1- The total optimization energy of
the thione tautomer (-1122.6683, -1122.5090 au) is more stable than the thiol one (-1122.6538, -1122.1976 au) according to the optimized studies using B3LYP/6-311G(d,p) and B3LYP/6-31++G(d,p) respectively.
2-
(dipole moment value of thione form in gas phase is greater than that of thiol form, and increases with increasing the dielectric constant of the solvent used) .
3- The calculated energy gap of thiol form is greater than that of thione.
which is the stable form ? and which is the reactive one?
from this data.
The final energies obtained from an electronic structure calculation are referenced to E = 0 corresponding to removing all of the electrons on the molecule to infinite separation, and then separating all of the atomic nuclei to infinite separation. Since the molecules are bound species, the total energy is a negative number. (A positive number would correspond to dissociation into separated electrons and atomic nuclei being energetically favored since the "products", i.e. separated electrons and ions, are at lower energy than the "reactant", i.e. the molecule.) Note that typically total energies are given in hartree because atomic units are used in the actual calculations. Chemically relevant energy differences are often given in kcal/mole (or kJ/mole) since these are values that are intuitive to chemists.
Total energies for two different species may not be compared directly. Total energies for two conformers or isomers of a given species are directly comparable.
To predict which species is more stable and which one is more reactive, one must consider the relevant reaction. Presumably instability means that the molecule (M) reacts with some other species (R) that is present, (such as oxygen in the air or even another molecule of the same kind) depleting the vessel of the molecule of interest and leading to an unwanted product (P).
M + R ---> P
If you wish to compare the thermodynamic stability of two species, M1 & M2, you must consider the two reaction,
M1 + R --> P1 (reaction 1)
M2 + R ---> P2 (reaction 2)
If E(X) denotes the total energy of molecule X, the energy changes for the two reactions are:
DE1 = E(P1) - E(M1) - E(R)
DE2 = E(P2) - E(M2) - E(R).
The reaction with the more positive value of DE has the more "stable" reactant, with respect to this specific reaction.
Note that the above assumes that the total internal energy change is a reasonable approximation of the enthalpy change, (which is typically true for reaction in open vessels where PV work is negligible). It also neglects the entropic contribution to the free energy change. Methods of computing the entropic contribution are available, but considerably more involved than simply computing energy differences.
If you wish to compare the kinetic stability of the two species, you need to compute the total energy of the transition state along each of the two reactions to determine which has the larger activation barrier, {E(TS) - E(M) - E(R)}.
Thank you very much for your answer.What is the best method to compute the transition state energy
Mrs/Miss Mahrous,
There are several methods available. The question is: What is the desirable accuracy and which method you shall use correlating between experimentnd theory?
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