# In collision theory why we multiply collision frequency by N/2?

In kinetic theory of gases we know that the average number of collisions, $$N_\mathrm{col}$$, is given by

$$N_\mathrm{col} = \sqrt{2\pi\sigma^2 \overline{v}_\mathrm{rel}}\ N$$

where $$N$$ is the number of particles, and $$\overline{v}_\mathrm{rel}$$ its average relative velocity.

If we consider bimolecular collisions we should have divided the total number of collisions $$N_\mathrm{col}$$ by $$2$$. But instead we multiply it by $$N/2$$. Why?

• I guess when you refer to bimolecular collisions you are really meaning a collision between like atoms or molecules (A + A) instead of disimilar species (A + B). You have to take into account that you have N molecules of each kind and ieach of this molecules can collide with the rest of the molecules in the sample.
– PAEP
Aug 6, 2023 at 16:33
• 2 objects can only collide 1 way: a-b. 3 objects can collide 3 ways: a-b, a-c and b-c. 4 objects can collide 6 ways: a-b, a-c, a-d, b-c, b-d, and c-d. The more objects, the more ways they can collide, and the more collisions, so N is a factor. Aug 6, 2023 at 19:43
• @paep I'm not sure I get the meaning of the angle signs. In MathJax They're set as relational operators. I'm confident that's not what you wanted to achieve. Aug 6, 2023 at 23:27
• @DrMoishePippik So this means that using the factor 'N' along with 1/2 gives us an accurate measurement of the number of collisions Aug 7, 2023 at 1:35
• Thanks for the tip @Martin-マーチン. I meant to write $\overline{v}_{rel}$.
– PAEP
Aug 7, 2023 at 10:33

I'm not sure if this answers your question, but in the special case where a collision/reaction combines two identical molecules $$\mathrm{A+A}$$ the number of indistinguishable combinations, i.e. the number of distinct A-A pairs of the two species must be calculated and this makes the bimolecular rate $$\displaystyle \sim \frac{N_A(N_A - 1)}{2!}$$ collisions. Normally, in dealing with chemical kinetics, $$N_A$$ is so vast that we can use $$\displaystyle N_A^2/2$$ without any error.
• This carries similarities to a single-rounded round-Robin tournament with $n$ participants, and $\frac{n(n-1)}{2}$ games (every participant plays only once with every other participant than him/herself). Note, the denominator I used here is $2$, not $2!$ (though indeed $2! = 2$ is true). Aug 7, 2023 at 8:45
• @Buttonwood only for interest, in a termolecular reaction the chance is $n(n-1)(n-2)/3!$, which is needed if a Monte Carlo method are used to calculate populations during reactions, rather than $n^3/6$, not that termolecular reactions actually ever occur except in a computer :) Aug 7, 2023 at 11:05
• @porphyrin Then (I now speculate) $2!$ is the more general, easier extendable approach to describe the situation with mathematics. Aug 8, 2023 at 7:02