Appendix B 

Essential Statistical Thermodynamics

Karl K. Irikura

Physical and Chemical Properties Division, 

National Institute of Standards and Technology, Gaithersburg, MD 20899

Some computational methods, particularly ab initio techniques, produce

detailed molecular information but no thermodynamic information

directly. Further calculations are needed to generate familiar, ideal-gas

quantities such as the standard molar entropy (SE), heat capacity (CpE),

and enthalpy change [HE(T)-HE(0)]. This Appendix details the necessary

procedures, including worked examples. Thermochemical calculations

can be extended to transition states of chemical reactions. Procedures are

provided for converting such information into rate constants. Tables are

also provided for unit conversions and physical constants.

Statistical thermodynamics calculations are necessary to compute properties as functions

of temperature. In some computations, such as ab initio electronic calculations of

molecular energy, the raw results do not even correspond to properties at absolute zero

temperature and must always be corrected. All the corrections are based upon molecular

spectroscopy, with temperature-dependence implicit in the molecular partition function,

Q. The partition function is used not only for theoretical predictions, but also to generate

most published thermochemical tables. Many data compilations include descriptions of

calculational procedures (1-3).

Corrections Unique to Ab Initio Predictions

By convention, energies from ab initio calculations are reported in hartrees, the atomic

unit of energy (1 hartree = 2625.5 kJ/mol = 627.51 kcal/mol = 219474.6 cm-1) (4). These

energies are negative, with the defined zero of energy being the fully-dissociated limit

(free electrons and bare nuclei). Ab initio models also invoke the approximation that the

atomic nuclei are stationary, with the electrons swarming about them. This is a good

approximation because nuclei are much heavier than electrons. Consequently, the
ZPE ' 1

2 j

3N&6

i'1

i (1)

H2%F2  2 HF (2)

resulting energies are for a hypothetical, non-vibrating molecule. Although oscillators

may be at rest in classical mechanics, real (quantum-mechanical) oscillators are always in

motion. The small residual motion at absolute zero temperature is the zero-point

vibrational energy, abbreviated ZPVE or ZPE. For a simple harmonic oscillator, the ZPE

equals one-half the vibrational frequency. Although all real molecular vibrations are at

least slightly anharmonic, they are usually approximated as harmonic. Thus, the

molecule’s ZPE may be taken as one-half the sum of the vibrational frequencies. In

equation 1, N is the number of atoms in the molecule and the i

 are the fundamental

vibrational frequencies. There are 3N-6 vibrations in a non-linear molecule and 3N-5 in

a linear molecule; equation 1 is for the more common non-linear case. The ZPE must be

added to the raw ab initio energy to obtain an energy corresponding to absolute zero

temperature, T = 0 K.

In practice, the ZPE correction is slightly complicated by the observation that ab

initio vibrational frequencies are often in error by +5% to +10%. To compensate for this

error, the computed frequencies are usually multiplied by empirical scaling factors. The

most recent recommendations are those of Scott and Radom (5). For example, they

suggest scaling HF/6-31G* frequencies by 0.8953 to predict vibrational spectra (i.e.,

fundamental frequencies), by 0.9135 for the computation of ZPEs, by 0.8905 to predict

enthalpy differences HE(298.15) - HE(0), and by 0.8978 to predict SE(298.15). The

methods for computing these quantities are described below. Common abbreviations and

acronyms of the ab initio literature are defined in the glossary (Appendix D) of this book.

In this Appendix, the degree sign (E) that indicates ideality and standard pressure (1 bar)

is omitted except where the thermal electron convention for ions is being emphasized (see

below).

Enthalpies of formation depend upon the thermodynamic conventions for

reference states of the elements. Since this information is not intrinsic to an isolated

molecule, an ab initio reaction energy (i.e., energies for at least two molecules) must be

combined with experimental data to compute an enthalpy of formation, f

HE. 

Example: fHE0 of hydrogen fluoride. There are many levels of approximation in ab

initio theory; several are described in the chapters of this book. For the present example,

we choose the CCSD(T)/aug-cc-pVTZ//B3LYP/6-31G(d) level. The notation indicates

that (1) molecular geometries are calculated at the density-functional B3LYP level using

the 6-31G(d) basis set and (2) molecular electronic energies are calculated at the high

CCSD(T) level of theory using the rather large aug-cc-pVTZ basis set. To compute an

enthalpy of formation for HF, we must also choose a balanced chemical reaction for

which to calculate an energy. We choose arbitrarily the reaction shown in equation 2.

