Gaussian Output
Gaussian Output |
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First look for the message that the optimization is through.
Optimization completed. -- Stationary point found. |
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These are internal coordinates of the molecule in terms of bond lengths and bond angles. Each atom is given a number. You can figure out the numbering by thinking about the atoms that are bonded and from your original input. For instance,
R(2,1) is the bond length between atoms 1 and 2. (1.184 Angstroms)
A(2,1,3) is the bond angle formed by the bonds between 2-1 and 1-3. (122.1 degrees)
D(2,4,1,3) is the dihedral angle between atoms 2, 4, 1, and 3.
The structure of the molecule also is supplied in Cartesian coordinates with the axes as shown above..
Standard orientation: ---------------------------------------------------------- Center Atomic Coordinates (Angstroms) Number Number X Y Z ---------------------------------------------------------- 1 6 .000000 .000000 -.519603 2 8 .000000 .000000 .664733 3 1 .000000 .924339 -1.100124 4 1 .000000 -.924339 -1.100124 |
The internal coordinate structure often is more useful than the Cartesian coordinates because we are primarily interested in bond lengths and bond angles.
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Look back in the output file about 25 - 30 lines before the "optimization completed" phrase. You should see the SCF energy of the optimized structure.
SCF Done: E(RHF) = -113.866331170 A.U. after 12 cycles
If you are doing an MP2 calculation, you will see the MP2 energy a few lines down in the output, written in exponential notation.
E2 = - 0.6318889865D+00 EUMP2 = -0.114498220156D+03
These energies are expressed in atomic units (Hartrees)
1 A. U. = 627.51 kcal/mol = 2,625.5 kJ/mol
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Directly below the Cartesian coordinates of the molecule you will find the rotational constants of the molecule.
Rotational constants (GHz): 293.4536205 40.1454922 35.3143626
These aren't much use to us for formaldehyde, which is nonlinear, but can be useful in analyzing the microwave spectra of linear molecules.
Close to the bottom of the output, you should find the predicted atomic charges and dipole moment of the molecule.
Total atomic charges:
1 C .134575
2 O -.415727
3 H .140576
4 H .140576
As we would expect from electronegativities, the oxygen atom is negatively charged. This produces a dipole moment (pointing toward the negative end) along the z-axis.
Dipole moment (Debye): X= .0000 Y= .0000 Z= -2.6663 Tot= 2.666
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In the last part of the job output from the frequency calculation you will find the predicted vibrational frequencies (cm-1) of the normal modes of the molecule. Also supplied are the predicted intensities of the IR and Raman bands corresponding to these normal modes.
1 2 3
B1 B2 A1
Frequencies -- 1335.5948 1383.4094 1679.4157
. . .
IR Inten -- .3711 23.1388 8.6198
Raman Activ -- .7628 4.5148 12.8466
. . .
4 5 6
A1 A1 B2
Frequencies -- 2027.8231 3160.8817 3232.9970
. . .
IR Inten -- 150.0938 49.6483 135.6548
Raman Activ -- 8.1210 137.6237 58.2266
Computational results usually have systematic errors. In the case of Hartree-Fock level calculations, for instance, it is known that calculated frequency values are almost always too high by 10% - 12%. To compensate for this systematic error, it is usual to multiply frequencies predicted at the HF/6-31G(d) level by an empirical factor of 0.893. Similarly, frequencies calculated at the MP2/6-31G(d) level are scaled by 0.943 8.
The predicted frequencies after applying the 0.8929 scale factor are listed below.
1 2 3
B1 B2 A1
Scaled Frequencies -- 1193 1235 1450
. . .
IR Inten -- .3711 23.1388 8.6198
Raman Activ -- .7628 4.5148 12.8466
. . .
4 5 6
A1 A1 B2
Scaled Frequencies -- 1811 2822 2887
. . .
IR Inten -- 150.0938 49.6483 135.6548
Raman Activ -- 8.1210 137.6237 58.2266
With scaling, the calculations predict two intense IR transitions, one at 1811
cm-1, and another at 2887 cm-1. These are around 4% higher than the experimental
values for gas-phase formaldehyde, 1746 cm-1 and 2782 cm-1, probably because we
have used only a medium-sized basis set.
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Temperature 298.150 Kelvin. Pressure 1.00000 Atm. Zero-point correction= .029201 (Hartree/Particle) Thermal correction to Energy= .032054 Thermal correction to Enthalpy= .032999 Thermal correction to Gibbs Free Energy= .008244 Sum of electronic and zero-point Energies= -113.837130 Sum of electronic and thermal Energies= -113.834277 Sum of electronic and thermal Enthalpies= -113.833333 Sum of electronic and thermal Free Energies= -113.858087
The sum of the electronic and thermal enthalpies is an estimate of the enthalpy
of the compound. To find the enthalpy of a reaction, do a frequency calculation
and determine the sum of the electronic energy and thermal enthalpies for each
compound. Then calculate the difference, products minus reactants, using the
these values. The enthalpies are expressed in atomic units (Hartrees)
1 A. U. = 627.51 kcal/mol = 2,625.5 kJ/mol
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The PCOL community acknowledges that partial support for this work was provided by the National Science Foundation's Division of Undergraduate Education through grant DUE #9950809. Additional support was provided by the Camille and Henry Dreyfus Foundation. PCOL faculty also acknowledge the National Science Teachers Association which awarded the PCOL Faculty Consortium the 1998 Gustav Ohaus Award for Innovation in College Science Teaching.
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This site created by David Whisnant
(whisnantdm@wofford.edu).
This page was last updated on March 9, 2005
llever@uscupstate.edu