 𝑁𝑘𝐵 + 𝑁𝑘𝐵 𝑙𝑛
+𝑁𝑘𝐵 𝑇
𝑁
𝜕𝑇
𝑉
𝑉
If we know q and T, then: E(T) and S(T) !!!
• 1 Mol
„Sum over states” Energy distribution of the molecules
Energy
Energy
𝑔𝑖 𝑒 −𝐸𝑖 /𝑘𝑏 𝑇
𝑃 𝑇 =
𝑔𝑖 𝑒 −𝐸𝑖 /𝑘𝑏 𝑇
E(T) and S(T)
+
microstates
How to get microstates?
Boltzmann distribution
H(T) = E(T) + RT
G(T) = H(T) – TS(T)
6
Microstates
• Combintation of Degree of freedom
– External: translation
– Internal: rotation, vibration and electronic
7
Translational states
What we need to know:
Molecular mass
1 molecule
𝑞𝑡𝑟𝑎𝑛𝑠
2𝜋𝑚𝑘𝐵 𝑇
=
ℎ2
3/2
𝑉
𝑞(𝑉, 𝑇)
𝜕𝑙𝑛𝑞
𝑆 𝑇 = 𝑁𝑘𝐵 + 𝑁𝑘𝐵 𝑙𝑛
+𝑁𝑘𝐵 𝑇
𝑁
𝜕𝑇
𝜕𝑙𝑛𝑞
𝐸(𝑇) = 𝑁𝑘𝐵 𝑇 2
𝜕𝑇 𝑉
1 mol
𝑉
Etrans(T)=3/2NkbT=3/2RT
Strans(T)=R[ln(qtrans)+1+3/2]
Etrans(T=298.15K)=3/2RT=3.7 kJ/mol
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https://www.youtube.com/watch?v=mXpfO9WhlPA
Rotational states
What we need to know:
Optimized geometry of the species → Rotational constants (Rigid rotor treatment)
Energy level diagram
≈2x
=1x
1 molecule 𝑞𝑟𝑜𝑡
𝜋
=
𝜎𝑟 2
𝐸(𝑇) = 𝑁𝑘𝐵 𝑇 2
1/2
𝜕𝑙𝑛𝑞
𝜕𝑇
𝑇 3/2
(Θ𝑟,𝑥 Θ𝑟,𝑦 Θ𝑟,𝑧 )1/2
𝑉
𝑞(𝑉, 𝑇)
𝜕𝑙𝑛𝑞
𝑆 𝑇 = 𝑁𝑘𝐵 + 𝑁𝑘𝐵 𝑙𝑛
+𝑁𝑘𝐵 𝑇
𝑁
𝜕𝑇
1 mol
Erot(T)=NkbT=RT
Srot(T)=R[ln(qrot)+1]
𝑉
Erot (T=298.15K)=RT=2.5 kJ/mol
9
Vibrational states
What we need to know:
Optimized geometry of the species
Force constants (k) → harmonic wavenumber (Harmonic oscillator approximation)
1 dimension (diatomics molecule)
1 molecule
𝑞𝑣𝑖𝑏 =
𝐾
1
1 − 𝑒 −Θ𝑣,𝐾 /𝑇
𝑞(𝑉, 𝑇)
𝜕𝑙𝑛𝑞
𝑆 𝑇 = 𝑁𝑘𝐵 + 𝑁𝑘𝐵 𝑙𝑛
+𝑁𝑘𝐵 𝑇
𝑁
𝜕𝑇
𝜕𝑙𝑛𝑞
𝐸(𝑇) = 𝑁𝑘𝐵 𝑇 2
𝜕𝑇 𝑉
𝐄𝐯𝐢𝐛 (𝐓) = 𝐑
1 mol
𝑲 Θ𝒗,𝑲
𝐒𝐯𝐢𝐛 (𝐓) = 𝐑
𝑲
𝟏
𝟐
𝑉
𝟏
+ Θ /𝑻 =𝐙𝐏𝐕𝐄 + 𝐑
𝒆 𝒗,𝑲 −𝟏
Θ𝒗,𝑲 /𝑻
+ 𝒍𝒏(𝟏 − 𝒆Θ𝒗,𝑲 /𝑻 )
𝒆Θ𝒗,𝑲 /𝑻 − 𝟏
𝑲
Θ𝒗,𝑲
𝒆Θ𝒗,𝑲 /𝑻 −𝟏
10
Scaling factors
The vibrational frequencies need adjustment (scale factor) to better match
experimental vibrational frequencies.
This scaling compensates errors from:
(1) Approximation in the solution of the electronic Schrödinger equation.
(2) Harmonic oscillator approach
How to get scaling?