Note that the ideal-gas energy and enthalpy are equal at 0 K, since H = E + PV = E +

nRT = E.
Q(T) ' j

i

exp(& i /kT) (3)

The optimized B3LYP/6-31G(d) bond lengths are 0.743, 1.404, and 0.934 D for

H2, F2, and HF respectively, in reasonably good agreement with the experimentally

derived values re = 0.741, 1.412, and 0.917 D respectively (6). The calculated (harmonic)

B3LYP/6-31G(d) vibrational frequencies are 4451, 1064, and 3978 cm-1 for H2, F2, and

HF respectively, in modest agreement with the experimentally derived harmonic

frequencies e = 4401, 917, and 4138 cm-1 respectively (6). Since 1 cm-1 equals only

0.01196 kJ/mol, small errors in ZPE will not cause significant errors in the final enthalpy

of formation. Scaling the calculated frequencies by 0.9806 (5) and substituting them into

equation 1 yields ZPE = 2182, 522, and 1950 cm-1 = 0.009943, 0.002377, and 0.008887

hartree for H2, F2, and HF respectively. At these optimized geometries, the

CCSD(T)/aug-cc-pVTZ energies are Ee = -1.172636, -199.313519, and -100.349402

hartree for H2, F2, and HF respectively. Adding the ZPEs thus leads to enthalpies (or

energies) at T = 0 K of E0 = -1.162693, -199.311142, and -100.340515 hartree for H2,

F2, and HF respectively. The calculated enthalpy change is then r

HE0(reaction 2) =

-0.207194 hartree = -544.0 kJ/mol. Using the experimental (defined, in these cases)

enthalpies of formation for H2 and F2 of 0 and 0 kJ/mol (1), we obtain f

HE0(HF) =

-272.0 kJ/mol. This is in good agreement with the experimental value of -272.5 ± 0.8

kJ/mol (1). All the ab initio calculations for this example were done on a personal

computer.

General Relationships of Statistical Thermodynamics

In the present context, statistical thermodynamics is meant to include the methods used

to convert molecular energy levels into macroscopic properties, especially enthalpies,

entropies, and heat capacities. Molecular energy levels arise from molecular translation

(i.e., motion through space), rotation, vibration, and electronic excitation. This

information constitutes the spectroscopy of the molecule of interest and can be obtained

experimentally or from calculations.

Partition Function. The molecular energy levels i

 are used to compute the molecular

partition function, usually denoted by the symbol Q, as shown in equation 3. The sum

extends over all energy levels. (Sometimes this sum is written only over all unique energy

levels, in which case a level degeneracy gi

 must be included in the sum.) However, for

very high temperatures at which the molecule becomes unstable, the extent of the sum

may be ambiguous. Tabulated thermochemical data must be used with caution under such

conditions; the values (1) may depend strongly upon the high-energy cutoff procedure

adopted and (2) may deviate implicitly from the ideal-gas model.

One typically chooses the lowest energy level to be the zero of energy, so that no

levels lie at negative energies. From equation 3 it follows that the largest contributions

to Q are from the lowest energy levels. Conversely, levels that lie far above kT (207 cm-1

at room temperature) have only a minor effect on Q and its derivative thermodynamic

quantities.
S ' Nk M

MT

(TlnQ)&lnN%1 (4)

Cv ' NkT M2

MT 2

(TlnQ) (5)

C

p ' Cv

%R (6)

H(T) ! H(0) ~ m

T

0

C

p

dT ' RT 2

Q

MQ

MT

%RT (7)

M

MT

(TlnQ) ' lnQ %

T

Q

MQ

MT (8)

M2

MT2

(TlnQ) ' 2

Q

MQ

MT

%

T

Q

M2Q

MT 2 & TQ 2

MQ

MT

2

(9)

MQ

MT ' 1

kT 2j

i

i exp(& i /kT) (10)

M2Q

MT 2 ' &2

T

MQ

MT

%

1

k 2T 4j

i

2

i exp(& i /kT) (11)

Thermodynamic Functions. Given the partition function, the usual molar

thermodynamic functions can be calculated based upon the following general equations.

Equation 4 is for the entropy, equation 5 for the heat capacity at constant volume,

equation 6 for the heat capacity at constant pressure, and equation 7 for the enthalpy

difference relative to absolute zero temperature. N is Avogadro’s number (6.022137 ×

1023), k is the Boltzmann constant (1.38066 × 1023 J/K), and the ideal-gas constant R /

Nk (4). The last two terms inside the brackets in equation 4 arise from the

indistinguishability of identical molecules, which requires a factor of (1/N!) in the partition

function for the ensemble. Expressions 4-7 may more easily be evaluated using equations

8-11 for the various derivatives.

Practical Calculations

A complete set of molecular energy levels is almost never available. To simplify the

problem, one usually adopts a model in which translation, rotation, vibration, and

electronic excitation are uncoupled. In other words, one makes the approximation that

the different types of motion are unaffected by each other and do not mix together. This

leads to a separability of Q into four factors that correspond to separate partition

functions for translation, rotation, vibration, and electronic excitation. This is shown in

equation 12, where the explicit dependence upon temperature has been dropped for
Q ' QtransQrotQvibQelec (12)

Qtrans ' (2 mkT)

3/2h &3 V (13)

Strans ' R[(3/2) ln(2 m/h 2

) % (5/2) ln(kT) & ln(p) % 5/2] (14)

C

p, trans ' (5/2) R (15)

simplicity. When electronically excited states are considered, one often assumes that the

translational, rotational, and vibrational spectra of the excited state are the same as those

of the ground electronic state. This is crude, but is convenient when no other information

is available. Moreover, if the excited state lies far above kT, the final results will not be

sensitive to such details.

