For the process moleene(l) → moleene(g) at 1 atm, ΔH°vap = 26.0 kJ/mol and ΔS°vap = 83.4 J/mol•K. Assuming these values are independent of T, what is the normal boiling point of moleene?

Question

Quizplus · Accepted Answer

The normal boiling point is when ΔG° = 0, which means ΔH°vap = TΔS°vap. Solving for T gives T = ΔH°vap / ΔS°vap = (26,000 J/mol) / (83.4 J/mol•K) = 311.6 K, which is approximately 39°C.

For the Process Moleene(l) → Moleene(g) at 1 Atm, ΔH°vap