甲醇与O[3P]反应机理的从头算及动力学  

Kinetics and ab initio study of the reaction methanol with O[~3P]

作　　者：刘坤辉[1] 冯文林[1] 李会英[1] 冀永强[1] 蒲敏[1] 

机构地区：[1]北京化工大学可控化学反应科学与技术基础教育部重点实验室,北京100029

出　　处：《北京化工大学学报（自然科学版）》2004年第3期82-86,共5页Journal of Beijing University of Chemical Technology(Natural Science Edition)

基　　金：国家教育部博士点基金 (1 9990 0 2 71 5 ) ;北京化工大学青年教师基金 (QN0 30 8)

摘　　要：采用UQCISD/ 6 31 1G (d ,p )从头算方法 ,优化甲醇和O [3 P ]的反应两个通道、反应物、过渡态和产物的几何构型。进一步运用G2方法进行单点能量校正 ,得出通道 (1 )和通道 (2 )的位垒分别是 4 8 86kJ/mol和 2 8 89kJ/mol。并指出通道 (1 )是吸热反应 ,而通道 (2 )是放热反应。在 30 0～ 32 0 0K温度范围内 ,采用传统过渡态理论计算两个反应通道各自的速率常数k1 和k2 ,由此采用非线性最小二乘法 ,得出这两个反应通道各自的速率方程为k1 =2 4 3× 1 0 -1 8×T2 2 3 ×exp(- 32 97/T)cm3 mol-1 s-1 (30 0K T 32 0 0K)k2 =6 1 2× 1 0 -1 8×T2 1 9×exp(- 1 396 /T)cm3 mol-1 s-1 (30 0K T 32 0 0K)通道 (2 )是主反应通道。讨论了通道 (2 )与通道 (1 )的速率常数比k2 /k1 对温度变化的依赖关系。计算得出CH3 OH和O[3 P]反应的总速率常数k1 + 2 。Ab initio UQCISD calculations, with 6 311G(d, p) basis sets were performed for the title reaction. The results show that the reaction has two product channels: .Both reactions are classical hydrogen abstraction. Using G2 theory, the calculated energy barriers were 48 86?kJ/mol and 28 89?kJ/mol respectively. In the range of temperature 300～3?200?K, the rate constants k 1 and k 2 were obtained by the transition state theory. Using the nonlinear least square method, the simulated rate equations were sequentially obtained for the both reactions k 1 =2 43×10 -18 × T 2 23 ×exp(-3?297/ T )?cm 3mol -1 s -1 (300?K T 3?200?K) k 2 =6 12×10 -18 × T 2 19 ×exp(-1?396/ T )?cm 3mol -1 s -1 (300?K T 3?200?K)Theoretical calculations suggest that the formation of CH 2OH is the dominant channel. The overall rate constants of the title reactions are consistent with the experimental values.

关 键 词：甲醇 三线态氧 从头算 速率常数 

分 类 号：O641[理学—物理化学]