A bromine clock reaction

wanna ask in the reaction:
5Br- + BrO3- + 6H+ ----- 3 Br2 + 3 H2O
which reactant is of first order, which is of second order

5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) ... (*)
Rate = k [Br-] [BrO3-] [H+]
The reaction is first order with respect to Br-(aq) and BrO3-(aq), and second order with respect to H+(aq).

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同埋in the reaction, 加 jor phenol. Gum 佢個function 係乜? Why 佢要very dilute
ask 多一條question, methyl orange is used as indicatior in this reaction. Whether it is neccessary to know how far the reaction has proceeded at the point where methyl orange is decolorized. Why?

The experiment is repeated for several times. In each experiment, a small fixed amount of phenol is added. Bromine formed in reaction (*) is removed by phenol (C6H5OH).
C6H5OH + 3Br2 → C6H2Br3OH + 3HBr
When phenol is completely reacted, the free bromine formed will decolorize the methyl orange indicator.

The phenol solution must be very dilute to ensure that the reaction of (*) has only proceeded to a small extent. Therefore, the changes in concentrations of all reactants are negligible, and thus the reaction rate is nearly constant which equals Δ[Br2]/t, whereΔ[Br2] is the change in concentration of Br­2 and t is the time taken for the decolorization of methyl orange.

It is not necessary to know how far the reaction has proceeded at the point where methyl orange is decolorized. However, it must ensure that the reaction has proceeded to the same point in each experiment. Since a fixed amount of phenol is added in each experiment, Δ[Br2] is the same when the methyl orange is decolorized. Since Rate =Δ[Br2]/t andΔ[Br2] is constant, Rate µ (1/t). Therefore, (1/t) can then be used to represent the reaction rate.