# Calculating internal energy of methane

I've recently found out my calculated value of internal energy of methane largely deviates from the ab initio output (at $$\pu{1000 K}$$ $$0.058~\text{Hartree} = \pu{152.3 kJ mol-1}$$). I used HF method, I know it's not highly-accurate, but I just need what internal energy should roughly be.

The way how I calculate:

$$H = U + PV = U + Nk_\mathrm{B}T$$

where $$U$$ is the internal energy, $$H$$ is the enthalpy. In this case, $$\Delta_\mathrm{f}H = \pu{-74.9 kJ mol-1}$$ (I found the enthalpy value of methane from heat of formation table on Wikipedia).

So, I simply substitute the temperature $$T = \pu{1000 K}$$ into the equation and I got $$\pu{-83.2 kJ mol-1}$$. First, the internal energy of a system cannot be negative. Second, even it's positive, why it deviates that much from ab initio calculations? Can someone give me a hint?

• I searched but I couldn't find about " ab initio output" what is it ? Jun 7, 2017 at 3:53
• It's from Gaussian09 calculation, by using HF/6-31G*. Jun 7, 2017 at 4:03

I believe the answer to your question is summarized well by the documentation of Psi4, an open source quantum chemistry package which can do similar calculations. I am fairly certain this is the problem you're running into but I could be wrong.

It is important to know that PSI4, like any other quantum chemistry program, does not compute the usual enthalpies, entropies, or Gibbs free energies of formation provided by most reference books. Instead, quantum chemistry programs compute “absolute” thermodynamic properties relative to infinitely separated nuclei and electrons, not “formation” values relative to elements in their standard states. If you are computing thermodynamic differences, like a reaction enthalpy computed as the enthalpy of the products minus the enthalpy of the reactants, then these “absolute” enthalpies are perfectly valid and usable. However, they cannot be mixed and matched with enthalpies of formation from reference books, since the zero of energy is not the same. Additionally, the “thermal energies” reported in kcal/mol are the finite-temperature corrections to the electronic total energy, and not the overall thermal energies themselves. If in doubt, use the reported Total Energies in Hartree/particle.

This is generally not the same as the output from Gaussian for a calculation on an isolated gas-phase atom. These values are referenced to the standard states of the elements. For example, the value for hydrogen atom is 1.01 kcal/mol. This is ($H_{H_2}^{\circ}(298K)- H_{H_2}^{\circ}(0K))/2$, and not ($H_{H}^{\circ}(298K)- H_{H}^{\circ}(0K))/2$ , which is what Gaussian calculates.