Question

The heat of vaporization (∆Hvap) of the element mercury (Hg) is 59.0 kJ/mol. If the vapor pressure of Hg is 0.0017 torr at 25°C, Calculate the normal boiling point of mercury.

Answer 1

Answer:

Tb Hg = 656.726 K

Explanation:

normal boiling point (Tb):

Clasius-Clapeyron's law:

• Tb = [(RLn(Po)/ΔHv) + (1/To)]∧(-1)

∴ R = 8.314 J/K.mol

∴To = 25°C ≅ 298 K

∴ Po = 0.0017 torr = 2.24 E-6 atm

∴ ΔHv = 59.0 KJ/mol = 59000 J/mol

⇒ Tb = [(8.314 J/K.mol)Ln(2.24 E-6))/(59000 J/mol)) + (1/298 K)]∧(-1)

⇒ Tb = [- 1.833 E-3 K-1 + 3.355 E-3 K-1 ]∧(-1)

⇒ Tb = [1.523 E-3]∧(-1)

⇒ Tb = 656.726 K