Abstract

The transition metal rhodium has been proved the effective catalyst to convert from NO(x) to N(2.) In the present work, we are mainly focused on the NO adsorption and decomposition reaction mechanism on the surface of the Rh(7)(+) cluster, and the calculated results suggest that the reaction can proceed via three steps. First, the NO can adsorb on the surface of the Rh(7)(+) cluster; second, the NO decomposes to N and O atoms; finally, the N atom reacts with the second adsorbed NO and reduces to a N(2) molecule. The N-O bond breaks to yield N and O atoms in the second step, which is the rate-limiting step of the whole catalytic cycle. This step goes over a relatively high barrier (TS(12)) of 39.6 kcal/mol and is strongly driven by a large exothermicity of 55.1 kcal/mol during the formation of stable compound 3, accompanied by the N and O atoms dispersed on the different Rh atoms of the Rh(7)(+) cluster. In addition, the last step is very complex due to the different possibilities of reaction mechanism. On the basis of the calculations, in contrast to the reaction path II that generates N(2) from two nitrogen atoms coupling, the reaction path I for the formation of intermediate N(2)O is found to be energetically more favorable. Present work would provide some valuable fundamental insights into the behavior of the nitric oxide adsorption and reduction reaction mechanism on the Rh(7)(+) cluster.

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