[Mimic]

We have the reaction
[tex]\ce{A}  \rightarrow \ce{B}[/tex]
in a discontinuous system with constant volume, the speed [itex]r[/itex] with this reaction reaches equilibrium is given by the formula
[tex] r = \dfrac{-\text{d[A]}}{\mathrm{d}t} [/tex]
Experimentally it has been observed that the speed of a reaction is a function of the temperature, through the kinetic constant [itex]k_r[/itex], and of the concentration of the reactants or of some of them, each high for a certain number obtained experimentally
[tex] r = k_r[\mathrm{A}]^x [/tex]
where [itex]x[/itex] is called the reaction order.
My question is this: since the reaction rate is defined by -d [A]/dt, where does the need to express it through the kinetic law?

[Babcock_Hall]

I would plug the definition of r into your second equation.  Then I would consider two cases, x = 1 and x = 2.  What is different about the time courses of these two cases?

[Mimic]

I would plug the definition of r into your second equation.  Then I would consider two cases, x = 1 and x = 2.  What is different about the time courses of these two cases?

With [itex]x = 1[/itex], r increases linearly with the A concentration
[tex] r = \dfrac{-\text{d[A]}}{\mathrm{d}t}  = k_r [\mathrm{A}] [/tex]
With [itex]x = 2[/itex], r will depend on the A concentration square
[tex] r = \dfrac{-\text{d[A]}}{\mathrm{d}t} = k_r [\mathrm{A}]^2 [/tex]
from which it can be deduced that a greater quantity of A will be required to increase r

[Babcock_Hall]

I don't think that I agree with your last sentence.  In any case, another way to think about your question is to integrate the two equations.  The case in which x = 1 produces a situation in which the half-life for [A] is a constant period.  The case in which x = 2 produces a different half-life.