8
E.G. Moschetta et al. / Journal of Catalysis 302 (2013) 1–9
specific P ligand [51]. For the five P ligands in our study, the BDEs
range from ꢃ108 to 148 kJ/mol, so it is unsurprising to learn that
ligand binding is enthalpy-driven from a theoretical perspective.
Their calculations do not account for the effect of solvent and their
model Pd structures are obviously different from ours, but these
values are a reasonable estimate for theoretical Pd–P binding ener-
gies. With the exception of P(OPh)3, our observed enthalpies are a
factor of 2–3 smaller for binding one equivalent of ligand to Pd,
which we attribute to the fact that the ligands and Pd complexes
are solvated. Additionally, their model Pd compound is negatively
charged, while ours is neutral.
We found that these metal–ligand interactions are enthalpy-
driven. These large, exothermic enthalpies indicate that solvent
reorganization likely contributed greatly to the observed enthal-
pies and played an active role in the metal–ligand equilibria. We
attributed the ability of the solvent molecules to interact with
the solutes as another contributor to the observed enthalpies of
binding. All ligands for which two equivalents bind to 1 had an in-
crease in entropy after the second ligand was bound, likely due to
displacement of a bound solvent molecule. The largest decrease in
observed entropy was seen for P(OPh)3 and was attributed to a loss
of translational and rigid-body rotational entropies.
The drastic variation in thermodynamic values obtained for
P(OPh)3 relative to the other P ligands warrants further discussion.
The other ligands in this study have rigid aryl groups connected di-
rectly to the P atom, whereas the phenyl groups in P(OPh)3 are
linked to the P atom by O atoms, which allows for more rotational
freedom when the ligand is unbound. All ligands will lose transla-
tional and rigid-body rotational entropy upon binding, but P(OPh)3
would seem to suffer the largest penalty. The enthalpies of dissolu-
tion for both ligands, PPh3 (26.21 0.04 kJ/mol) and P(OPh)3
(6.86 0.01 kJ/mol), are endothermic. In both cases, there are
intermolecular forces that must be overcome in order for dissolu-
tion to occur. The enthalpy of solution for PPh3 is larger in magni-
tude because its lattice energy must be overcome in order for it to
dissolve (it is a solid) while P(OPh)3 is a liquid and has no such en-
ergy. If the enthalpy of fusion for PPh3 (19.69 kJ/mol) [52] is taken
into account as a crude approximation of its lattice energy, then
dissolution of a hypothetical PPh3 liquid would be the difference
between the enthalpy of dissolution and the enthalpy of fusion,
which is 6.52 kJ/mol, slightly smaller in magnitude than the en-
thalpy of solution of P(OPh)3, which is 6.86 kJ/mol. Since the en-
thalpy of dissolution for P(OPh)3 is positive, overcoming its own
intermolecular forces and breaking up solvent–solvent intermolec-
ular forces are larger in combined magnitude than solute–solvent
interactions. The inability of P(OPh)3 to overcome these forces
which, along with its enormous rotational entropy penalty, ex-
plains why P(OPh)3 has such a large, negative entropy of binding
[53]. Searle and Williams estimate an entropy loss of 8.8–44.8 kJ/
Acknowledgments
The work was supported by The Pennsylvania State University
and The Penn State Institutes of Energy and Environment (PSIEE)
through start-up funds provided to RMR and a 3M Non-Tenured
Faculty Grant. K.G. acknowledges support from the NASA WISER
program for undergraduate research.
Appendix A. Supplementary material
Supplementary data associated with this article can be found, in
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