# Estimating Vapor Pressure and Vapor Mole Fraction

• Chemistry
• DoctorBim
In summary, the question involves finding the vapor pressure and molar fraction of benzene in a benzene-toluene liquid solution at 120C, with a molar fraction of benzene at 0.68. Using Raoult's law and the given experimental values, the estimated values are 2.38 atm for vapor pressure and 0.79 for the molar fraction of benzene in the vapor. Some approximations may have been made and help is needed in understanding how to use Trouton's rule.
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

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?

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

## 1. What is vapor pressure?

Vapor pressure is the pressure exerted by the gaseous molecules of a substance in equilibrium with its liquid or solid state at a specific temperature.

## 2. How is vapor pressure measured?

Vapor pressure can be measured using various methods such as the Antoine equation, which relates vapor pressure to temperature, or using a vapor pressure apparatus like a bomb calorimeter.

## 3. What factors affect vapor pressure?

Temperature, intermolecular forces, and molecular weight are factors that can affect vapor pressure. Generally, as temperature increases, so does vapor pressure. Stronger intermolecular forces and higher molecular weight can decrease vapor pressure.

## 4. What is vapor mole fraction?

Vapor mole fraction is the ratio of the number of moles of vapor present in a system to the total number of moles of all components in the system. It is used to describe the composition of a gaseous mixture.

## 5. How is vapor mole fraction calculated?

Vapor mole fraction can be calculated by dividing the moles of vapor by the total moles in the system. This can be determined experimentally or by using the ideal gas law and knowing the pressure, volume, and temperature of the system.

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