Different energy units are used conventionally in the fields of molecular

spectroscopy, quantum chemistry, and thermochemistry. To provide some feeling for

magnitudes, the values of the thermal energy kT, at “room temperature” (298.15 K) and

at 1000 K, are listed in Table I in several units. In this Appendix, all units are of the SI

(Système International: kg, m, s, Pa, K) unless otherwise indicated.

Table I. Thermal energy (kT) at two temperatures, expressed in various units

Unit “Room

temperature” 1000 K

kelvin (K) 298.15 1000

wavenumber (cm-1) 207.2 695.0

Hertz (s-1) 6.212 × 1012 2.084 × 1013

kJ/mol 2.479 8.314

kcal/mol 0.592 1.987

electron volt (eV) 0.0257 0.0862

hartree (atomic unit) 0.000944 0.003167

Translational Partition Function. Rigorously, Qtrans must be calculated from a sum

over all the translational energy levels that are available to a molecule confined to a cubic

box of volume V = RT/p (molar volume of an ideal gas at temperature T and pressure p).

This is seldom done. Instead, the sum is approximated as an integral to obtain equations

13-16. This approximation is good as long as m3/2T 5/2p-1 » h3(2 )

-3/2k-5/2 (3). At the

standard pressure p = 1 bar = 105

 Pa = 0.986923 atm, this condition is met for sufficiently

heavy molecules, m (in amu) » 11.4 T -5/3, and for sufficiently high temperatures, T » 4.31

m-3/5 (m expressed in amu). Fortunately, this covers the conditions of common chemical

interest. For an atomic ideal gas, there is no vibrational or rotational motion.
[H(T)&H(0)]trans ' (5/2) RT (16)

Qrot ' 8 2

h 3 (2 kT)

3/2 (IA IBIC)1/2 ' (kT/h)3/2 (ABC)&1/2 1/2 &1 (17)

Q linear

rot ' 8 2 IkT

h 2 ' kT/( hB) (18)

As an example, we can calculate the standard entropy for neon ideal gas at T =

298.15 K. The atomic mass is converted to SI units using the equivalence NA amu =

0.001 kg, where NA = 6.022 × 1023 mol-1 is the Avogadro constant. Thus for 20Ne (m =

19.992 amu), m = 3.320 × 10-26 kg. The values of the physical constants are h = 6.626

× 10-34 J s, k = 1.381 × 10-23 J K-1, and R = kNA = 8.3145 J mol-1 K-1 (7,4). The standard

pressure is p = 105

 Pa. Substituting these values into equation 14 yields S298.15(

20Ne) =

146.21 J mol-1 K-1. For 21Ne (m = 20.994 amu), S298.15(

21Ne) = 146.82 J mol-1 K-1, and for

22Ne (m = 21.991 amu), S298.15(22Ne) = 147.40 J mol-1 K-1. Averaging these values using

the natural abundances of 90.48%, 0.27%, and 9.25%, respectively (7,8), we find S298.15

= 146.32 J mol-1 K-1 for the naturally occurring isotopic distribution. This agrees well

with the accepted value of 146.33 J mol-1 K-1 (9).

Rotational Partition Function. The free rotation of a rigid molecule is also quantized

(the angular momentum and its projection are integer multiples of h/2 ), so the rotational

energy is restricted to certain discrete levels. Rotational spectra are characterized by the

constants A, B, and C, where A / h/(8 2IA) and likewise for B and C. The quantities IA,B,C

are the principal moments of inertia of the molecule, with the convention IA # IB # IC (or

A $ B $ C). Many computer programs, including ab initio packages, report the rotational

constants when provided with a molecular geometry. The moments can also be calculated

manually as the eigenvalues of the inertial tensor, which has elements like

I and , where the index i runs over all atoms in

xy ' &j mi xi yi Ixx ' %j mi (y

2

i %z

2

i )

the molecule and the coordinate origin is at the center of mass. Linear molecules (IA = 0)

are described by a single rotational constant, B, and a single moment of inertia, I. Details

may be found in textbooks of molecular spectroscopy. 

Fortunately, at high enough temperatures (kT » hA), the sum can be replaced by

an integral as it is for translation. In the general case, the rotational partition function is

given by equation 17. For linear molecules, equation 18 should be used instead. In these

and subsequent equations, the symbol denotes the “rotational symmetry number” or

“external symmetry number” for the molecule. This is the number of unique orientations

of the rigid molecule that only interchange identical atoms. It preserves parity restrictions

on the interchange of identical nuclei when summation is replaced by integration.

Identifying the correct symmetry number is a common point of difficulty; it is discussed

further below.

