Question

Compound Normal Bpt (oC) Hvap (kJ/mol)

Methanol 64.6 35.3

iso-Propanol 82.3 45.4

Cyclohexane 80.7 33.0

Ethanol has a normal boiling point of 78.3 oC. Why is the normal boiling point higher than methanol but lower than iso-propanol listed above? Would you expect the heat of vaporization of ethanol to follow the same trend? Explain.

Calculate the heat of vaporization of cyclohexane using the normal boiling point of 80.7 oC. The vapor pressure of cyclohexane at 25.0 oC is 0.132 atm.

Answer 1

Let us show the molecules

CH3-OH methanol

C2H5-OH ethanol

C3H7OH isopropanol

as you can see, the mass of each component increases, therefore, you should expect properties of ethanol to be

methanol < ETHANOL < Propanol

thats why the normal BP is between those

you will expect to have the next trend in enthalpy of vaporization

methanol < ETHANOL < Propanol

since we require MORE energy to evaporate as more mass is pressent

B)

for heat vaporization of cyclohexane

T1 = 80.7°C = 353.7 P1 = 1 atm (since normal means 1 atm)

T2 = 25°C = 298K P2 = 0.132 atm

apply clasisu equation

ln(P2/P1) = H/R*(1/T1-1/T2)

ln(0.132/1) = H/8.314*(1/353.7 - 1/298)

H = ln(0.132/1)*8.314/((1/353.7 - 1/298) ) = 31858.1954868 J

H = 31.85 kJ/mol of cyclohexane