Problem 57 - Chemistry An Atoms First Approach

Question

Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) and hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) form an ideal solution. At \(25^{\circ} \mathrm{C}\) the vapor pressures of pentane and hexane are 511 and \(150 .\) torr, respectively. A solution is prepared by mixing \(25 \mathrm{mL} \text { pentane (density, } 0.63 \mathrm{g} / \mathrm{mL})\) with \(45 \mathrm{mL}\) hexane (density, 0.66 g/mL). a. What is the vapor pressure of the resulting solution? b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?

Step-by-Step Solution

Answer

a. The vapor pressure of the resulting solution is \(290.12 \, \mathrm{torr}\). b. The composition by mole fraction of pentane in the vapor that is in equilibrium with this solution is \(0.678\).

Step 1: Calculate the number of moles of pentane and hexane in the solution

We are given the volume and density of each component. We can find the mass of each component and then calculate the number of moles. The molar masses of pentane and hexane are \(72.15 \, \mathrm{g/mol}\) and \(86.18 \, \mathrm{g/mol}\), respectively. Mass of pentane \(= (25 \, \mathrm{mL})(0.63 \, \mathrm{g/mL}) = 15.75 \, \mathrm{g}\) Moles of pentane \(= \cfrac{15.75\, \mathrm{g}}{72.15\, \mathrm{g/mol}} = 0.218\, \mathrm{mol}\) Mass of hexane \(= (45\, \mathrm{mL})(0.66\, \mathrm{g/mL}) = 29.7\, \mathrm{g}\) Moles of hexane \(= \cfrac{29.7\, \mathrm{g}}{86.18\, \mathrm{g/mol}} = 0.345\, \mathrm{mol}\)

Step 2: Calculate the mole fractions of pentane and hexane in the solution

Mole fraction (\(χ_i\)) is the ratio of the number of moles of a component to the total number of moles of all components in the mixture. Mole fraction of pentane (\(χ_\mathrm{pentane}\)) \(= \cfrac{\text{moles of pentane}}{\text{total moles}} = \cfrac{0.218}{0.218 + 0.345} = 0.387\) Mole fraction of hexane (\(χ_\mathrm{hexane}\)) \(= \cfrac{\text{moles of hexane}}{\text{total moles}} = \cfrac{0.345}{0.218 + 0.345} = 0.613\)

Step 3: Calculate the vapor pressure of the solution using Raoult's law

Raoult's law states that the partial pressure of a component in an ideal solution is equal to the product of its mole fraction and its vapor pressure as a pure liquid. The total vapor pressure of the solution is the sum of the partial pressures of all components. Vapor pressure of the solution \(= P_\mathrm{total} = χ_\mathrm{pentane} P^\circ_\mathrm{pentane} + χ_\mathrm{hexane} P^\circ_\mathrm{hexane}\) \(= (0.387)(511\, \mathrm{torr}) + (0.613)(150\, \mathrm{torr}) = 198.167 + 91.95 = 290.12 \, \mathrm{torr}\)

Step 4: Determine the mole fraction of pentane in the vapor in equilibrium with the solution

Let's denote the mole fraction of pentane in the vapor as \(y_\mathrm{pentane}\). We can find this by using the ratio of the partial pressure of pentane in the solution to the total vapor pressure. Mole fraction of pentane in the vapor \(y_\mathrm{pentane} = \cfrac{χ_\mathrm{pentane} P^\circ_\mathrm{pentane}}{P_\mathrm{total}} = \cfrac{(0.387)(511\, \mathrm{torr})}{290.12\, \mathrm{torr}} = 0.678\) a. The vapor pressure of the resulting solution is \(290.12 \, \mathrm{torr}\). b. The composition by mole fraction of pentane in the vapor that is in equilibrium with this solution is \(0.678\).

Key Concepts

Vapor Pressure of Solutions

Understanding how the vapor pressure of solutions is determined is essential for many fields of chemistry, especially when dealing with mixtures. Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) form. It's a key concept when predicting how a solution will behave under various conditions.

In the context of Raoult's law, which applies to ideal solutions, the vapor pressure of a solution can be calculated as the sum of the partial pressures of each of its components. The partial pressure of each component is, in turn, the product of the mole fraction of the component in the liquid phase and its pure vapor pressure. This law simplifies the description of vapor-liquid equilibrium for solutions by making the assumption that the intermolecular forces between unlike molecules are similar to those between like molecules.

To calculate the vapor pressure of the solution, we first need to determine the mole fraction of each component (as shown in the given exercise steps). Then, we multiply these mole fractions by the respective pure vapor pressures and sum them up. The reasoning behind this is that each component of an ideal solution contributes to the total vapor pressure proportionally to its amount (mole fraction) and its inherent vapor pressure when pure.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of one particular component to the total number of moles of all components present in the mixture. Mathematically, the mole fraction, denoted as \( χ_i \), of component \( i \) is given by \( χ_i = \frac{n_i}{n_{total}} \), where \( n_i \) is the number of moles of that component, and \( n_{total} \) is the sum of moles of all components.

This dimensionless quantity plays a crucial role when we apply Raoult's law to find the vapor pressure of a solution. In the exercise provided, the mole fractions of pentane and hexane in the liquid phase are important for determining not only the vapor pressure of the solution but also the composition of the vapor phase. The mole fraction is key in various thermodynamic calculations, as it relates to the composition of both phases when a liquid is in equilibrium with its vapor.

Ideal Solution

An ideal solution is a hypothetical solution in which the enthalpy of mixing is zero and the volume of mixing is also considered to be negligible. In other words, there are no heat changes or volume changes when the components are mixed to form the solution. These solutions follow Raoult's law at all concentrations for all the components in the mixture.

An ideal solution assumes that the intermolecular forces between dissimilar molecules are equivalent to those between similar molecules. Due to this assumption, the properties of the solution can be predicted from the properties of the pure components. For example, pentane and hexane form an ideal solution as described in the exercise, because the size and type of intermolecular forces (dispersion forces) in pentane and hexane are similar.

However, in reality, many solutions deviate from ideal behavior due to interactions between molecules of different substances being different from those within the same substance. These are called non-ideal solutions and often require additional factors or corrections to be applied to Raoult's law to accurately describe their behavior.