For the typical case (equation 17), the thermodynamic functions are given by

equations 19-21. For linear molecules (equation 18), equations 22-24 are used instead.
Srot ' R[ln(8 2

/ ) % (3/2) ln(2 kT/h 2) % (1/2) ln(IAIB IC) % 3/2]

' R[(3/2) ln(kT/h) & (1/2) ln(ABC/ ) & ln( ) % 3/2] (19)

C

p, rot ' (3/2) R (20)

[H(T)&H(0)]rot ' (3/2) RT (21)

S linear

rot ' R[ln(8 2

IkT/ h 2) %1] ' R[ln(kT/ hB) %1] (22)

Clinear

p, rot ' R (23)

[H(T)&H(0)]linear

rot ' RT (24)

External Symmetry Number. Some computer programs, such as many ab initio

packages, determine the molecular symmetry and external symmetry number ( )

automatically. If such a program is unavailable, may be determined by hand. With

practice, this becomes very fast. 

If you are familiar enough with group theory to identify the molecule’s point

group (10), then can be determined from Table II (11). Without identifying the point

group, one can count manually the number of orientations of the rigid molecule that

interchange only identical atoms.

Table II. Symmetry numbers corresponding to symmetry point groups

Group Group Group Group

C1, Ci, Cs, C4v 1 D4h 2 T, Td 12 Oh 24

Ih 60 Sn n/2 Cn, Cnv, Cnh n Dn, Dnh, Dnd 2n

For example, the benzene molecule (C6H6) belongs to the D6h point group. From

Table II, = 12. Alternatively, one can draw the molecule as a hexagon with numbered

vertices. Rotating the drawing by n × 60E, where n runs from 0 to 5, generates six

different orientations that are distinguished only by the artificial numbering of the vertices.

Each of these six orientations can be flipped over to generate another orientation, for a

total of 12 unique orientations, = 12.

Another example is methyl chloride, CH3Cl. This belongs to the C3v point group,

so = 3. Alternatively, one can artificially number the hydrogen atoms and see that there

are three unique orientations, related by rotations of n × 120E (n = 0-2) around the C-Cl

bond axis.
Qvib ' k

i

1&e &h i

/kT &1

(25)

Svib ' &R j

i

ln 1&e &h i

/kT

% R j

i

h i

kT

e &h i

/kT

1&e &h i

/kT (26)

C

p, vib ' Rj

i

h i

kT

2

e &h i

/kT

1&e &h i

/kT 2 (27)

[H(T)&H(0)]vib ' RT j

i

h i

kT

e &h i

/kT

1&e &h i

/kT (28)

Chlorobenzene, C6H5Cl, belongs to the C2v point group, so = 2. Alternatively,

one can again number the hydrogen atoms and see that there are two unique orientations,

related by rotations of n × 180E (n = 0-1) around the C-Cl bond axis. In contrast, toluene

(C6H5CH3) belongs to the Cs point group, so = 1. There are no ways to rotate or flip

the molecule rigidly that will leave it unchanged. Allowing the methyl group to rotate

leads to an internal symmetry number which is discussed below, following the section on

internal rotation.

Vibrational Partition Function. To complete the simple rigid-rotator/harmonic

oscillator (RRHO) model, one must consider the molecular vibrations. As indicated in

the discussion of ZPE (equation 1), a molecule that contains N atoms has 3N-6 vibrational

frequencies (3N-5 for linear molecules). The partition function is given in equation 25,

where the product runs over all vibrational frequencies i. The corresponding

thermodynamic functions are given by equations 26-28.

Example: Hydrogen Fluoride. Earlier we used the results of ab initio

calculations to obtain a value for fHE0(HF). The other equations above permit us to

compute ab initio thermodynamic functions, which will provide an enthalpy of formation

at the more useful temperature of 298.15 K. Results are summarized in Table III. For

simplicity, we will neglect the naturally occurring heavy isotopes of hydrogen. The

molecular weight of 1

H19F is 20.006 amu. Using equation 14, as done above for neon,

leads to Strans = 146.22 J mol-1 K-1. HF is a linear molecule, so we use equation 22 to

calculate Srot. The ab initio calculation reports a rotational constant B = 605.64 GHz =

6.0564 × 1011 s-1 based upon the calculated B3LYP/6-31G(d) equilibrium geometry and

the most common isotopes. This molecule belongs to the C4v point group ( = 1); there

are no identical nuclei that can be interchanged by any rotation. Hence Srot = 27.67 J mol-1

K-1. For the vibrational contribution, we scale the B3LYP/6-31G(d) frequency of 3987

cm-1 by 1.0015 as suggested for entropies (5) to obtain = 3993 cm-1. This is multiplied

by the speed of light, c = 2.998 × 1010 cm s-1 (7,4), to convert wavenumbers to SI

frequency units, = 1.197 × 1014 s-1. Thus h /kT = 19.27 and equation 26 yields Svib =

7.22 × 10-7 J mol-1 K-1. The total entropy is S298.15 = Strans + Srot + Svib = 173.89 J mol-1 K-1,

in good agreement with the accepted value of 173.78 J mol-1 K-1 (9). 
For enthalpy and heat capacity, the B3LYP/6-31G(d) frequency is scaled by

0.9989 (5) to obtain = 3983 cm-1. The heat capacity Cp(HF) is calculated using

equations 15, 23, and 27, leading to Cp = Cp, trans + Cp, rot + Cp, vib = (5/2)R + R + 1.38 × 10-

5

 J mol-1 K-1 = 29.10 J mol-1 K-1. This compares well with the accepted value of 29.14

J mol-1 K-1 (1). Finally, the enthalpy difference can be computed using equations 16, 24,

and 28 to be [H(298.15)-H(0)] = (5/2) RT + RT + 2.14 × 10-4 J mol-1 = 8.68 kJ mol-1.

