-1
$\begingroup$

Do atoms combine with the nearest atoms possible because of lower delta E (less energy needed to form a stable nucleus)?

$\endgroup$

closed as unclear what you're asking by bon, Loong, M.A.R., Geoff Hutchison, LDC3 Mar 23 '15 at 3:54

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ If atoms combined with the nearest atom, then the air would be composed of $\ce {N2, O2, NO, NO2, ...}$. There are reasons that determine which compounds are preferred over other compounds. $\endgroup$ – LDC3 Mar 22 '15 at 16:52
  • $\begingroup$ Nucleus?? Are you talking about nuclear reactions? $\endgroup$ – Mithoron Mar 22 '15 at 17:15
  • $\begingroup$ "Combining with the nearest atom"? How do you know they will? Also, is this about a chemical or nuclear reaction? $\endgroup$ – M.A.R. Mar 22 '15 at 17:16
1
$\begingroup$

Let's say we have an atomic soup that contains atoms (what follows also applies to molecules) $\ce{A, B}$ and $\ce{C}$. From these reactants it would be possible to form products $\ce{A-B, A-C}$ and $\ce{B-C}$. Just which of these molecules, if any, will form depends on several factors including the $\Delta E$ that you mention.

Background

We usually talk about a reaction coordinate, like the one shown below.

enter image description here

image source

We have our reactants on one side and products on the other, and a barrier connecting the two. The height of the barrier is referred to as the activation energy ($E_{a}$). The Arrhenius equation is often used to describe the rate (k) for this process, where A is the pre-exponential factor (more on this in a minute), R is a constant and T is the temperature (°K)

$k = Ae^{-E_a/(RT)}$

Back to the question

Just because atoms collide doesn't necessarily mean they will react to produce a product. Sometimes they must collide from a certain direction. Think of molecules with complex shapes, for them to react we need the other reactant to collide with a certain part of the larger molecule, maybe too from a certain angle or direction. These geometrical factors are contained in the pre-exponential factor. Also the collision must have enough energy to get the reactants over the energy barrier.

So just because molecules are close doesn't guarantee that they will react. More important is how often molecules collide (concentration), the geometry of the collision (pre-exponential factor) and the energy (E) they carry.

$\endgroup$

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