(1) Do it yourself
(2) Find it
(a) http://cccbdb.nist.gov/
(b) literature
Experimental vibrational frequencies (νi),
Theoretical vibrational frequencies (ωi).
Scaling factor (c) and its relative uncertainty (𝒖𝒓 𝟐 ):
ν
(ωi2 (𝑐 − ωi )2 )
(νi ωi)
i
𝑐=
𝑢𝑟 2 =
2
2
(ωi )
(ωi )
c=0.9608
for B3LYP/6-31G(d)
11
Electronic states
Usually it is not considered, but it can be important:
- Spectroscopy
- Species having low lying excitations
Energy
Units
S0/S1/T1
1.223 Å/1.238 Å/1.230Å
1.393 Å/1.408 Å/1.400Å
1.223 Å/1.239 Å/1.230Å
1.478 Å/1.434 Å/1.453Å
1.390 Å/1.409 Å/1.401Å
kJ/mol eV
nm
S1
T1
303.9 3.15
393.6
287.9 2.98
415.5
S0
0
1.386 Å/1.371 Å/1.376Å
1.389 Å/1.370 Å/1.375Å
1.401 Å/1.433 Å/1.426Å
1.397 Å/1.430 Å/1.425Å
1.486 Å/1.393 Å/1.406Å
1.207 Å/1.261 Å/1.299Å
1.112 Å/1.111 Å/1.096Å
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Energy
Jablonski diagram
Jablonski diagram
Potential energy diagram
Energy
Electronic excited state (e.g. S1)
Rotationally, vibrationally and electronically excited state
Rotational ground state, vibrationally and electronically excited state
Rotationally excited, vivrational ground state and electronically excited state
Rotational and vibrational ground state, electronically excited state
microstate
Elecronic ground state (e.g. S0)
Rotationally vibrationally excited state and elecronic ground state
Rotational ground state, vibrationally excited state and elecronic ground state
Rotationally excited, vibrational and elecronic ground state
Rotational, vibrational and elecronic ground state (first molecular state!)
Interatomic distance
13
Thermodynamics terminology
X in energy dimension
E0≡Etot+ZPVE
E°(T)≡Etot+Ethermal
H°(T)=E°(T)+RT
G°(T)≡H°(T)-TS°(T)
Microstates
Zero-point corrected energy
Thermal-corrected energy
Standard enthalpy (pV=nRT!)
Standard Gibbs free energy
Macroscopic properties:
Etot
P(T)
H°(rmin)(T)
ZPVE
P(0K)
Minimum of potential energy surface (Etot(rmin))
E°(rmin)(T)
E0(rmin) (first molecular state!)
G°(rmin)(T)
Interatomic distance
Ethermal(T=0K)=ZPVE
14
Ethermal(T)=Etrans(T)+Erot(T)+Evib(T) (+Eelec(T))
Thermodynamics terminology
X in energy dimension
E0=Etot+ZPVE
E°(T)=Etot+Ecorr
H°(T)=Etot+Hcorr
G°(T)= Etot+Gcorr
Zero-point correction=
Thermal correction to Energy=
Thermal correction to Enthalpy=
Thermal correction to Gibbs Free Energy=
Sum of electronic and zero-point Energies=
Sum of electronic and thermal Energies=
Sum of electronic and thermal Enthalpies=
Sum of electronic and thermal Free Energies=
Total
E (Thermal)
KCal/Mol
537.139
0.799551 ZPVE
0.855985 Ecorr
0.856929 Hcorr
0.699092 Gcorr
-2392.526502
-2392.470067
-2392.469123
-2392.626960
CV
Cal/Mol-Kelvin
206.443
Etot+ZPVE
Etot+Ecorr
Etot+Hcorr
Etot+Gcorr
S
Cal/Mol-Kelvin
332.196
Etot
H°(rmin)(T)
E°(rmin)(T)
Hcorr Ecorr
Gcorr
Minimum of potential energy surface (Etot(rmin))
G°(rmin)(T)
Interatomic distance
15
Reference state?
Different definition of reference state in experiment and theory, conversion needed
Experimental
E(kJ/mol)
Ref (Theory):
Theory
E(Hartree)
0 Hartree
9C6+(g)+8H+(g)+2Cl17+(g) +3O8+ (g)+120e- (g)
93C(g)+82H(g)+22Cl(g) +33O(g) 9184.246kJ/mol
-1490.021129 Hartree
=9·716.68+
8·217.998+
2·121.301+
3·249.18
Ref (Experiment):
9Cgrafit+4H2(g)+Cl2 (g) +1.5O2 (g)
e.g. B3LYP/6-31G(d)
=9·-37.843920+
8·-0.497912+
2·-460.133882+
3·-75.058263
0 kJ/mol
fH°
C9H8Cl2O3
-404.5 kJ/mol
-1493.673273 Hartree
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