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

Answer:

44.5 kJ/mol

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.

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