The molar enthalpy of vaporization of hexane (C₆H₁₄) is 28.9 kJ/mol, and its normal boiling point is 68.73^$\circ$C. What is the vapor pressure of hexane at 25^$\circ$C A) 4.44 torr B) 117 torr C) 171 torr D) 759 torr E) 3370 torr

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation to solve this problem:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

where P1 is the vapor pressure at the boiling point (68.73°C = 341.88 K), ΔHvap is the molar enthalpy of vaporization (28.9 kJ/mol), R is the gas constant (8.314 J/molK), T1 is the boiling point in Kelvin (341.88 K), and T2 is the temperature we want to find the vapor pressure at (25°C = 298.15 K).

Solving for P2, we get:
P2 = P1 * e^(-(ΔHvap/R) * (1/T2 - 1/T1))

Plugging in the values, we get:
P2 = 760 torr * e^(-(28.9*10^3 J/mol) / (8.314 J/molK * 298.15 K)) = 171 torr

Therefore, the vapor pressure of hexane at 25°C is approximately 171 torr. The correct answer is choice C.

The Molar Enthalpy of Vaporization of Hexane (C6H14) Is 28 ∘\circ∘

The Molar Enthalpy of Vaporization of Hexane (C₆H₁₄) Is 28 $\circ$