Is there a relationship between viscosity and enthalpy of vaporization?
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that the term $(v^V-v^L)$ becomes increasingly decisive as you go above 1 bar, where the assumption $v^V \gg v^L$ is no longer valid.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.