Chap27 solutions

Physical Chemistry

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27 Molecular reaction dynamics Solutions to exercises Discussion questions E27.1(b) A reaction in solution can be regarded as the outcome of two stages: one is the encounter of two reactant species, which is followed by their reaction, the second stage, if they acquire their activation energy. If the rate-determining step is the former, then the reaction is said to be diffusion-controlled. If the rate-determining step is the latter, then the reaction is activation controlled. For a reaction of the form A + B P that obeys the second-order rate law ν = k 2 [A][B], in the diffusion-controlled regime, k 2 = 4 πR DN A where D is the sum of the diffusion coef±cients of the two reactant species and R is the distance at which reaction occurs. A further approximation is that each molecule obeys the Stokes–Einstein relation and Stokes’ law, and then k 2 8 RT 3 η where η is the viscosity of the medium. The result suggests that k 2 is independent of the radii of the reactants. E27.2(b) In the kinetic salt effect, the rate of a reaction in solution is changed by modi±cation of the ionic strength of the medium. If the reactant ions have the same sign of charge (as in cation/cation or anion/anion reactions), then an increase in ionic strength increases the rate constant. If the reactant ions have opposite signs (as in cation/anion reactions), then an increase in ionic strength decreases the rate constant. In the former case, the effect can be traced to the denser ionic atmosphere (see the Debye–Huckel theory) that forms round the newly formed and highly charged ion that constitutes the activated complex and the stronger interaction of that ion with the atmosphere. In the latter case, the ion corresponding to the activated complex has a lower charge than the reactants and hence it has a more diffuse ionic atmosphere and interacts with it more weakly. In the limit of low ionic strength the rate constant can be expected to follow the relation log k = log k + 2 Az A z B I 1 / 2 E27.3(b) Refer to Figs 27.21 and 27.22 of the text. The ±rst of these ±gures shows an attractive potential energy surface, the second, a repulsive surface. (a) Consider Fig. 27.21. If the original molecule is vibrationally excited, then a collision with an incoming molecule takes the system along the ²oor of the potential energy valley (trajectory C). This path is bottled up in the region of the reactants, and does not take the system to the saddle point. If, however, the same amount of energy is present solely as translational kinetic energy, then the system moves along a successful encounter trajectory C and travels smoothly over the saddle point into products. We can therefore conclude that reactions with attractive potential energy surfaces proceed more ef±ciently if the energy is in relative translational motion. Moreover, the potential surface shows that once past the saddle point the trajectory runs up the steep wall of the product valley, and then rolls from side to side as it falls to the foot of the valley as the products separate. In other words, the products emerge in a vibrationally excited state.
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Chap27 solutions - 27 Molecular reaction dynamics Solutions...

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