Answer to Question #231523 in General Chemistry for Savannah

Question

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 410 torr?

Answer 1

• The external pressure of the benzene is:

P_b = 410 torr = 54662.20 P_b

• The molar enthalpy of vaporization of benzene is:

ΔH_a =30.72 Kj/mol = 30720 J/mol

• The normal boiling temperature of the benzene is:

T_a = 80.10^oC = 353.10K

The temperature of benzene to boil at 410 torr is T_b

atmospheric pressure is 101325 P_a

The benzene boil at normal atmospheric pressure where boiling temperature is 80.10

The expression by using the Clausius-Clapeyron equation to evaluate the temperature of benzene to boil at pressure of 410 torr is

• ln( 101325 /54662.20 ) = (30720 / 8.314). (1/T_b - 1/ 353.10 )
• (0.61716×8.314 / 30720 ) = 1/ T_b - 1/353.10 )
• 1/T_b = 1.670 x 10^-4 + 1/353.10
• 1/T_b = 1.670 x 10^-4 + 0.002832
• 1/T_b = 0.002999
• T_b = 1/0.002999
• T_b = 333.44 K
• T_b = 60.44^oC

Thus the boiling temperature of benzene at pressure of 410 torr is 60.44^oC

∘