Rogers Computational Chemistry Using the PC

The experimental enthalpy of isomerization of cyclopropane is isomH298 ¼8:0 0:2 kcal mol 1. Why is this enthalpy of isomerization negative? Evidently, there is not much difference between computed values of isomE0 and isomH298 as obtained from the thermodynamic equations

isomE0 ¼ [G2 (0 K)(propene)] [G2 (0 K)(cyclopropane)]

isomH298 ¼ [G2 Enthalpy(propene)] [G2 Enthalpy

(cyclopropane)]

We shall discuss this difference in the section on thermodynamic functions below.

The GAMESS Implementation

High-level molecular orbital calculations can be carried out with the freeware program GAMESS [General Atomic and Molecular Electronic Structure System

(Schmidt et al. 1993, 1998)]. Input files can be written from the save command of PCMODEL just as GAUSSIAN input files are. Input files are copied to the working filename INPUT before a run, and output files are designated filename.out. We ran our programs from a separate GAMESS directory. GAMESS is written for professionals, so it is not quite as user friendly as the commercial program GAUSSIAN, but it is by no means beyond the student level.

Exercise 10-9

Run a single point STO-3G calculation of the total energy of H2O at the MM3 geometry in the GAMESS implementation. Compare your result with the identical calculation in the GAUSSIAN implementation. Repeat the calculation using the double zeta valence (DZV) and triple zeta valence (TZV) basis sets in the GAMESS implementations. Comment on the relative energies calculated by single, double, and triple zeta basis sets.

Solution 10-9

Using PCMODEL, draw H2O. Minimize using the MM3 force field. Save to the filename water.inp (or some such) in the GAMESS format. Copy to your GAMESS directory. Copy to filename INPUT and be sure that PUNCH has been renamed to PUNCH.OLD or has been erased entirely. Run using GAMESS.EXE > FILENAME.OUT. The INPUT file

$CONTRL SCFTYP ¼ RHF COORD ¼ CART $END $BASIS GBASIS ¼ STO NGAUSS ¼ 3 $END

$DATA water

leads to

FINAL ENERGY IS 74.9610273642 AFTER 13 ITERATIONS

Repeat in the GAUSSIAN implementation. The .gif file

Double zeta valence or triple zeta valence calculations can be carried out by putting DZV or TZV in place of STO NGAUSS ¼ 3 in the second line of the INPUT file in the GAMESS implementation. The calculated energies become progressively lower (better) for double and triple zeta basis sets

FINAL ENERGY IS 76.00923

FINAL ENERGY IS 76.02007

but they approach a limit.

COMPUTER PROJECT 10-3 j The Bonding Energy Curve of H2: GAMESS

Plot the curve of the bond energy of H2 vs. internuclear distance for the H2 molecule using the STO-3G, double zeta valence (DZV), and triple zeta valence (TZV) basis sets in the GAMESS implementation.

Procedure. Carry out the STO-3G single point calculations on H2 in a way

˚

similar to that of Exercise 10-9 at interatomic distances of 0.4 to 1.2 A at intervals

˚

of 0.1 A. This is best done in the z-matrix format, for example,

$CONTRL SCFTYP ¼ RHF COORD ¼ ZMT $END $BASIS GBASIS ¼ STO NGAUSS ¼ 3 $END

$DATA

hydrogen Cn 2

H1.0

˚

for a bond distance of 1.0 A, where z-matrix format is signified in the input as COORD ¼ ZMT. This input file results in

TOTAL ENERGY ¼ 1.0661086701

as part of the output data block. Double and triple zeta basis sets are obtained by replacing STO NGAUSS ¼ 3 with DZV or TZV.

The Thermodynamic Functions

The output of a G3(MP2) calculation for propene, for example,

G3MP2(0 K) ¼ 117.672791

which we have called E0[G3(MP2)], is the energy of the molecule in the ground state and in the gas phase at 0 K relative to isolated nuclei and electrons. From a theoretical point of view, the problem is solved, but for practical purposes, we would like to have the enthalpy of formation in the standard state f H and the Gibbs free energy of formation in the standard state f G at some other temperature, most importantly, 298 K.

