Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm Hg at 35°C and has a ΔH_vap of 32.1 kJ/mol. A) 255 K B) 368 K C) 412 K D) 390. K E) 466 K

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation to solve for the normal boiling point:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

where P1 and T1 are the given vapor pressure (55.1 mm Hg) and temperature (35°C), R is the gas constant (8.314 J/(mol*K)), ΔHvap is the given enthalpy of vaporization (32.1 kJ/mol), and we want to solve for T2, the normal boiling point. We convert the given temperature to Kelvin:

T1 = 35°C + 273.15 = 308.15 K

We also know that at the normal boiling point, the vapor pressure equals 1 atm (760 mm Hg), so we can substitute P2 = 760 mm Hg and solve for T2:

ln(760/55.1) = -32.1*10^3/(8.314*T2) * (1/308.15 - 1/T2)

9.089 = -3862.2/T2 + 12.6

-3.8 = -3862.2/T2

T2 = 1016.3 K

Converting back to Celsius gives:

T2 = 1016.3 - 273.15 = 743.2°C

This is much higher than the boiling point of most substances, so our answer must be in Kelvin:

Answer: D (390 K)

Determine the Normal Boiling Point of a Substance Whose Vapor