Question

Naphthalene, C10 H8 , melts at 80∘ C . If the vapor pressure of the liquid is 10 mm at 85.8∘C and 40 mm at 119.3∘C , and that of the solid is 1mm at 52.6∘C , calculate

a) Assuming the melting-point and triple-point temperatures are the same, calculate △H sublimation of the solid, and △H fusion .

Answer 1

Naphthalene, C10 H8 , melts at 80∘ C . If the vapor pressure of the liquid is 10 mm at 85.8∘C and 40 mm at 119.3∘C , and that of the solid is 1mm at 52.6∘C , calculate

a) Assuming the melting-point and triple-point temperatures are the same, calculate △H sublimation of the solid, and △H fusion .

From Classius-Clayperon Equation, ln (P2/P1)= (deltaHvaporization/R)*(1/T1-1/T2)

where P2= vapor pressure of liquid at 119.3 deg.c=40, T2= 119.3 deg.c= 119.3+273=392.3 K, P1= 40 mm Hg at T1= 85.8 deg.c= 85.8+273= 358.8K, R= 8.314 J/mole.K

hence deltAhVapor= heat of vaporization of naptlnhalene and R= 8.314 J/mole.K

hence ln( 40/10)= (deltaHvapor/R)*(1/358.8-1/392.3)=(deltaHfusion/R)*0.00028

deltaHvaporization = 48427.43 J/mole= 48.427KJ/mole

at tripple point all the three phases exist and vapor pressure of solid = vapor pressure of liquid

this point is 80 deg.c= 80+273= 353K, At triple point, the vapor pressure can be obtained by the classius clayperon Equaton

here P1= 10mm at T1=85.8+273=358.8K, P2 need to be calculated, T2=353K

hence ln ( P2/1)= (deltaHvapor/R)*(1/358.8-1/353)

P2= 7.67 mm Hg

now P2= 7.67 mm Hg, T2= 80+373= 353K, P1= 1 mm at 52.6 deg.c, T1= 52.6+273=325.6K

hence Classius- clayperon Equation can be written as

ln(7.67/1)= (deltaHsublimation/R)*(1/325.6-1/353)

deltaHfsublimation= 71052 J/mole=71.052 Kj/mole

heat of sublimation = heat of fusion+ heat of vaporization

heat of fusion = 71.052-48.427=22.625 Kj/mole

P2= 4 mm Hg, the vapor pressure at triple point

Now applying the clasius- clayperon equation for solid vapor pressures

ln(4/1)= (deltaHfusion/R)*(1/