1
$\begingroup$

Many non-covalent interactions (e.g. electrostatic interactions) result in decreased entropy and are driven by negative enthalpy. If the heat released in the reaction has its physical basis in an increase in the average kinetic energy of the degrees of freedom contributing to the heat capacity of the solution (i.e. surrounding water molecules move faster), how exactly does binding exert a force on these degrees of freedom? How does binding cause increased thermal motion of solution molecules? Is it simply that the two attractive moieties (e.g. a lysine and a glutamate residue) are accelerated towards each other as they convert potential energy into kinetic, and then release heat via collisions with solution molecules? Is all exothermic heat release the result of potential energy being converted into kinetic energy in reaction products and the subsequent collisions with solution?

$\endgroup$
3
$\begingroup$

I think that you are on the right track. I try to explain how I see things below.

As a bond forms the reactants must absorb some energy from their surroundings so that the activation barrier between reactants and products may be surmounted. (By the Boltzmann distribution there are always a few collisions with far more energy than the average). At the top of the barrier is the transition state (TS), for this to be crossed the excess energy has to be taken away from the TS more rapidly that it may cross back to reactants.

The transition state is effectively a very hot species; it has far more energy than the surrounding medium. This can be thought of as a species having a large number of vibrational and rotational quanta excited. As far as is known$^*$ a transition state lasts for only a few tens of femtoseconds to a few picoseconds because vibrational/rotational energy is lost very rapidly to the surroundings. This excess energy is carried away also as kinetic energy of the products if, say, two molecules of product are formed. As time progresses this energy excites more and more vibrational and rotational modes in the surrounding medium, which may be a solvent or protein's backbone and residues. Eventually this energy diffuses away from the reaction site (via anharmonic vibrational couplings) and the product eventually has vibrational and rotational modes excited with quanta in accord with the prevailing temperature.

$*$ Although transition states are universal there is a paucity of actual, i.e. time resolved spectroscopic data on their properties.

(A more detailed approach has been given by Kramers (see M. Daune ' Molecular Biophysics for a discussion). In the condensed phase 'friction' or 'drag' on the reaction coordinate slows the reaction (compare to gas phase) and the reactants 'diffuse' towards and across the transition state. 'Friction' here means vibrational and rotational energy affecting the reaction coordinate, by impeding or promoting, by turns, the progress of the reaction. The Kramers approach allows a consistent approach to gas phase reactions from isolated molecules, then in the presence various pressures of inert gas right through to reactions in viscous solution.)

$\endgroup$
  • $\begingroup$ That clarifies a lot. I'm still unclear as to how a weak interaction such as a hydrogen-bond or electrostatic interaction could release heat. Is it just released via the acceleration of the species and collisional relaxation? Protein folding is said to involve many such electrostatic interactions, which must release heat to the surroundings in order to proceed (i.e. they must have negative enthalpy) because they necessarily decrease entropy by limiting conformational sampling (a source: ncbi.nlm.nih.gov/books/NBK22567). $\endgroup$ – Jory Aug 5 '16 at 2:43
  • $\begingroup$ Or in the case of an electrostatic interaction, does the heat come from the photons that are released? $\endgroup$ – Jory Aug 5 '16 at 6:48
  • $\begingroup$ @jory, no photons need be involved. In any interaction (h bond, dispersion, dipole-dipole etc ) it is the potential energy that holds the two parts together. So when a bond breaks this energy has to go somewhere, usually it gets taken up by a combination of vibrational , rotational quanta and translation of the species. These constitute the heat released. Thus then diffuses away into the surroundings, (primarily via collisions) raising the temperature by a tiny bit. $\endgroup$ – porphyrin Aug 5 '16 at 13:27

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.