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氟原子与甲醇的反应:CHO/CHOH的产率以及CHO + CHO和CHO + HO反应的速率常数。

The reaction of fluorine atoms with methanol: yield of CHO/CHOH and rate constant of the reactions CHO + CHO and CHO + HO.

作者信息

Assaf Emmanuel, Schoemaecker Coralie, Vereecken Luc, Fittschen Christa

机构信息

Université Lille, CNRS, UMR 8522 - PC2A - Physicochimie des Processus de Combustion et de l'Atmosphère, F-59000 Lille, France.

出版信息

Phys Chem Chem Phys. 2018 Apr 25;20(16):10660-10670. doi: 10.1039/c7cp05770a.

Abstract

Xenondifluoride, XeF2, has been photolysed in the presence of methanol, CH3OH. Two reaction pathways are possible: F + CH3OH → CH2OH + HF and F + CH3OH → CH3O + HF. Both products, CH2OH and CH3O, will be converted to HO2 in the presence of O2. The rate constants for the reaction of both radicals with O2 differ by more than 3 orders of magnitude, which allows an unequivocal distinction between the two reactions when measuring HO2 concentrations in the presence of different O2 concentrations. The following yields have then been determined from time-resolved HO2 profiles: φCH2OH = (0.497 ± 0.013) and φCH3O = (0.503 ± 0.013). Experiments under low O2 concentrations lead to reaction mixtures containing nearly equal amounts of HO2 (converted from the first reaction) and CH3O (from the second reaction). The subsequent HO2 decays are very sensitive to the rate constants of the reaction between these two radicals and the following rate constants have been obtained: k(CH3O + CH3O) = (7.0 ± 1.4) × 10-11 cm3 s-1 and k(CH3O + HO2) = (1.1 ± 0.2) × 10-10 cm3 s-1. The latter reaction has also been theoretically investigated on the CCSD(T)//M06-2X/aug-cc-pVTZ level of theory and CH3OH + O2 have been identified as the main products. Using μVTST, a virtually pressure independent rate constant of k(CH3O + HO2) = 4.7 × 10-11 cm3 s-1 has been obtained, in good agreement with the experiment.

摘要

二氟化氙(XeF₂)在甲醇(CH₃OH)存在的情况下发生了光解。可能存在两条反应途径:F + CH₃OH → CH₂OH + HF 和 F + CH₃OH → CH₃O + HF。在有O₂存在的情况下,两种产物CH₂OH和CH₃O都会转化为HO₂。两种自由基与O₂反应的速率常数相差超过3个数量级,这使得在测量不同O₂浓度下的HO₂浓度时,能够明确区分这两个反应。然后根据时间分辨的HO₂分布确定了以下产率:φCH₂OH = (0.497 ± 0.013) 以及 φCH₃O = (0.503 ± 0.013)。低O₂浓度下的实验导致反应混合物中含有几乎等量的HO₂(由第一个反应转化而来)和CH₃O(来自第二个反应)。随后HO₂的衰减对这两种自由基之间反应的速率常数非常敏感,并获得了以下速率常数:k(CH₃O + CH₃O) = (7.0 ± 1.4) × 10⁻¹¹ cm³ s⁻¹ 和 k(CH₃O + HO₂) = (1.1 ± 0.2) × 10⁻¹⁰ cm³ s⁻¹。还在CCSD(T)//M06 - 2X/aug - cc - pVTZ理论水平上对后一个反应进行了理论研究,并确定CH₃OH + O₂为主要产物。使用μVTST,获得了实际上与压力无关的速率常数k(CH₃O + HO₂) = 4.7 × 10⁻¹¹ cm³ s⁻¹,与实验结果吻合良好。

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