The normal boiling point of methanol (CH₃OH)is 64.6°C.Given that the vapor pressure of methanol is 75.0 torr at 15.2°C, calculate the molar enthalpy of vaporization of methanol. A)0.383 kJ/mol B)3.00 kJ/mol C)27.5 kJ/mol D)38.0 kJ/mol E)74.7 kJ/mol

Question

Quizplus · Accepted Answer

Using the Clausius-Clapeyron Equation, we can calculate the molar enthalpy of vaporization:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

where P1 = 75.0 torr, T1 = 15.2°C + 273.15 = 288.35 K, P2 = 760 torr (1 atm), and T2 = 64.6°C + 273.15 = 337.75 K.

Substituting the values and solving for ΔHvap:

ln(760/75.0) = -ΔHvap/8.314 J/mol*K * (1/337.75 K - 1/288.35 K)
ΔHvap = -(8.314 J/mol*K * ln(760/75.0)) / (1/337.75 K - 1/288.35 K)
ΔHvap = -38.0 kJ/mol

Therefore, the molar enthalpy of vaporization of methanol is 38.0 kJ/mol, which is choice D.

The Normal Boiling Point of Methanol (CH3OH)is 64

The Normal Boiling Point of Methanol (CH₃OH)is 64