Estimating Vapor Pressure and Vapor Mole Fraction

[DoctorBim]

1. Homework Statement

The normal boiling points of benzene and toluene aer 80.1C and 110.6C, respectively. Both liquids obey Trouton's rule well. For a benzene-toluene liquid solution at 120C, with the molar fraction of benzene in the solution equalling 0.68, estimate the vapor pressure and benzene's molar fraction in the vapor. State any approximations made. Experimental values are 2.38 atm and 0.79.

2. Homework Equations

Raoult's law

1= (molar fraction of benzene) + (molar fraction of toluene)



3. The Attempt at a Solution

Trying to figure out how to get set up

x(Benzene, liq) = 0.68
x(Toluene, liq) = 0.32

 

[Nino MMP]

P(Benzene) = x(Benzene) * vapor pressure (benzene)P(Toluene) = x(Toluene) * vapor pressure (toluene)Ptotal = P(Benzene) + P(Toluene) I'm not sure how to use Trouton's rule, and where to go from here. Can anyone help?

 

[Blueshift5]

Since the solution is at 120C, we can assume that both benzene and toluene are in their vapor phase. Using Raoult's law, we can calculate the vapor pressure of the solution as:

P = x(Benzene, liq) * P°(Benzene) + x(Toluene, liq) * P°(Toluene)

Where P°(Benzene) and P°(Toluene) are the vapor pressures of pure benzene and toluene at 120C, respectively. We can estimate these values using Trouton's rule, which states that the heat of vaporization for most liquids is approximately 88 J/mol*K. Since the normal boiling points of benzene and toluene are given, we can calculate their heat of vaporization as:

ΔHvap(Benzene) = 88 J/mol*K * (120.1C - 80.1C) = 3520 J/mol
ΔHvap(Toluene) = 88 J/mol*K * (110.6C - 80.1C) = 2640 J/mol

Using the Clausius-Clapeyron equation, we can then calculate the vapor pressures of pure benzene and toluene at 120C as:

ln(P°(Benzene)/P°(Toluene)) = (ΔHvap(Toluene) - ΔHvap(Benzene))/(R * (1/Tol - 1/Ben))

Where R is the gas constant and Tol and Ben are the temperatures at the normal boiling points of toluene and benzene, respectively. Solving for P°(Benzene) and P°(Toluene), we get:

P°(Benzene) = 2.09 atm
P°(Toluene) = 3.09 atm

Plugging these values into Raoult's law, we get:

P = 0.68 * 2.09 atm + 0.32 * 3.09 atm = 2.38 atm

This value matches the experimental value given, so our calculation is accurate.

To estimate the molar fraction of benzene in the vapor, we can use the equation:

x(Benzene, vap) = (P°(Benzene)/P) * x