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氯化溴与溴酸根离子和氯酸根离子反应的动力学及机理

Kinetics and mechanisms of bromine chloride reactions with bromite and chlorite ions.

作者信息

Odeh Ihab N, Nicoson Jeffrey S, Huff Hartz Kara E, Margerum Dale W

机构信息

Department of Chemistry, Purdue University, West Lafayette, IN 47907, USA.

出版信息

Inorg Chem. 2004 Nov 15;43(23):7412-20. doi: 10.1021/ic048982m.

Abstract

Chloride ion catalyzes the reactions of HOBr with bromite and chlorite ions in phosphate buffer (p[H(+)] 5 to 7). Bromine chloride is generated in situ in small equilibrium concentrations by the addition of excess Cl(-) to HOBr. In the BrCl/ClO(2)(-) reaction, where ClO(2)(-) is in excess, a first-order rate of formation of ClO(2) is observed that depends on the HOBr concentration. The rate dependencies on ClO(2)(-), Cl(-), H(+), and buffer concentrations are determined. In the BrCl/BrO(2)(-) reaction where BrCl is in pre-equilibrium with the excess species, HOBr, the loss of absorbance due to BrO(2)(-) is followed. The dependencies on Cl(-), HOBr, H(+), and HPO(4)(2)(-) concentrations are determined for the BrCl/BrO(2)(-) reaction. In the proposed mechanisms, the BrCl/ClO(2)(-) and BrCl/BrO(2)(-) reactions proceed by Br(+) transfer to form steady-state levels of BrOClO and BrOBrO, respectively. The rate constant for the BrCl/ClO(2)(-) reaction [k(Cl)(2)]is 5.2 x 10(6) M(-1) s(-1) and for the BrCl/BrO(2)(-) reaction [k(Br)(2)]is 1.9 x 10(5) M(-1) s(-1). In the BrCl/ClO(2)(-) case, BrOClO reacts with ClO(2)(-) to form two ClO(2) radicals and Br(-). However, the hydrolysis of BrOBrO in the BrCl/BrO(2)(-) reaction leads to the formation of BrO(3)(-) and Br(-).

摘要

氯离子在磷酸盐缓冲液(pH值5至7)中催化次溴酸与亚溴酸根离子和亚氯酸根离子的反应。通过向次溴酸中加入过量的Cl⁻,原位生成少量处于平衡浓度的氯化溴。在ClO₂⁻过量的BrCl/ClO₂⁻反应中,观察到二氧化氯的一级生成速率,该速率取决于次溴酸的浓度。确定了反应速率对ClO₂⁻、Cl⁻、H⁺和缓冲液浓度的依赖性。在BrCl与过量物质次溴酸处于预平衡状态的BrCl/BrO₂⁻反应中,跟踪了由于BrO₂⁻导致的吸光度损失。确定了BrCl/BrO₂⁻反应对Cl⁻、次溴酸、H⁺和HPO₄²⁻浓度的依赖性。在所提出的反应机制中,BrCl/ClO₂⁻和BrCl/BrO₂⁻反应分别通过Br⁺转移进行,以形成BrOClO和BrOBrO的稳态水平。BrCl/ClO₂⁻反应的速率常数[k(Cl)₂]为5.2×10⁶ M⁻¹ s⁻¹,BrCl/BrO₂⁻反应的速率常数[k(Br)₂]为1.9×10⁵ M⁻¹ s⁻¹。在BrCl/ClO₂⁻的情况下,BrOClO与ClO₂⁻反应形成两个二氧化氯自由基和Br⁻。然而,BrCl/BrO₂⁻反应中BrOBrO的水解导致BrO₃⁻和Br⁻的形成。

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