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?

731 K ( 458 ºC )

Explanation:

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

TΔSºvap = ΔHº

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 10⁴ J/mol

Thus boiling point is:

Tb = 5.42 x 10⁴ J/mol / 74.1 J/mol K = 731 K = (731-273)ºC = 458 ºC

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