【作者】 王娟娟；

【导师】 庞先勇；

【作者基本信息】 太原理工大学 ， 物理化学， 2012， 硕士

【摘要】 本文在第一原理量子力学计算基础上,采用广义梯度近似(GGA)的密度泛函理论(DFT)并结合平板模型,研究了丙胺在不同氧覆盖度下,氧化产物不同的原因,乙醇在低氧覆盖的Au(111)表面的氧化反应细节与反应机理,及助剂M(M=CS、Ru、Rh、Pd、Pt、O)吸附在Ag(111)表面对乙烯环氧化选择性的影响。通过计算丙胺、乙醇、乙烯及其在氧化反应中可能出现中间体的吸附构型和最稳定吸附位,及反应可能出现的过渡态构型,反应的活化能,并从电子结构及能量分解方面对其结果进行分析,得到一些变化趋势及特征,为以后的研究提供理论参考。具体结论如下：1.通过密度泛函理论,研究了在清洁及有氧修饰的Au(111)表面,丙胺的选择性氧化。在不同的氧覆盖度下,丙胺氧化的反应机理及反应产物也不同。因此,我们建立了单层(ML)氧覆盖度是0、2/9、2/3ML的Au(111)表面,从电子结构及能量分解分析,得出结论：Au(111)表面上氧覆盖度越高,金的活性越小,底物的变形越大,吸附分子在表面的吸附更加不稳定,导致丙胺和丙氰的吸附困难,吸附能降低。2.在不同的氧覆盖度下：丙胺经历N-H键和C-H键的断裂,在低氧覆盖度(00=2/9ML)下得到丙氰和水,并且在高氧覆盖度(θo=2/3ML)下产生丙氰、丙醛和水。丙胺氧化第一步CH3CH2CH2NH2→CH3CH2CH2NH的活化能是0.16eV(θo=2/9ML)和0.38eV(θo=2/3ML),第二步CH3CH2CH2NH2→CH3CH2CH2NH的活化能是0.16eV和0.25eV。C-H键的断裂,第一步CH3CH2CH2N→CH3CH2CHN活化能在低氧覆盖度下是0.20eV(0o=2/9ML)和0.25eV(0o=2/3ML)；第二步CH3CH2CHN→CH3CH2CN在高氧覆盖度下是0.15eV(θo=2/9ML)和0.26eV(θo=2/3ML)(CH3CH2CHN→CH3CH2CN)。另一个反应CH3CH2CHN→CH3CH2CHO发生在高氧覆盖度下,并且反应的活化能是0.41eV。计算结果表明由于低的活化能,丙胺在室温下就可被氧化。而且,低氧覆盖度下反应的活化能普遍比高覆盖度下的活化能低,这与实验结果一致。通过研究氧覆盖度0,1/9MLAu(111)表面上,乙醇氧化的反应机理,发现在氧覆盖度是清洁,1/9ML的表面,乙醇的吸附能分别是0.12eV与-0.02eV。随着表面氧覆盖度的增加,吸附能改变的趋势同前面得到结论是一致的：随着氧覆盖度的增加,反应中各个吸附物种的吸附能减小了。由于乙醇有三种氢键：O-H键(CH3CH2OH),Cα-H键(CH3CH2OH),Cp-H键(GH3CH2OH)。因此,在有氧覆盖的Au(111)表面,乙醇第一步反应中有三种脱氢方法。Path a1:乙醇和氧反应生成乙氧基CH3CH2OH+O→CH3CH2O+OH,活化能是0.03eV；Path a2:乙醇Cα_H键(CH3CH2OH)的断裂(CH3CH2OH+O→CH3CHOH+OH),活化能是0.39eV; Path a3:乙醇的Cβ-H键(CH3CH2OH)断裂生成CH2CH2OH和OH(CH3CH2OH+O→CH2CH2OH+OH),活化能是1.04eV。比较得CH3CH2OH+O→CH3CH2O+OH的活化能最低,选其作为反应的第一步。接着是第二步乙氧基的进一步氧化脱氢,乙氧基也包含两种氢键(CH3CH2O中的Cβ-H键与CH3CH2OH中的Cα-H键)。同样也设计并模拟了两个途径：Path b1:CH3CH2O+OH→CH3CHO+H2O,活化能为0.40eV。Path b2:乙氧基的Cβ-H键断裂生成CH2CH2O和H2O,活化能为1.2eV。比较知Path b,反应产生乙醛需要的能量最小,因此反应倾向于断裂Cα-H键。在有氧覆盖的Au(111)表面(θo=1/9ML),乙醇经历O-H和Cα-H键的断裂最后生成了乙醛和水。3.通过研究在Ag(111)清洁表面和Cs、Rh、Pb、Pt、O原子改性的Ag(111)表面,乙烯(C2H4)和氧(O)反应生成环乙烷(C2H4O)和乙醛(CH3CHO)的过程,确立了反应物、过渡态和产物的几何构型和稳定吸附位。同清洁的Au(111)表面相比,Cs、Rh、Pb、Pt、O都通过一定的方式提高了对环氧化物的选择性。Cs吸附在Ag(111)面时形成了粒子域,稳定了C2H4O过渡态的生成,提高了C2H4O的选择性。Ru、Rh、Pd、Pt分别吸附在Ag(111)表面影响了氧的酸碱性,提高了反应的选择性。综上所述,在助剂氧的作用下,环乙烷的产率提高了。结果证明氧覆盖度的增加有利于生成环乙烷。更多还原