First, we would like to change the reference state from the isolated nuclei and electrons to the elements in their standard states, C(graphite) and H2(g) at 298 K. This leads to the energy of formation at 0 K f E0, which is identical to the enthalpy of formation f H0 at 0 K. The energy and enthalpy are identical only at 0 K. Next we would like to know the enthalpy change on heating propene from 0 to 298 K so as to obtain the enthalpy of formation from the isolated nuclei and electrons elements H298. This we will convert to f H298 from the elements in their standard states at 298 K. From that, with the absolute entropy S of propene at 298, we arrive at the standard Gibbs free energy of formation f G298, which leads to the equilibrium constant at 298 K for reactions involving propene. Classical thermodynamic equations permit these conversions to be carried out in a straightforward way, but because the heat capacities CP and absolute entropies S are usually not known over a wide range of temperatures, the power of statistical thermodynamics is also brought to bear.

The entire procedure can be carried out in steps. We find the ground-state energy of formation of propene at 0 K from C and H atoms in the gaseous state

CðgÞ þ 3HðgÞ ! C3H6ðgÞ

by subtracting the energy of formation of C(g) and H(g) in the ground state at 0 K relative to their constituent isolated nuclei and electrons, from the energy of formation of propene(g) in the ground state relative to its constituent isolated nuclei and electrons (Fig. 10-4).

The result of this calculation is 1:29373 hartrees ¼ 811:82851 kcal mol 1. To this we add the energy of formation of C(g) and H(g) from the elements in the standard state, C(graphite) and H2(g) (Fig. 10-5).

We now know the energy of the propene thermodynamic state {propene(g)} relative to the state {3 C(g) and 6 H(g)} and the energy of the thermodynamic standard state of the elements relative to the same state {3 C(g) and 6 H(g)}, which is opposite in sign to the summed energies of formation of 3 C(g) and 6 H(g). The energy difference between these thermodynamic states is

811:82851 þ 509:94 þ 309:78 ¼ 7:89 kcal mol 1

which is the energy of formation at 0 K f E0 of propene(g) (Fig. 10-6).

3C(g) + 6H(g)

The same series of calculations starting with

G3MP2 Enthalpy ¼ 117.667683

lead to f H ¼ f H298, except that the enthalpies of formation of gaseous atoms are 169.73 kcal mol 1 from C(graphite) and 101.14 kcal mol 1 for H2(gas) at 298.15 K. These give

f H298ðpropeneðgÞÞ ¼ 4:29 ¼ kcal mol 1

as compared to the experimental value of 4:78 0:19 kcal mol 1.

The remaining question is how we got from G3MP2(0 K) ¼ 117.672791 to G3MP2 Enthalpy ¼ 117.667683. This is not a textbook of classical thermodynamics (see Klotz and Rosenberg, 2000) or statistical thermodynamics (see McQuarrie, 1997 or Maczek, 1998), so we shall use a few equations from these fields opportunistically, without explanation. The definition of heat capacity of an ideal gas

qE

CV ¼ qT V

leads to

To evaluate this integral, we must know CV as a function of temperature, and usually this is not known.

Statistical thermodynamics tells us that CV is made up of four parts, translational, rotational, vibrational, and electronic. Generally, the last part is zero over the range 0 to 298 K and the first two parts sum to 5/2 R, where R is the gas constant. This leaves us only the vibrational part to worry about. The vibrational contribution to the heat capacity is

x Evib ¼ RT ex 1

where

x ¼ Evib

kBT

We don’t know Evib but we can approximate it from the vibrational spacing of the bond vibrations in the harmonic oscillator approximation.

