Question

12.62

Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1 ∘C.

At what temperature does benzene boil when the external pressure is 405 torr ?

Answer 1

At normal boiling point pressure is 760 torr
T1= 80.1 oC = (273.15 + 80.1)K = 353.25 K
P1 = 760 torr (anything boils when pressure equals atmospheric pressure)

T2=?
P2 = 405 torr

delta H=30.72 KJ/mol = 30720 J/mol

use:
ln(P2/P1) = (- delta H/R)*(1/T2 - 1/T1)
ln(405/760) = (-30720 /8.314)* (1/T2 - 1/353.25 )
-0.63 = -3695*(1/T2 - 1/353.25 )
1.71*10^-4 = (1/T2 - 1/353.25 )
1/T2 = 3*10^-3
T2 = 333 K
      = 333-273.15 oC
      = 59.85 oC
Answer: 59.85 oC