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Kinetic law

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**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**:

--- 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 [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.

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