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I understand that in an exothermic reaction energy is transferred to the surroundings. More energy is produced from making bonds than absorbed to break bonds.

I would imagine that the products have lower chemical energy (bond energy perhaps?) but have higher kinetic energy as measuring the temperature of the surroundings is effectively measuring the kinetic energy of the specific molecules? I only doubt myself because on enthalpy change graphs the products are always lower than the reactants in an exothermic reaction, so increasing KE doesn't seem likely. If someone could clear up my confusion that'd be great!

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    $\begingroup$ The energy diagrams that you see solely plot chemical potential energy. Any kinetic energy that the particles possess is not reflected in the diagram. $\endgroup$ – orthocresol Nov 21 '16 at 12:39
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In line with the comment of orthocresol, these potential energy diagrams only display the change in the chemical potential as a function of the reaction coordinate.

To clarify: Let's consider the theoretical situation of a reaction mixture in complete vacuum that undergoes a reaction towards a product state. If we assume that the reaction is exothermic and there is no loss of energy due to irradiation, then this means that the products are heated up by an amount of energy that is gained from the loss in chemical potential (i.e. the exothermic reaction). In other words, if you would plot the chemical potential + the kinetic energy (the two terms combined) as a function of the reaction coordinate, you would obtain a straight line due to the law of conservation of energy, because chemical energy is converted to kinetic energy (heat).

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There are four main ways that a molecule can store energy: translation, rotation, vibration, and electronic energy. The first three of these are kinetic energy in one way or another, and the fourth is less commonly important when looking at the products of a reaction, so forget about that.

From the outset we can say yes, of course, the products have more kinetic energy because energy must be conserved. I wanna go into more detail though to hopefully demonstrate some of the intricacies of where the product energy is stored.

velocity flux contour map http://202.141.40.218/wiki/images/Velocityfluxcontourmap.jpg

The picture above is a miracle of experimental chemistry.

This plot shows you all observed trajectories of the $\ce{DF}$ molecule in the reaction,$$\ce{F + D_2 -> DF + D}$$

I can hear the naysayers now, "who would possibly care about such a useless reaction?" Well, there are many interesting things to be learned from very simple systems.

The plot is made by preparing all incoming product molecules with a total kinetic energy of, in this case, $1.82\ \frac{kcal}{mol}$. After the reaction takes place, that incoming kinetic energy, which was all in the form of translational energy (it need not be but here it was) must be stored in either vibrations, rotations, or translation.

Each of the dashed lines represents a vibrational state of the product $\ce{DF}$ molecule. Each of the contours represents a rotational state. All the rest of the energy is then stored in translational kinetic energy.

There are many details that can be gathered from this one picture, but in this case I will highlight a detail that might change how you think about the products of a reaction. In this case, the products are most often in the $v=3$ vibrational state. That is, when this reaction happens and you look at the products, you'll find a molecule meandering along pretty slowly, but vibrating quite vigorously. You can see that $v=3$ is most populated by looking at the density of the contour lines.

Also, notice that the reaction almost never proceeds and ends up in the $v=1$ vibrational state. How strange that the molecules have a preference!!

So, to tie this back to something we're more familiar with such as exothermic reactions taking place in a beaker. When a reaction happens, and there is excess energy to be stored in the products, it will almost always go into some combination of vibration, rotation, and translation. In the experiment above, after the reaction happens, nothing else is around to take that energy so we can see what the true product state looks like. In a beaker, a molecule that is made rotationally excited quickly collides with something else and gives away some of its energy. Eventually, it is quite likely everything will settle down to just plain old translational kinetic energy and some rotations, but initially, it is quite feasible the products start out vibrationally excited. Hopefully that answers your question, but also gives you some other things to think about.


The figure was found online somewhere, but it comes from this paper.

Neumark, D. M., Wodtke, A. M., Robinson, G. N., Hayden, C. C., Shobatake, K., Sparks, R. K., ... & Lee, Y. T. (1985). Molecular beam studies of the F+ D2 and F+ HD reactions. The Journal of chemical physics, 82(7), 3067-3077.

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