I am learning about chemical kinetics and dynamics and as I understand for a general chemical reaction $$\ce{aA + bB -> cC + dD} $$ whose reaction rate, r, can be described by an elementary rate law can be written as follows: $$r= k[\ce{A]^a[\ce{B}]^b} $$

What I do not seem to conceptually understand is if you look at a chemical reaction from a collision theory standpoint, would it not be true that as one reactant that is in a higher stoichiometric ratio relative to the other, the probabilty that the reactants collide (assuming any collision gives a reaction) would be inversely proportional to the stoichiometry of the reactant. For example in the following: $$\ce {A + 2B -> C} $$

You basically need 2 mols of B for every mol of A for a successful reaction to produce C, so if the probability that A successfully collides with a mol of B is $\propto$ to $1/z$, then the probability that A successfully collides with TWO mols of B $\propto 1/z^2$ which is lower, so why does a reaction rate, which is $\propto$ to the probability of a collision, with an elementary rate law $\propto$ to a concentration raised to an exponent?

  • 1
    $\begingroup$ Are you acquainted with the Gillespie algorithm? $\endgroup$ Jun 14, 2018 at 6:43
  • $\begingroup$ $k$ is different for different reactions, see my comment to one of the answers. $\endgroup$
    – Karsten
    Aug 11, 2023 at 8:10

1 Answer 1


Let's divide the reaction volume into V little boxes and randomly throw ${\rm N}_A$ molecules of A into it and ${\rm N}_B$ molecules of B. The probability $p_A$ that a given box has a molecule of A in it is ${\rm N}_A$/V, in the limit that $V >> {\rm N}_A$, i.e. an ideal dilute solution, and likewise the probability $p_B$ that it has a molecule of B in it is ${\rm N}_B/V$. The probability it has two molecules of B in it is ${p_B}^2$, again in the limit that $V >> {\rm N}_B$ and ${\rm N}_B >> 1$.

Consequently, the probability all three atoms are close to one another is $({\rm N}_A/V)({\rm N}_B/V)^2$, which is proportional to the terms in your rate law. (This isn't a derivation, by the way, but only a rationalization I hope is appealing to intuition. I've swept all kinds of detail and qualification under the rug.)

Where your intuition may be leading you astray is in the assumption that squaring the molarity of B makes the RHS of your rate law bigger, and that doesn't match up with your intuition that it should be smaller if it involves a less likely event (a trimolecular collision). But that is not the case. Squaring numbers only makes them bigger if they are greater than 1, below 1 it makes them smaller. And of course the prefactor will be different, too. What the squaring really does is change the dependence of the rate law on changes in B concentration. It says the rate law is much more sensitive to changes in B concentration than A concentration. And, given that you need two B molecules to collide with each A molecules, that should be intuitively compelling.

  • 1
    $\begingroup$ If all concentrations are 1 M, you might think reactions have the same rate no matter what the rate law. However, the rate constant $k$ depends on the reaction. For termolecular reactions (three particles colliding), the expectation is that the rate constant is “supersmall” compared to dimolecular reactions. $\endgroup$
    – Karsten
    Aug 11, 2023 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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