This can be used to compute f

HE298.15(HF) = fHE0(HF) + [H(298.15)-H(0)]HF -

[H(298.15)-H(0)]elements. Taking the ab initio value f

HE0(HF) = -272.0 kJ/mol from

above, the calculated enthalpy difference of 8.68 kJ/mol for HF, and the accepted

enthalpy differences of (8.47)/2 and (8.83)/2 kJ/mol for 1⁄2H2 and 1⁄2F2 (9), we obtain

f

HE298.15(HF) = -272.0 kJ/mol, in agreement with the accepted value of -273.3 ± 0.7 kJ

mol-1 (9) [-272.5 ± 0.8 kJ/mol is listed in older ref (1)].

Table III. Results for Hydrogen Fluoride Example

Contribution S, J/(mol@K) Cp, J/(mol@K) [H(298.15)-H(0)], kJ/mol

Translation 146.22 20.79 6.20

Rotation 27.67 8.31 2.48

Vibration 7 × 10-7 1 × 10-5 2 × 10-4

Total 173.89 29.10 8.68

Electronic Partition Function. Although they may not have low-lying electronic excited

states, some molecules have degenerate electronic ground states. Free radicals are a

common example. They may have unpaired electrons in their electronic ground states and

a net electron spin of S = nunpaired/2, where nunpaired is the number of unpaired electrons.

(Beware not to confuse the spin quantum number S with the entropy.) The multiplicity,

or degeneracy g, of such a state is g = (2S+1). Using degeneracy numbers is equivalent

to an explicit count of all states, including degenerate ones. Thus, Qelec = g is a constant

and only affects the entropy: Selec = R ln(g) and Cp, elec = [H(T)-H(0)]elec = 0. Since most

free radicals have only a single unpaired electron, the usual effect is to increase the

entropy by Rln(2). In addition to spin degeneracies, some states have spatial

degeneracies. This situation is most common for diatomic molecules. Linear molecules

with a spatial symmetry other than G (e.g., or ) have a spatial degeneracy of 2. For

example, the OH radical has a 2 ground state, so its degeneracy is g = 2 (spin) × 2

(spatial) = 4. If there are both spin and spatial degeneracies, spin-orbit coupling lifts the

degeneracy, often significantly. In the example of OH, the 4-fold degenerate ground state

is split into two doubly-degenerate levels separated by 139.2 cm-1 (6). In such a case the

low-lying excited states should be included in the calculation of thermodynamic quantities.

The partition function is given by equation 29, where i and gi are the excitation energies
Qelec ' j gi exp(& i /kT) (29)

Selec ' Rln j gi

e & i

/kT

% R j gi

( i/kT)e & i

/kT

j gi

e & i

/kT (30)

C

p, elec ' R j gi

( i/kT)

2 e & i

/kT

j gi

e & i

/kT & R j gi

( i/kT) e & i

/kT

j gi

e & i

/kT

2

(31)

[H(T)&H(0)]elec ' RT j gi

( i/kT) e & i

/kT

j gi

e & i

/kT (32)

(spectroscopic T0) and degeneracies of the excited states, g0 and 0 / 0 are for the ground

state, and the sum runs over all the electronic states being considered, including the

ground state. The contributions to the thermal functions are given by equations 30-32.

This treatment assumes, rather crudely, that the rotations and vibrations are unaffected

by electronic excitation.

Example: Entropy of Methyl Radical. For simplicity, we again neglect minor

isotopes. Results are summarized in Table IV. The molecular weight of CH3 is then m

= 15.023 amu, so Strans = 142.65 J mol-1 K-1. This is a flat, triangular molecule that

belongs to the D3h point group, = 6. The experimental bond length is re = 1.0767 D =

1.0767 × 10-10 m (12-14). The moments of inertia can be evaluated using the symmetry

of this oblate top, or more generally by diagonalizing the inertial tensor. We place the

molecule in the yz plane with one hydrogen atom on the z axis. The center of mass

coincides with the carbon atom. The cartesian coordinates then lead to an inertial tensor

with components Ixx = 3mHre

2

, Ixy = Iyx = 0, Ixz = Izx = 0, Iyy = (3/2)mHre

2

, Iyz = Izy = 0, and

Izz = (3/2)mHre

2

 = 2.910 × 10-47 kg m2. This is already diagonal, with eigenvalues IA =

IB = 2.910 × 10-47 kg m2 and IC = 5.820 kg m2 so that Srot = 43.50 J mol-1 K-1 (equation

