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From what I know in a closed system the potential energy + the kinetic energy is constant since no external work is done on the system. In my textbook it says that enthalpy is equal to the sum of the KE + PE meaning total energy. My question is how can the energy be lost to the environment after the collision of two particles. Since as the particle approach each other KE is turned into PE until activated complex is reached. Once it is reached they break up into smaller particles, repel each other and convert the PE back to KE. In conclusion all energy in the system is constant and no energy is lost. I don't quite understand why we have Enthalpy vs Time graphs. Since the energy in the system was never changed it simply converted between forms. For my homework when I pretended that enthalpy was simply PE and that endothermic = increase in PE and exothermic = decrease in PE I usually got it corrected(Such as question below). Is it not true that if you decrease PE by a amount KE would increase by that amount hence no change?

For instance a question from my homework "After a reaction, the product have less KE than original reactant molecules. Is the reaction endothermic or exothermic?"

Im thinking to myself that if KE has decreased PE must increase hence no change in energy so neither endothermic or exothermic. But the answer is endothermic.

In conclusion my main question is can I ignore the definition the textbook gave me and define enthalpy as total PE and delta enthalpy = change in PE rather than total energy.

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  • $\begingroup$ No, potential energy is internal energy. Enthalpy is a different thing. $\endgroup$ – Ivan Neretin Jul 7 '17 at 6:23
  • $\begingroup$ If enthalpy is different what is the difference? $\endgroup$ – coderhk Jul 7 '17 at 6:36
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    $\begingroup$ $$\underbrace{\hskip 0.75em H\hskip 0.75em}_\text{enthalpy}\;=\;\underbrace{\hskip 0.75em U\hskip 0.75em}_\text{energy}\;+\;\underbrace{\hskip 2em pV\hskip 2em}_\text{that's the difference}$$ $\endgroup$ – Ivan Neretin Jul 7 '17 at 7:16
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    $\begingroup$ @IvanNeretin Those some godly TeX skills, sir. Nice $\endgroup$ – Stian Yttervik Jul 7 '17 at 14:35
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In any way, ignoring the definitions given in a textbook is a bad idea. As for the question, the main problem in your thoughts is that you mixed to many things from different fields of science. Your example (two particles, kinetic energy, potential energy etc) looks like classical mechanics. So if you want to explain it, please, use tools offered by classical mechanics. Enthalpy is a measure of total energy in thermodynamics. It is used to explain processes taking place in large systems which can be characterized with volume, pressure, temperature etc. It is not used to describe one particle. Maybe you can take it into account and revice your question. Make it clear what exactly you want to understant: change in the kinetic and potential energy in interactions of two particles or change in enthalpy in exothermic and endothermic reactions.

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I know in thermodynamics, enthalpy specifically relates to changes in temperature. However, there is also another aspect that supplies energy into a system which is pressure.

The actual formula would look like

(Change in Energy) = q + w

Where q is the enthalpy, and w is work done by pressure.

In classical mechanics, if the system conserves energy, then its P+K=0, however, in real systems, energy is most likely lost to friction, so you are forgetting that in actuality it is

P+k=w

Where w is work.

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  • $\begingroup$ That is not an answer $\endgroup$ – Greg Jul 8 '17 at 4:52

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