Chapter 6, Problem 6.25E

Phosphorus exists as several allotropes that have varying properties. The enthalpy of transition from white P to red P, ${\Delta }_{\text{trans}}H$, is $-1\text{8 kJ/mol}$. The densities of white and red phosphorus are $1.823$ and $2.27{\text{0 g/cm}}^{\text{3}}$, respectively. At what temperature does white phosphorus become the stable phase at $62\text{5 atm}$ of pressure? Assume the formula for phosphorus is ${\text{P}}_{\text{4}}$.

Interpretation Introduction

Interpretation:

The temperature at which the white phosphorus becomes the stable phase at $62\text{5 atm}$ of pressure is to be calculated.

Concept introduction:

The clausius-clapeyron equation states the relation between vapor pressure and the absolute temperature.

The alternate form of Clausius-Clapeyron equation is,

$P_2-P_1=\frac{\Delta \overline{H}}{\Delta \overline{V}}\ln \frac{T_2}{T_1}$

Answer to Problem 6.25E

The temperature at which the white phosphorus become the stable phase at $62\text{5 atm}$ of pressure is $312.\text{304 K}$.

Explanation of Solution

Given

The density of white phosphorus is $\text{1}{\text{.823 g/cm}}^{\text{3}}$.

The density of red phosphorus is $2.27{\text{0 g/cm}}^{\text{3}}$.

Molar mass of phosphorus is $31\text{ g/mol}$.

The enthalpy of transition from white P to red P, ${\Delta }_{\text{trans}}H$, is $-1\text{8 kJ/mol}$.

The alternate form of Clausius-Clapeyron equation is,

$P_2-P_1=\frac{\Delta \overline{H}}{\Delta \overline{V}}\ln \frac{T_2}{T_1}$ (1)

The final temperature is calculated by the equation,

$\begin{array}{rll}\ln \frac{T_2}{T_1}& =& \frac{\left(P_2-P_1\right)\Delta \overline{V}}{\Delta \overline{H}}\\ T_2& =& T_1{e}^{\frac{\left(P_2-P_1\right)\left(V_2-V_1\right)}{\Delta \overline{H}}}\end{array}$ (2)

First, the molar volumes of each phosphorus atom are calculated with the help of their densities. The molar volumes are calculated by the formula,

$\Delta \overline{V}=\frac{\text{Mass}}{\text{Density}}$ (3)

Substitute the values of molar mass and density of white phosphorus, ${\text{P}}_{\text{4}}$ in equation (3) to calculate its molar volume.

$\begin{array}{rll}\Delta \overline{V}& =& \frac{31\text{ g/mol}\times 4}{\text{1}{\text{.823 g/cm}}^{\text{3}}}\times \left(\frac{{\text{1 m}}^3}{{10}^6{\text{ cm}}^{\text{3}}}\right)\\ & =& 68.02{\text{ cm}}^{\text{3}}\text{/mol}\times \left(\frac{{\text{1 m}}^3}{{10}^6{\text{ cm}}^{\text{3}}}\right)\\ & =& 68.02\times {10}^{-6}{\text{ m}}^{\text{3}}\text{/mol}\end{array}$

Similarly, the molar volume of red phosphorus is calculated as,

$\begin{array}{rll}\Delta \overline{V}& =& \frac{31\text{ g/mol}\times 4}{2.27{\text{0 g/cm}}^{\text{3}}}\times \left(\frac{{\text{1 m}}^3}{{10}^6{\text{ cm}}^{\text{3}}}\right)\\ & =& {\text{ cm}}^{\text{3}}\text{/mol}\times \left(\frac{{\text{1 m}}^3}{{10}^6{\text{ cm}}^{\text{3}}}\right)\\ & =& 54.62\times {10}^{-6}{\text{ m}}^{\text{3}}\text{/mol}\end{array}$

The initial temperature, $T_1$ is $\text{25}°\text{C}$

The conversion of Celsius to Kelvin is done as,

$\text{0}°\text{C}=2\text{73 K}$

Therefore, the conversion of $\text{25}°\text{C}$ to Kelvin is done as,

$\begin{array}{c}\text{25}°\text{C}=25+2\text{73 K}\\ =298\text{ K}\end{array}$

The initial pressure, $P_1$ is $\text{1 atm}=1013\text{25 Pa}$.

The final pressure is

$\begin{array}{rll}62\text{5 atm}& =& 62\text{5}\times 1013\text{25 Pa}\\ & =& 6332812\text{5 Pa}\end{array}$

Substitute the values of the enthalpy of transition, change in molar volume, pressure and initial temperature in equation (2) to calculate the final temperature.

$\begin{array}{rll}{T}_2& =& 298\text{ K}\times e^{\frac{\left(6332812\text{5 Pa}-1013\text{25 Pa}\right)\left(\left(54.62\times {10}^{-6}-68.02\times {10}^{-6}\right){\text{m}}^{\text{3}}\text{/mol}\right)}{-1\text{8 }\times 100\text{0 J/mol}}}\\ & =& 298\text{ K}\times e^{\frac{\left(63226800\text{ Pa}\right)\left(\left(-1.34\times {10}^{-5}\right){\text{m}}^{\text{3}}\text{/mol}\right)}{-1\text{8 }\times 100\text{0 J/mol}}}\\ & =& 298\text{ K}\times e^{0.0470}\\ & =& 298\text{ K}\times 1.048\end{array}$

Simplify the above expression,

$T_2=312.\text{304 K}$

Hence, the temperature at which the white phosphorus become the stable phase at $62\text{5 atm}$ of pressure is $312.\text{304 K}$.

Conclusion

The temperature at which the white phosphorus become the stable phase at $62\text{5 atm}$ of pressure is $312.\text{304 K}$.