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

Respuesta :

Answer:

Tnbp = 731.44 K

Explanation:

Trouton's law:

  • ΔS°vap = ΔH°vap / Tnbp

∎ Tnbp: temperature at normal boiling point

∮ ΔH°vap = 54.2 KJ/mol

∮ ΔS°vap = (74.1 J/mol.K)×(KJ/1000 J) = 0.0741 KJ/mol.K

⇒ Tnbp = ΔH°vap / ΔS°vap

⇒ Tnbp = (54.2 KJ/mol) / (0.0741 KJ/mol.K)

⇒ Tnbp = 731.44 K

baraltoa

Answer:

731 K ( 458 ÂșC )

Explanation:

For a change of phase, in this case vaporization, we have:

TΔSÂșvap = ΔHÂș

assuming ∆H°vap, and ∆S°vap are independent of T.

So we can determine the boiling point by:

Tb =  ∆H°vap/ ∆S°vap

First we will need to convert ∆H°vap to J/mol to be consistent in the units.

∆H°vap = 54.2 kJ x 1000 J / kJ = 5.42 x 10 J/mol

Thus boiling point is:

Tb = 5.42 x 10⁎ J/mol / 74.1 J/mol K = 731 K = (731-273)ÂșC = 458 ÂșC

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