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The collision theory of chemical kinetics

  1. Jul 2, 2012 #1
    I am studying the activation energy of a reaction, and I notice that it says when the products are more stable than the reactants the reaction is exothermic, and endothermic when the products are less stable than the reactants.

    It doesn't say why, but my supposition is that when the products are less stable than the reactants, they need to absorb heat from the environment to get enough energy for the reaction to happen.

    As for when the products are more stable than the reactants, there is an excess of energy, so it releases that energy as heat.

    I just want to know if this is correct.

    Also, they show a potential energy vs. reaction progress graph for an exothermic and endothermic reaction, and the exothermic graph has the products with less potential energy than the products, and the endothermic graph has the products with more potential energy than the reactants. Does having higher stability mean less potential energy?
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  3. Jul 2, 2012 #2


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    Yes. This comes from the conservation of energy. Exothermic reactions convert chemical potential energy into thermal energy whereas endothermic reactions convert thermal energy into chemical potential energy.

    Yes, that is correct (although in chemistry stability can have other meanings other than the thermodynamic stability you're talking about here).
  4. Jul 2, 2012 #3
    So this begs the question, why does the activation energy increase for both cases? I would think in the exothermic case, the Activation energy isn't even needed because wouldn't the reaction happen on its own since it wishes to be at a lower potential energy? Yet the graph shows an activation energy that is positive, albeit the activation energy is less than the endothermic case.

    If the reactants are more stable (less potential energy), they should need the extra heat(endothermic) for the reaction to occur. I can see the activation energy as the heat absorbed.

    If the reactants are less stable (more potential energy), they need to expend that extra energy to be more stable. Where is the activation energy?

    On the graphs, for the exothermic case there is a small jump in potential energy, and then a huge dive in it after the activation energy portion. Is this dive in potential the potential being turned into heat?

    On the endothermic case, there is a large jump in potential energy, and then a small dive after the activation energy. Is this small dive the potential being turned into heat as well, but there is just a lot less heat given off than the exothermic, so overall it absorbed more potential energy as heat than it released, and for the exothermic case there is a small jump in potential then a huge dive after, is all that huge dive after the potential being turned into heat, but before there was a small increase, but the net is that it released more heat than it absorbed?

    Basically, does an exothermic reaction exclusively give off heat, or does it take in a little heat and give off a lot more heat, so the net effect is more heat given off than taken in, so we call it exothermic and vice versa for endothermic.
    Last edited: Jul 2, 2012
  5. Jul 3, 2012 #4
    So when I studied all of this stuff (literally just last term), it helped me to consider thermodynamics and kinetics as two separate entities. The fundamental difference between the two is the dependence on reaction pathway. While thermo is path-independent, kinetics is path-dependent.

    The activation energy for a given reaction is almost always positive. There have been some reaction pathways with documented negative EA, but those are really more of an exception.

    Collision theory is based on the premise that some of the participating molecules are colliding with one another to yield a product (which goes back to the idea of rate constants for given rxns). In order for a molecule to form a product it needs to have a necessary amount of kinetic energy to break & form new bonds. This necessary amount of kinetic energy is known as activation energy.

    Thus, to answer your question, even though an exothermic reaction be can spontaneous, the reactant molecules still need enough kinetic energy to drive the reaction forward to products.
  6. Jul 3, 2012 #5


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    Here's where the other definition of stability comes in. If molecule A reacts exothermically to form molecule B and there is no activation energy for the reaction, then any molecule A that exists will instantaneously convert to molecule B. In other words, you could never isolate any molecule A to study the reaction in the first place.

    Therefore, in addition to considering the thermodynamic stability of a molecule (how low is its chemical potential energy), it is also important to consider the kinetic stability of a molecule as well. Molecules whose reactions require larger activations energies are more kinetically stable than molecules whose reactions require smaller activation energies.

    As an example, consider graphite and diamond. At standard temperatures and pressures, graphite has a lower chemical potential energy than diamond. Therefore, diamond is thermodynamically unstable, and it is thermodynamically favorable to convert diamond into graphite (i.e. the reaction is exothermic). However, the activation energy for this conversion is huge such that at standard temperatures and pressures, we never see this reaction occur. Thus, the fact that "diamonds are forever" is a result of their kinetic stability, not their thermodynamic stability.

    You've got the right idea. An exothermic reaction needs a little bit of energy to get started, but it gives off more than enough energy to compensate for the initial input of energy. A good example here is starting a fire. If you put gasoline in open air, it is not going to spontaneously combust. You need to provide activation energy in the form of a spark. Once the spark lights the fire, however, you get back much more energy than the energy of the spark.
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