I seem to be unsure about a discrepancy in energy conservation utilizing the equipartition theorem. Let's say I have a molecule within a thermal reservoir. For example, I will use a molecule of $NH_3$. I will assume that the temperature of reservoir is sufficiently high enough that the high temperature limit can be assumed for the thermodynamical ensembles and molecule's internal energy will be share equally among its degrees of freedom. Specifically, I will focus on its translational, vibrational, and rotational motions.

According to the equipartition theorem, $NH_3$ should have three translational degrees of freedom (each of which gives $\displaystyle \frac{1}{2} k_B T$

$$U_{tr}=\frac{3}{2} k_B T$$

Similarly for rotational energy

$$U_{rot}=\frac{3}{2} k_B T$$

And for a polyatomic molecule such as $NH_3$, there are $3N-6$ vibrational degrees of freedom each of which give energy $k_B T$ which gives

$$U_{vib}=6 k_B T$$

Therefore the total energy of the $NH_3$ molecule is

$$U_{tot}=9 k_B T$$

Now what if the $NH_3$ molecule completely dissociates so that

$$NH_3 \rightarrow N + 3H$$

Now we only have 3 translational degrees of freedom for each of the 4 atoms. This gives a total internal energy of the system to be

$$U_{tot}=6 k_B T$$

Of course due to energy conservation this energy must have gone somewhere. My question therefore is, where did the extra $3 k_B T$ worth of energy go? My best guess would be that it either was lost as heat to the surroundings when the bonds were broken or it went into the actual dissociation of the molecule. Would I be correct in any of these assumptions?


Consider the simpler case of a diatomic molecule where there is just one degree of freedom. The potential energy as a function of bond length will look something like this:


Near the minimum the potential is approximately quadratic, that is:

$$ V(x) = kx^2 $$

for some effective force constant $x$. So if we want to make our molecule vibrate we have to (1) give the atoms some kinetic energy and (2) give them some extra potential energy. For any potential $V(r)=ar^n$ the kinetic energy $T$ and potential energy $V$ are related by the virial theorem:

$$ 2T = nV $$

so for a quadratic potential where $n=2$ we end up with $T=V$. And this is why the energy associated with a vibrational mode is $kT$ not $\tfrac{1}{2}kT$, because for every $\tfrac{1}{2}kT$ we add as kinetic energy we have to add another $\tfrac{1}{2}kT$ as potential energy.

However as we climb out of the potential well to large interatomic distances the potential flattens out and eventual becomes independent of distance, i.e. $n=0$ in our equation above, so now we just need to add kinetic energy without needing to add any potential energy at the same time, and the energy becomes just $\tfrac{1}{2}kT$ i.e. the same as a free particle.

So that extra $\tfrac{1}{2}kT$ of energy went into climbing out of the potential well.

  • $\begingroup$ I'm not convinced by the last sentence. What does $kT/2$ have to do with the height of the potential well? Can't it be arbitrarily higher or lower? $\endgroup$ – knzhou Aug 6 '16 at 15:53

The 6$k_BT$ for the whole body (NH$_3$) and for the separated atoms is the same. In calculating this value for ammonia (translation + rotation) one considers only that the atoms move as a group. Only when using 3N-6 does the internal dynamics of a molecule come in, i.e. vibrating bonds. Thus the difference in energy (3k$_B$T) is that needed to break the bonds, i.e. to take the molecule from its potential well to separate atoms. Each of the three normal vibrational modes account for k$_B$T, (1/2)k$_B$T each for potential & kinetic energy, which make up the difference. (Each squared term in the energy (potential & kinetic) equates to (1/2)k$_B$T via the Virial Theorem)

(Even if the potential energy is not harmonic and thus a different relationship between kinetic and potential energy is given by the virial theorem, when the energy from 3N-6 modes was calculated a harmonic potential was assumed, because the energy is (1/2)k$_B$T, thus it is ok to assume this when bonds are broken. Clearly if a different potential was used the energy would be different, but must still cancel when the bond is broken as the potential has to be the same.)


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