1
$\begingroup$

I used Gaussian software to determine the single point energy (SPE) of 3 molecules namely ethane, ethene and phenol.

The SPE for the molecules are -79.83, -78.59 and for phenol it is quite large at -307. However, I am not able to explain the trend.

I am thinking does SPE have anything to do with the molecular weight of the molecule for example, larger mass = more negative SPE?? I cant think of a better explanation. Can someone please help me out?

thank you

$\endgroup$
5
$\begingroup$

The energies you're seeing are those resulting from solving for the eigenfunctions of the Coulomb Hamiltonian under the Born-Oppenheimer approximation, where the Hamiltonian can be broken down into the following 5 components:

nuclear kinetic energy operator

electronic kinetic energy operator

electron-nuclear attraction operator

electron-electron repulsion operator

nuclear-nuclear repulsion operator

The most important thing to take away from these expressions is that they all contain summations over nuclei, electrons, or both; combined with the presence of the set of nuclear charges $\{Z\}$, there is a non-linear dependence of the molecular energy on more than just the nuclear masses.

This is one reason why it usually isn't instructive to compare absolute energies of unrelated molecules.

$\endgroup$
0
$\begingroup$

Although you did not provided the method used to calculate the energies, it is most probably HF or DFT. These methods use the electronic Hamiltonian, as provided in answer by pentavalentcarbon. The energy corresponds to formation of the compound from bare nuclei and electrons. Such value is of little use on itself.

You probably expect the result to be Heat of Formation, i.e. formation from atoms at their standard state. To obtain such a number, you'd need to calculate the electronic energies of corresponding atoms and formulate the corresponding chemical equations.

Alternatively, semi-empirical methods (AM1, PM3, PM6 in Gaussian) give the Heat of Formation directly.

To overcome the difficulty, one often uses the isodesmic reaction to obtain the relative energies, which are of better use.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.