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I am confused about the energies we are talking about when speaking in regards to the Maxwell Boltzmann Distribution and activation energies.

  1. Is there an relationship between kinetic energy and free energy for ideal gases? I am wondering if it’s possible to plot the x-axis of a Maxwell Boltzmann Distribution in terms of free energy as opposed to kinetic energy or velocity.
  2. Is the activation energy the difference in the free energy of the activated complex and the average free energy of the gas (or any reactant)?
  3. From the answer here, the activation energy mark on Maxwell Boltzmann Distributions like this one refers to a minimum kinetic energy level. Since the activated complex/transition state is a state, I do not understand why having a minimum kinetic energy (and right collision orientation) means reactants will reach the transition state. How come there is no free energy requirement?
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  • $\begingroup$ @Crafter0800 I fail to see how removing "confusion" from the title made it better. It is still as unspecific as it can get. Unearthing a three year old question for this is probably not worth it. $\endgroup$ – Martin - マーチン Jun 19 '17 at 8:28
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My initial thoughts on this are that you are viewing the Maxwell-boltzmann $velocity$ distribution the wrong way. Let me explain,

In nature there are three main distributions that "things" follow:

1) Boltzmann

2) Bose-Einstein

3) Fermi-Dirac

The first one is a classical result while the latter two are the result of quantum mechanical effects (symmetry of the wavefunction with respect to permutation). These three distribution functions describe the "way" energy is distributed among the energy levels of a molecule/atom/generic particle.

Perhaps a better way to phrase it would be that the macrosystem can be separated into lots of tiny microsystems. Each of these tiny microsystems can have their own particular "state". The individual microstates will combine in a collection or $ensemble$ to give the overall macrostate observed on our scale.

At this point we can vary our definition of how to define the ensemble and indeed there are quite a few buzz terms to learn. A $canonical$ ensemble is one in which the number of particles in each microstate ($n_i$), the volumes of the microstates, and their temperatures are all the same. If you choose the $micro-canonical$ ensemble then you would have $n_i$, $V_i$ and the energy of the $i$th state as being kept constant but not the temperature. A $grand-canonical$ ensemble has constant $n_i$, $V_i$ and chemical potential $\mu _i$.

Now depending on what you choose you will get slightly different equations, but they are all variations on a general unifying theme:

"The frequency distribution of a particular state or outcome out of all those states or outcomes possible, is proportional to \begin{equation} e^{-\frac{E}{KT}} \end{equation} Which is know as the Boltzmann factor. $E$ is the energy of the state, which varies from state to state."

You have already come across this factor in kinetic theory. !!AS A PARTICULAR EXAMPLE OF A NATURAL PHENOMENA THAT OBEYS THE BOLTZMANN DISTRIBUTION!!.

You can clearly see that the velocity distribution function, the frequency of having a particular velocity, is proportional to the Boltzmann factor, in this case the particular energy we are considering is the kinetic energy, so let $E=\frac{1}{2}mv^2$. \begin{equation} f(v)=\sqrt{\frac{m}{2\pi KT}}e^{-\frac{mv^2}{2KT}} \end{equation}

The Maxwell velocity distribution is just a special case of a Boltzmann distribution.

With this in mind I'm sure you can see that many different phenomena will be described by this distribution. If you wanted to put the Gibbs energy in there (indeed as we do in transition state theory (a personal favourite of mine)) then we can.

This brings me to your next question which is a little easier to answer. The Activation energy can readily be described by a Gibbs potential as indeed it usually is. I think this inherently answers your third question.

If you have any questions I will try to answer them when I can! :)

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  • $\begingroup$ The Boltzmann factor basically says that any energy term (u, h, g, KE etc.) for ideal gases follows the same frequency distribution? My confusion lies in fact that KE is not a state function (I am likely mixing things up here), but E above is described as energy of the state. $\endgroup$ – Yandle Oct 16 '14 at 19:08
  • $\begingroup$ Just to be sure I understood 2 and 3 correctly, activation energy is the difference in free energy between the transition state and reactant. Only particles with free energy above the transition state can react, actual velocity/KE of collision doesn't matter? $\endgroup$ – Yandle Oct 16 '14 at 19:23
  • $\begingroup$ Not really ... You use the Boltzmann factor to obtain the thermodynamic potentials/state functions /.... its all about relating the micro properties to the macro ... the molecular details that give rise to the classical thermodynamic quantities ... perhaps I mislead you with the transition state comment ... that is quite an advanced topic which mixes kinetics with thermodynamics which is indeed its own question :) $\endgroup$ – AngusTheMan Oct 16 '14 at 19:59
  • $\begingroup$ That is why you can mix a bit of Gibbs in there ... but in general you want to have the energy of the state in the Boltzmann factor ... $\endgroup$ – AngusTheMan Oct 16 '14 at 20:00
  • $\begingroup$ I was referencing this diagram that shows activation energy as a difference in free energy between transition state and reactant average energy, is this correct? Is gas KE here basically a part of the free energy and internal energy (it is microscopic, not the system KE as in first law)? In simple terms, is it correct to think that a reaction cannot happen unless the reactants has free energy above the transition state (but even if this is satisfied other factors like KE, orientation also plays a role). $\endgroup$ – Yandle Oct 16 '14 at 20:23

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