Rotational Correction

Interpretation of THERM rotational correction

References

This summary is based mainly on two papers:

• Hydrogen Atom Bond Increments for Calculation of Thermodynamic Properties of Hydrocarbon Radical Species, Tsan H. Lay, Joseph W. Bozzelli, Anthony M. Dean, and Edward R. Ritter, J. Phys. Chem. 1995, 99, 39, 14514–14527: This paper outlines the original implementation of THERM.

• Energy Levels and Thermodynamic Functions for Molecules with InternalRotation I. Rigid Frame with Attached Tops, Kenneth S. Pitzer and William D. Gwinn, J. Chem. Phys. 10, 428 (1942): This paper used for the table for conversion of Moments of Inertia (Ir) and Torsion Barriers (V) to entropy.

Information and Summary found in THERM paper

This summary follows the example in the THERM paper.

In THERM and THERGAS, to calculate the thermodynamics of a radical, they calculate the parent molecule and then examine the loss/gain (among other things) due to the formation of the radical. If you look at Table 5 of Bozzelli's THERM, you can see the corrections due to the formation of the radical (he collects all these corrections in one HBI term --- these are in Table 5).

For calculation of CCJ, see Table 5, the first set, there is a loss of the CH3-CH3 rotor and a gain of the CH3-CH2. rotor. The Ir and V value should be used to calculate the correction to the Entropy (and heat capacities):

• rotor - CH3-CH3:: -3.864 cal/(mol K)

• rotor - CH3-CH2: 4.756 cal/(mol K)

From the THERM paer (pg 14518 - Section 'Hindered Internal Rotations'):

The differences in the entropy of internal rotation between parent (molecule) and daughter (radical) resulting from changes in rotational barriers and moments of inertia are usually the major components of the entropy increment in HBI groups. The method and tables of Pitzer and Ginn are used to calculate the contribution of hindered internal rotations to the thermodynamic functions SO298 and Cp(T);

From the THERM paper (pg 14518 - Section 'Hindered Internal Rotations'):

The barriers of hindered internal rotations for stable molecules and free radicals considered in this work are listed in Table 4.

From Table 4: Moments of Inertia (Ir) and Torsion Barriers (V) for the hindered rotations about single bonds:

CH3-CH3 Ir=1.6 V=2.9

RCH2-CH3 Ir=3.0 V=3.3

CH3-CH2 Ir= 1.2 V=0.1

RCH2-CH2 Ir=0.1 V=2

From the Appendix of the THERM paper (pg 14526):

There is no simple equation to express S(int-rot) and Cp(int-rot). These parameters are calculated using the method and tables of Pitzer and Gwinn . Two parameters are needed to utilize the tables of Pitzer and Gwinn in this calculation of S(int-rot) and CP(int-rot): V (actually V/RT) and l/Qf. V is the potential barrier, R is the ideal gas constant, T is temperature, and Qf is the partition function of free rotation and is given by

Q, = (8 (pi)**3 Ir k T)**(1/2)/(hn)

where Z, is the reduced moment of inertia, k is Boltzmann's constant, and n is the number of potential minima per revolution. The calculation of Qf and the interpolation of Pitzer and Gwinn's tables to obtain the entropy and heat capacity values are implemented (automated) by a computer code. Table IV (in ref 32 for entropy decrease from classical free rotation is incorporated into the code and used to calculate the entropy values. Table VI therein is used to calculate the heat capacities.

In Benson's book, in the notes of Table A.16, the formula for Q is given as:

Q = 3.6/(sigma)* (Ir T/100)**2

For CH3-CH3 and CH3-CH2 this is 2.62 and 2.26, respectively. This give 1/Qf values of 0.38 and 0.29, respectively.

'Reproduction' of example given in the Appendix of THERM paper

From the Appendix of THERM paper (pg 145526):

Using CH3-CH3 as an example, Ir of this symmetric coaxial CH3 rotor can be calculated from

lr = 1/I1 + 1/I2

where I1 and I2 are the moments of inertia of each methyl top. Ir is 1.6 amu A*, and Qf is then calculated as 2.62. With the rotational barrier equal to 2.9 kcal/mol, the entropy contribution is calculated, from use of the tables, as 3.86 cal/(mol K) at 298K. The reduced moment of inertia for rotation of the CH2 group in the ethyl radical is calculated as 1.2 amu A2, and with the barrier of 0.1 kcal/mol, the internal rotational entropy contribution at 298 K is obtained as 4.76 cal/(mol K).

