The molar enthalpy of vaporization of carbon disulfide is 26.74 kJ/mol, and its normal boiling point is 46°C. What is the vapor pressure of CS₂ at 0°C? A)447 torr B)4160 torr C)313 torr D)139 torr E)5.47 torr

Question

Quizplus · Accepted Answer

To solve this problem, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

where P1 is the vapor pressure at the boiling point (46°C or 319 K), P2 is the vapor pressure at 0°C (273 K), ΔHvap is the molar enthalpy of vaporization (26.74 kJ/mol), and R is the gas constant (8.31 J/mol*K).

Plugging in the values:

ln(P2/760 torr) = -26.74 kJ/mol / (8.31 J/mol*K) * (1/273 K - 1/319 K)

Solving for P2:

P2 = 760 torr * e^(-26.74 kJ/mol / (8.31 J/mol*K) * (1/273 K - 1/319 K))

P2 ≈ 139 torr

Therefore, the answer is D (139 torr).

The Molar Enthalpy of Vaporization of Carbon Disulfide Is 26