Production and magic numbers of large hydrated ... - ACS Publications

Production and magic numbers of large hydrated anion clusters X-(H2O)N=0-59 (X = hydroxyl, oxygen atom or molecule, and ozone) under thermal condition...
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J . Phys. Chem. 1990, 94, 8500-8502

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where Mi, are determined by the set of quadratic equations

parts of the molecule. As would be seen below, this factor slightly renormalizes F,/. Ideas similar to those used for the eqs 3-5 derivation were proposed by Flory' in the theory of polymers. So this approach could be called Flory-like approximation. It should be said that the assumption eq 4 is rather a strong one: as the formation of an H bond reduces the number of degrees of freedom of the molecule it cannot but influence the formation of other H bonds with the same molecule. However, this assumption can be justified in two cases: (i) for molecules with a small number of H bonds per molecule, and (ii) for big flexible molecules. I n other cases it can be used for the rough estimation of z. Using eqs 3-5 in eq 2 one obtains in the thermodynamic limit

Equations 7 and 8 give the excess free energy due to H bonds for the discussed lattice model. It should be noted that the lattice coordinate number z enters these equations in the form Fu - kBT In (zl,) only. As the free energy Fij contains the term kBT In z , eqs 7 and 8 are in fact independent of the lattice properties (I am grateful to the reviewer for this remark). As an example of using eqs 7 and 8 let us consider the binary mixture with H-donor molecules A and H-acceptor molecules B: d, = aB = 1, dB = aA = 0. If NA >> NB (donor sites are in excess) then one obtains from eqs 7 and 8 FH = kBTNB In

(9)

where y = (NE - MAB)/NB is a probability for a molecule B to be not involved in H bonding. The value of y is determined by the following equations:

Y = 1/(1 + K N A / N ) K = Z ~ A B~ ~ P ( - F A B / ~ B T )

These expressions are consistent with those of Huyskens2 In the case of a water solution of an inert substance, d A = aA = 2 , d B = a B = 0. Substituting these values in eqs 7 and 8 one obtains after some algebra the chemical potentials of the components. They are also consistent with the expressions of Huyskens2 in both limiting cases NA/NB >> 1 and NA/NB