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Kinetic law

(1/1)

Mimic:
We have the reaction
$$\ce{A} \rightarrow \ce{B}$$
in a discontinuous system with constant volume, the speed $r$ with this reaction reaches equilibrium is given by the formula
$$r = \dfrac{-\text{d[A]}}{\mathrm{d}t}$$
Experimentally it has been observed that the speed of a reaction is a function of the temperature, through the kinetic constant $k_r$, and of the concentration of the reactants or of some of them, each high for a certain number obtained experimentally
$$r = k_r[\mathrm{A}]^x$$
where $x$ 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:

--- Quote from: Babcock_Hall on July 02, 2022, 07:50:05 AM ---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?

--- End quote ---

With $x = 1$, r increases linearly with the A concentration
$$r = \dfrac{-\text{d[A]}}{\mathrm{d}t} = k_r [\mathrm{A}]$$
With $x = 2$, r will depend on the A concentration square
$$r = \dfrac{-\text{d[A]}}{\mathrm{d}t} = k_r [\mathrm{A}]^2$$
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.