Abstract
AbstractThe photolysis of pure N2O, N2O and N2, and N2O and C3H6 mixtures at 1470 Å and room temperature has been studied to determine the relative importance of the primary processes. The results are equation image where ϕ{O(1D)} = 0.515 represents both the O(1D) produced in the primary act and that produced by collisional quenching of O(1S); ϕ{N2(3Σ)} = 0.084 represents only that portion of N2(3ϕ) which dissociates N2O on deactivation; and ϕ{O(1S)} = 0.38 – ±{N(2D)} represents only that portion of O(1S) which enters into chemical reaction with N2O. If the reaction of O(1S) with N2O yields only N2 and O2 as products, which seems likely from potential‐energy curve considerations then ±{O(1S)} = 0.135 ± 0.06 and ϕ{N(2D)} = 0.245 ± 0.06. Young and coworkers [4] have found from spectroscopic observations that the total quantum yield of O(1S) is about 0.5. Thus it can be concluded that collisional removal of O(1S) by N2O yields mainly O(1D) with chemical reaction being less important. Furthermore, most of the O(1D) is produced this way, and the true primary yield of O(1D) is about 0.15. The metastable N(2D) is not deactivated by N2O, but is removed by chemical reaction to produce N2 and NO. The results further indicate that N2(3Σ) dissociates N2O at least 80% of the time during quenching. The relative efficiency of N2O compared to N2 is about 2 for the removal of O(1D). O(1S) is removed about 90 times as efficiently by C3H6 as by N2O.
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