Predicting sign of enthalpy of allotropic transition of phosphorus How to predict the sign of enthalpy during the transition
$$\ce{P(s, white) -&gt; P(s, red)}?$$
$\ce{P(s, red)}$ is more stable than $\ce{P(s, white)}$ due to release in angle strain.
Since during formation of red phosphorus from white phosphorus bond breaking and bond formation both takes place, how to predict sign of enthalpy?
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.