The reaction of O( 1 D ) with CH 4

The room‐temperature photolysis of N 2 O (10–100 torr) at 2139 Å to produce O( 1 D ) has been studied in the presence of CH 4 (10–891 torr). The reactions of O( 1 D ) with CH 4 were found to be\documentclass{article}\pagestyle{empty}\begin{document}$$\begin{array}{*{20}c} {(4)} & {{\rm O(}^{\rm 1} D{\rm)} + {\rm CH}_{\rm 4} } & \to & {{\rm HO + CH}_{\rm 3} } & {0.95 \pm 0.05} \\ {(8)} & {} & \to & {{\rm O(}^{\rm 3} P{\rm)} + {\rm CH}_{\rm 4} } & {{\rm 0}{\rm .05} \pm {\rm 0}{\rm .05}} \\ {} & {} & \to & {{\rm CH}_{\rm 2} + {\rm H}_{\rm 2} {\rm O}} & {{\rm < 0}{\rm .03}} \\ {} & {} & \to & {{\rm CH}_{\rm 3} {\rm OH}} & {{\rm < 0}{\rm .01}} \\ {} & {} & \to & {{\rm CH}_{\rm 2} {\rm O + H}_{\rm 2}} & {{\rm < 0}{\rm .002}} \\\end{array}$$\end{document} The method of chemical difference was used to measure the rate constant ratio k 4 /( k 2 + k 3 ), where reactions (2) and (3) are\documentclass{article}\pagestyle{empty}\begin{document}$$\begin{array}{*{20}c} {(2)} & {{\rm O(}^{\rm 1} D{\rm)} + {\rm N}_{\rm 2} {\rm O}} & { \to {\rm N}_{\rm 2} + {\rm O}_{\rm 2} } \\ {(3)} & {} & { \to 2{\rm NO}} \\\end{array}$$\end{document} The CH 3 radicals produced in reaction (4) react with the O 2 and NO produced in reactions (2) and (3). Thus, near the endpoint of the internal titration, ϕ{C 2 H 6 } gives an accurate measure of k 4 /( k 2 + k 3 ). For the translationally energetic O( 1 D ) atoms produced in the photolysis, k 4 /( k 2 + k 3 ) = 2.28 ± 0.20. However, if He is added to remove the excess translational energy, then k 4 /( k 2 + k 3 ) drops to 1.35 ± 0.3.