【Abstract】 In this thesis, the reaction mechanism for selective oxidation of propylamine and ethanol on clean and oxygen-covered gold surface has been studied by the first principle quantum mechinism calculations at the level of generalized gradient approximation (GGA) with slab model. In addition, adsorption and oxidation of ethylene on clean and M-Promoted (M=Cs, Ru, Rh, Pd, Pt, O) Ag(111) surfaces have been investigated by the same way. On the basis of above systemical studies, the configuration of possible intermediates, the most stable adsorption and transition state structure in the reaction, adsorption energy, and the activation energy of reaction have been obtained. By analyzing the above date, the electronic strecture of configurations and decompositing adsorption energy, the conclusions are summarized as follows:(1) The calculation results on reaction mechanism for selective oxidation of propylamine on oxygen-covered gold have been indicated that the adsorption energy of propylamine decrease with the increasing oxygen coverage, that is,-0.38,-0.20and-0.10eV on clean,2/9monolayer (ML) and2/3monolayer (ML) oxygen, respectively. The adsorption energies of the intermediates also have the trend of the gradual lower. The present work also indicated that the final product distribution depends on the oxygen coverage:Propylamine undergoes N-H bond and C-H bond cleavage to produce propionitrile and water at low-oxygen-coverage (θ0=2/9ML), and to yield propionitrile, propionaldehyde and water at high-oxygen-coverage (θ0=2/3ML).The energy barrier of first step of propyamine oxidation (CH3CH2CH2NH2→CH3CH2CH2NH) is0.16eV (θ0=2/9ML) and0.38eV (θ0=2/3ML). On the second step, the barrier energy is0.16eV and0.25eV of CH3CH2CH2NH→CH3CH2CH2N, the next both of C-H breakage and the barrier energy is0.20eV (CH3CH2CH2N→CH3CH2CHN) and0.25eV (CH3CH2CHN→CH3CH2CN) eV on low oxygen coverage, and0.15eV (CH3CH2CH2N→CH3CH2CHN) and0.26eV eV(CH3CH2CHN→CH3CH2CN) on the high oxygen coverage. The additional reaction step of CH3CH2CHN→CH3CH2CHO is occurred on the high oxygen coverage, and the associated barrier is0.41eV. The calculation results show that the oxidation of propylamine can occur at room temperature due to the lower energy barrier. Furthermore, it was found that the energy barrier for the possible reaction steps at the low oxygen coverage is general smaller than that on high oxygen coverage, which agrees with the experimental results.(2) The mechanism of ethanol oxidation on O-adsorbed Au(111) surface is study. The oxygen coverage on Au(111) surface is zero,1/9monolayer (ML). On clean,1/9ML oxygen coverage, the adsorption energy of ethanol respectively are-0.12,-0.02eV. The chance rule of the adsorption energy with the increase of oxygen coverage is meet with the conclusion above:the adsorption energy of adsorption species reduce. Ethanol has three hydrogen bonding:O-H bond(CH3CH2OH), C-H bond(CH3CH2OH), C-H bond(CH3CH2OH). So ethanol oxidation on Au(111) surface of oxygen coverage, the way of dehydrogenation have three on the frist step. So there are three pathway designed.(1), reaction of ethanol and oxygen atom produce ethoxyl (CH3CH2OH+O→CH3CH2O+OH). The energy barrier is0.03eV.(2) The another way is Cα-H bond (CH3CH2OH) breaking of ethanol (CH3CH2OH+O→CH3CHOH+OH) and the energy barrier is0.39eV.(3), Cβ-H bond(CH3CH2OH) cleavage of ethanol lead to CH2CH2OH and OH (CH3CH2OH+O→CH2CH2OH+OH) and the activation barrier is1.04eV. From the calculation results that the energy barrier of CH3CH2OH+O→CH3CH2O+OH is lowest, so it is best reaction pathway. The second step is oxidation of the produced ethoxyl also include two C-H (Ca-H bond CH3CH2O, Cβ-H bond of CH3CH2O). Two pathway is designed here.(1), CH3CH2O+OH→CH3CHO+H2O need the energy barrier of0.40eV.(2), the path is breakage of Cβ-H bond of ethoxyl (CH3CH2O+OH→CH2CH2O+H2O) and the activation barrier is1.2eV. Generation of acetaldehyde need minimum energy, so reaction tend to break Ca-H bond. Ethanol undergoes O-H bond and Ca-H bond cleavage to produce acetaldehyde and water at oxygen-coverage (θ0=1/9ML) Au(111) surfaces.(3) On clean and M-promoted Ag(111) surfaces, the oxidation reaction of ethylene (C2H4) and oxygen (O) produce ethylene oxide (C2H4O) and acetaldehyde (CH3CHO). The reactants, transition state, product geometric configurations and the stablest state adsorption site are established on clean and M-promoted Ag(111) surfaces. Further the analysis, as compared to clean Au(111) surface, the presence of Cs、Rh、 Pb、Pt and O on Au(111) surfaces improved selectivity of C2H4O. The Cs atom is adsorbed on the Au(111) surface form electric field, then stabilize transition state, leading to C2H4O. The Ru、Rh、Pd、Pt atoms respectively adsorbed on Ag(111) surfaces affect acid and alkaline of the oxygen, so the selectivity enhancement of ethylene oxide. When the oxygen atom promote in ethylene epoxidation on Ag(111), yield of the ethylene oxide improved. The results proof that the increase of oxygen coverage is beneficial to product ethylene oxide.更多还原