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CH_4和CO是两种主要的温室效应气体和空气污染物,催化氧化是最有效的消除CH_4和CO的方法.研发不含贵金属的金属氧化物催化剂或者减少催化剂中贵金属用量为该领域研究热点.SnO_2是一种重要的宽禁带n型半导体材料,广泛应用于气敏器件、锂离子电池以及光电设备.SnO_2表面富含活泼的缺位氧且具有良好的热稳定性,因此其在催化方面的性能近年来逐渐受到人们关注.在过去的5年中,本团队深入研究了SnO_2材料在空气污染治理和绿色能源生产等领域的应用及其催化性质.发现通过其它阳离子如Fe~(3+),Cr~(3+),Ta~(5+),Ce~(4+)和Nb~(5+)等的掺杂,替换晶格中部分Sn~(4+),形成金红石型SnO_2固溶体结构,显著提高了催化剂氧物种的流动性、活性和催化剂本身的热稳定性.固溶体材料是一类重要的催化剂,受到广泛关注.一个典型的例子是铈锆固溶体,其作为储氧材料已广泛应用于汽车尾气净化器.形成固溶体结构后,氧化铈的储氧能力和热稳定性得到显著提高.为有效形成固溶体,两个阳离子需要具有相似的离子半径和电负性.以往,人们基于结构中金属阳离子和氧阴离子的离子半径提出了容忍因子的判别方法,以此来判断固溶体是否能有效形成及所生成固溶体的稳定性.我们在前期工作中,以Sn-Nb固溶体为例,提出了简单的X射线衍射(XRD)外推法来计算固溶体晶格容量,即形成稳定固溶体时客体阳离子取代主体晶格阳离子的最大值.作为延续工作,本文采用共沉淀法制备了一系列Sn/M(M=Mn,Zr,Ti,Pb)摩尔比为9/1的SnO_2基催化剂,并用于CH_4和CO催化氧化.结果表明,Mn~(3+),Zr~(4+),Ti~(4+)和Pb~(4+)均可以掺杂进四方金红石型SnO_2晶格中,形成稳定的固溶体结构.其中Sn-Mn-O固溶体表现出最高活性.为了深入研究Mn_2O_3在SnO_2中的晶格容量及最优催化剂配比,采用共沉淀法制备了一系列不同Sn/Mn摩尔比的样品,采用XRD,N_2-BET,H_2-TPR,SEM和XPS等手段对其物理化学性能进行了表征,并考察了对CH_4的催化氧化性能.通过XRD外推法测定了Mn~(3+)离子在SnO_2中的晶格容量为0.135g Mn_2O_3/g SnO_2,相当于Sn/Mn摩尔比为79/21.这表明形成稳定的固溶体后,SnO_2晶格中最多只有21%Sn~(4+)可以被Mn~(3+)替代;当Mn~(3+)含量超过晶格容量时,过量的Mn~(3+)在催化剂表面形成Mn_2O_3,对催化剂活性不利.类似于Sn-Nb-O固溶体,在Sn-Mn-O催化剂体系中亦观察到明显的晶格容量效应.纯相的Sn-Mn-O固溶体比含过量Mn_2O_3晶相的Sn-Mn-O催化剂具有更高活性.
CH_4 and CO are the two major greenhouse gases and air pollutants, and catalytic oxidation is the most effective method to eliminate CH_4 and CO. It is a hot topic in this field to develop a metal oxide catalyst without precious metal or to reduce the amount of precious metal in the catalyst. SnO 2 is an important kind of wide-bandgap n-type semiconductor material, which is widely used in gas sensing devices, lithium ion batteries and optoelectronic devices. The SnO 2 surface is rich in active oxygen vacancies and has good thermal stability. Therefore, Has been attracting more and more attentions in recent years.For the past 5 years, the team has deeply studied the application and catalytic properties of SnO_2 in air pollution control and green energy production, and found that through the other cations such as Fe 3+ ), Cr ~ (3 +), Ta ~ (5 +), Ce ~ (4+) and Nb ~ (5+) were used to replace some Sn ~ (4+) in the crystal lattice to form rutile SnO_2 The solid solution structure significantly increases the mobility and activity of the catalyst oxygen species and the thermal stability of the catalyst itself.Solid solution materials are an important class of catalysts and have received widespread attention.A typical example is the cerium-zirconium solid solution, which has been used as an oxygen storage material Widely used in cars Air purifier.After the formation of solid solution structure, the oxygen storage capacity and thermal stability of cerium oxide has been significantly improved.For the effective formation of solid solution, the two cations need to have similar ion radius and electronegativity.In the past, people based on the structure of the metal cation And oxygen ion radius of the ion tolerance of the proposed discriminant method to determine whether the solid solution can be effectively formed and the stability of the resulting solid solution in our previous work to Sn-Nb solid solution, for example, put forward a simple X Ray diffraction (XRD) extrapolation method to calculate the solid solution lattice capacity, that is, the formation of stable solid solution guest cationic substitution host lattice cation maximum .As a continuation of work, this paper prepared by coprecipitation a series of Sn / M (M = Mn, Zr, Ti, Pb) molar ratio of 9: 1 and was used for the catalytic oxidation of CH_4 and CO. The results showed that Mn_3 +, Zr_4 +, Ti_4 + And Pb 4+ can be doped into the tetragonal rutile SnO 2 lattice to form a stable solid solution structure, in which the Sn-Mn-O solid solution exhibits the highest activity.In order to further study the lattice capacity of Mn 2 O 3 in SnO 2 and Optimum catalyst ratio, prepared by coprecipitation A series of samples with different molar ratios of Sn / Mn were characterized by XRD, N_2-BET, H_2-TPR, SEM and XPS, respectively. The catalytic oxidation of CH_4 was also investigated. The lattice capacity of Mn ~ (3+) ions in SnO_2 was 0.135g Mn_2O_3 / g SnO_2, which was equivalent to the Sn / Mn molar ratio of 79/21, indicating that the most lattice of SnO_2 was formed after the formation of stable solid solution Only 21% of Sn 4+ could be replaced by Mn 3+. When the content of Mn 3+ exceeded the capacity of the lattice, Mn 3+ formed Mn 2 O 3 on the surface of the catalyst, In the Sn-Mn-O catalyst system, a significant lattice-volume effect was also observed, similar to the Sn-Nb-O solid solution.The pure Sn-Mn-O solid solution exhibited better lattice- O catalyst has higher activity.