For the elementary reaction $\ce{2O -> O2}$, when calculating the rate of concentration change of atomic oxygen as $\frac{d[O]}{dt}$, would I use the standard rate coefficient $k_f$ or would I have to multiply by 2 since 2 oxygen molecules are being consumed?

$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -2 \cdot k_f \cdot [O] \cdot [O]$


$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -k_f \cdot [O] \cdot [O]$

Or would the rate constant define the dependency on atomic oxygen?

  • 2
    $\begingroup$ Given that equation, the rate is for production of $\ce{O2}$. So you do need to factor for the stoichiometry of $\ce{O}$ relative to $\ce{O2}$. $\endgroup$
    – Zhe
    Oct 16, 2018 at 17:43

2 Answers 2


In general if the reaction is $\ce{aA + bB \to cC +dD}$ the the rate is written as $$r=\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$$.

If the reaction is indeed second order in oxygen atoms as stated by the OP, then the first equation given by the OP is correct, but unusual (you could just leave the factor two in the denominator on the left hand side instead of multiplying by two on the right hand side).


When we measure the rate of the reactant concentration change (rate of disappearance of the reactant) or the rate of formation of the product experimentally, we neglect the coefficient and equation, but we take the balance equation into consideration when calculating the rate of disappearance or formation of each species in the reaction, so: rate of disappearance of oxygen atoms = double the rate of formation of oxygen molecules.

$\ce{\frac{d[O]}{dt} = - k \cdot [O]^2}=2 \cdot{\ce{\frac{d[O2]}{dt}}}$


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.