Question

1 7. The initial rate of the reaction: BrOz (aq) + 5 Br(aq) + 8 H+(aq) + 3 Brz(1) + H2O(1) Has been measured at the reactant concentrations shown (in mol/L) Experiment [Br03-] [Br] [H] Initial rate (mol/(L:s)) 0.10 0.10 0.10 8.0 x 10-4 2 0.20 0.10 0.10 1.6 x 10-3 3 0.10 0.20 0.10 1.6 x 10-3 4 0.10 0.10 0.20 3.2 x 10-3 a) what is the rate law expression and the rate constant k? [3] b) what is the rate when [BrO3 ]=[Br]=[H"]=0.20 mol/L? [1]

Answer 1

Answer: BrOz (aq) + 5 Br" (aq) + 8 H+ (aq) + 3 Br2 (1) + H20 (1) case i & ii By increasing (BrO3] by 2 times the rate increases by 2 times order with respect to BrO3 = 1 Case i and iii By increasing Br" by 2 times rate increases by 2 times order with respect to Br = 1 Case i and iV By increasing H by 2 times the rate increases by 4 times order of the reaction with respect to H* = 2 overall the rate of the reaction can be expressed as rate = K [BrO3 ] [Br] [H]? Case i rate = K [BrO3] [Br] [H]? 8.0 x 104 = K (0.10) (0.10) (0.10)2 8.0 x 104 = K 10-4 K = 8.0 x 10-1 = 8.0 Conclusion: Equilibrium constant = 0.8 rate = K [BrO3 ] [Br] [H*]? b) Answer: rate = 8.0 (0-20) (0.20) (0.20)2 = 8.0 x 0.0016 = 0.0128 = 12.8 x 10-3