Rate order when the slow step only contains intermediates?

[![enter image description here](https://i.stack.imgur.com/bJKJA.png)](https://i.stack.imgur.com/bJKJA.png)


I feel a bit sheepish asking this, but I was told that the stoichiometry of the slow step of a mechanism indicates the order of a reaction with respect to a reactant. However, in the slow step of this reaction, there is no reactant (of the overall equation) reacting. So how does one determine the rate order? Would it be in terms of the intermediate.


I don't see how the rate could be $k[C\_4H\_{11}O^+]$ as that is not a reactant in the overall equation


Image source: <https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(McMurry)/10%3A_Organohalides/10.05%3A_Preparing_Alkyl_Halides_from_Alcohols>




Step 1 is fast and reversible so that in effect an equilibrium amount of $\mathrm{C\_4H\_9OHH^+}$ is formed which reacts slowly in step 2. Once the $\mathrm{C\_4H\_9^+}$ is formed it reacts rapidly in step 3 so the reversible part of step 2 never occurs to any great extent. Thus the limiting step is step 2 with first order reaction rate $-k\_2[\mathrm{C\_4H\_9OHH^+}]$ and the product forms at minus this rate.


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