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)38.0 kJ/mol D)27.5 kJ/mol E)74.7 kJ/mol

Question

Quizplus · Accepted Answer

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

ln(P2/P1) = (-ΔHvap/R) x (1/T2 - 1/T1)

where P1 is the vapor pressure at the normal boiling point (1 atm), P2 is the vapor pressure at 15.2°C (converted to atm), R is the gas constant (0.0821 L·atm/mol·K), T1 is the normal boiling point in Kelvin (337.8 K) and T2 is the temperature in Kelvin at which the vapor pressure is measured (288.35 K):

ln(75.0 torr/1 atm) = (-ΔHvap/0.0821) x (1/288.35 K - 1/337.8 K)

Simplifying and solving for ΔHvap, we get:

ΔHvap = -ln(75.0 torr/1 atm) x 0.0821 x (1/288.35 K - 1/337.8 K)

ΔHvap = 38.0 kJ/mol

Therefore, the answer is C: 38.0 kJ/mol.

The Normal Boiling Point of Methanol (CH3OH)is 64

The Normal Boiling Point of Methanol (CH₃OH)is 64