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How to predict the sign of enthalpy during the transition

$$\ce{P(s, white) -> P(s, red)}?$$

$\ce{P(s, red)}$ 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?

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    $\begingroup$ Does the downvoter care to give a chance to improve the question? $\endgroup$
    – user69608
    Aug 19 '20 at 14:39
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    $\begingroup$ white P has 4 P atoms in a tetrahedral arrangement, so all bond angles are 60degrees. Like just about any three-membered ring, these structures are highly strained. Conversion to red phosphorus breaks two of the three membered rings open, relieving some of the strain with the entropic cost of polymerizing. Very similar to polymerization of ethylene oxide. $\endgroup$
    – Andrew
    Aug 20 '20 at 13:15
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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 $\Delta U_m<0$. 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 $\Delta H<0$). Interestingly, even though white phosphorus is less thermodynamically stable it is considered the standard state of P.

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  • $\begingroup$ It is assumed that strain in fact is a decisive determinant of the change in energy, although this is not something I know about, and irrelevant for the calculations, but if true it allows the logical conclusion that $\Delta H<0$. $\endgroup$
    – Buck Thorn
    Aug 19 '20 at 16:33

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