Question

Naphthalene, C10H8, melts at 80.0 ˚C; the vapor pressure of the solid is 1.00 torr at 52.6 ˚C, and that of the liquid is 10.0 torr at 85.8 ˚C and 40.0 torr at 119.3 ˚C. Calculate: (a) the enthalpy of vaporization, the normal boiling point, and the entropy of vaporization at the boiling point; (b) the vapor pressure at the melting point (Hint: liquid and solid are in equilibrium there); (a) the enthalpy of sublimation and enthalpy of fusion of the solid (assume that the melting point and the triple point are the same); (d) the (highest) temperature at which the vapor pressure will be (less than or equal to) 1.0x10-5 torr.

Answer 1

Answering only Answer 1 and 2:

(a) The enthalpy of vaporization of Naphthalene, C10H8:

H_Vap for the compound having p1 vapor pressure at Temp T1 and p2 vapor pressure at Temp T2.

Naphthalene has vapor pressure of 10.0 torr at 85.8 ˚C (358.8 K) and 40.0 torr at 119.3 ˚C (392.3 K).

As per the Clausius-Clapeyron Equation,

ln (P2/P1) = H_Vap/R * (1/T₁ - 1/T₂)

ln (40/10) = (H_Vap / 8.31447) (1/358.8 - 1/392.3)

1.386 = (H_Vap / 8.31447) (0.0002379)

H_Vap = 1.386* 8.31447 / 0.0002379 = 48439 J = 48.439 KJ mol^-1

Boiling point of Naphthalene: At boiling point the vapor pressure of the liquid will be equal to the atmostpheric pressure 760 mm Hg or 760 torr.

So, P1= 40 torr, T1 =392.3 K

P2= 760 torr, T2 =??

Again putting the values in Clausius-Clapeyron Equation,

ln (P2/P1) = H_Vap/R * (1/T₁ - 1/T₂)

ln(760/40) = (48439 Jmol^-1/8.31447) (1/392.3 - 1/T2)

2.94 = 5825.86 (1/392.3 - 1/T2)

(T2-392.3)/392.3*T2= .000504

T2-392.3 = .19797* T2

T2= ~350.2 K

Entropy of vaporization at the boiling point:

G_Vap =H_Vap -T* S_Vap

At boiling point, liquid and solid are in equilibrium so  G_Vap = 0

So , H_Vap -T_b.p* S_Vap =0

H_Vap =T_bp* S_Vap

S_{Vap =} H_{Vap /} T_{bp =} 48439 J mol^-1 / 350.2 = 138.3 J mol⁻¹ K⁻¹.