Oxygen atom transfer from trans-L(H2O)RhOOH2+ {L = [14]aneN(4) (L-1), meso-Me-6[14]aneN(4) (L-2), and (NH3)(4)} to iodide takes place according to the rate law -d[L(H2O)RhOOH2+]/dt = k(i)[L(H2O)RhOOH2+][I-][H+]. At 0.10 M ionic strength and 25 degreesC, the rate constant k(l)/M-2 s(-1) has values of 8.8 x 10(3) [L = (NH3)(4)], 536 (L-1), and 530 (L-2). The final products are LRh(H2O)(2)(3+) and I-2/I-3(-). The (NH3)(4)(H2O)RhOOH2+/Br- reaction also exhibits mixed third-order kinetics with k(Br) approximate to 1.8 M-2 s(-1) at high concentrations of acid (close to 1 M) and bromide (close to 0.1 M) and an ionic strength of 1.0 M. Under these conditions, Br-2/Br-3(-) is produced in stoichiometric amounts. As the concentrations of acid and bromide decrease, the reaction begins to generate O-2 at the expense of Br-2, until the limit at which [H+] less than or equal to 0.10 M and [Br-] less than or equal to 0.010 M, when Br-2/Br-3(-) is no longer observed and O-2 is produced quantitatively. At this limit, the loss of (NH3)(4)(H2O)RhOOH2+ is about twice as fast as it is at the high [H+] and [Br-] extreme, and the stoichiometry is 2(NH3)(4)(H2O)RhOOH2+ --> 2(NH3)(4)(H2O)RhOH2+ + O-2; i.e., the reaction has turned into the bromide-catalyzed disproportionation of coordinated hydroperoxide. In the proposed mechanism, the hydrolysis of the initially formed Br-2 produces HOBr, the active oxidant for the second equivalent of (NH3)(4)(H2O)RhOOH2+. The rate constant k(HOBr) for the HOBr/(NH3)(4)(H2O)RhOOH2+ reaction is 2.9 x 10(8) M-1 s(-1).