The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.370 atm? (∆Hvap = 28.5 kJ/mol)

Answer 1

Consider the Clausius Clapeyron equation:

ln(P2/P1) = -∆Hvap / R (1/T2 - 1/T1)

P1 = 1 atm (for the normal boiling point)

P2 = 0.370 atm

∆Hvap = 28.5 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (convert so units agree with those of ∆H)

T1 = 282ºC + 273 = 555K

T2 = ?

ln(0.370/1) = -28.5/0.008314 (1/T2 - 1/555)

-0.994 = -3428(1/T2 - 0.00180)

-0.994 = -3428/T2 + 6.17

-7.16 = -3428/T2

T2 = 479K

T2 = 479K - 273 = 206ºC