超声/K2S2O8体系降解左氧氟沙星过程中活性自由基产生机制

以左氧氟沙星水溶液为实验材料，比较了超声、K₂S₂O₈和超声/K₂S₂O₈ 3种体系对左氧氟沙星的去除效果，并通过K₂S₂O₈浓度、超声功率、pH值及添加异丙醇或叔丁醇的对照实验，分析了超声/K₂S₂O₈体系去除左氧氟沙星过程中活性自由基产生机制及变化规律.结果表明：在左氧氟沙星浓度30 mg·L^-1、过硫酸盐浓度20 mmol·L^-1、超声功率360 W、pH=9的条件下，反应180 min，超声/K₂S₂O₈体系中左氧氟沙星的去除率达到90%，超声体系为13.89%，K₂S₂O₈体系只有5.82%；超声/K₂S₂O₈体系存在协同作用，产生了更多的SO₄^-·和·OH；超声/K₂S₂O₈体系中·OH的产生有超声和链式反应两种途径，其中，链式反应的贡献较大，K₂S₂O₈的存在是·OH产生的必要条件，链式反应在自由基浓度较高条件下会发生猝灭，K₂S₂O₈浓度对SO₄^-·的产生有重要影响，但K₂S₂O₈浓度过高对SO₄^-·的产生无益；超声/K₂S₂O₈体系降解左氧氟沙星过程中SO₄^-·与·OH是通过链式反应共同增减的，·OH在反应中占主导地位；反应初期·OH作用较大，随反应时间延长逐渐减小，SO₄^-·作用在反应初期较小，随着反应进行作用逐渐增大.

Generation mechanism of active free radicals in the processes of levofloxacin degradation in US/K2S2O8 system

In this study, the degradation of levofloxacin in aqueous solution with Ultrasonic (US), K₂S₂O₈ and US/K₂S₂O₈ were investigated. Production of active free radicals in US/K₂S₂O₈ system were studied based on the variation of K₂S₂O₈ concentration, ultrasonic power, pH value and the addition of isopropanol or tert butanol. The results showed that under the conditions of levofloxacin concentration 30 mg·L^-1, the ultrasonic power 360 W, K₂S₂O₈ concentration 20 mmol·L^-1 and pH=9, the removal rate of levofloxacin in US/K₂S₂O₈, K₂S₂O₈ and US system after 180 min was 90%, 5.82% and 13.89%, respectively. The synergistic effect in US/K₂S₂O₈ system facilitated the production of more SO₄^-·and·OH free radicals. Moreover, both ultrasonic and chain reaction were capable to produce·OH in US/K₂S₂O₈ system, and the chain reaction was the major contributor. The existence of K₂S₂O₈ was also the necessary condition for·OH production. The chain reaction would be quenched when the concentration of free radicals at high level. The concentration of K₂S₂O₈ had a significant effect on the production of SO₄^-·, but the high concentration of K₂S₂O₈ would not benefit the production of SO₄^-·. In the processes of levofloxacin degradation in US/K₂S₂O₈ system SO₄^-·and·OH free radicals were increasing and decreasing together by chain reaction, and·OH played a dominant role in the reaction. It shows that·OH plays a major role in the early stage of the reaction, and gradually weakens with the extension of the reaction time. SO₄^-·has little contribution in the early stage of the reaction, and gradually increases with the proceeding of the reaction.