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((P_2)/(P_1))=(\Delta H_(vap))/(R)[(1)/(T_1)-(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((140.5mmHg)/(462.7mmHg))=(\Delta H_(vap))/(8.314J/mol.K)[(1)/(252)-(1)/(232)]\\\\\Delta H_(vap)=28966.6J/mol=28.97kJ/mol`

Therefore, the value of
`\Delta H_(vap)` is 28.97 kJ/mol