What are the molar heat of vaporization in (KJ/mol) and the normal boiling point

Question

What are the molar heat of vaporization in (KJ/mol) and the normal boiling point of a liquid that has a vapor pressure of 254 mm Hg at 25 deg C and a vapor pressure of 648 mm Hg at 45 deg C?

So I converted:
254 mm Hg to 0.3342 atm
648 mm Hg to 0.8526 atm
25C to 298.15K
45C to 318.15K

Then I used ln(P1/P2)=DeltaH/R (1/T1 - 1/T2)

and got Delta H = 36929.9012 J/mol = 37 KJ/mol

Now I have no idea how to solve for the normal boiling point of this liquid, can someone please help me step by step? Thanks!!!!

Explanation / Answer

Boiling point is when the vapor pressure overcomes the atmospheric pressure, which is 1 atm, or 760 torr.

Using the Clasius-Clapeyron equation:

ln(P1/P2) = (?Hvap/R)(1/T2 - 1/T1)
?Hvap = Rln(P1/P2)/(1/T2 - 1/T1)
= (8.314472 * 10^-3 kJ/mol-K)ln(760 torr/40 torr)/(1/(273.15 + 20 K) - 1/(273.15 + 98.4 K))
? 34.0116996 kJ/mol

Rounding to one sig fig, the molar heat of vaporization of n-heptane is about 3.0 * 10^1 kJ/mol.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.