Oxidation of formate ion by peroxydisulfate in water solution: Why are there "strange" kinetics? As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: 
$r=k\ce{[HCOO^{−}]}^{1/2} \ce{[S2O8^{2−}]}$. 
The first reaction is very slow, so most of the peroxydisulfate is consumed in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.

WHAT does this mean?
So, why is the rate constant NOT according to the slowest step? 

Why it is not just $k\ce{[S2O8^{2-}]}$?
The rate of the third reaction affects the rate of the first reaction, i.e. they are competitive reactions and in that case the rds method can't be applied.

The rate of the first reaction is $r=-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t$, but the change of the concentration of the peroxydisulfate depends on the third reaction, too:
$$-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t=k_1\ce{[S_2O_8^{2-}]} + k_3 \ce{[CO_2^-]}\ce{[S_2O_8^{2-}]}$$