19). The observed vibrational frequencies of CH3 are 3004.4, 606.5, 3160.8, and 1396

cm-1 for 1, 2, 3, and 4 respectively (15). Since 3 and 4 are both doubly degenerate

(eN symmetry), they are counted twice and we have the correct number of vibrations, 3N6 = 6. Converting to SI units leads to Svib = 6.51 × 10-5 + 1.84 + 6.42 × 10-5 + 0.15 =

1.99 J mol-1 K-1. In this case, this is a radical with one unpaired electron, S = 1⁄2

(electronic ground state is 2A2O), so the degeneracy g = 2 and Selec = Rln(2). Adding the

four contributions to the entropy gives S298.15 = 193.9 J mol-1 K-1, in agreement with the

literature value of 194.2 ± 1.3 J mol-1 K-1 (1).

Example: Entropy of Hydroxyl Radical. For simplicity, we again neglect

minor isotopes. Results are summarized in Table IV. The molecular weight of OH is

then m = 17.003 amu, so Strans = 144.19 J mol-1 K-1. Again using the simple RRHO model,

the observed bond length is re = 0.96966 D (6) and the symmetry number = 1 (C4v point
Qfree rotor ' ( int h)

&1 8 3

Iint kT 1/2 (33)

Iint ' Itop&I

2

top

2

IA

%

2

IB

%

2

IC

(34)

group). Hence I = 1.480 × 10-47 kg m2

 and Srot = 28.22 J mol-1 K-1. Using the observed

vibrational fundamental for the vibrational frequency, = 0 = 3568 cm-1 (6) leads to Svib

= 5.04 × 10-6 J mol-1 K-1. If spin-orbit splitting is ignored, Qelec = 4 as explained above,

so Selec = 11.53 J mol-1 K-1. Combining these four contributions yields S = 183.9 J mol-1

K-1. If instead the spin-orbit splitting is included, so that g0 = g1 = 2 and 1 = 139.2 cm-1

= 2.765 × 10-21 J, then equation 30 yields Selec = 11.08 J mol-1 K-1 and so the total S =

183.5 J mol-1 K-1. The literature value is 183.71 ± 0.04 J mol-1 K-1 (1).

Table IV. Methyl and Hydroxyl Examples, SE298, J/(mol@K)

Contribution Methyl Hydroxyla Hydroxylb

Translation 142.65 144.19 144.19

Rotation 43.50 28.22 28.22

Vibration 1.99 5 × 10-6 5 × 10-6

Electronic 5.76 11.53 11.08

Total 193.9 183.9 183.5

a

Spin-orbit splitting ignored. bSpin-orbit splitting included.

Internal Rotation. This refers to torsional motion, most commonly involving methyl

groups. There are three ways to treat such a rotor, depending upon its barrier to rotation.

The free and hindered rotor models require that an internal symmetry number, int, be

included. int equals the number of minima (or maxima) in the torsional potential energy

curve. The harmonic oscillator model does not require int because it ignores all but one

of the energy minima. For intermediate barrier heights (hindered rotor), the

appropriateness of an internal symmetry number may be confusing. In such cases, avoid

over- or under-counting states by ensuring that the limiting case of infinite barriers

(harmonic oscillator model, no int) moves smoothly into the limiting case of zero barrier

(free rotor model, int needed) as the barrier height decreases. Note that the vibrational

frequency corresponding to the torsion must be deleted if the torsion is treated as a free

or hindered rotation.

Free Rotor. If the barrier to rotation is much less than kT, then the rotor may be

considered freely rotating. For a symmetric rotor such as a methyl group, the partition

function is given by equation 33, where Iint is the reduced moment of inertia for the

internal rotation and is given by equation 34 (3). Asymmetric rotors can be treated using
Sfree rotor ' R[(1/2) ln(8 3

Iint kT) & ln( int h) %(1/2)] (35)

C

p, free rotor ' (1/2) R (36)

[H(T)&H(0)]free rotor ' (1/2) RT (37)

an appropriate formula for the reduced moment (3). In equation 34, Itop is the moment

of inertia of the rotating fragment about the axis of internal rotation. This is expressed

as , where the mi

 are atomic masses, ri I is the distance of atom i from the top ' j mi r

2

i

axis of internal rotation, and the sum runs over all atoms in the rotating fragment. The

quantities , , and are the cosines of the angles formed between the internal rotation

axis and the principal axes of the overall molecule that correspond to IA, IB, and IC,

respectively. Contributions to the thermodynamic functions are given in equations 35-37.

Harmonic Oscillator. If the barrier to internal rotation is much greater than kT,

one can consider the torsion to be a non-rotating, harmonic oscillator. Treatment is the

same as for other vibrations.

Hindered Rotor. This is the common, intermediate case, when the torsional

barrier V is comparable to kT. If the torsional potential is assumed to have the simple

form , then the tables of Pitzer and Gwinn are usually used to U( ) ' V(1&cos int )/2

compute the contribution of the hindered rotor to the thermodynamic functions (16,17).

