What's the derivation of the 'Kinetic Equation' in Chemical kinetics?

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The discussion revolves around the derivation and understanding of the chemical kinetics formula v = k[A]^m[B]^n, which describes the reaction rate based on the concentrations of reactants A and B. The initial inquiry seeks a rigorous derivation or logical interpretation of this formula, particularly how it relates to reaction orders. It is explained that for first-order reactions, the rate is proportional to the concentration of a single species, while for second-order reactions, the rate depends on the interaction frequency between two species, correlating to the product of their concentrations. The conversation confirms that the formula was initially hypothesized and later validated through experimental evidence.
Ale_Rodo
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Hi,

I'm following an introduction course to chemistry and I am reviewing the chapter on Chemical kinetics.
It's shown that the reaction speed for a certain component of a general chemical equation such as aA +bB <-> cC + dD , might be expressed as v = k[A]m[ B]m.

I was wondering where it does come from. It's just plain curiosity, I don't really need to know this for the upcoming exam but I would really appreciate if someone could give a rigorous derivation or a 'sense-full' logic interpretation of said formula.

Thank you in advance.
 
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It goes something like this: For a first order reaction involving a single molecular species, the reaction rate has to be proportional to the amount of the species, as represented by its concentration. For a 2nd order reaction, involving interaction of two molecular species, the rate of the reaction depends on the frequency of encounters between the two species, which is proportional to the product of their concentrations.
 
Was the formula first supposed like that and the experimentally proved?
 
Ale_Rodo said:
Was the formula first supposed like that and the experimentally proved?
If I understand your question correctly, then, pretty much yes.
 
Chestermiller said:
If I understand your question correctly, then, pretty much yes.
Fair enough this time, thanks!
 
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