3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think that the equation of $P = N/N*$ lacks a ^(-1) exponent in the RHS. $\endgroup$ – user1420303 Sep 18 '16 at 15:11
  • $\begingroup$ @user1420303 I think you might be right, silly typo I'll fix it $\endgroup$ – getafix Sep 18 '16 at 15:15
  • $\begingroup$ Yes, that always will work. Although this does not answer your question, as you stated "after singling out this critical bond, we have $n−n^∗$ quanta left, and $s−1$ oscillators". $\endgroup$ – user1420303 Sep 18 '16 at 17:23
3
$\begingroup$

Your equations for N and $N^*$ are correct.

The first N is the total number of ways to distribute n quanta among s oscillators in energy range E to $E+dE$.

To calculate $N^*$ we need the total number of ways to reach a distribution that leads to reaction with $n^*$ number of quanta in the critical oscillator. This is given by the total number of arrangements (quanta into oscillators) having energy 0 to E, which equals $$N^*=(k+s)!/(k!s!)$$ with k quanta and s oscillators. To obtain this last equation ( $N^*$) we sum $N=...$ equations i.e. $$N^* =\Sigma _{m=0}^k \frac{(m+s-1)!}{m!(s-1)!}$$ and is most easily seen by direct counting, say, with say $n=4$ quanta and $s=3$ oscillators. You can confirm the equation for N in the same way.

For a maximum of $k=n-n^*$ quanta in the $s-1$ remaining oscillators your $N^*$ is produced.

$\endgroup$
  • $\begingroup$ Hi! Thanks for your answer. I am sorry but I don't follow completely, I wish to know why the expression for $N^*$ is what it is instead of $N^* = \frac{(n-n^*+s-2)!}{(n-n^*)!(s-2)!}$ (which is what I thought initially). Any insight would be appreciated. Am I missing something? Should I pick up a book on combinatorics instead? haha $\endgroup$ – getafix Sep 18 '16 at 10:28
  • 1
    $\begingroup$ Hi, I've added more explanation, hopefully this will clarify things for you. The important point is that $n^*$ is the total number of arrangements from energy 0. $\endgroup$ – porphyrin Sep 18 '16 at 15:47
  • $\begingroup$ Indeed, I think things are a lot clearer now. I edited and updated my question a little while ago, with a direct counting exercise and the result make sense to me now. Perhaps you can take some time corroborate that? $\endgroup$ – getafix Sep 18 '16 at 15:55
  • 1
    $\begingroup$ Direct counting works just like putting any number of balls in to bags, say 4 balls into 3 bags in all possible arrangements. This gives you N. Add up all those similar arrangements from n equal to zero to 4 to get $N^*$. If you try with large numbers it becomes impossible. Alternatively sum algebraically to confirm $N^*$ $\endgroup$ – porphyrin Sep 18 '16 at 16:02
1
$\begingroup$

I find some flaws in the reasoning.

As a secondary problem, braking a bond give rise to two fragments, each of which must be treated with their correct number of normal vibrational modes.

Imagine one molecule with 11 atoms and $3N-6=27$ vibrational modes. After a bond break you can have for example: a chemical specie with 6 and other with 5 atoms, with $3*6-6=12$ and $3*5-6=9$ vibrational modes. So neither $s-1=27-1=26$ or $s-2$ make sense.

It is more important conceptually that as you are interested in the rate constant, you must concentrate the analysis in the "present" time. It is not a comparison of actual and future states. Molecules that already reacted can not be used in this context to deduce how the energy distribution in the molecules that did not react yet influence them in relation of velocity.

That is, it is still used $s-1$ because you are trying to get

The probability of of energy concentrating into one oscillator as described above

for reactive molecules.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.