4
$\begingroup$

Suppose $\ce{Na^+}$ and $\ce{Cl^-}$ form an ionic bond. In doing so they will come closer by attraction. Now as they come closer their speed will increase and hence the kinetic energy increases. So basically the electrostatic potential energy will be controverted into kinetic energy.

  • My question is why do these processes that involve attraction cause a release of energy from the system? And after this release why is the system said to be more stable?

My understanding is there might be two concepts involved.

First, as the ions approach each other their speed increase, but as soon as they touch each other they collide and reflect back. So the ions keep on oscillating. These oscillations are what we know as heat and the temperature of $\ce{NaCl}$ will be higher than its surroundings so it will release its energy in the form of heat.

Second, it might be because any accelerating charge releases electromagnetic waves, and so as long as the ions oscillate they'll keep releasing energy in the form of EM waves which may be trapped by other ions or may escape.

After the energy will be released from the system(NaCl ion here) our system must become more stable. Stable in the sense that NaCl crystal will not convert back into ionic form separated from other ions because for coming back to that state the system must be provided the lost energy back by some external agency. This is the way how I understand the term stabilty; I don't what does this term mean in chemistry. I also don't know why " every system in nature wants to be more stable" -- This is the phrasing that I heard in school days from chemistry teacher. This fact must be related to some thermodynamic fact of kind that every system wants to release energy, but why?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ I'm afraid it doesn't have much to do with chemistry. $\endgroup$
    – Mithoron
    Jul 23, 2015 at 14:12
  • 2
    $\begingroup$ @Mithoron This is what I was told by many chemistry teachers when I was a student. They also told be that lesser energy --> More stability. And every thing in nature wants to be more stable. Although I never quite got their 'nature' arguments, like XYZ chemical reacts in a particular way because of its nature etc etc. $\endgroup$
    – user31782
    Jul 23, 2015 at 14:17
  • $\begingroup$ It's very basic physis problem - in chemistry it's get as given. Enthalpy should decrease entropy increase. $\endgroup$
    – Mithoron
    Jul 23, 2015 at 14:41
  • 1
    $\begingroup$ @Mithoron That's exactly what my question is. Why should Enthalpy decrease and entropy increase? $\endgroup$
    – user31782
    Jul 23, 2015 at 14:46
  • $\begingroup$ chemistry.stackexchange.com/questions/19150/… - your q. is more basic than that? $\endgroup$
    – Mithoron
    Jul 23, 2015 at 15:10

1 Answer 1

Reset to default
4
$\begingroup$

Attraction is a shorthand way of saying "potential energy gets lower as the distance gets lower". The opposite is true of repulsion. The potential energy of two like-charged particles, e.g. two $\ce{Na+}$ atoms, will increase as they are brought closer together.

Attraction (and its opposite, repulsion), are forces with units (in SI) of Newtons, or $\frac{\rm kg\cdot m}{\rm s^2}= \rm N$.

Energy has units of Joules, or $\frac{\rm kg\cdot m^2}{\rm s^2} = {\rm N\cdot m} $.

The relationship between a force $F$ between two particles separated by a distance $r$ and the potential energy $E$ of those particles is $F = -\frac{dE}{dr}$.

If the energy gets lower ($dE$ is negative) when two particles get closer together ($dr$ is negative), then the force $F$ is attractive. If the energy gets higher ($dE$ is positive) as the distance gets lower ($dr$ is negative), then the force is repulsive.

Force and energy are two different mathematical ways of looking at the same thing. That's why they are always related.$%edit$

Improve this answer
$\endgroup$
8
  • $\begingroup$ Is $E$ total energy here or only the potential energy. Is $F$ magnitude of force? How can we prove $ F=−dE/dr$ in this case? As far as I know Work done = $\vec F.\cdot \vec{dr}$, where work is being done on a particle and $|\vec{dr}|$ is the net distance moved not the distance between two particles. And lastly in which form the energy lost will appear? K.E or heat? $\endgroup$
    – user31782
    Jul 23, 2015 at 17:56
  • $\begingroup$ $E$ is potential energy; I'll edit my post to make that clear. The way in which energy is lost depends on the system. For a sodium ion and a chloride ion in a vaccum, the "lost" potential energy is converted to kinetic energy as the particles accelerate towards each other. For a sodium ion and a chloride ion in a viscous (uncharged) solvent, the potential energy is lost as heat through viscous dissipation (i.e. the friction of the ion moving through the solvent). $\endgroup$
    – Curt F.
    Jul 23, 2015 at 18:17
  • $\begingroup$ And yes the vector form of the relationship is $dE = \overrightarrow{\bf F}\cdot \overrightarrow{d{\bf r}}$ $\endgroup$
    – Curt F.
    Jul 23, 2015 at 18:19
  • $\begingroup$ I think I didn't phrase my question well. I know that P.E. will be converted into K.E. I need to edit my question. $\endgroup$
    – user31782
    Jul 23, 2015 at 18:19
  • $\begingroup$ I think your answer isn't rigorous. You might mean to take one $\ce{Na^+}$ atom as the origin. "when two particles get closer together (dr is negative)" But $dr=r_2-r_1$ so if two particles get closer then $dr=$postive. The equation you are using, $F=−dE/dr$ seems to come from vector calculus but it is missing vector notations. $\endgroup$
    – user31782
    Jul 23, 2015 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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