The vapor pressure of benzene at 25^$\omicron$C is 94.6 Torr and its enthalpy of vaporization is 30.8 kJ . mol ^- ¹ . Estimate the normal boiling point of benzene.Assume the enthalpy of vaporization is independent of temperature.

Question

Quizplus · Accepted Answer

Using the Clausius-Clapeyron equation:
ln(P2/P1) = (-∆Hvap/R)((1/T2) - (1/T1))
where P1 is the vapor pressure at the known temperature T1, and P2 is the unknown vapor pressure at the boiling point temperature T2.

Solving for T2:
ln(P2/94.6 Torr) = (-30.8 kJ/mol)/(8.314 J/mol-K) * ((1/T2) - (1/298 K))
ln(P2/94.6 Torr) = (-3742.84 K) * ((1/T2) - (0.003355));

Using the given value of the normal boiling point temperature, Tb, we know that:
P2 = 1 atm = 760 Torr;
T2 = Tb = 353 K.

Therefore, we can solve for the natural log of the ratio of the vapor pressures.
ln(8.037/94.6) = (-3742.84 K) * ((1/353 K) - (0.003355));
-4.7181 = (-3742.84) * (0.002836);
-4.7181/(3742.84 * 0.002836) = 358.1 K

Since the boiling point temperature is usually reported to the nearest degree, the best answer is B) 358 K.

The Vapor Pressure of Benzene at 25 ο\omicronο C Is 94

The Vapor Pressure of Benzene at 25^$\omicron$C Is 94