In most of the cases the reactions are exothermic because all systems want to reduce their net energy and one way to do that is by loosing the $KE$ of the molecules to the surrounding molecules by collisions.

Now suppose that that the molecules of our system cannot emit any radiation so the only way for them to reduce their net energy is by giving their $KE$ to other molecules which are not a part of the system.

Now if I make water with Hydrogen and Oxygen, it results in an exothermic reaction. My question is that hasn’t the $KE$ of the molecules increased when we have mixed Hydrogen and Oxygen together before they pass their $KE’s$ to the molecules of the air? Or in other words does every exothermic system becomes unstable ( because of the increase in the $KE$ of the molecules and hence in net increase in energy ) before it can pass out it’s energy to the surroundings ? And is it that exothermic systems will tend to increase their energy if they cannot pass on their energy to the surroundings ?

  • 2
    $\begingroup$ I think you are talking about a transition state. $\endgroup$ – Zhe Jan 4 '17 at 14:50
  • $\begingroup$ @Zhe , wikipedia states that transition state is related to potential energy but here I am referring to Kinetic energy! Moreover in a video at youtube she said that the transition state theory assumes that the reactants and products are in equilibrium... . But the the question which I have addressed does not mention that too! So I think that transition state is somewhat related to the distance between the atoms ( potential energy ) but $KE$ is obviously related to the change in that length per unit time. $\endgroup$ – Aaryan Dewan Jan 4 '17 at 15:32
  • $\begingroup$ If we're nitpicking, then the first statement in your question is wrong. Systems don't care to lose energy. The universe just needs to increase its entropy. Based on your comment, it's no longer clear what you're actually asking. What do you mean by "unstable"? $\endgroup$ – Zhe Jan 4 '17 at 15:57
  • $\begingroup$ The KE of molecules does not somehow increase as they already have a distribution of energies (Maxwell-Boltzmann) and distribution of potential energy (Boltzmann) which both vary with temperature. These energies interconvert and occasionally there is enough energy to overcome an activation energy and cross a transition state towards products (and vice versa but with lower chance if reaction is exothermic). (btw. In the TS model of reactions there is equilibrium between reactants and the TS and between products and the TS). $\endgroup$ – porphyrin Jan 7 '17 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.