Naturally, this type of thermodynamic problem calls for a formula. In this case, we need the Clausius-Clapeyron equation which is shown in the attached picture. Take note of the given units shown in the legend.
Tb = 6 + 273 = 279 K
T0 = -25 + 273 = 248 K
P0 = 429 torr * 101.325 kPa/706 torr = 61.57 kPa
Applying the equation,
279 = {[8.314(ln61.57 - ln101.325)/ΔHvap] + 1/248}⁻¹
Use a scientific calculator to find ΔHvap.
ΔHvap = 9,244.25 J/mol
