Abstract
The nascent internal state distributions of NH(ND) radicals produced in the reaction of N(2DJ) with HD were measured and compared with those for analogous reactions with H2 and D2. There was no leaving-atom isotope effect in the rotational distributions of NH(v′′=0) and ND(v′′=0), except at highly rotationally excited levels near the thermochemical limits. The observed isotope effect at high rotational levels can be explained by the conservation of both energy and angular momentum. The absence of isotope effect at low and medium rotational levels suggests that the transition state is a bent H(D)–N–H(D) type where the H(D)–H(D) bond is broken.
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