CHM 5414 Lecture Notes
3 October 1996

Fireworks seem the perfect example of stochastic energy eruptions, but they're actually painstakingly choreographed. When you labor in a fireworks factory, you leave nothing to chance!

OK...I pulled a fast one on you last lecture. It's not that we produced something wrong so much as we cheated by considering an intramolecular reaction A = B like the isomerization of methyl cyanide (CH₃CN) and methyl isocyanate (CH₃NC) which bypassed worrying about the numbers of moles of things created and destroyed. N = 0 so features dependent on N cancelled out.

Let's rectify that by including N explicitly so we needn't confine ourselves arbitrarily to N=0 in future!

If we let z be the partition function for a molecule (the count of state thermally available to it) and Z be that for N such molecules (N_A if a mole), what's the relation of Z to the N little z's? Given that one molecule's occupancy of some state does not, in principle, influence another's molecule of any of the states available to it, the overall number of states available to any two molecules A and B would be expected to be Z = z(A)z(B), the product. (This is because the probabilities of independent events multiply.)

So for N such molecules, each having the same z, we'd expect Z=z^N. But we have a problem; all the molecules had the same z because the were identical. And our original W=N!/n! count presumed they were distinguishable in the first instance so we could count N! different permutations! The correction we'll have to supply is to recognize that we couldn't recognize a different configuration brought about by swapping any two or more identical molecules;
so Z = z^N/N!

Notice that we didn't say W'=W/N!; that would give W'<1. Intolerable. Why then doesn't Z=z^N/N! suffer the same fate? Because the number of states available to molecules at normal T and P far exceeds N. There are, for example, some 6 orders of magnitude more translational wave functions available to a gas at STP than there are molecules in the gas! That is because _TRANS = (h²/8mL²)(n_X² + n_Y² + n_Z²) <<< kT for n_(X,Y,Z) = N_A.

Hence since z >> N, it must be the case that z^N >>> N!

Now I want to pull another fast one on you.

We saw last lecture how (molecular) thermodynamic energies were proportional to ln(z). Take the energy . Since it is the sum of the translational and internal energies, we find that the partition function, z, can be divided up into a product of z_trans z_int. This is because any internal energy (rotation, vibration, electronic) can be paired up with any translational state (n_XYZ), so each
exp{-(_trans + _int)} partitions into two exponential factors
exp{-_trans} times exp{-_int}.

That's like having a sum of terms X_iY_j over all i and j. Every cross term is represented. That means that it's really (X_i) times (Y_j). Hence z = z _trans z _int as a product.

BS alert: if the internal energy is the sum of rotational, vibrational, and electronic energies, then z _int = z _rot z _vib z_el, making the overall z the product of all four energy component z's.

What makes this bogus science (that is what BS stands for, isn't it?) is that the rotational and vibrational energy levels depend on parameters which differ for each electronic state! No one molecule can be in electronic state 1 but vibrating as if it were in electronic state 2, now can it?! Of course not. That ruins the statement about "every cross term" being represented. So internal energy doesn't separate as advertised!

How can we claim otherwise? Simple. Just as translational energy levels are vanishingly small compared to kT, scattering their occupancy out over enormous numbers of such states, excited electronic energy levels are almost always enormous compared to kT; so only the ground state need be considered. Indeed, z _el = g _o, the degeneracy of the electronic ground state. That factor separates out of the sum, leaving rotation and vibration to decide if they're independent.

Here we're on shakier ground. Vibrations flex the bond distances and thus moments of inertia which in turn influence rotational levels. That's how skaters spin faster by pulling their arms in to their bodies. Fortunately, excited vibrational levels too well exceed kT; so ~99% of the occupancy is of the ground vibrational state. Since it doesn't flex much, it is quasi-separable from rotation, and z _int ~ g _o z _rot z _vib.

Why have I worked SO hard to convince you that z is a separable product? Because that makes ln(z) a separable sum! Since all the thermo energy-like critters will depend on ln(z), we can expect an E _trans, E _rot, E _vib, and a (pretty dull) E _el. The same will hold for entropy, but equilibrium constant, K, being a product of z's will split up into a product of K's.

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Now combine what we've just learned. Each z is a product. And for N molecules, Z=z^N/N! So do we use that 1/N! for each z? No. Once we've determined that Z _trans=z^N _trans/N!, the molecules are no longer indistinguishable! They each have a different translational state; so even though they aren't sitting still in a gas, we could, in principle, tag them with their translational state index! So now that they are distinguishable, we can leave the 1/N! off z_int! This works because there're more than enough translational states to go around.

Thus, Z=(z^N_trans/N!) z^N_rot z^N_vib z^N_el. Not only does ln(Z) fragment that into a sum, those ^N exponents simplify to coefficients, ln(z^N) = N ln(z).
And the ubiquitous k's, as in S = k lnW, will become kN_A=R, the gas constant.

So let's do one:

E = _{i i}
= (N/z)_i exp(-_i) where z = exp(-_i)
= - (N/z) dz/d since d{exp(-_i)}/d = - _i exp(-_i)
= - d{ln(z^N)/d

= - d{ln(Z)}/d (distinguishable molecules or not since d(1/N!)/d = 0)

= -[{ln(Z)}/]_V (at fixed V otherwise varying L would change _trans)
= -[{ln(Z)}/T]_V [T/d]_V (but =1/kT so set T/=1/[/dT and voila)
= -[{ln(Z)}/T]_V (-kT ²)

E = + kT ² [{ln(Z)}/T]_V (but Z=z^N and ln[z^N]=N ln[z] so)
= + RT ² [{ln(z)}/T]_V (note the z not Z)

And the other thermodynamic energies follow similarly. We'll do S next lecture.

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

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