The molar enthalpy of vaporization of hexane (C₆H₁₄) is 28.9 kJ/mol, and its normal boiling point is 68.73°C. What is the vapor pressure of hexane at 25°C? (R = 8.314 J/K • mol) A) 171 torr B) 4.44 torr C) 117 torr D) 3370 torr E) 759 torr

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The Clausius-Clapeyron equation can be used to calculate the vapor pressure at a different temperature given the enthalpy of vaporization and the normal boiling point. The equation is ln(P1/P2) = ΔHvap/R * (1/T2 - 1/T1), where ΔHvap is the molar enthalpy of vaporization, R is the gas constant, T1 and T2 are the initial and final temperatures in Kelvin, and P1 and P2 are the initial and final pressures. Given that the normal boiling point of hexane is 68.73°C (341.88 K) and the enthalpy of vaporization is 28.9 kJ/mol, we can calculate the vapor pressure at 25°C (298.15 K) knowing that the vapor pressure at the boiling point (T1) is 1 atm (760 torr). Rearranging and solving for P2 (the vapor pressure at 25°C) involves converting ΔHvap to J/mol (since R is given in J/K•mol), and using the temperatures in Kelvin. Solving this equation yields a vapor pressure that is closest to the value given in choice A, 171 torr.

The Molar Enthalpy of Vaporization of Hexane (C6H14) Is 28

The Molar Enthalpy of Vaporization of Hexane (C₆H₁₄) Is 28