N2 emission-channel change in NO reduction over stepped Pd(211) by angle-resolved desorption

Abstract

A sharp change in the N 2 emission channel from N 2 O(a)→N 2 (g) +O(a) to N(a)+N(a)→N 2 (g) has been found at around 500 K in a steady-state NO+D 2 reaction over stepped Pd(211)= [(S)3(111)×(100)] by means of angle-resolved desorption. The desorbing N 2 is highly collimated at around 30° off normal toward the step-down direction below about 500 K due to the intermediate N 2 O decomposition, whereas, above 500 K, the near normally directed desorption due to the recombination of N(a) is relatively enhanced. The N 2 O decomposition channel is promoted when the reaction is carried out with hydrogen (deuterium) and the channel change is accelerated by quick changes of the amounts of surface hydrogen and oxygen (or NO(a)) into the opposite directions, and enhanced nitrogen removal as ammonia on the resultant hydrogen-rich surface. In the steady-state NO+CO reaction, the N 2 emission channel gradually changes above 500 K toward recombination. A model for the off-normal N 2 emission is briefly described.

Insight into the Reduction of NO by H2 on the Stepped Pd(211) Surface

“…In addition, the experiment shows that N 2 O is an active and important intermediate for the formation of N 2 , , and the production of N 2 O and N 2 involved the same intermediates, trans -N 2 O 2 . N 2 O is also decomposed to N 2 on Rh(110) and Rh(100) surfaces. , The dimer path is feasible and N 2 emission from the decomposition of N 2 O takes place below 500 K on the Pd(211) surface. …”

“…The adsorption, dissociation, and desorption of NO on Pd(111), (100), and (311) surfaces indicated that the stepped Pd(311) surface was active for the thermal dissociation of NO due to the formation of N 2 and N 2 O, while NO molecule desorbed from Pd(111) and Pd(100) surfaces . In addition, N 2 O easily dissociated on the Pd(211) surface leading to the emission of N 2 in a steady-state NO+CO+D 2 reaction, while N 2 O is as the product desorbing from the Pd(111) surface . The experiment shows that the catalytic activity of Pd(211) for the reaction of NO-H 2 depends on the ratio of the area between the (100) steps and the (111) terraces and the stepped Pd­(21l) surface is more active for NO decomposition than the Pd­(1l1) surface. , The DFT results also demonstrate that the dissociation of NO on Pd(211) is facilitated compared to that on Pd(111) .…”

An Experimental Approach to Transition States of Surface Reactions; Energy Partitioning in Repulsive Desorption

“…Hence, the DFT-GGA calculations have predicted that, in the course of dissociation, the N 2 O tilts toward the nearest step in the step-up direction and bends with the terminal O toward the bridge site at the step edge. The N-O bond concomitantly elongates [38] and the terminal nitrogen atom of the dissociating molecule moves from the stable bridge site onto Pd atoms on the (111) terrace [38,40]. A very similar dissociation was first proposed on stepped Pt(211) by Burch [41].…”

“…This model predicts larger collimation angles with stronger repulsive forces operative toward the nascent N 2 from the counter-oxygen atom. Thus, the collimation angle on Pd(110) may be smaller than that on Rh(110) or Rh(100) because the repulsion is expected to be larger on rhodium than on palladium, considering smaller work functions [37] and smaller lattice constants on rhodium; i.e., more polarized or closer oxygen is formed on Rh surfaces [38].…”

“…N 2 desorption is sharply collimated at [25][26][27][28][29][30][31] • off normal toward the step-down direction in either the thermal decomposition of N 2 O(a) at around 110 K [39] or a steady-state NO+D 2 (or NO+CO) reaction below about 550 K [38]. In the latter, of course, the intermediate N 2 O(a) emits the product N 2 .…”