So I agree with the other posts that it's probably not that important. However, I can show you a quick derivation that may or may not help:
For first order:
where the rate is dependent on reactant A:
Rate=-d[A]/dt=k[A]
right, because, rate is a change in concentration over time and since it's a reactant the concentration of [A] will be decreasing to that means that to get the proper rate you need to make it negative. Anyway, so that equals k[A] because we said that rate depends on concentration of A, and k is just a constant.
Ok, so now we rearrange to get all the [A] on one side and all the t on the other side:
-d[A]/[A]=kdt
if you integrate both sides you get -ln[A]=kt
actually, you should integrate from Ao to A but that will basically give you the same thing, I'm just trying to show you why it's negative ln
ok, for second order:
it's the same thing except this time it will depend on the concentration of two reactants (which might be two molecules of A or a molecule of A and another reactant or w/e). We'll just use two molecules of A for simplicity.
Rate=-d[A]/dt=k[A]^2
Ok, so same thing, rearrange and you get
-d[A]/[A]^2=kdt
now integrate and you get 1/[A]=kdt
again you should actually do it from Ao to A and t(0) to t but this shows you where the 1/[A] comes from.
oh, actually, so here's how you're gonna remember it:
the relationship comes from the integral of -1/x^n where n is the order of the reaction. for n=1, it's the integral of -1/x which is -ln(x). when n=2, it's the integral of -1/x^2 which is 1/x. hope that helps.
