Was reviewing some of my physical chemistry lectures and attempted a a derivation of the unimolecular gas phase reaction using a calculation based on the RRK model (Rice-Ramsperger-Kassel), and ran into some trouble half way into the derivation.
My Derivation
I have a unimolecular gas phase reaction that takes place as follows:
$$\ce{A}+ \ce{A} \rightleftharpoons^{k_a} \ce{A^*} + \ce{A}$$ $$ \ce{A^*}\rightarrow^{k_b} \ce{P}$$
Where, $\ce{A}, \ce{A^*}, \ce{P}$ represent the reactant, energetically excited reactant and product respectively.
Now, the way I understand it is that the RRK model is premised on the fact that the even though a molecule may have enough energy to react, said energy is distributed over all available modes of motion, and it has to be funnelled into a particular mode, (say particular vibrational mode to cause bond cleavage) for the reaction to proceed.
So I suppose that my reactant system has $s$ identical harmonic oscillators, and as an approximation all of them have the same frequency $\nu$. These vibrational modes are excited by $E = n h\nu$. The number of ways of distributing is $$N = \frac{(n+s-1)!}{n!(s-1)!}$$
The rationale behind this is that I have $n$ quanta that need to be distributed over $s$ oscillators. I think of these as $s$ containers, created by using $s-1$ partitions. The numerator is the total no. of arrangements of $(n+s-1)!$ and since the $n!$ and $(s-1)!$ arrangements of the objects are indistinguishable we divide by the terms in the denominator.
This is the reasoning that was presented to us in our course, and that I have attempted to reproduce
Now, let's single out a bond that will break if it is given energy (at least) $E^*= n^*h \nu$
So, after singling out this critical bond, we have $n-n^*$ quanta left, and $s-1$ oscillators (i.e $s-2$ partitions, if I am to follow the rationale described previously).
However, this is where I believe I run into a problem, and my derivation disagrees with what was presented in class, namely
$$N^* = \frac{(n-n^*+s-1)!}{(n-n^*)!(s-1)!}$$
My question is that shouldn't there be an $(s-2)$ term instead of an $(s-1)$ in the expression for $N^*$. It's been a few months, and I don't remember the reasoning presented in class, and since we have summer break right now, I can't exactly ask my professor. Am I missing something?
Also, I find this derivation a little bit convoluted, if someone can provide a alternative, or suggest a modification.
Anyway, using the expression for $N^*$ given above, I do eventually end up with the correct result.
The probability of of energy concentrating into one oscillator as described above is $$P = \frac{N^*}{N} = \frac{n! (n-n^*+s-1)!}{(n-n^*)!(n+s-1)!}$$, and assuming $ (s-1) \ll n-n^*$,
$$ P \approxeq \left (\frac{n-n^*}{n} \right )^{s-1} = \left (1 -\frac{E^*}{E} \right )^{s-1} $$
and thus, the Kassel form of the unimolecular rate constant is
$$k_b (E) = \left (1 -\frac{E^*}{E} \right )^{s-1} k_b \qquad \text{for} \quad E\geq E^*$$
UPDATE I may/may not have answered my own question. I spent some time reading about combinatorics this evening and came up with this little stars and bars exercise to work out my problem:
Say we have 20 quanta ($n$) available, and we have 6 ($s$) oscillators. Using stars to represent the quanta, and 5 bars ($s-1$) to represent the partitions, I can draw something like this:
$\text{**|***|***|*****|*****|**}$
The no. of ways of arranging this collection is $N$ as described above.
Now, I assume that the disassociation requires at least 6 ($n^*$) quanta:
$\text{******} \quad \text{|*|***|**|****|****}$
The left most collection is the critical oscillator (say we use exactly 6 quanta), and we hold that fixed. What is left on the right is $n-n^* = 20-6 = 14$ quanta and the 5 bars (still, $s-1$ partitions), and the total no. of arrangements is $N^*$ above. Similarly, now if we have focus 7 quanta in our critical oscillator (we still need a min. of 6,so we have one extra), I can draw something like:
$\text{******} \quad \text{*|*|***|**|****|***}$
Again, the leftmost collection is the critical oscillator. And our combinatorial problem is still of arranging $n-n^* = 14$ quanta and $s-1 = 5$ partitions. I feel this works; could someone please tell me otherwise if this not the case.