When these four (or three) contributions are summed for a molecule such as propene, we have the thermal correction to the energy G3MP2(0 K). The result is G3MP2 Energy in the G3(MP2) output block. To this is added PV, which is equal to RT for an ideal gas, in accordance with the classical definition of the enthalpy

H ¼ E þ PV ¼ E þ RT

The sum of the energy correction for heating the molecule from 0 to 298 K plus RT is called the thermal enthalpy correction (TCH) and yields

G3MP2 Enthalpy ¼ 117.667683

in the output block of the G3(MP2) calculation

G3MP2(0 K) ¼ 117.672791

G3MP2 Energy ¼ 117.668627

G3MP2 Enthalpy ¼ 117.667683

G3MP2 Free Energy ¼ 117.697857

One can obtain THC for the G2 or G3 family of calculations by taking G3MP2 Enthalpy G3MP2(0 K).

From the third law of thermodynamics, the entropy S ¼ 0 at 0 K makes it possible to calculate S at any temperature from statistical thermodynamics within the harmonic oscillator approximation (Maczek, 1998). From this, S of formation can be found, leading to f G298 and the equilibrium constant of any reaction at 298 K for which the algebraic sum of f G298 for all of the constituents is known. A detailed knowledge of S, which we already have, leads to Keq at any temperature. Variation in pressure on a reacting system can also be handled by classical thermodynamic methods.

One can now see why there is not much difference between computed values ofisomE0 and isomH298 as obtained from the thermodynamic equations in Computer Project 10-2

isom ¼ [G2 (0 K)(propene)] [G2 (0 K)(cyclopropane)]isomH298 ¼ [G2 Enthalpy(propene)]

[G2 Enthalpy(cyclopropane)]

The difference between the energy of a molecule at 0 K and its enthalpy at 298 depends on the thermal contribution due to vibration at the two temperatures. If the molecule in question is rigid, with few vibrational degrees of freedom, this contribution will be small, as it is for propene and cyclopropane. For larger molecules with a good deal of vibrational freedom, the difference will be correspondingly larger.

Koopmans’s Theorem and Photoelectron Spectra

In studying molecular orbital theory, it is difficult to avoid the question of how ‘‘real’’ orbitals are. Are they ‘‘mere’’ mathematical abstractions? The question of reality in quantum mechanics has a long and contentious history that we shall not pretend to settle here but Koopmans’s theorem and photoelectron spectra must certainly be taken into account by anyone who does.

Koopmans proposed that the orbital structure of a cation Mþ ought to be nearly the same as that of the molecule that engenders it, so that the amount of energy necessary to remove electrons from a stable molecule by hitting it with high-energy photons

hn þ M ! Mþ þ e

ought to be equal and opposite to the energy of the orbitals they come from. (Note that we have been using Koopmans’s theorem implicitly in our thermochemical calculations.) If more than enough energy is supplied to a molecule to drive electrons from one or more molecular orbitals, different excess energies Eexcess should be imparted to them according to the binding energy they had in their orbitals

Ein Eorb ¼ Eexcess

The excess energies can be measured for a known Ein by essentially a stopping potential method, giving a spectrum. This spectrum is then matched with calculated orbital energies (eigenvalues) derived from molecular orbital calculations.

Exercise 10-10

The measured energy spectrum of ethylene is shown in Fig. 10-7.

Electron

Emission

Intensity

The occupied eigenvalues of ethylene according to a 6-31G calculation are

Alpha occ. eigenvalues -- 11.23781 11.23620 1.03132

0.77980 0.63720

Alpha occ. eigenvalues -- 0.57707 0.49601 0.37194

Match the energies of the three highest orbitals with the peaks in Fig. 10-7.

Solution 10-10

The highest eigenvalues have the smallest (negative) energies in the third line of occupied eigenvalues. Converted to electron volts (conversion factor 27.21, with a change in sign), they are 15.7, 13.5, 10.1 eV, respectively. Quantitatively, the match isn’t as good as we might wish. Nevertheless, we have sound evidence of three molecular orbitals with energies in the vicinity of the three highest 6-31G eigenfunctions. Remember that the orbital structure of the cation is not really the same as that of the neutral molecule; that is an approximation.