My Calculations for CH3-CH3

In my own calculations for CH3-CH3, Ir=1.6, which means Q=2.62 (same as in the above in the THERM paper). 1/Q is 0.38.

V is 2.9 (from Table 4 of THERM paper). V/RT=2900/(1.9869*298)=4.90

Using Table III (TABLE III. Entropy, S (cal. per degree mole). ) in Gwinn and Pitzer to translate to S value. Extrapolating between the table values, we get (S-Sf)=0.657:

• Q=0.40 V/RT=5.0 (S-Sf)=1.235

• Q=0.35 V/RT=5.0 (S-Sf)=1.281

• Q=0.40 V/RT=4.5 (S-Sf)=1.116

• Q=0.35 V/RT=4.5 (S-Sf)=1.153

• 1.235 - (1.281-1.235)*( (0.40-0.38)/(0.40-0.35))=1.263

• 1.116 - (1.153-1.116)*( (0.40-0.38)/(0.40-0.351.234))=1.138

• 1.263 - (1.263-1.138)*( (5.0-4.9)/(5.0-4.5) )=

• 1.236*3=3.70

However, this does come close to the value of 3.86 (from THERM paper: the entropy contribution is calculated, from use of the tables, as 3.86 cal/(mol K) at 298K.), but it is not exact. In addition, the number 3.86/3=1.29 is outside the range of numbers possible for the values of 1/Qf=0.38 and V/RT=4.90.

Same Calculation for CH3-CH2

For CH3-CH2 Ir=1.2 which means:

Q=Q = 3.6/(sigma)* (Ir T/100)**(1/2) = 3.6/6*(1.2 *298/100)**(1/2) = 1.135,

Q=Q = 3.6/(sigma)* (Ir T/100)**(1/2) = 3.6/2*(1.2 *298/100)**(1/2) = 3.404,

giving 1/Qf=0.88 (or 0.293 for sigma=2).

V is 0.1 (from Table 4 of THERM paper).

V/RT=100/(1.987*298)=0.169

The value from the table for 1/Q=0.88 V=0.169 looks like it is close to zero (even when multiplied by 6).

(The value from the table for 1/Q=0.293 V=0.169

This is not close to the value given in Table 5 of the THERM paper of 4.756. The value close to zero is closer to the THERGAS assumption that the only contribution is loss of the rotation from the parent.

Calculation of RCH2CH3

For RCH2-CH3 Ir=3.0 which means:

Q=Q = 3.6/(sigma)* (Ir T/100)**(1/2) = 3.6/3*(3.0 *298/100)**(1/2) = 3.588 ,

giving 1/Qf= 0.279

V is 3.3 (from Table 4 of THERM paper).

V/RT=3300/(1.987*298)=5.573

The point is: 1/Q=0.279 V=5.573

From the table

1/Q=0.4 V=6.0 E=1.444 1.444*3 = 4.332

1/Q=0.35 V=5.0 E=1.218 1.218*3 = 3.654

This is sort of close to the value given in Table 5 of the THERM paper of 4.283.

Backtracking

From the Table 5 of THERM paper

CH3CH3 Erot = 3.864 sigma=3

CH3CH2 Erot= 4.756 sigma=6

RCH2CH3 Erot = 4.283 sigma = 3

RCH2CH2 Erot = 5.183 sigma =2

This means from the (V/RT,Qf) tables there should be a value of:

CH3CH3 3.864/3=1.288

CH3CH2 4.756/6=0.7927

Sf=4.6+R/2 ln(Ir*T/(n**2 300) from table A.18 in Benson

CH3CH3 Ir=1.6 V=2.9 n=3 Sf= 3.468

CH3CH2 Ir=1.2 V=0.1 n=6 Sf= 3.386

RCH2CH3 Ir=3.0 V=3.3 n=3 Sf= 3.502

RCH2CH2 Ir=1.8 V=0.1 n=2 Sf= 3.800