Abstract
The kinetics and mechanism of the chlorine(III)-HOBr reaction were studied by the stopped-flow method under acidic conditions, pH 1.0-3.0, in 1.0 M NaClO(4) and at 25.0 degrees C. The overall redox process occurs in two consecutive steps via the formation of the BrClO(2) intermediate. The electron transfer reactions are coupled with bromine hydrolysis, the formation of the tribromide ion, and the protolytic equilibrium of chlorine(III). On the basis of simultaneous evaluation of the kinetic traces, the following rate constants were obtained for the redox steps: HClO(2) + HOBr right harpoon over left harpoon BrClO(2) + H(2)O, k(3) = (3.34 +/- 0.02) x 10(4) M(-1) s(-1), k(-3) = (3.5 +/- 1.3) x 10(3) s(-1); BrClO(2) + ClO(2)(-)<==>2ClO(2) + Br(-), k(4) = (2.9 +/- 1.0) x 10(7) M(-1) s(-1). The second step was practically irreversible under the conditions applied, and the value of k(-4) could not be determined. The equilibrium constant for the formation of BrClO(2), K(3) = 9.5 M(-1), was calculated from the kinetic results, and it was confirmed that this species is a very powerful oxidant. The redox potential was also estimated for the BrClO(2) + e(-) = Br(-) + ClO(2) reaction: epsilon(0) approximately 1.70 V.
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