However, we still need to integrate this over all values of ps that correspond to enough energy ps /2u to access the TSs energy, which we denote E*. Moreover, we have to account for the fact that it may be that not all trajectories with kinetic energy equal to e or greater pass on to form product molecules, some trajectories may pass through the Ts but later recross the TS and return to produce reactants. Moreover, it may be that some trajectories with kinetic energy along the reaction coordinate less than e* can react by tunneling through the barrier The way we account for the facts that a reactive trajectory must have at least E* in energy along s and that not all trajectories with this energy will react is to integrate over only values of ps greater than (2ue")and to include in the integral a so-called transmission coefficient k that specifies the fraction of trajectories crossing the ts that eventually proceed onward to products. Putting all of these pieces together, we carry out the integration over ps just described to obtain SS(/h)K exp(-P32/2ukT(p/u8 ) ds dp where the momentum is integrated from ps =(2uE*)to oo and the s-coordinate is integrated only over the small region 8s. If the transmission coefficient is factored out of the integral(treating it as a multiplicative factor), the integral over ps can be done and yields the following (kT/h)exp(E*/kT) 66 However, we still need to integrate this over all values of ps that correspond to enough energy ps 2 /2m to access the TS’s energy, which we denote E*. Moreover, we have to account for the fact that it may be that not all trajectories with kinetic energy equal to E* or greater pass on to form product molecules; some trajectories may pass through the TS but later recross the TS and return to produce reactants. Moreover, it may be that some trajectories with kinetic energy along the reaction coordinate less than E* can react by tunneling through the barrier. The way we account for the facts that a reactive trajectory must have at least E* in energy along s and that not all trajectories with this energy will react is to integrate over only values of ps greater than (2mE*)1/2 and to include in the integral a so-called transmission coefficient k that specifies the fraction of trajectories crossing the TS that eventually proceed onward to products. Putting all of these pieces together, we carry out the integration over ps just described to obtain: ò ò (1/h) k exp(-ps 2 /2mkT) (ps /mds ) ds dps where the momentum is integrated from ps = (2mE*)1/2 to ¥ and the s-coordinate is integrated only over the small region ds. If the transmission coefficient is factored out of the integral (treating it as a multiplicative factor), the integral over ps can be done and yields the following: k (kT/h) exp(-E*/kT)