The normal boiling point of methanol (CH₃OH) is 64.6^$\circ$C. Given that the vapor pressure of methanol is 75.0 torr at 15.2^$\circ$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

We can use the Clausius-Clapeyron equation to solve for the molar enthalpy of vaporization:

ln(P2/P1) = (-ΔHvap/R)(1/T2 - 1/T1)

where P1 and T1 are the given vapor pressure and temperature respectively, and P2 and T2 are the vapor pressure and normal boiling point respectively.

Plugging in the given values, we get:

ln(75.0 torr/760 torr) = (-ΔHvap/8.314 J/mol·K)(1/(337.75 K) - 1/(337.75 K))

Solving for ΔHvap, we get:

ΔHvap = -(ln(75.0/760)·8.314·337.75)/(1/337.75 - 1/337.75)

ΔHvap = 38.0 kJ/mol

Therefore, the best choice is D, 38.0 kJ/mol.

The Normal Boiling Point of Methanol (CH3OH) Is 64 ∘\circ∘ C Given That the Vapor Pressure of Methanol Is 75

The Normal Boiling Point of Methanol (CH₃OH) Is 64 $\circ$ C Given That the Vapor Pressure of Methanol Is 75