Photodecomposition reaction of ethyl bromide at 147 nm

Abstract

A vacuum ultraviolet photolysis of ethyl bromide at 147 nm was studied over the pressure range of 0.5-50 Torr at room temperature, using a Xenon resonance lamp. The pressure effect with and without He as an additive was investigated. As a collision partner $C_2H_5Br$ was found to be more effective than He in energy transfer. A scavenger effect of the reaction was also observed by adding NO as a radical scavenger. The extinction coefficient of $C_2H_5Br$ at 147 nm and 298 K was found to be $712 \pm 7 atm^{-1} cm^{-1}$. The principal reaction products were $CH_4,\; C_2H_2,\; C_2H_4$, and $C_2H_6$. Increasing pressure, the product quantum yields, $\phi$, of $CH_4 (\phi=0.019 \pm 0.004)$ and $C_2H_4 (\phi = 0.57 \pm 0.01)$ remain constant, while that of $C_2H_2$ ($\phi=0.03$ to 0) was found to be weak negative pressure dependence, and that of $C_2H_6$ ($\phi$=0.39 to 0.43) was weak positive pressure dependence. Addition of NO completely suppressed the formation of $C_2H_2$ and $C_2H_6$, and partially reduced that of $C_2H_4 (\phi = 0.50 \pm 0.01)$. These results were interpreted in terms of two channel competition between the molecular eliminations and the formation of radicals. And their contributions to the decomposition are approximately half-and-half. From the aforementioned results, it seems to be reasonable to assume that there exist two electronically excited states. One state which is initially formed plays a role as the main source of the molecular elimination products. And the other state resulted from the first excited state by collisional cross over goes on the radical decomposition modes.