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哈尔滨理工大学学士学位论文 Fe(II,III)催化H2O2氧化降解水中甲基橙的研究

摘 要

染料行业的迅猛发展导致了严重的水污染，染料废水浓度高、毒性大，对环境及与环境息息相关的人类都具有重大的威胁，所以染料废水是当前水污染问题中急需解决的重要问题。偶氮染料甲基橙溶解在水中对人类具有很强的致癌性，因此对这种染料废水必须采取非常有效的方法进行治理，治理后达到排放标准后才能排入水体。

本文采用的是用Fe(II,III)催化过氧化氢(H2O2)氧化降解的方式，即Fenton反应与类-Fenton反应降解甲基橙。以甲基橙作为目标污染物，以Fe(II,III)作为催化剂，用H2O2作为氧化剂，来降解甲基橙。通过控制变量法控制变量，研究Fenton(Fe(II)/H2O2)反应与类-Fenton(Fe(III)/H2O2)反应氧化降解效能，做出考察pH值、Fe(II,III)浓度、H2O2浓度、水中存在阴离子(硝酸根离子(NO3-)、硫酸根离子(SO42-)、氯离子(Cl-))、抑制剂(甲醇、叔丁醇)等影响因素对Fe(II,III)催化H2O2氧化降解甲基橙效能的研究。可以得出如下结论：

(1)反应条件需要在酸性条件下，pH越小，反应降解甲基橙速率越快，降解效果越好。反应最佳pH值为3，即pH=3时，甲基橙降解效果越好，降解速率越快。初始溶液中H2O2的浓度越大，甲基橙的降解速率越快，降解效果越好。在Fe(III)催化H2O2降解甲基橙的过程中，H2O2浓度的增加，使得Fe(III)向 Fe(II)的转化速率越快，同时也使得羟基自由基的产生速率加快，从而加速了反应的进行。初始溶液中Fe(III)浓度越高，甲基橙降解效率越高，甲基橙降解效果越好。由此可见，Fe(III)浓度的增加会加快Fe(III)催化H2O2氧化降解甲基橙的反应速率，因为Fe(III)浓度增加了，随之而来的Fe(II)浓度也会增大，羟基自由基的产生速率也随之而加快，使得整个反应速度变快。最佳的硝酸铁浓度为1.6mmol/L。

(2)阴离子主要考察硝酸根离子，硫酸根离子，氯离子。其中硝酸根离子对甲基橙降解率无明显影响；硫酸根离子浓度越大，甲基橙的降解越慢，降解效果越差；氯离子浓度越大，甲基橙的降解越慢，降解效果越差。由上面的章节可知三种阴离子在相同浓度下对甲基橙降解率：硝酸根离子>硫酸根离子>氯离子。

(3)本文只考察两种抑制剂，甲醇和叔丁醇。抑制剂能非常明显的起到在反应体系中对甲基橙的抑制降解效果，由上文可知甲醇对甲基橙的抑制降解效果比叔丁醇好。

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哈尔滨理工大学学士学位论文

关键词 羟基自由基(·OH)；Fe(II,III)；H2O2；甲基橙

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哈尔滨理工大学学士学位论文 Fe(II,III) catalytic hydrogen peroxide oxidation degradation of methyl orange in aqueous solution

Atract

The rapid development of dye industry has led to severe water pollution. The environment and humans have a significant threat that is high concentrations and toxicity of dye waste water, so the dye waste water is an important issue in the current water pollution problems needing to be resolved. Methyl orange dissolved in water has highly carcinogenic to humans, therefore the water of Methyl orange is discharged into lakes after met emission standards by a very effective method of treatment.

Methyl orange is degraded by peroxide (H2O2) that is catalyzed by Fe(II,III) in this passage, that is Fenton and like-Fenton. Methyl orange is degraded when Fe(II,III) is catalyst, and H2O2 is oxidizer. Record time when reaction starts, measure the Aorbance in different time and calculate C/C0. Through controlling variables, research effectiveness that Methyl orange is degraded in Fenton (Fe(II)/H2O2) and like-Fenton (Fe(III)/H2O2) and the factors are pH,concentration of Fe(II,III), H2O2, anion in water (NO3-, SO42-, Cl-), inhibitors (methanol、tert-butanol). It can be concluded:

(1)Fenton and like-Fenton reaction must be carried out under acidic conditions, and with the acidity is stronger and as pH is more smaller, the reaction is more faster. When pH is 3, Methyl orange degraded is more and fast. The higher the concentration of H2O2, the faster the degradation rate of methyl orange in Fenton and like-Fenton. But when the concentration of H2O2 is exorbitant, H2O2 will inhibit Fenton and like-Fenton. The higher the concentration of Fe(II) and Fe(III), the faster the degradation rate of methyl

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哈尔滨理工大学学士学位论文 orange in Fenton and like-Fenton. But when the concentration of Fe(II) and Fe(III) is exorbitant, Fe(II) and Fe(III) has no effect in Fenton and like-Fenton.

(2)The factors are NO3-, SO42-, Cl- in water. And NO3- has no effect; the higher the concentration of SO42- and Cl-, the slower the degradation rate of methyl orange in Fenton and like-Fenton. Three kinds of anion degradation rate of methyl orange at the same concentration: NO3-> SO42-> Cl-.

(3)Two inhibitors are methanol and tert-butanol in this passage. Methanol inhibiting degradation of methyl orange is better than t-butanol in Fenton and like-Fenton.

Keywords Hydroxyl radical (·OH); Fe (II, III); Hydrogen peroxide; Methyl orange

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