find boiling point

Question

For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. Calculate the boiling point of benzene (in °C) when the atmospheric pressure is 83.4 kPa.

Answer 1

We can use the Clausius-Clapeyron equation:

ln(P1/P2) = ∆Hvap/R (1/T2 - 1/T1)

P1 = 1 atm

T1 = 80.1ºC + 273 = 353K

P2 = 83.4 kPa x 1 atm/101.325 kPa = 0.823 atm

T2 = ?

R = 8.3145 J mol^-1 K^-1

∆Hvap = 30.8 kJ/mol = 30,800 J/mol (use J to be consistent with units of R)

ln(1/0.823) = 30,800/8.3145 (1/T2 - 1/353)

0.1948 = 3704 (1/T2 - 0.002833)

0.1948 = 3704/T2 - 10.49

10.60 = 3704/T2

T2 = 349K

ºC = 349 - 273 = 76.4ºC