Question

Benzene has a heat of vaporization of 30.72kJmol−1 and a normal boiling point of 80.1 ∘C....

Benzene has a heat of vaporization of 30.72kJmol−1 and a normal boiling point of 80.1 ∘C.

At what temperature does benzene boil when the external pressure is 599 mbar ?

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Answer #1

Solution :

Use the Clausius-Clapeyron Equation

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

deltaHvap = 30.72kJ/mol

T1 = 80.1 +273 = 353.1K

T2 = need to find out

P1 = is the presssure normal boiling point = 1atm

P2 =599 mbar = 0.59atm ( as 1 atm = 1013mbar)

R = 8.314*10-3 kJ/K*mol

ln ( 0.59atm /1atm) = - 30.72kJ/mol / 8.314*10-3 kJ/K*mol ( 1/T2-1/353.1K)

-0.53 = -3695( 1/T2- 0.00283)

-0.53 /-3695 =( 1/T2- 0.00283)

1.43*10-4 = 1/T2- 0.00283

1/T2 = 1.43*10-4 +0.00283

1/T2 = 2.97*10-3

T2 = 336K

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