Is there a relationship between viscosity and enthalpy of vaporization? It would be useful to be able to predict relative evaporation rates of liquids.  For small reduced temperatures the heat of vaporization is relatively constant, but it is not one of the more common properties cited for substances.

Is there any sort of functional relationship between enthalpy of vaporization and common properties like viscosity, melting point, and boiling point?
Yes. The phase envelope (or equilibrium curve) of most liquids have a similar shape and can be approximated, to a reasonable degree, by using the Clausius-Clapeyron equation:

$$
\Delta h_v=T(v^V-v^L)\frac{dP^s}{dT}
$$

which requires finding ${dP^s}/{dT}$ from e.g. a vapor-pressure-temperature correlation like the Antoine equation, and also a separate estimate for $\Delta v$ before $\Delta h_v$ can be obtained. This is an exact thermodynamic relationship, but is often simplified by using the ideal gas law for $v^V$ and neglecting the liquid volume $v^L$ i.e. $v^V \gg v^L$, giving

$$
\Delta h_v=\frac{RT^2}{P^s}\frac{dP^s}{dT}
$$

However, it is very important to note that the term $(v^V-v^L)$ becomes increasingly decisive as you go above 1 bar, where the assumption $v^V \gg v^L$ is no longer valid. In such a case, separate correlations for $v^L$ and $v^V$ are required before the correct curvature is obtained. $v^L$, for instance, can be calculated from e.g. a density correlation like the Rackett equation, and $v^v$ from an equation of state like Peng-Robinson.