Question

Q1: The normal boiling point of liquid methanol is 338 K. Assuming that its molar heat of vaporization is constant at 36.6 kJ/mol, the boiling point of CH₃OH when the external pressure is 1.24 atm is  K.

Q2: From the following vapor pressure data for diethyl ether, an estimate of the molar heat of vaporization of C₂H₅OC₂H₅ is  kJ/mol.

P, mm Hg	T, Kelvins
100	262
400	291

NOTE: please explain the steps. I am not 100% on the process. I keep calculating wrong. Thank you.

Answer 1

Q.1 At boiling point the vapor pressure due to liquid becomes equal to the atmospheric pressure.

Hence at temperature, T1 = 338 K, Pressure, P1 = 1 atm

Pressure, P2 = 1.24 atm, temperature, T2 = ?

Hv = 36.6 KJ/mol = 36600 J/mol

Applying Clausius-Clapeyron Equation

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

=> ln(1.24 atm / 1 atm) = (36600 Jmol-1/8.314 JK-1mol-1) x (1/338K - 1/T2)

=> 0.215 = 4404.33 K x (1/338K - 1/T2)

=>1/338K - 1/T2 = 4.88 x 10^-5 K-1

=> T2 = 348 K

Hence when the external pressure is 1.24 atm, the boiling point of CH3OH will be 348 K (answer)

Q.2: Given

P1 = 100 mm Hg, T1 = 262 K

P2 = 400 mm Hg, T2 = 291 K

Hv = ?

Applying Clausius-Clapeyron Equation

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

=> ln(400 mmHg/ 100 mmHg) = (Hv/8.314 JK-1mol-1) x (1/262K - 1/291)

=>  ln(400 mmHg/ 100 mmHg) x 8.314 JK-1mol-1 =  Hv x (29 / 262 x 291)

=> 11.526 = Hv x (29 / 262 x 291)

=>  Hv = 11.526 * 262 x 291 / 29 = 30301 J/mol = 30.3 KJ/mol (answer)