A liquid has an enthalpy of vaporization of 30.3 kJ/mol.At 262 K it has a vapor pressure of 121 mmHg.What is the normal boiling point of this liquid? (R = 8.314 J/(K· mol))
A) 278 K
B) 302 K
C) 262 K
D) 248 K
E) 231 K

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation:

ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)

where P1 and T1 are the given values, P2 is the pressure at the normal boiling point (1 atm), T2 is the normal boiling point (what we are solving for), ΔHvap is the enthalpy of vaporization, and R is the gas constant.

Plugging in the values we get:

ln(760/121) = (-30300 J/mol / (8.314 J/(K·mol))) * (1/T2 - 1/262K)

Simplifying:

ln(6.28) = (-3642.7) * (1/T2 - 0.0038)

ln(6.28)/(-3642.7) = 1/T2 - 0.0038

1/T2 = ln(6.28)/(-3642.7) + 0.0038

T2 = 302 K

Therefore, the answer is B: 302 K.

A Liquid Has an Enthalpy of Vaporization of 30