在甲醇合成过程中Cu/ZnO/ZrO催化剂的可逆和不可逆的结构变化
Lucas Warmuth1, Matthias Steurer1, Dieter Schild2
1Institute of Catalysis Research and Technology (IKFT), Karlsruhe Institute of Technology (KIT), 76344 Eggenstein-Leopoldshafen, Germany.
ACS applied materials & interfaces
|February 9, 2024
概括
甲醇合成中的催化剂失活主要是由铜烧结和氧化重组引起的. 重新激活工作部分恢复了催化剂结构,突出了的促进作用.
科学领域:
- 材料科学 材料科学 材料科学
- 催化科学 催化科学
- 化学工程是化学工程的重要组成部分.
背景情况:
- 异质催化剂的运行稳定性与它们的结构和化学状态密切相关.
- 对这些关系的准确理解对于推动工业过程中的催化剂开发至关重要.
研究的目的:
- 研究用于甲醇合成的Cu/ZnO/ZrO2催化剂的失活机制.
- 分析催化剂老化过程中发生的依赖时间的结构和化学变化.
- 评价降低反应对老化催化剂的有效性.
主要方法:
- 使用并行设置进行时间解析的实验.
- 包括X射线吸收光谱 (XAS),X射线光电子光谱 (XPS),N2物理吸收和传输电子显微镜 (TEM) 在内的表征技术.
- 分析了不同时间 (0,50,935小时在流中) 的催化剂样本.
主要成果:
- 催化剂失活主要归因于铜 (Cu0) 域的烧结和氧化 (ZnO) 域内的重组.
- 使用H2/N2的还原反应部分恢复了催化剂的结构完整性.
- (ZrO2) 起到促进作用,在初始减少后观察到的重新分散.
结论:
- 烧结和重组是Cu/ZnO/ZrO2甲醇合成催化剂的关键失活路径.
- 有限的结构性复苏是通过降低性反应来实现的.
- ZrO2的促进作用及其再分散是催化剂行为的重要方面.
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