Question

The vapor pressures of SO₂(s) and SO₂(l) in the vicinity of the triple point are given by the equations, Ln (P/torr) = 23.9812 - 4244.7/T and Ln (P/torr) = 19.2453 - 3298.4/T, respectively, where T is absolute temperature. (i)Determine from these equations the three phase change enthalpies, ∆subH°, ∆vapH° and ∆fusH°. (ii)Show your determination of the temperature and pressure at the triple point of sulfur dioxide. (iii)Assume that the enthalpy of vaporization remains at the value determined here, and show your work to predict the standard boiling point temperature of SO₂.

Answer 1

at the triple point, all the thee phases co-exist, Hence at the triple point, the vapor presssre of liquid SO2= vapor pressure of solid SO2

23.9812- 4244.7/T= 19.2453-3298.4/T

1/T*(4244.7-3298.4)= 23.9812-19.2453

T= 199.8 K

hence at the triple point, lnP=23.9812-4244.7.4/199.8, P(bar)= 15.43 Torr = 15.34/760 atm =0.020 atm

since solid will have lesser vapor pressure and liquid will have more vapor pressure, choose any temperature and calculate the vapor pressure, let us say 100K, vapor pressure calculated from lnP= 23.9812-4244.7/T gives less vapor pressure than the vapor presure calculated from lnP =19.2453-3298.4/T.

Hence, lnP= 23.9812-4244.7/T is the vapor pressure of solid SO2 and lnP= 19.2453-3298.4/T refers to liquid vapor pressure.

at two different temperatures less than 199.8K, let us say 100 and 150K

lnP2= 23.9812- 4244.7/100 and lnP1= 23.9812-4244.7/150

ln(P1/P2)= 4244.7*(1/100-1/150)

but ln(P1/P2)= deltaH/R*(1/T2-1/T1), when the enthalpy of fusion remains constant

hence deltaH fsubimation/R= 4244.7, deltaHfus= 4244.7*8.314 J/mole =35290.44 J/mole

similarly, from the vapor pressure of liquid SO2, deltaH Vapr= 3298.4*8.314=27432 J/mole

enthalpy of sublimation = enthalpy of fusion + enthalpy of vaporization

enthalpy of fusion = 35290.44-27432 J/mole =7867.54 J/mole