E.s.r. studies of the reactions of atomic oxygen and hydrogen with simple hydrocarbons
E.s.r. studies of the reactions of atomic oxygen and hydrogen with simple hydrocarbons
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E.s.r.
DOI：
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发表时间：
1967
期刊：
影响因子：
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通讯作者：
B. Thrush
中科院分区：
文献类型：
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作者：
J. Brown;B. Thrush
电子自旋共振已用于研究氮载体中的氧原子和氩载体中的氢原子与简单碳氢化合物在总压力约为 2 mm Hg 的快速流动系统中的反应。氧原子和乙炔之间的反应主要通过以下机理发生： O+C2H2= CH2+CO (3) O+CH2= CO+2H (6) CH2+C2H2= C3H4(12) 其中 k3=(9.2 ± 0.4)× 1010 cm3 摩尔–1 sec–1 和 k6=(2.7 ± 1.0)k12 均在 298°K 下。氧原子与甲基乙炔反应的第一步是O+CH3。 CCH=CH3。 CH : + CO (13)，其中 k13=(4 ± 1)× 1011 cm3 摩尔–1 秒–1，298°K；氧原子消失的总速率常数大 3-4 倍。
O+C2H4 和 O+CH4 反应的机理已从它们的总体化学计量和氢原子产率中推导出来。对于初始步骤 O+C2H4= CH3+HCO (15) O+CH4= CH3+OH (23)k15=(3.2 ± 0.4)× 1011 cm3 摩尔–1 秒–1 在 298°K 和 k23=(7 ± 2)× 1012 exp (–7700 ± 300/RT) cm3 摩尔–1 秒–1 在 450 和600°K。氢原子向甲基乙炔和乙烯的加成是二阶过程，H+C3H4= C3H5(4) H+C2H4= C2H5(17)，其中 k4=(2.4 ± 0.3)× 1011 cm3 mol–1 sec-1，k17=(8.8 ± 0.4)× 1010 cm3 mol–1 sec-1，温度为 298°K。
Electron spin resonance has been used to study the reaction of oxygen atoms in a nitrogen carrier and of hydrogen atoms in an argon carrier with simple hydrocarbons in a fast flow system at total pressures around 2 mm Hg. The reaction between oxygen atoms and acetylene occurs predominantly by the mechanism: O+C2H2= CH2+CO (3) O+CH2= CO+2H (6) CH2+C2H2= C3H4(12) where k3=(9.2 ± 0.4)× 1010 cm3 mole–1 sec–1 and k6=(2.7 ± 1.0)k12 both at 298°K. The initial step in the reaction of oxygen atoms with methyl acetylene is O+CH3 . C CH = CH3 . CH : + CO (13) with k13=(4 ± 1)× 1011 cm3 mole–1 sec–1 at 298°K; the total rate constant for the disappearance of oxygen atoms being 3–4 times greater.
The mechanisms of the O+C2H4 and O+CH4 reactions have been deduced from their overall stoichiometries and the hydrogen atom yields. For the initial steps O+C2H4= CH3+HCO (15) O+CH4= CH3+OH (23)k15=(3.2 ± 0.4)× 1011 cm3 mole–1 sec–1 at 298°K and k23=(7 ± 2)× 1012 exp (–7700 ± 300/RT) cm3 mole–1 sec–1 between 450 and 600°K. The addition of hydrogen atoms to methyl acetylene and to ethylene were second order processes, H+C3H4= C3H5(4) H+C2H4= C2H5(17) with k4=(2.4 ± 0.3)× 1011 cm3 mole–1 sec–1 and k17=(8.8 ± 0.4)× 1010 cm3 mole–1 sec–1 at 298°K.