Their tables are in terms of the dimensionless variables x and y, where and x'V/(kT)

and Iint y' is defined as for a free rotor (see above). int h(8 3Iint kT)

&1/2

Example: Entropy of Ethane at T = 184 K [adapted from ref (17)]. Results

are summarized in Table V. Ignoring minor isotopes as before, for C2H6 we have m =

30.047 amu, so Strans = 141.26 J mol-1 K-1. The experimental geometry is staggered (D3d

point group, = 6) and defined by rCC = 1.535 D, rCH = 1.094 D, and CCH = 111.2E (12-

14). If we choose coordinates so that the origin is at the center of mass, the carbon atoms

lie on the z-axis, and the yz plane is a reflection plane of symmetry, then the elements of

the inertial tensor are, in (amu D2), Ixx = Iyy = 25.46, Ixy = Iyx = Ixz = Izx = Iyz = Izy = 0, and

Izz = 6.291 amu D2

. This is already diagonal, with eigenvalues IA = 1.045 × 10-46 kg m2

and IB = IC = 4.228 × 10-46 kg m2 so that Srot = 62.17 J mol-1 K-1 (equation 19). The

observed vibrational frequencies (18) are 2954, 1388, 995, 289 (torsion), 2896, and 1379

cm-1 (non-degenerate), and 2969, 1468, 1190, 2985, 1469, and 822 cm-1 (doubly

degenerate), for a total of 3N-6 = 18 vibrational modes and Svib = 3.36 J mol-1 K-1 (0.25

J mol-1 K-1 excluding the torsional mode). The total entropy in the RRHO model is thus

SE184 = 206.8 J mol-1 K-1, which is below the experimental value S = 207.7 ± 0.6 J mol-1

K-1 (17).

If we consider the torsion to be a free, unhindered rotor, then we require the

corresponding reduced moment of inertia. In this case, in the rotor axis is aligned with

the A axis of the molecule, so that = 1 and = = 0 in equation 34. This gives Iint = Itop

- Itop

2

/IA. The symmetry of this molecule requires that IA = 2Itop, because the moment of
f HEthermal electron convention ' f Hion convention % (5/2)qRT (38)

inertia of the whole molecule around the A axis (viz., the C-C bond axis) is twice that of

a single methyl group. Thus Iint = 2.613 × 10-47 kg m2

. The internal symmetry number is

int = 3, since there are three equivalent values of the torsion angle (0E, 120E, and 240E).

Equation 35 yields Sfree rotor = 10.09 J mol-1 K-1, for a total entropy of SE184 = 213.8 J mol-1

K-1 in the free-rotor model, which is higher than the experimental value.

To apply the hindered-rotor model we need a value for the torsional barrier height.

This can be estimated from the observed torsional vibrational frequency (in s-1) as V .

8 2

Iint

2

/ int

2

 = 1.720 × 10-20 J (or 10.4 kJ/mol). Thus the parameters are x = 6.77 and y

= 0.490. Interpolating within the standard tables (16,17), Shindered rotor = 3.99 J mol-1 K-1,

so that the total entropy is S = 207.7 J mol-1 K-1, in agreement with the experimental

value.

Table V. Ethane Example, SE184 (J mol-1 K-1)

Contribution Harmonic Rotor Free Rotor Hindered Rotor

Translation 141.26 141.26 141.26

Rotation 62.17 62.17 62.17

Vibration 0.25 0.25 0.25

Torsion 3.11 10.09 3.99

Total 206.8 213.8 207.7

Charged Molecules: Two Conventions. The balanced chemical equation describing

an ionization process involves at least one free electron. There are two major conventions

for the thermodynamic properties of the electron. Most compilations of thermochemical

data adopt the thermal electron convention. In this convention, the free electron is

treated as a chemical element, so that its ideal-gas enthalpy of formation is zero at all

temperatures. In contrast, most of the literature in mass spectrometry and ion chemistry

adopts the ion convention, sometimes also called the stationary electron convention. In

this convention, the enthalpy content of the electron is ignored. At absolute zero

temperature there is no difference between the two conventions, but in general enthalpies

of formation under the two conventions are related by equation 38, where q is the

(signed) charge on the ion in question (±1 in most cases). A thorough discussion is

provided in the introduction to the GIANT tables (19). When reporting the

thermochemistry of ions, it is important always to indicate which convention is being

used. Especially beware not to combine enthalpies of formation that were derived using

different conventions.
r(T) ' kT

h exp & G ‡RT

(39)

Ea ' RT 2 M

MT

(lnr) ' H ‡ % MRT (40)

A ' r(T) exp(%Ea /RT) (41)

Chemical Kinetics

The equilibrium constant for a reaction is Keq = exp(- G/RT), where G = H - T S and

the differences are between the reactants and products, e.g., S = Sproducts - Sreactants.

