I have been reading through a case study about calculating the order of reaction for the following reaction:

$$\ce{H2O2 + 3 I- + 2 H+ -> I3- + 2 H2O}$$

The paper uses an iodine clock reaction with vitamin C to calculate its order of reactions, experimentally. However, when writing out the rate law, the author writes it as such:

$$\mathrm{rate} = k[\ce{H2O2}]^x[\ce{I-}]^y$$

How come it is not written as follows:

$$\mathrm{rate} = k[\ce{H2O2}]^x[\ce{I-}]^y[\ce{H+}]^z\;?$$

For what reason is the order of reaction of hydrogen omitted? My thinking was that it has an order of reaction of $z = 0$ and therefore it does not affect the rate at all. However, I was unable to find a source to verify this.

  • 2
    $\begingroup$ H+ isn't in the rate determining step so can be ignored. $\endgroup$
    – Mithoron
    Commented Nov 6, 2019 at 22:34
  • $\begingroup$ That would explain it, but would you perhaps have a source for this statement? $\endgroup$
    – Liam
    Commented Nov 7, 2019 at 6:45
  • 3
    $\begingroup$ Another possibility is that if you have vitamin C (the acid) in a large excess, then $\ce{[H+]}$ might not change significantly. If this is the case, then you can actually use an effective rate coefficient defined by $k = k' \ce{[H+]}^z$ . I am not sure if this is the case here, but if it is, then this reaction is a so-called pseudo-first order reaction. $\endgroup$
    – user23638
    Commented Nov 7, 2019 at 12:38
  • 1
    $\begingroup$ But according to my experiment, HCl concentration can influence the rate of the iodine clock reaction... $\endgroup$
    – user129034
    Commented Nov 19, 2022 at 10:53
  • $\begingroup$ I assume [H+] is kept constant – being in excess – and can be present implicitly in the value of k=f([H+]). Similarly as a reaction with general kinetics of the 2nd order can be considered of the pseudo-first order when one reactant is in excess. $\endgroup$
    – Poutnik
    Commented Nov 19, 2022 at 11:27

2 Answers 2


Imagine you are doing a process which takes 'n' steps. Label the steps : 1,2,3..n, now consider the time it takes to do step 1 , step -2 and so on. Now, wouldn't you agree that the time taken to complete process would be entirely dependent on the step which takes the most time(slowest step) to complete?

So, in your example the hydrogen ions are in one of the faster steps. So, if the concentration of hydrogen is 'enough' such that once you come to the step that their ample hydrogens to fuel the reaction, then it would seem that rate is independent of hydrogen concentration ( I say seem because there may be minute effects of it on the rate)


Although the comment by Ezze is a valid and helpful answer, according to chegg.com, the order of reaction with respect to the $\ce{H+}$ ion is $\ce{0}$ and therefore it can be omitted from the rate expression as $\ce{[H+]^0= 1}$.


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