The vapour pressure of pure aniline is 1314 Pa at 350 K and 13372 Pa at 400 K. Determine the enthalpy of vaporization of aniline.

Question

Quizplus · Accepted Answer

Using the Clausius-Clapeyron equation:

ln(P2/P1) = (-ΔHvap/R)(1/T2 - 1/T1)

where:
P1 = 1314 Pa (vapour pressure at T1 = 350 K)
P2 = 13372 Pa (vapour pressure at T2 = 400 K)
R = 8.314 J/molK (gas constant)

Rearranging the equation and plugging in the values:

- ln(P1/P2) = (ΔHvap/R)(1/T2 - 1/T1)
- ln(1314/13372) = (ΔHvap/8.314)(1/400 - 1/350)
- -2.970 = ΔHvap/8.314(-0.001429)
- ΔHvap = (-2.970)(8.314)(-700.2)
- ΔHvap = 182.32 kJ/mol

Rounding off to one decimal place, the enthalpy of vaporization of aniline is 53.9 kJ mol^-1. Therefore, the correct choice is C.

The Vapour Pressure of Pure Aniline Is 1314 Pa at 350