Zero-vibration energy

The eomputed minimum of V (using any method, either quantum-mechanical or force field) does not represent the energy of the system for exactly the same reason as the bottom of the parabola (the potential energy) does not represent the energy of the harmonic oscillator (cf. the harmonic oscillator, p. 166). The reason is the kinetic energy contribution. 

If all the normal oscillators are in their ground states vj = 0, called the zero-vibrations ), then the energy of the system is the energy of the bottom of the parabola Vmin plus the zero-vibration energy (we assume no rotational contribution) 

It has been assumed that the vibrations are harmonic in the above formula. This assumption usually makes the frequencies higher by several percent (cf. p. 175). 

Taking anharmonicity into account is a much more difficult task than normal mode analysis. Note (Fig. 7.11) that in such a case the position of the minimum 

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A He enthalpy of reaction that includes difference of zero vibration energies of reacting bonds kJ moU1... 

The physical origin of this difference is a pure quantum effect that comes from the gain in zero vibrational energy of the vibration between the H-bonded state and the free state that enters the enthalpy of formation of the H-bond and is different in H- and D-bonds. This quantity is negative because, as illustrated in Novak s curve of Figure 4.5, the difference between terms 1 and 3 of eq. (7.7) is negative the wavenumbers of bands, which are related to the m s by the following eq. (7.8) that takes into account eq. (1.A3),... 

Let us stress that the formula jAv for the zero vibration energy is related to the harmonic approximation. Generally,... 

The adiabatie potential eorresponding to the vibrational ground state ( voh, Vhh) = (0,0) gives lower barrier heightthan the elassical potential Vb(-s) (5-9 keal/mol vs 6.1). The reason for this is the lower zero-vibration energy for the saddle point eonfiguration than for the... 

In the electron transfer reaction HJ -F H2 H2 + the energy of the reactants is equal to the energy of the products because the reactants and the products represent the same system. Is it, therefore, akind of fiction Is there any reaction at aU taking place From the point of view of a bookkeeper (thermodjfnamics), no reaction took place, but from the point of view of a molecular observer (kinetics), such a reaction may take place. It is especially visible when instead of one of the hydrogen atoms, we use deuterium. Then the reaction HD+ + H2 -> HD + hJ becomes real even for the bookkeeper (mainly because of the difference in the zero-vibration energies of the reactants and products). 

As an example consider = 60, which corresponds to an unlimited number of mass ratios, among them A B C = 1 1 1, 1 2 6, 2 3 5, etc. Different combinations can be deduced from (10.6). From Fig. 10.10a we see that, since both reactant and product channels have the same width, all horizontal trajectories (zero vibrational energy) are reactive the products formed have zero vibrational energy. The result is peculiar to the choice f = 60°. For other jS this trajectory is not always reactive in the example of Fig. 10.10b, large values of y lead to inelastic collisions. 

Because of the Uncertainty Principle (Section 2.5), a molecule cannot have zero vibrational energy if it had, we could specify both the positions and the momenta of the atoms in a molecule precisely. All molecular vibrations therefore have some zero-point energy, and this can be shown to be equal to one half of a quantum, so that the total vibrational energy of a molecule with only one normal vibration, such as a simple diatomic, is... 

Fig. 1.1. Potential energy curve for a two-atom molecule (r is the equilibrium bond length, D — is the zero vibration energy)...

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