Question

Construct the phase diagram (in Excel, or some digital format) for benzene near its triple point at 36 torr and 5.50 degree C using the following data: ∆fusH = 10.6 kJ/mol, ∆vapH = 30.8 kJ/mol, density (solid) = 0.891g/cm−3 , and density (liquid) = 0.879g/cm−3 . (You should be able to calculate the phase boundaries based on known slopes from the triple point).

Answer 1

T = 5.50oC = 278.65 K

--> First we have to find the slope of the vaporization line near the triple point.

C6H6(l) = C6H6(g)

(dP/dT)1 = dvapH*P/RT2

= (30.8e003 J mol-1)(36 Torr)(1atm/760 Torr)/(8.314 J K-1 mol-1)(278.65 K)2

= 2.26e-003 atm/K

--> Now to find the slope of the fusion line near the triple point.

C6H6(s) = C6H6(l)

dVfus = Vl - Vs = m/pl - m/ps = m(ps-pl)/pspl

= (78 g mol-1)((0.891 - 0.879)g cm-3)/((0.891 g cm-3)(0.879 g cm-3)

= 1.1968e-006 m3 mol-1

dSfus = dHfus/T = 10.6e003 J mol-1/2787.65 K

= 38.04 J K-1 mol-1

(dp/dT)2 = dSfus/dVfus --> Clausius-Clapeyron equation

= (10.6e003 J mol-1)/(1.1968e-006 m3 mol-1) * (1 atm/1.01325e005 J m-3)

= 3.14e005 atm K-1

--> Finally to determine the slope for the line of sublimation.

C6H6(s) = C6H6(g)

(dp/dT)3 = dsubH*P/RT2

= (41.4e003 J mol-1)(36 Torr)(1 atm/760 Torr)/(8.314 J K-1 mol-1)(278.65 K)2

= 3.04e-003 atm K-1