Question

Propanol has a normal boiling point of 97.8°C. At 400 torr, it has a boiling point of 82.0°C. What is the heat of vaporization?24.9 kJ/mol38.7 kJ/mol12.3 kJ/mol44.4 kJ/mol52.7 kJ/mol

Answer 1

44.5 kJ/mol

Step-by-step explanation:

Propanol has a normal boiling point of T₁ = 97.8°C = 371.0 K. "Normal" refers to a pressure P₁ = 1 atm = 760 torr. At 400 torr (P₂), it has a boiling point of T₂ = 82.0°C = 355.2 K. We can find the heat of vaporization (ΔHvap) using the two point form of the Clausius-Clapeyron equation.

`ln((P_(2))/(P_(1)) )=(-\Delta H_(vap))/(R) .((1)/(T_(2))-(1)/(T_(1)))\\ln((400torr)/(760torr))=(-\Delta H_(vap))/(8.314J/K.mol) .((1)/(355.2K)-(1)/(371.0K))\\\Delta H_(vap)=4.45 * 10^(4) J/mol=44.5 kJ/mol`