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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 $k\ce{[S2O8^{2-}]}$?

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    $\begingroup$ I think you already have got an answer to your question here. $\endgroup$ – Philipp Oct 8 '14 at 21:27
  • $\begingroup$ @Philipp No but I am m=not understanding the meaning of that line $\endgroup$ – DSinghvi Oct 9 '14 at 12:12
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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: $$-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t=k_1\ce{[S_2O_8^{2-}]} + k_3 \ce{[CO_2^-]}\ce{[S_2O_8^{2-}]}$$

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  • $\begingroup$ Could you please elaborate $\endgroup$ – DSinghvi Nov 12 '14 at 14:02
  • $\begingroup$ but that's only possible if the 3rd step is slow not fast $\endgroup$ – DSinghvi Nov 12 '14 at 14:55
  • $\begingroup$ Why do you think so? Regardless of the rate of the third reaction, it will affect the concentration of the peroxydisulfate, even more so if it has a big rate. $\endgroup$ – RBW Nov 12 '14 at 14:59
  • $\begingroup$ O ya right..but than there are many - many step reactions where first step is slow and has a rate constant accordingly so why does it have a different rate costants $\endgroup$ – DSinghvi Nov 12 '14 at 15:20
  • $\begingroup$ Because the reactant of the slowest, in this case the first, step participates in some other step, too. In most introductory examples in kinetics, that's not the case and there you can simply apply rds. $\endgroup$ – RBW Nov 12 '14 at 15:40
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In this answer, I will attempt to explain using chemical principles, rather than detailed mathematics.

As has been said already, reactions 1 and 3 both consume $\ce{S2O8^{2-}}$. Now, for the rate determining step to dominate the decay of this species, there would necessarily be very little consumption of it in any other steps. IE, there would be little consumption of it in reaction 3, implying that there is very little supply of the $\ce{CO2^-}$ radical.

Now, unfortunately for the complexity of the kinetics, there is a rapid interconversion of $\ce{SO4^-}$ to $\ce{CO2^-}$ by reaction with formate, which in turn rapidly reacts with more $\ce{S2O8^{2-}}$ to reform the $\ce{SO4^-}$ which can react again with formate, you get the picture. So, although there will only be a very small concentration of $\ce{SO4^-}$ and $\ce{CO2^-}$ present (by virtue of the slow reaction in step 1), the sheer relative speed of formate induced interconversion makes these other steps important. Of course, the more formate we have, the more $\ce{CO2^-}$ we can liberate (to a point, hence the square root), and the faster we can remove $\ce{S2O8^{2-}}$. This is, in fact, a catalytic cycle within the greater reaction scheme.

This is what is meant by the steady state approximation. As the concentrations of $\ce{SO4^-}$ and $\ce{CO2^-}$ are always very small, we can assume that their total rate of change across all reactions in the scheme is (extremely close to) zero. This allows us to write a set of simultaneous equations from which we can try to determine the (often quite complex) rate law by estimating the "steady state concentrations" of the low concentration species in terms of the other species of interest.

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  • $\begingroup$ I have updated your use of mathematical/ chemical environments. The way you tried to use it, it sometimes breaks at the end of the lines, making it very hard to understand. $\endgroup$ – Martin - マーチン Nov 18 '14 at 6:04

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