I understand that that energy is required to overcome intermolecular forces holding solute particles together in the crystal, but I don't understand how this relates to lattice energies and respective solubilities in water.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
When an ionic compound dissolves in water, the water molecules surround the individual ions. Since water is very polar, when water surrounds an ion there is a decrease in overall energy. However, to pull the ions apart in the first place, there has to be an energy increase to overcome the lattice energy. If the lattice energy (cost) is larger than the energy of solvation of the ions (gain) the compound won't dissociate, and it will be insoluble.
There is a large grey area where lattice energy and solvation energy are close, and for those compounds temperature has a big influence on solubility.
$\begingroup$
$\endgroup$
You can break the process of dissolution down into sub-processes and use Hess’ law to calculate the overall process’ enthalpy. Specifically, the processes you need to consider are:
- destruction of the lattice, loss of $\Delta_\mathrm{L}H^0$
- solvation of the cation, gain of $\Delta_\mathrm{solv,1}H^0$
- solvation of the anion, gain of $\Delta_\mathrm{solv,2}H^0$
According to Hess’ law, the total enthalpy of dissolution $\Delta_\mathrm{D}H^0$ is thus:
$$\Delta_\mathrm{D}H^0 = \Delta_\mathrm{solv,1}H^0 + \Delta_\mathrm{solv,2}H^0 - \Delta_\mathrm{L}H^0$$
Thus, a higher lattice enthalpy will decrease the value of $\Delta_\mathrm{D}H^0$, i.e. decrease the enthalpy gained from dissolution, i.e. make dissolution ‘harder’.
There is also a kinetic effect associated: even if the overall process is favourable, it may just be a very slow reaction for the lattice to break down. This may initially be perceived as a lower solubility. However, given time (and potentially heat to increase molecular motion), you will arrive at a thermodynamic solubility which will be higher than the initially perceived one.
