For a reaction: $\ce{aA->product}$

The rate law is $\text{Rate}=-\dfrac{\mathrm d[A]}{a\,\mathrm dt} = k$


  • $k$ is the rate constant
  • $[\ce{A}]$ is the concentration of the reactant $\ce{A}$ in the reaction
  • $a$ is the stoichiometric coefficient of $\ce{A}$

If we integrate the relation, we get:

$$[\ce{A}]-[\ce{A}]_0=-akt $$

However, my instructor wrote the relation on board as below;


Why is $a$ not needed in the integrated rate law?

  • $\begingroup$ Because what you are referring to (with the inverse of the stoichiometric coefficient multiplied with d[A]/dt)is the reaction rate(r). This rate is related to the concentrations of the reactants via the rate equation(r= f([A]) which can only be determined experimentally. If it is zero order, you already know that d[A]/dt= [A]^(0). Hence there is no role of a here $\endgroup$ – YUSUF HASAN Dec 15 '18 at 14:38
  • $\begingroup$ I think you take the rate expression of the reaction as $\frac{1}{a}\times(\text{ the rate of disappearance of}\mathrm{A}$), and your instructor take the rate expression of the reaction as equal to (the rate of disappearance $\mathrm{A}$) . Both cases accurately describe the kinetics, but different in defining the rate and $\mathrm{k}$. $\endgroup$ – Adnan AL-Amleh Dec 16 '18 at 0:20

You are right, for consistency the stoichiometric coefficient should be retained. Keeping track of the stoichiometric coefficients is essential to convert between rates for different species.

If you want to compare the rate of formation of reagent A and product, you write

$$\text{Rate}=-\frac{\mathrm d[A]}{a\,\mathrm dt} = \frac{\mathrm d[product]}{\,\mathrm dt}= k$$

and so:

$$\ce{[A]-[A]_0=-a*k*t} $$





However you would say that the reagent A is consumed with an effective rate constant $\ce{k' = a*k}$.


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