The reaction 4 HBr(a) + O2(g) --> 2 H70(9) + 2 Brz(o) proceeds through the following mechanism: HBr(9) + O2(9) --> HOOB(9) HOOB (g) + HBr(g) -> 2 HOB (g) HOBr() + HBr(9) - Br (9) + H20(0) (a) The first step of this mechanism is rate-determining (slow). What is the rate law for this reaction? O Rate = k [HBr] [0,1 O Rate = k [HBr? [0] Rate - k [Br] [0,12 O Rate = k [Br]1/2 [02] O Rate = k [HB] [0,142 O Rate = k [HBO] O Rate -k (HBr? (0,12 (b) What would the rate law be if the third step of this mechanism were rate-determining? O Rate -k (HBr] [0,1 O Rate = k [HBr]? [02] O Rate - k [Br] [0,12 O Rate = k [Br]1/2 [02] O Rate = k [HB] [O]'? O Rate = k [HBO]2 O Rate = k [HB-]> [0,11/2 O Rate-k (HBr]