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Photolysis of azomethane–propane mixtures. Arrhenius parameters for the hydrogen abstraction reactions of methyl with azomethane and with propane
Author(s) -
Durban Philip C.,
Marshall Roger M.
Publication year - 1980
Publication title -
international journal of chemical kinetics
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.341
H-Index - 68
eISSN - 1097-4601
pISSN - 0538-8066
DOI - 10.1002/kin.550121212
Subject(s) - chemistry , isobutane , propane , arrhenius equation , propene , reaction rate constant , photochemistry , photodissociation , hydrogen atom abstraction , butane , quantum yield , hydrogen , methane , kinetics , organic chemistry , catalysis , activation energy , physics , quantum mechanics , fluorescence
Mixtures of up to 14% azomethane in propane have been photolyzed using mainly 366 nm radiation in the ranges of 323–453 K and 25–200 torr. Detailed measurements were made of the yields of nitrogen, methane, and ethane. Other products observed were isobutane, n ‐butane, ethene, and propene. A detailed mechanism is proposed and shown to account for the observed variation of product yields with experimental conditions. The quantum yield of the molecular process\documentclass{article}\pagestyle{empty}\begin{document}$$ {\rm CH}_3 {\rm N}_2 {\rm CH}_3 + h\nu \to {\rm C}_2 {\rm H}_6 + {\rm N}_2 $$\end{document}is found to be given by the temperature‐independent equation\documentclass{article}\pagestyle{empty}\begin{document}$$ \phi _2 = 0.0092 \pm 0.0005] $$\end{document}The values of rate constants obtained are\documentclass{article}\pagestyle{empty}\begin{document}$$ {\rm log} k_3 ({\rm cm}^3 /{\rm mol}\,{\rm sec}) = 12.03 \pm 0.15 - 40.8 \pm 1.1{\rm kJ}/{\rm mol}/(2.3RT) $$\end{document}\documentclass{article}\pagestyle{empty}\begin{document}$$ {\rm log}k_4 ({\rm cm}^3 /{\rm mol}\,{\rm sec}) = 11.42 \pm 0.15 - 40.7 \pm 1.1{\rm kJ}/{\rm mol}/(2.3RT) $$\end{document}where the reactions areand it is assumed that the rate constant for the reactionis given by\documentclass{article}\pagestyle{empty}\begin{document}$$ {\rm log}k_5 ({\rm cm}^3 /{\rm mol}\,{\rm sec}) = 13.4 $$\end{document}

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