The normal boiling point of bromine is 58.8°C. Given that the vapor pressure of bromine is 75.0 torr at 2.5°C, calculate the molar enthalpy of vaporization of bromine.
A)3.76 kJ/mol
B)31.3 kJ/mol
C)2.90 kJ/mol
D)3.57 kJ/mol
E)29.7 kJ/mol

Question

Quizplus · Accepted Answer

The Clausius-Clapeyron equation relates vapor pressure and temperature to the enthalpy of vaporization:
ln(P2/P1) = -ΔHvap/R (1/T2-1/T1)
P1 = vapor pressure at T1 = 75.0 torr
T1 = 2.5°C + 273.15 = 275.65 K
T2 = normal boiling point = 58.8°C + 273.15 = 332.95 K
R = 8.314 J/(mol*K)
Substituting and solving for ΔHvap:
ln(P2/P1) = ln(1) = 0
ΔHvap = -ln(1)/((1/332.95) - (1/275.65)) * (8.314 J/mol*K) / 1000 J/kJ/mol
ΔHvap = 31.3 kJ/mol. 
Therefore, the answer is B, and the molar enthalpy of vaporization of bromine is 31.3 kJ/mol.

The Normal Boiling Point of Bromine Is 58