Kinetic proof of law of mass action Suppose we have a chemical reaction of the form
$$n_1 \mathrm{A}_1 + \cdots + n_r \mathrm{A}_r \rightleftharpoons m_1 \mathrm{B}_1 + \cdots + m_s \mathrm{B}_s$$
where $\mathrm{A}_i$ and $\mathrm{B}_i$ are molecules, and the $n_i$ and $m_i$ are the integer coefficients of the equation. Let $[\mathrm{X}]$ be the concentration of $\mathrm{X}$, that is, the amount of moles per unit volume. The law of mass action says that at chemical equilibrium, the following holds:
$$
\frac{[\mathrm{A}_1]^{n_1}\cdots[\mathrm{A}_r]^{n_r}}{[\mathrm{B}_1]^{m_1}\cdots[\mathrm{B}_s]^{m_s}} = K(T)
$$
In Fermi's Thermodynamics, he gives a proof that uses kinetic theory (slight edition and emphasis by me):

A reaction from left to right can occur as a result of a multiple collision involving $n_1$ molecules $\mathrm{A}_1$, $n_2$ molecules $\mathrm{A}_2,\,\dots,\,n_r$ molecules $\mathrm{A}_r$. The frequency of such multiple collisions is obviously proportional to the $n_i$th power of $[\mathrm{A}_i]$, that is, to the product:
$$ [\mathrm{A}_1]^{n_1}\cdots[\mathrm{A}_r]^{n_r}$$

That is not obvious at all for me. Could someone go into a little more detail explaining why the frequency of collisions should be proportional to $[\mathrm{A}_i]^{n_i}$?
 A: Think of the volume as divided into many unit cells. The probability that a molecule of type $\mathrm{A}_i$ is present in the cell is proportional (in the leading term) to the number of molecules of type $\mathrm{A}_i$. The combined probability of having $n_1$ molecules of type $\mathrm{A}_1$, etc, is the product of the individual probabilities (assuming independent events). This boils down to concentration products.
A: This may be obvious to Fermi, but not to the rest of us. 
One has to keep in mind the context of the derivation,  Fermi is talking here about ideal gases.  In other words, gases which are very dilute, such that the interparticle distance is much much larger than the range of the interparticle interaction.    This means that particles don't see each other very often, and that the probability of a reaction occurring in the time interval that it takes a particle to travel a distance equal to the interparticle spacing is $\ll 1 $.  
To calculate the frequency of collisions (that is the number of collisions per unit volume per unit time) let us find the probability that all collision reactants are in a given volume $\Delta V$ within a certain time interval $\Delta t$. 
The probability of finding one particle of reactant $A_{i}$ in the volume $\Delta V$ and during time interval $\Delta t$ is given by 
$$ P_{i} = [A_{i}] \Delta V \Delta t / \tau$$ 
where $[A_{i}]$ is the concentration of the reactant and $\tau$ is the average time between collisions.    Notice that this expression is only valid if: 


*

*$\Delta V \ll V$, the reaction volume we consider is much smaller than the total volume of the system

*$\Delta t \ll \tau$, the reaction time is much smaller than the average time between particle collisions.  


In other words, the expression above for the probability is only valid when you get a very small probability out of it ($P_{i} \ll 1$).   
Now, let's consider the probability of finding all the reactants in the same volume element $\Delta V$ at the same time interval $\Delta t$.   Here is where the stoichiometric coefficients come in.   As an example let's consider the following reaction 
$$ 3 A_{1} + A_{2} \rightarrow B $$ 
The probability of finding 3 particles of $A_{1}$ and 1 of $A_{2}$ in a given volume $\Delta V$ and time interval $\Delta t$ is equal to
$$ P = [A_{1}]\Delta V \Delta t / \tau \times \left( [A_{1}] - \frac{1}{V} \right) \Delta V \Delta t /\tau \times \left([A_{1}] - \frac{2}{V} \right) \Delta V \Delta t/\tau \times [A_{2}] \Delta V \Delta t/ \tau $$ 
Notice that the probabilities of the 2nd and 3rd particles of $A_{i}$ are conditional probabilities, conditioned on the fact there are already particles of $A_{i}$ in the volume element. Since the gas is dilute, those conditional probabilities can be estimated by using the the effective concentration of $A_{1}$, which is only the original concentration reduced by the number of particles already present in the volume element.
As long as the stoichiometric coefficients are much smaller than the number of particles in the total volume $V$, that is $n_{i} \ll [A_{i}]V$, we are safe in making the approximation 
$$ P \approx  ([A_{1}] \Delta V \Delta t/\tau)^{3} \times [A_{2}] \Delta V \Delta t /\tau$$ 
In the general case, we see that the collision rate, $K$, will be proportional to the probability $P$ per unit volume and unit time: 
$$ K \propto  \frac{ \prod_{i} ( [A_{i}]^{n_{i}} \Delta V \Delta t/\tau \,)}{ \Delta V \Delta t  } $$  
if we absorb all of the volume elements and the time intervals and $\tau$'s into a proportionality constant, $K'(T)$, which is specific to the particular reaction (and, as kinetic theory tells us, depends on temperature), then we have the result:  
The frequency of reactions from left to right is given by  
$$ K'(T)  \prod_{i} [A_{i}]^{n_{i}} $$ 
Is that it? 
In one of the comments somebody asked the following question:   

But when calculating the frequency of collisions, wouldn't you have to take into account the probability of having a number of molecules not only equal to $n_{i}$ but also greater? Since, after all, the reaction should occur when you have at least $n_{i}$ molecules of type $A_{i}$, not exactly $n_{i}$.

The answer to that question is that such probability is very small and therefore it can be neglected.    In fact let's look at the ratio of the probability of having $n_{i}+1$ particles of $A_{i}$  to the probability of having $n_{i}$ particles in the same volume element and at the same time interval:  
$$ \frac{ ([A_{i}] \Delta V \Delta t / \tau )^{n_{i}+1} } {([A_{i}] \Delta V \Delta t /\tau )^{n_{i}} } =  [A_{i}] \Delta V \Delta t /\tau \ll 1 $$ 
The probability of having more than $n_{i}+1$ particles of reactant $A_{i}$ in the volume element is so small compared to the probability of having just $n_{i}$ that it simply does not matter.  
All of this was very obvious for the great Fermi. 
A: As known, collision frequency is the number of collisions among the reacting species that takes place per second per unit volume of the reaction mixture.  
For instance, under ordinary conditions of temperature and pressure, in a gaseous system, the collision frequency of binary collisions is of the order of $10^{25}$ to $10^{28}$. But all the collisions among the reacting species at a given temperature are not effective in bringing about the chemical reaction.  
It is to be noted that rate of reaction is proportional to collision frequency, greater the number of collisions, greater is the probability of formation of the products in a short time. In other words, collision frequency is proportional to rate of the reaction. From law of mass action we know that rate of reaction is proportional to $[A_i]^{n_i}$. Therefore, it is obvious that collision frequency is proportional to $[A_i]^{n_i}$.  

Credits: Modern's ABC of Chemistry-XII-2013 Edition-Page.No.352 *Some modification has been done.