Simple transition-state theory for chemical kinetics assumes that the reaction rate is

limited by formation of a transient transition state, which is the point of maximum energy

along the path from reactants to products. The transition state is considered to be in

quasi-equilibrium with the reactants. If differences between reactants and the transition

state are denoted with a double dagger, e.g., S‡

 = STS - Sreactants, then the rate constant

(denoted r here to avoid confusion with the Boltzmann constant) is given by equation 39.

As for stable species, G‡

 = H‡ - T S‡. Thus rate constants can be calculated easily

from the “thermochemistry” for transition states. In such calculations, the imaginary

vibrational frequency is ignored, so that there are only 3N-7 molecular vibrations in the

transition structure (3N-6 if linear). If all internal and external symmetry numbers are

included in the rotational partition functions, then any reaction path degeneracy will

usually be included automatically. Occasionally, however, stereochemical factors are also

needed (20).

Experimental, temperature-dependent rate constants are often presented as an

Arrhenius plot of r(T) vs. 1/T. This is motivated by the observation that such plots are

nearly linear, r(T) = A exp(-Ea/RT), where the pre-exponential factor A is usually called

simply the A-factor and Ea is the phenomenological activation energy. It is often

desirable to report A and Ea in computational studies, for comparison with the values

derived from experimental data. They may be determined using equations 40 and 41,

where M is the molecularity of the reaction (e.g., M = 1 for a unimolecular and M = 2 for

a bimolecular reaction). The derived A and Ea are weakly temperature-dependent. This

is consistent with experimental results, which are often fitted using the three-parameter

modified Arrhenius expression . This functional form leads to r(T)'A ) T n exp(&E )

a /RT)

a better fit than the ordinary Arrhenius expression, but the parameters may have little

physical interpretation. If the RRHO approximation is accepted, then the three

parameters are given by equations 42-44, where the yi in equation 42 are the reduced

vibrational frequencies, yi = h i/kT. However, in practice it is often best to determine the

parameters AN, n, and by fitting calculated rate constants to the modified Arrhenius E )

a

expression.
n ' 1 % j

TS

y ‡

i e &y ‡

i 1%e &y ‡i

1&e y ‡

i

2 & j

reactants

yie &yi 1%e &yi

1&e yi 2 (42)

E )

a ' RT 2 M

MT

(lnr) & nRT ' H ‡ % (M&n)RT (43)

A ) ' r(T) T &n exp(%E )

a /RT) (44)

Ftunnel ' 1%

1

24

h ‡/kT 2

(45)

Ab initio energies are now precise enough that it is becoming common to use

kinetic theories more sophisticated than simple transition-state theory. When the reaction

coordinate is dominated by motion of a hydrogen atom, corrections for quantummechanical tunneling are often made (21). The simplest is the Wigner correction, which

requires only the imaginary vibrational frequency ‡

i associated with the reaction

coordinate. To apply this correction, the calculated rate is multiplied by Ftunnel (equation

45). Better results may be obtained by fitting the energy profile of the reaction to an

Eckart potential function (see the chapter by Petersson in this book). The location of the

transition state can also be refined. Variational transition-state theory defines the

transition state as the maximum in the free energy along the reaction path, instead of the

maximum along the vibrationless potential energy curve. Such a definition is essential for

reactions such as simple bond cleavage, which usually has no barrier in excess of the bond

energy.

Units and Constants

In actual calculations, many practical difficulties involve incompatible units. In addition

to the standard units of the SI, many others are in use, usually for historical reasons.

Conversion factors among selected units are provided in Table VI. For convenience, the

values of commonly-used constants are collected in Table VII. Detailed information is

available on-line at http://www.physics.nist.gov/PhysRefData/contents.html#SI.
Table VI. Unit Conversions

Quantity Unit Conversiona SI Unit

energy hartree (atomic unit) 2 625.500 kJ/mol

energy cal 4.184 J

energy cm-1 (wavenumber) 0.011 962 66 kJ/mol

energy eV 96.485 31 kJ/mol

energy K (temperature) 8.314 511 × 10-3 kJ/mol

distance D 10-10 m

distance bohr (atomic unit) 5.291 772 × 10-11 m

mass amu or u 1.660 540 × 10-27 kg

pressure bar 105 Pa

pressure atm 101 325 Pa

pressure Torr or mm-Hg 133.322 37 Pa

pressure (density) cm-3 (at 298.15 K; ideal gas) 4.166 43 × 10-15 Pa

pressure (density) cm-3 (arb. temp.; ideal gas) 106 kT Pa

pressure (density) M or mol/L (ideal gas) 103 RT Pa

dipole moment atomic unit 8.478 358 × 10-30 C@m

dipole moment D (debye) 3.335 641 × 10-30 C@m

a

Multiply the quantity expressed in the units of column 2 by the conversion factor in

column 3 to obtain the quantity expressed in units of column 4 (SI units).

Table VII. Physical Constants

Quantity Value

k 1.380 66 × 10-23 J K-1

NA 6.022 137 × 1023 mol-1

R = kNA 8.314 510 J mol-1 K-1

h 6.626 076 × 10-34 J s

c 299 792 458 m s-1
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