Question

Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enthalpy of vaporization of sulfur dioxide? (R = 8.314 J/K⋅mol)

Answer 1

Answer : The value of $\Delta H_{vap}$ is 28.97 kJ/mol

Explanation :

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $-21.0^oC$ = 462.7 mmHg

$P_2$ = vapor pressure at temperature $-44.0^oC$ = 140.5 mmHg

$\Delta H_{vap}$ = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $-21.0^oC=[-21.0+273]K=252K$

$T_2$ = final temperature = $45^oC=[-41.0+273]K=232K$

Putting values in above equation, we get:

$\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol$

Therefore, the value of $\Delta H_{vap}$ is 28.97 kJ/mol