Predicting sign of enthalpy of allotropic transition of phosphorus
You can use the fundamental relation between energy and enthalpy. At constant temperature and pressure:
$$\Delta H_m = \Delta U_m + P\Delta V_m = \Delta U_m + PM (\frac{1}{\rho_{red}}-\frac{1}{\rho_{white}})$$
where the subindex m labels molar properties and M is the molar mass. Taking values of the density posted in the Wikipedia indicates that red phosphorus has a significantly higher density than white (~2.25 vs 1.82 g/mL). This means the volume decreases and the surroundings do work on the system during the reaction. From this data one can approximate the value of $P\Delta V$ at standard pressure (1 bar):
$$\Delta H_m=   \Delta U_m -\pu{0.32 J/mol} $$
In addition the difference in strain would indicate that $\Delta U_m&lt;0$. This would mean that the enthalpy change should be negative, as can be verified by looking up thermodynamic tables. The Wikipedia page describing the allotropes of P (or alternately the CRC handbook) provides a value  of $\Delta H_m=\pu{-17.6 kJ/mol}$, so the correction for the volume contraction can be seen to be negligible (and does not influence the conclusion that $\Delta H&lt;0$). Interestingly, even though white phosphorus is less thermodynamically stable it is considered the standard state of P.