Preparation of Ce-0.6-Cu-60/Al-40-[O] catalyst and role of CeO2/CuO in simultaneous removal of H2S and PH3
Preparation of Ce-0.6-Cu-60/Al-40-[O] catalyst and role of CeO2/CuO in simultaneous removal of H2S and PH3
复制标题
Ce-0.6-Cu-60/Al-40-[O]催化剂的制备及其CeO2/CuO同时脱除H2S和PH3的作用
DOI：
10.1016/j.jtice.2018.03.007
复制
发表时间：
2018
影响因子：
5.7
通讯作者：
Li Kai
中科院分区：
文献类型：
--
作者：
Wang Yingwu;Lin Qiang;Ning Ping;Wang Chi;Sun Xin;Li Kai
工业排放的H2S和PH 3由于其毒性而对环境和人类有害。本研究在低温下同时脱除H2S和PH 3。结果表明，Ce0.6-Cu 60/Al 40-[O]催化剂的脱硫率最高，可达9.12mol（H2S）/g和8.09mol（PH 3）/g。采用HRTEM、SEM、XRD、BET、H2-TPR、XPS和原位FTIR等手段研究了CeO 2和CuO在Al 2 O3上的负载作用。结果表明，CuO作为一种“晶体魔术师”促进了催化剂的性能，同时破坏了Al 2 O3的规则结构，从而对催化剂的比表面积、孔容和平均孔径产生了负面影响。CeO 2起到了“仙女”的作用，在一定程度上修复了CuO的破坏，并通过调整晶粒尺寸促进了CuO的活性。因此，CeO 2和CuO成功地击败了“敌人”-H2S和PH 3。H2S与CuO反应生成CuS和H2O，而PH 3和O2在载体上生成P2 O 5。然后，由于H2O的存在，P2 O 5进一步转化为HPO 3。CeO 2促进了HPO 3向H3 PO 4的转化。最后，“水晶魔术师”被牺牲，以获得预期的结果。
H2S and PH3, emitted from industries, are harmful to the environment and humans because of their toxicity. In this study, H2S and PH3were simultaneously removed at low temperature. It was found that the Ce0.6–Cu60/Al40-[O] catalyst showed the best results could reach 9.12 mol (H2S)/g and 8.09 mol (PH3)/g. The role of CeO2and CuO loaded on Al2O3was investigated by HRTEM, SEM, XRD, BET, H2-TPR, XPS and in situ FTIR. The results revealed that CuO acted as a ‘crystal magician’ to promote the power of the catalysts, while destroying the regular structure of Al2O3, which in turn negatively affected the specific surface area, pore volume, and average pore size of the catalyst. CeO2acted as ‘fairy’ repairing the destruction to some extent and promoting the activity of the magician CuO by adjusting the grain size. Thus, CeO2and CuO succeeded in defeating the ‘enemy’- H2S and PH3. H2S reacted with CuO to form CuS and H2O, and PH3and O2generated P2O5on the carrier. Then, P2O5was further converted to HPO3due to the presence of H2O. CeO2accelerated the conversion of HPO3to H3PO4. Finally, the “crystal magician” was sacrificed to obtain the desired results.