N <sub> 2 </ Sub> O Catalyst Rh Surface Adsorption Mechanism

With the development of the city and the increase of the motor vehicles,there are more and more pollution in the air,and the proportion of air pollution is increasing,so it damages the environment seriously.Automotive three-way catalytic(TWC)converters consist of various combinations of Pt,Pd and Rh for the simultaneous.Among these metals,rhodium is key component.The pollution of NO_x contains NO and N₂O.As the direct pollutants,NO has been researched mostly.As the reaction intermediates,N₂O has been researched relatively little,but also the pollution of it can not be neglected.The process of catalysis is multi-steps process,in which both the constiunet and the topology of the surface of the catalyzer can affect these processes.So the catalysis is very complex,and there are questions like these:which kind of the structure of the adsorbate is favourable? What's the median product? How does the median product transform into product? These problems have to be explored further.In this paper,on the basis of experimental and theoretical studies on the catalytic reaction of N₂O in the TWC,we caculated the chemidorption and Transition state of N₂O on Rh(111),Rh(100)surface with CASTEP program on the level of DFT(Density Functional Theory)which is usually used in theoretical studies on the system in transition metal catalysts.The PW91 function is used as the exchang-correlation function.In order to ensure which adsorption position is the most stable position.We consider two conditions, one is N adsorption,another is O adsorption.For the(111)surface,we have considered the top,twofold bridge,threefold hollow fee and hcp sites.For the(100)surface,we have considered the top,twofold bridge,fourfold hollow sites.Three-layer(3x3)slab model on a periodic system is used to simulate the metal surface.The calculation results are presented as following:(1)Thealculation results of the adsorption of N₂O on Rh(111)surface are presented as following:compare with the length of the bond of N-N,N-O,Rh-N and Rh-O on the both of N adsorption and O adsorption,we found that the more stable adsorption is N adsorption.Through the difference of the lengh of the bond and the adsorptive energy,we found atop of N adsorption is the most stable adsorption.This result is similar to the result of the study of J.M.Ricart.et.al,who calculated the atop adsorptive energy of N₂O on Rh(111)surface which is(6x6)slab model.Their results is atop of N adsorption is the most stable adsorption.The energy of it is-0.40ev.(2)The lculation results of the transition state of N₂O on Rh(111)surface are presented as following:The results of the active energy is 345.04kJ/mol.On Rh(111)surface N₂O is active when it have enough energy.The atom of O is probably on twofold bridge of Rh(111).The length of N-O is 0.1759nm longer than adsoption state.The bond of N-N is enhanced,but the bond of Rh-N is shorter than adsoption state.It enhances the adsorption of N on Rh(111)surface,then we get the transition state of Rh...N₂...O.From the transition state to the product,the bond of N_c-O is dis- connected,and then the atom of O is released.At last it is formed as the adsorption state of N-Rh and O.N₂(a)may be desorbed as N₂(g).If the energy of the transition state is reduced,the molecular of N-2O is released when the bond of N_t-Rh is discon-nected.(3)The alculation results of the adsorption of N₂O on Rh(100)surface are presented as following:compare with the length of the bond and the adsorptive energy of N-N,N-O, Rh-N and Rh-O on the both of N adsorption and O adsorption,we found atop of N adsorption is the most stable adsorption.Otherwise,form the adsorptive energy we found the energy of twofold bridge is also small,which is -0.5036ev,and form the change of the bond we found the bond of Rh-N is also short,so the twofold bridge of adsorption is also possible.But the the bond of N-N is so long,and the analysis mechanism is different from atop of N adsorption.(4)The lculation results of the transition state of N₂O on Rh(100)surface are presented as following:The results of the active energy is 351.73kJ/mol.Atop of N adsorption have enough energy,then it turns into the transition state.On Rh(100)surface the atom of N is adsorbed on the atop of Rh(100),but the atom of O is probably on fourfold site of Rh(100).The length fo N-O is 0.0994nm longer than adsoption state.The bond of N-N is enhanced,but the bond of Rh-N is shorter than adsoption state.It enhances the adsorption of N on Rh(111)surface,then we get the transition state of Rh...N₂...O.If the transition state changes to the product,the bond of N_c-O is disconnectedand then the atom of O(a)is released.Then they turn into N₂(a)and O(a).At last N₂(a)turns into N₂(g).If the energy of the transition state is reduced,the molecular of N₂O is released when the bond of N_t-Rh is disconnected.(5)From the calculation results we can found atop of N adsorption is the most stable adsorption on both of Rh(111)surface and Rh(100)surface.The results of this adsorption is similar to the A.V.Zeigami mechanism and L-H mechanism,which is N₂O_gas(?)N₂O_ads. From the length of the bond and adsorptive energy,we found the twofold bridge of adsorption is also possible.But the the bond of N-N is too long,and the analysis mechanism is different from atop of N adsorption.