Conceptual question about rate laws

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Integrated rate laws are typically derived from simple reactions like A -> B, where a first-order reaction is expressed as v = -d[A]/dt = k[A]. When considering reactions with stoichiometric coefficients, such as aA -> bB, the rate law remains first order with respect to A, but the integrated form must account for the stoichiometry. This leads to the conclusion that ln[A]/[A]i = -akt for a reaction with coefficient 'a', suggesting different integrated rate laws for varying stoichiometric factors. However, the fundamental form of the rate law does not change; it is adjusted for stoichiometry to avoid confusion between reaction order and stoichiometry. The choice to simplify the presentation of rate laws in textbooks is often to enhance clarity in teaching and experimental analysis.
minivanhighwa
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So anytime I've seen textbooks explain integrated rate laws, they usually start with a reaction of the form A -> B and then from there say, if we know the reaction is first order with respect to [A] then:

v = -d[A]/dt = k[A]

And then subsequently integrate this to find ln[A]/[A]i = -kt

I get that. My issue is this. From what I always thought, the integrated rate laws are universal, and now that I have this equation I can use it any time I know that a reaction is first order with respect to A. But what if I have an equation of the form:

aA -> bB

where there is a stoichiometric coefficient in front of A. I was always under the impression that v = - (1/a) d[A]/dt = (1/b) d/dt... If experimental data still tells us that the rate law is first order with respect to A, can I just use the same integrated rate law that I found above? or would I have to say the following:

v = - (1/a) d[A]/dt = k[A]

in which case you would get:
ln[A]/[A]i = -(a)kt

So for a reaction 2A -> B

You would find ln[A]/[A]i = -2kt

Assuming the rate was always first order with respect to [A], wouldn't you get a different integrated rate law for every instance that you have a different stoichiometric factor (a) in front of your reactant? If so why isn't the general form of the equation given as

ln[A]/[A]i = -akt

Hope this makes sense - not sure where my reasoning is off
 
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I suspect this is a matter of definitions and pedagogy.

When I think of a "different rate law," I view the zero-order rate law

[A] = [A]0 - kt

to be different than the first-order rate law

[A] = [A]0*exp(-kt).

Adjusting a rate law by a constant to correct for stoichiometry does not substantively change the form of the rate law.

I suspect the reason that a more general rate law is not introduced for a few reasons, most of which has to do with avoiding confusion between order of a reaction and stoichiometry (I used to notice this happening quite a bit while a teaching assistant for introductory general chemistry) and since - if feasible - one can pick what to measure in an experiment to make your subsequent analysis of the reaction kinetics & mechanism easier.
 
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