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The Production of Synthesis Gas by the Redox of Cerium Oxide

机译:氧化铈氧化还原法生产合成气

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摘要

The oxidation of CH_4 with CeO_2 in the absence of gaseous oxygen (Step1) and subsequent reduction of CO_2 to CO or of H_2O to H_2(Step2) by the reduced cerium oxide, CeO_(2-x), have been studied under stmospheric pressure at 673 to 1073K. The reaction of Step1 occurred at >= 873K producing H_2 and CO with the ratio of 2. This synthesis gas was strongly suggested to be formed directly from CH_4. Addition of Pt black (1 wt percent) to CeO_2 remarkably enhanced this reaction. The reaction of Step2 proceeded at >=673K for both CO_2 and H_2O. When the degree of reduction of the oxide in Step1 had been adjusted to ca. 10 percent, both CO_2 and H_2O were stoichiometrically converted to CO and H_2, respectively. If the degree of the reduction exceeded 10 percent in Step1, the carbon deposited on the CeO_(2-x) strongly retarded the formation of CO from CO_2 in Step2. However, the decomposition of H_2O was not affected. Thus, it is suggested that the reoxidation of the reduced oxide (Step2) should be operated with H_2O.
机译:在大气压力下,于室温下研究了在不存在气态氧的情况下用CeO_2氧化CH_4的氧化(步骤1),然后通过还原的氧化铈CeO_(2-x)将CO_2还原为CO或将H_2O还原为H_2(步骤2)。 673至1073K。步骤1的反应在> = 873K处发生,以2的比例生成H_2和CO。强烈建议该合成气直接由CH_4形成。在CeO_2中加入Pt黑(1 wt%)显着增强了该反应。对于CO_2和H_2O,步骤2的反应均在> = 673K​​下进行。当将步骤1中的氧化物的还原度调节至约1。 10%的CO_2和H_2O分别化学计量转化为CO和H_2。如果在步骤1中还原度超过10%,则在步骤2中,沉积在CeO_(2-x)上的碳会强烈阻碍CO_2生成CO。但是,H_2O的分解不受影响。因此,建议还原的氧化物的再氧化(步骤2)应以H_2O进行。

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