K/Zr助剂对Fe基催化剂CO加氢性能的影响

Effects of K/Zr Promoter on Iron-based Catalyst for CO Hydrogenation

摘要:

研究了K、Zr助剂对沉淀铁基催化剂在颗粒形貌、晶相组成、还原和碳化的影响，利用X射线粉末衍射、扫描电镜、拉曼光谱、H2-TPR和CO-TPR对催化剂进行了表征，并考察了催化剂对CO加氢制低碳烯烃反应的催化性能。结果表明：K助剂有利于α-Fe2O3生成100～150nm的多面体颗粒，Zr助剂有利于α-Fe2O3颗粒分散生成10～50nm的不规则颗粒，K和Zr助剂协同有利于生成20～30nm的规则球形颗粒；K和Zr助剂使催化剂H2低温还原峰温度分别提高85℃和14℃，同时使CO低温还原/碳化峰温度分别提高68℃和12℃；反应过程中K有助于生成活性Fe5C2物相，使CO转化率提高24.7%，CH4含量降低10.37%，烯烷比提高2.24，而Zr对碳化铁活性物相和FTO反应无显著影响；K和Zr助剂存在显著的协同效应，FTO反应过程中Fe3O4晶粒尺寸由反应前的26.4nm减小至反应后的21.2nm，且晶体趋向于无定形态，CO转化率达到90%以上，产物分布中C2=~C4=达到30.51mol%，烯烷比达到3.59，C2=~C4=收率达到28.65%，且Fe基催化剂具有较高的稳定性。

Abstract:

Effects of K and Zr promoter on precipitated iron-based catalyst for CO hydrogenation were investigated. The bulk structure, surface morphology, bonded state of surface elements, and chemical properties of the catalysts were characterized by means of X-ray diffraction(XRD), scanning electron microscopy(SEM), Raman spectroscopy, H2/CO-temperature-programmed reduction and catalytic activity tests for the selective formation of light olefins. Results showed that K was benefit to format 100～150nm solid particles of α-Fe2O3, and Zr contributed to the dispersion of α-Fe2O3 to the particles with 10～50 nm diameter, while the cooperation of K and Zr could helpful to generate spherical particles with diameter of 20～30 nm. K and Zr could improve the low H2 reduction temperature of 85℃ and 14℃ respectively, meanwhile improve the low CO reduction/carbonization temperature of 68℃ and 12℃. K in reaction process could generate active Fe5C2 phase, improve the CO conversion rate from 42.3% to 67.0%, lower the content of CH4 by 10.37% and increased the O/P by 2.24, but Zr exhibited no such function. The catalyst promoted by K and Zr show a remarkable synergic effect which leaded to the reduction of Fe3O4 crystal from 26.4nm of fresh catalyst to 21.2nm after it was used, tending to amorphous type. Meanwhile, the synergic effect of K and Zr increased CO conversion to above 90%, C2=~C4= distribution to 30.51mol%, O/P to 3.59 and the yield of C2=~C4= got about 28.65%.