# Kinetics of molecule reacting with another of itself

For the following reaction:

$$\ce{O3 + O3 -> 3O2}$$

The rule (which makes sense for combination of two different moluecules) states that the rate constant R should be:

$$R \propto \ce{[O3][O3]} = \ce{[O_3]^2}$$

But in the case of a combination of the same molecule this seems counter-intuitive. If you had double the amount of $\ce{O3}$, wouldn't the rate simply double (instead of increasing as the square)?

If you had double the amount of $$\ce{O3}$$, wouldn't the rate simply double (instead of increasing as the square)?
$$\ce{A + B -> C}$$
We know that: $$\ce{\frac{d[\ce{P}]}{dt}} \propto \ce{[A][B]}$$
and that doubling $$\ce{[A]}$$ doubles $$\ce{\frac{d[\ce{P}]}{dt}}$$ because when we substitute $$[2\ce{A}]$$ for $$\ce{[A]}$$, we get: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A][B]}=2\ce{[A][B]}$$
If $$\ce{A}=\ce{B}$$, however, then it follows that we must also substitute $$[2\ce{A}]$$ for $$\ce{[B]}$$, which then gives us: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A}][2\ce{A}]=4\ce{[A][A]}=4\ce{[A][B]}$$