4
$\begingroup$

Consider a chemical reaction where two different particles form another one

$$\ce{O + O_2 -> O_3}$$

I find it confusing how this can be an exothermic process. How to picture the release of $+\Delta H$ to the surroundings.

And am I right in viewing this energy value only as the net energy coming from an integration of the particle potentials as they get closer along the reaction coordinate?

$\endgroup$
  • $\begingroup$ Can you provide a source for the thermodynamic data? I'd have thought it was an endothermic reaction. $\endgroup$ – CHM Aug 8 '12 at 15:55
  • $\begingroup$ @CHM: I thought this is a exothermic one, but that doesn't really matter - the question goes for any exothermic reaction. (One could argue that the way I wrote it down doesn't even claim that this particular reaction is exothermic, but it's of course somewhat implied.) $\endgroup$ – Nikolaj-K Aug 8 '12 at 16:16
2
$\begingroup$

Essentially yes - if you consider the two particles colliding into a potential well, the kinetic energy associated with each will (depending on the angular specifics) be apportioned into rotational, vibrational, and even occasionally electronic excitations. So, the two particles collide, and are then vibrating and/or rotating about each other with all the translational energy converted in the collision.

Non-reaction collisions with other ambient particles allow relaxation into lower-energy states, transmitting energy to those particles. After some collisions the new species formed is in equilibrium, transferring between low-energy rotational and vibrational states as it collides further, and the change in potential energy has been dissipated as rotational, vibrational, and translational energy in other particles.

(I guess I could really use some sort of animation for this.)

$\endgroup$
  • $\begingroup$ Mhm, my understanding of the energy package given to the outside, the so called system, hasn't really improved. $\endgroup$ – Nikolaj-K Aug 8 '12 at 18:26
2
$\begingroup$

The "packet" of energy exchanged fall into one of three forms: heat, work, or electromagnetic radiation. The first two are kinetic energy phenomena.

Expanding on Aesin's answer about potential and kinetic energy.

Reactions that are exothermic convert potential energy into kinetic energy. The potential energy is the energy stored in the chemical bonds (electrons), and the kinetic energy is the molecular motion (rotation, translation, and vibration). Consider your "system" to be the electrons and the "surroundings" to be the nuclei of the atoms.

According to thermochemistry data from the NIST chemistry webbook for ozone and atomic oxygen, this reaction is exothermic (the heat of formation of molecular oxygen is by definition zero):

$$\Delta H ^0_{rxn}=\sum \Delta H ^0_{f}(products)-\sum \Delta H ^0_{f}(reactants) $$ $$\Delta H ^0_{rxn}= \Delta H ^0_{f}(\text{O}_3 )- \Delta H ^0_{f}(\text{O}) $$ $$\Delta H ^0_{rxn}= 142.67~~\text{kJ/mol}-249.18~~\text{kJ/mol}=-106.51~~\text{kJ/mol} $$

In this case the potential energy of the electrons in ozone (total bond order of 3) is lower than the potential energy of a molecule of oxygen and an atom of oxygen (total bond order of 2). That extra potential energy is needed to keep O and O2 apart. When that energy is no longer needed, it has to become some other kind of energy (First Law of Thermodynamics: Conservation of Energy). In all cases of exothermic reactions, the extra potential energy becomes kinetic energy. If this kinetic energy increase occurs without a change in volume, then the temperature increases, and we say that "heat" has been exchanged. If the volume is allowed to change, then the increase in kinetic energy is converted to work.

Conversely, for an endothermic reaction, kinetic energy (work or heat) is converted into potential energy (less stable electronic arrangements).

Electromagnetic radiation is a little different. Instead of increasing the kinetic energy of the nuclei, an exothermic reaction can release a photon of light. In the case of the ozone reaction, the photon would have a wavelenth of 1.12 micrometers, in the near-IR.

$\endgroup$

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.