Order of reaction for hydrogen ions being ignored

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.

• H+ isn't in the rate determining step so can be ignored. Nov 6, 2019 at 22:34
• That would explain it, but would you perhaps have a source for this statement?
– Liam
Nov 7, 2019 at 6:45
• 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.
– user23638
Nov 7, 2019 at 12:38

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 $$\ce{[H+]^0= 1}$$.