A liquid has an enthalpy of vaporization of 30.4 kJ/mol.At 269 K it has a vapor pressure of 102 mmHg.What is the normal boiling point of this liquid? (R = 8.314 J/(K· mol))
A) 287 K
B) 316 K
C) 269 K
D) 253 K
E) 234 K

Question

Quizplus · Accepted Answer

The Clausius-Clapeyron equation relates the vapor pressure of a liquid with its enthalpy of vaporization and temperature:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

where P1 and T1 are known values, and P2 and T2 are the vapor pressure and temperature that we need to find, respectively. R is the gas constant.

We can solve for T2 by rearranging the equation:

1/T2 = 1/T1 - (1/R)(ΔHvap/ln(P2/P1))

Substituting the given values, we obtain:

1/T2 = 1/269 K - (1/8.314 J/(K·mol))(30.4 kJ/mol / ln(102 mmHg/760 mmHg))

Solving for T2, we get:

T2 = 316 K

Therefore, the normal boiling point of this liquid is 316 K, or 43 degrees Celsius. The answer is (B).

A Liquid Has an Enthalpy of Vaporization of 30