For the process moleene(l) $\rightarrow$ moleene(g) at 1 atm, $\Delta$H$\degree$vap = 30.0 kJ/mol and $\Delta$S$\degree$vap = 78.4 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 found using the equation ΔG = ΔH - TΔS = 0. Solving for T gives T = ΔH/ΔS = (30,000 J/mol) / (78.4 J/mol•K) = 382.65 K, which is approximately 110°C.

For the Process Moleene(l) →\rightarrow→ Moleene(g) at 1 Atm Δ\DeltaΔ H °\degree° Vap = 300 KJ/mol And Moleene(l)

For the Process Moleene(l) $\rightarrow$ Moleene(g) at 1 Atm $\Delta$ H $\degree$ Vap = 300 KJ/mol And Moleene(l)