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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.

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    $\begingroup$ H+ isn't in the rate determining step so can be ignored. $\endgroup$ – Mithoron Nov 6 '19 at 22:34
  • $\begingroup$ That would explain it, but would you perhaps have a source for this statement? $\endgroup$ – Liam Nov 7 '19 at 6:45
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    $\begingroup$ 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. $\endgroup$ – Ezze Nov 7 '19 at 12:38
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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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