[Big-Daddy]

If I have a zero-order reaction A → Products with the rate (rate of loss in concentration of A) shown by rate=k (where k is the rate constant for the rate of loss in concentration of A; since it is zero-order, k is equal to the rate), and I integrate to find kt=A₀-A, is it true to say that, if the reaction were instead aA → Products, the integrated law would still be kt=A₀-A? Or would it now be akt=A₀-A? And how do I find k given the value of a, from a graph of A concentration (y-axis) against time (x-axis)? If a=1, then gradient=-k, but what if a does not =1?

[curiouscat]

Those are two different reactions. You cannot predict one rate const. from the other.

If for both [ rate  = rate of loss in concentration of A ] then kt=A0-A holds for both. Depends entirely on how you define k.

In any case k'=ak so it's all about constants.

k = Slope /a ; doesn't matter if or not a = 1.