Chapter 6, Problem 6.61E

At the triple point of a substance, the vapor pressures of the solid and the liquid phase are the same. Using the Clausius-Clapeyron equation for each combination of phases, one can show that the temperature of the triple point, $T_{\text{tp}}$, is

$T_{\text{tp}}=-\frac{{\Delta }_{\text{fus}}H}{R\ln \frac{p_{\text{liq}}}{p_{\text{sol}}}+\frac{{\Delta }_{\text{vap}}H}{T_{\text{liq}}}-\frac{{\Delta }_{\text{sub}}H}{T_{\text{sol}}}}$

where the data points ( $p_{\text{liq}}$, $T_{\text{liq}}$) and ( $p_{\text{sol}}$, $T_{\text{sol}}$) are the vapor pressure and temperature data for the liquid and solid phases, respectively.

(a) Given these data, calculate $T_{\text{tp}}$ for iodine, ${\text{I}}_{\text{2}}$.

$\begin{array}{l}{\Delta }_{\text{fus}}H15.52\text{kJ/mol}\\ {\Delta }_{\text{sub}}H57.09\text{kJ/mol}\\ {\Delta }_{\text{vap}}H41.57\text{kJ/mol}\end{array}$

$\begin{array}{l}{p}_{\text{sol}}\text{at}\text{364}\text{K}\text{10}\text{.00}\text{torr}\\ p_{\text{liq}}\text{at}\text{410}\text{K}20\text{0}\text{.0}\text{torr}\end{array}$

(b) Calculate the vapor pressure of ${\text{I}}_{\text{2}}$ at the triple point using the Clausius-Clapeyron equation and the fact that the normal boiling point of ${\text{I}}_{\text{2}}$ is $457.4\text{K}$.