Theoretical study of the effect of reagent rotation on the reaction of F+H2(ν=0,J)
- 22 November 1995
- journal article
- research article
- Published by AIP Publishing in The Journal of Chemical Physics
- Vol. 103 (20) , 8884-8890
- https://doi.org/10.1063/1.470077
Abstract
Quasiclassical calculations on the Muckerman V potential energy surface were carried out on the reaction F+H2(ν=0, J)→FH+H at a relative energy of 2 kcal/mol for J=0 to 10. This surface is, to use the classification of Levine and co‐workers, very oblate; for a given distance Rc.m. from F to the center of mass of H2 the potential energy is much lower for the collinear configuration (χ=0) than for the perpendicular configuration (χ=90 deg). The goal of the work was to understand the effect of molecular rotation on such an extremely oblate surface. It proved useful to decompose the reactive cross section QR(J) into the product of a hitting cross section Q≠hit(J) for F hitting H2 times the probability PR(J) of reaction occurring once F hits H2. Both Q≠hit(J) and PR(J) go through minima at J≊4–5. We determined that Q≠hit(J=0) is increased by about a factor of 2 by ‘‘reorientation’’ of the H2 molecule towards a linear configuration by the F atom as it approaches. For J≳0 Q≠hit(J) declines due both to loss of this reorientation effect as well as to the more oblique approach of the trajectory to the reactant valley. Many trajectories bounce off the repulsive wall near χ=90 deg before the F atom can hit H2; this effect has been discussed by other authors. The initial decline of PR(J) with J is due to a relatively unusual feature of the potential surface, whereby rotation of the H2 molecule away from a linear F–H–H configuration can easily switch the system from the product region back to the reactant region of the system. Both Q≠hit(J) and PR(J) increase above J=5 because the H2 molecule now has enough rotational energy to rotate through the barrier at χ=90 deg rather than bounce off it.Keywords
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