For the process of a certain liquid vaporizing at 1 atm,$\Delta$H°_vap= 54.2 kJ/mol and $\Delta$S°_vap= 74.1 J/mol K.Assuming these values are independent of T,what is the normal boiling point of this liquid?

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation to solve for the normal boiling point.
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
At the normal boiling point, P2 = 1 atm and T2 = Tbp.
We can assume P1 = 0, since we are dealing with vaporization at 1 atm.
Therefore:
ln(1/0) = (54.2 kJ/mol / 8.314 J/mol K) * (1/298 K - 1/Tbp)
Tbp = 458 °C

For the Process of a Certain Liquid Vaporizing at 1 Δ\DeltaΔ

For the Process of a Certain Liquid Vaporizing at 1 $\Delta$