In zero order why we don't need the coefficient of reactant in integration of rate law?

Question

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

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

Where:

• $k$ 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;

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

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

Answer 1

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:

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

or

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

and

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

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