Sulfur dioxide has a vapor pressure of 462.7 mm Hg at -21.0 $\circ$ C and a vapor pressure of 140.5 mm Hg at -44.0 $\circ$ C.What is the enthalpy of vaporization of sulfur dioxide? (R = 8.314 J/K.mol)

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation to calculate the enthalpy of vaporization:

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

where P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, R is the gas constant (8.314 J/K*mol), and ΔHvap is the enthalpy of vaporization.

We can plug in the values given in the problem:

ln(140.5/462.7) = -(ΔHvap/8.314) * (1/(-44+273) - 1/(-21+273))

Simplifying:

-1.2385 = -(ΔHvap/8.314) * (0.004636 - 0.007948)

-1.2385 = -(ΔHvap/8.314) * (-0.003312)

ΔHvap = (-1.2385 / -0.003312) * 8.314 = 24.9 kJ/mol

Therefore, the answer is (D) 24.9 kJ/mol.

Sulfur Dioxide Has a Vapor Pressure of 462 ∘\circ∘ C and a Vapor Pressure of 140

Sulfur Dioxide Has a Vapor Pressure of 462 $\circ$ C and a Vapor Pressure of 140