reaction order (A > B) 
A(t) 
linear graph 
t _{1/2} 
O^{th} 
A =  k t + A_{o}  A vs. time

[A]_{o} / (2k) 
1^{st}  ln A =  k t + ln A_{o} 
ln A vs. time

ln 2 / k 
2^{nd}  (1 / A) = k t + (1 / A_{o})  (1 / A) vs. time

1 / (k [A]_{o}) 
Derive equation for 1^{st} and 2^{nd} order reaction with a single reactant [having more than a single reactant, raises the complexity, so it won’t be shown]
For a 1^{st} order reaction: A > B,
[1]
where describes the slope / derivative of the curve in a graph of [A] vs. time and is referred to as a differential equation. Given the expression for , our goal is to determine A(t), which is a function that describes the graph plotting [A] versus time, t. To "solve" the above differential equation, “take the integral” of [1]:
ln A =  k t + C
at t = 0, [to evaluate the value of C]
ln A_{o} = C
substitue
ln A =  kt + ln A_{o}
ln A  ln A_{o} =  k t
or A = A _{o} e^{ k t } [see below for an alternative equation]
For the 2^{nd} order reaction: A > B,
where describes the slope / derivative of the curve in a graph of [A] vs. time. As above, solving this differential equation:
at t = 0,
substitue,
A similar method would be used to treat a 0^{th} order reaction: A > B [not shown]
Notice that A(t) for a 1^{st}& 2^{nd} order reaction is on the "ap equation sheet".
noncalculus derivation of rate law.
derive rate law from Reaction mechanism  steady state approximation method applied to a biology example. [file in pdf format]
a slightly more realistic treatment for the reaction A <> B, which unfortunately has a somewhat complicated analysis; mathematica simulation.
Alternative equation describing the solution to the above 1^{st} order differential equation (supplied by Su Tran, Ms. Saucedo, & Mr. Ring; 2006)
Using “my equation”
where
At the halflife,
substituting [2] into [1],