# Reaction rate coefficient constant multiplier for equation with multiple atoms consumed?

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 $$\frac{d[O]}{dt} = -2 \cdot k_f \cdot [O] \cdot [O]$$

or

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

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

• 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 $\ce{O2}$. – Zhe Oct 16 '18 at 17:43

In general if the reaction is $$\ce{aA + bB \to cC +dD}$$ the the rate is written as $$r=\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=+\frac{1}{c}\frac{d[C]}{dt} = k[A]^\alpha[B]^\beta]\cdots$$ and the reaction order is $$n=\alpha + \beta +\cdots$$.

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.

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