# Why are the derivatives of the concentrations divided by stoichiometric coefficients?

Consider this simple dioxygen to ozone chemical equation

$$\ce{3 O2 -> 2 O3}$$

with the average rate of the reaction

$$r = -\frac{1}{3}\frac{\Delta[\ce{O2}]}{\Delta t} = \frac{1}{2}\frac{\Delta[\ce{O3}]}{\Delta t}.$$

So, the rate of change of concentration of the reactant $$\ce{O2}$$ must be divided by its stoichiometric coefficient, i.e., by $$3$$ and so does that of $$\ce O_3$$, i.e., by $$2$$.

But what I think after doing this is that by doing this division, we are no longer referring to this particular reaction as it requires $$3$$ molecules of $$\ce{O2}$$ and to form $$2$$ molecules of $$\ce{O3}$$. Why should the reaction rate, if in terms of reactants here, be divided by the stoichiometric coefficient $$3$$ if there are $$3$$ molecules taking part in the reaction and thus $$3$$ molecules contributing to the rate?

• I've taken a liberty to correct technicalities, but the title needs to be rewritten to reflect the question, which I'm afraid I don't quite understand. You appear to apply the rate expression by its definition, whereas the mechanism is a bit more complex (see also Chapman mechanism). This might possible the source of confusion. Jun 1, 2023 at 4:45
• Rate is the property of the reaction, and not of O2 or O3. There is no point in treating them separately. Jun 1, 2023 at 6:29

For $$\ce{3 O2 -> 2 O3}$$ and

$$r = -\frac{1}{3}\frac{\mathrm{d}[\ce{O2}]}{\mathrm{d} t} = \frac{1}{2}\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d} t},$$

you can consider the reaction rate as the rate of formal elementary operations of taking 3 molecules of $$\ce{O2}$$ to form 2 molecules of $$\ce{O3}$$. The reaction rate $$\pu{1 mol/L/s}$$ is then formally $$\pu{1 mol/s}$$ of such operations in $$\ce{1 L}$$. You should consider that whatever component concentration rate you take, it has to lead to the same reaction rate, as the reaction rate does not depend on what component you choose.

1 operation takes $$\ce{3 O2}$$ molecules to make $$\ce{2 O3}$$ molecules. $$\pu{1 mol}$$ of operations takes $$\pu{3 mol}$$ of $$\ce{O2}$$ to make $$\ce{2 mol}$$ of $$\ce{O3}$$.

Now all what remains is making rates from them, as above.

It is closely related to $$\ce{3 O2 -> 2 O3}$$ reaction enthalpy $$\Delta H_\mathrm{r} [\pu{kJ/mol}]$$,
which is enthalpy change for "$$\pu{1 mol} \text{ of reaction operations}"$$, leading from $$\ce{\pu{3 mol} O2}$$ to $$\ce{\pu{2 mol} O3}$$.

Be aware it is not an elementary reaction and the rates for O2 and O3 do not exactly fit to each other with 3:2 ratio.