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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 -> B

where there is a stoichiometric coefficient in front of A. 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