5
$\begingroup$

I might have a bit of a conceptual misunderstanding here, but how do we account for the entropy of a system when constructing a Hamiltonian? All entropy should be accounted, because the Hamiltonian includes the total energy, including the entropic contribution. However, when considering e.g. $\ce{H2+}$, the fact that the dissociated state is more entropically favoured doesn't go from the Hamiltonian.

Any suggestion that would clean up my understanding of the relationship between entropy and energy in this and other cases would be very helpful. Thanks!

$\endgroup$
  • 3
    $\begingroup$ The calculation only involves solving for total energy as a function of internuclear separation and so bonding and antibonding potential energy curves are produced. If you want now to dissociate the molecule say from its lowest energy state then energy has to be added, say from a photon or collision and this is where the entropy enters. Similarly to get atoms from infinity to the lowest molecular energy level involves removing energy and changing entropy. $\endgroup$ – porphyrin Jan 29 '17 at 14:52
  • $\begingroup$ @porphyrin Thank you! (and sorry for a delay in my reply) But when we dissociate a molecule, we account for the electrostatic interaction between its two nuclei and the electrons, rather that for the probability of $H$ and $H^+$ being found close together (this is how I understand entropy). Does electrostatic interaction already include this? $\endgroup$ – GingerBadger Feb 4 '17 at 14:51
  • 1
    $\begingroup$ I don't fully understand you last question; you seem confused so my answer is to define entropy and perhaps this will help. Entropy is best defined from statistical mechanics as $s=k\ln(\Omega)$ where k is the Boltzmann constant and $\Omega$ the number of different ways any energy levels can be occupied. In a molecule, there is entropy associated with vibrations and rotations. These are calculated via the partition function; see most phys. chem. texts for details. $\endgroup$ – porphyrin Feb 6 '17 at 14:35
  • $\begingroup$ @porphyrin thank you! I'm still a bit confused but I think it is clearer to me now! $\endgroup$ – GingerBadger Feb 7 '17 at 22:49
4
$\begingroup$

One definition of entropy is that the number of microstates that makes up a macrostate influences the probability of observing the macrostate. So your wavefunction needs to be multi-configurational and include both $\mathrm{H}_2^+$ and $\mathrm{H}^+ + \mathrm{H}$. The usual approach is to invoke the Born-Oppenheimer approximation, which makes the wavefunction depend on a single static set of nuclear coordinates, i.e. single-configurational. This approach will just give you the energy and you must compute reaction probability using statistical mechanics.

Multi-configurational SCF does include entropy in the following sense. Let's say you have a molecule with a HOMO-LUMO gap that is sufficiently small to allow for significant mixing. You run a CASSCF(2,2) to determine the LUMO contribution and is 5%. Now suppose you have the same system, same HOMO-LUMO gap, but the LUMO is doubly degenerate. Your CASSCF(2,3) would now have a larger LUMO contribution, though I am not sure it would be 10%.

$\endgroup$

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.