Larger Molecules I: Isodesmic Reactions

Granting that absolute energy calculations may be very accurate by G2 and G3 methods, they are also very demanding of computer resources. Long ago Warren Hehre (1970) suggested a method for determining relative energies by using lowerlevel molecular orbital calculations in such a way that the error cancels across an isodesmic reaction. An isodesmic reaction is a reaction in which the number of bonds and bond types are the same on either side of the reaction but their arrangement is different. This permits determination of the enthalpy of reaction and thus the enthalpy of formation of one component of the reaction provided that all the others are known.

Because the f H298 values of many small molecules are known (Pedley, 1986) to within 0.1 or 0.2 kcal mol 1, one need not start with atoms in the hypothetical formation reaction as in the sections on energies of atomization and ionization and the thermodynamic function above. One can build up the target molecule from smaller molecules rather than from atoms. Suppose again, for illustrative purposes, that we make believe we don’t know f H298 of propene but we do know f H298 of the simpler C2 hydrocarbons ethene and ethane along with f H298 of methane. An isodesmic reaction containing these enthalpies is

Note that, for thermochemical purposes, there is no requirement that we can actually carry out the reaction. Systematic computational errors will, in some measure, cancel between the right and left sides of isodesmic reactions (10-20), giving an estimate of the f H298. GAMESS calculations at the STO-3G level lead to total energies of

77:073955 þ ð 78:306180Þ ! 115:660299 þ ð 39:726864Þ

obtained from the ENERGY COMPONENTS section of the output file. The calculated enthalpy of reaction is

where np and nr are the appropriate stoichiometric coefficients. The calculatedrH298 above and the experimental f H298 for CH4(g), CH2 CH2(g), and CH3 CH3(g), which are 17.90, 12.54, and 20.08 kcal mol 1; respectively (Pedley, 1986), leave only the unknown f H298 (propene) in the equation

4:41 ¼ f H298ðpropeneÞ 17:90 ð12:54 20:08Þf H298ðpropeneÞ ¼ 5:95 kcal mol 1

This compares with the G3(MP2) value of f H298ðpropeneÞ ¼ 4:29 kcal mol 1 and the corresponding experimental value of f H298ðpropeneÞ ¼ 4:78 0:19 kcal mol 1.

Exercise 10-11

Carry out a calculation of f H298 (propene) at the 6-31G MP2 level of theory in the GAMESS implementation.

Solution 10-11

The required TOTAL ENERGY entries are

117:29857 40:27913 ð 78:18420 þ ð 79:38560ÞÞ ¼ 0:00790 h ¼ 4:96 kcal mol 1

f H298ðpropeneÞ ¼ 4:96 10:36 ¼ 5:40 kcal mol 1

as contrasted to the G3(MP2) value of f H298ðpropeneÞ ¼ 4:29 kcal mol 1 and the experimental value of f H298ðpropeneÞ ¼ 4:78 0:19 kcal mol 1.

A second issue that arises in relation to isodesmic reaction enthalpies is why they should exist at all. If all we are doing is rearranging bonds, shouldn’t the summed bond energies be the same on either side of the reaction? Not really. A negative 6-31G MP2 enthalpy of 5 kcal mol 1 for the reaction

CH2 CH2ðgÞ þ CH3 CH3ðgÞ ! CH3CH CH2ðgÞ þ CH4ðgÞ

tells us that the thermodynamic state on the right containing propene is more stable than the thermodynamic state consisting of the simpler molecules on the left. We recognize this enthalpy difference as the ‘‘hyperconjugation’’ stabilization that a methyl group exerts on an a double bond, and we find that it is about 5 kcal mol 1 in the 6-31G MP2 model chemistry. At this level of accuracy, distinctions among the terms energy, enthalpy, and free energy are usually not made and they are used