For reaction II:
5Br− + BrO3− + 6H+ → 3Br2 + 3H2O,
The overall reaction rate is:
r=−51ΔtΔ[Br−]=−31ΔtΔ[Br2].
Given ΔtΔ[Br−]=2×10−4 mol L−1 min−1
The reaction rate is r=52×10−4=4×10−5 mol L−1 min−1.
Since the rates of both reactions are equal at 10.10 am and reaction I is first-order:
r=k[A], so 4×10−5=k×10−2, giving k=4×10−3 min−1.