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A gas containing nitrogen, benzene, and toluene is in equilibrium with a liquid mixture of 40 mole \(\%\) benzene-60 mole\% toluene at 100^'C and 10 atm. Estimate the gas-phase composition (mole fractions) using Raoult's law. State your assumptions. Why would you have confidence in the accuracy of Raoult's law?

Short Answer

Expert verified
The mole fractions of each component in the gas phase can be calculated using Raoult's law. This result is highly confident due to the similar intermolecular interactions between benzene and toluene.

Step by step solution

01

Understanding Raoult’s law

Raoult's law states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. The formula can be written as, \( P_{i} \) = \( X_{i} \) * \( P_{i}^{*} \), where \( P_{i} \) is the partial pressure of component i, \( X_{i} \) is the mole fraction of component i in the solution, and \( P_{i}^{*} \) is the vapor pressure of pure component i at the given temperature.
02

Identifying the mole fractions

From the problem, we know that the liquid mixture contains 40% benzene and 60% toluene. Thus, the mole fractions of benzene (\( X_{B} \)) and toluene (\( X_{T} \)) in the liquid phase are 0.4 and 0.6 respectively.
03

Calculating the partial pressures

Assuming that the vapor pressures of pure benzene and toluene at 100°C are known (let's denote them as \( P_{B}^{*} \) and \( P_{T}^{*} \)), we can calculate the partial pressures of benzene and toluene in the gas phase using Raoult's law. For benzene: \( P_{B} \) = \( X_{B} \) * \( P_{B}^{*} \), and for toluene: \( P_{T} \) = \( X_{T} \) * \( P_{T}^{*} \)
04

Calculating the mole fractions in the gas phase

The total pressure of the gas mixture is the sum of the partial pressures of all three components. As nitrogen is an inert gas, it does not interact with benzene or toluene, so its partial pressure equals the total pressure minus the sum of the partial pressures of benzene and toluene. Thus,\( P_{N} \) = \( P_{total} \) - \( P_{B} \) - \( P_{T} \). Then, the mole fractions of each component in the gas phase can be calculated by dividing its partial pressure by the total pressure. For nitrogen: \( Y_{N} \)= \( P_{N} / P_{total} \), for benzene: \( Y_{B} \)= \( P_{B} / P_{total} \), and for toluene: \( Y_{T} \) = \( P_{T} / P_{total} \).
05

Confidence in the accuracy of Raoult's law

Raoult’s law is accurate for ideal solutions where interactions between molecules of different components are similar to those between molecules of the same component. Since benzene and toluene are both hydrocarbon compounds, they are likely to exhibit similar intermolecular interactions, which give us confidence in the accuracy of Raoult's law for this specific case

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vapor-Liquid Equilibrium
In a solution, **vapor-liquid equilibrium** occurs when the rate of evaporation of the liquid matches the rate at which the gas condenses back into the liquid. This balance between phases at a given temperature and pressure is crucial for explaining the behavior of mixtures, particularly when applying Raoult's law. In our case, the mixture of benzene and toluene reaches a state of equilibrium. This allows us to predict the composition of the gas phase using known properties of the liquid phase. The equilibrium ensures that each component distributes between the liquid and vapor phases according to its own partial pressure over its pure form, multiplied by its mole fraction in the solution.
Understanding vapor-liquid equilibrium helps in determining how components distribute themselves between the liquid and gas phases, providing a foundation for further calculations using Raoult's Law.
Partial Pressure Calculation
**Partial pressure calculation** is a key step in analyzing vapor-liquid equilibrium. Each component in a mixture contributes to the total pressure within the gas phase. Raoult's law facilitates the computation of partial pressures by relating them to mole fractions and the vapor pressures of the pure components.
- For benzene and toluene, you need the vapor pressures at the given temperature. Let's use hypothetical values as an example. If benzene's vapor pressure at 100°C is, let's say, 100 kPa and toluene's is 200 kPa, the partial pressures would be calculated as follows:
- Benzene's partial pressure: \[ P_{B} = X_{B} \times P_{B}^{*} \] where \( X_{B} \) is benzene's mole fraction and \( P_{B}^{*} \) is benzene's vapor pressure.
- Similarly, for toluene: \[ P_{T} = X_{T} \times P_{T}^{*} \] where \( X_{T} \) is toluene's mole fraction and \( P_{T}^{*} \) is toluene's vapor pressure.
The sum of these partial pressures provides insight into how each component contributes to the overall behavior of the gas mixture, crucial for determining gas-phase compositions.
Ideal Solutions
**Ideal solutions** are those in which the interactions between different molecules are similar to those within the same component. This concept is essential when applying Raoult's law. In an ideal solution, each component's contribution to the total vapor pressure directly correlates to its concentration in the liquid phase and its pure component vapor pressure.
It's assumed that the molecules do not experience additional attractions or repulsions when mixed together, making calculations straightforward. Benzene and toluene, both being hydrocarbons, tend to follow ideal behavior fairly closely. This is why the accuracy of Raoult’s law in predicting their vapor pressures and partial pressures is enhanced. Using Raoult’s law under these conditions generally yields reliable results, as intermolecular forces don't significantly deviate from ideal predictions.

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Most popular questions from this chapter

The solubility of sodium bicarbonate in water is \(11.1 \mathrm{g} \mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(30^{\circ} \mathrm{C}\) and \(16.4 \mathrm{g}\) \(\mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(60^{\circ} \mathrm{C} .\) If a saturated solution of \(\mathrm{NaHCO}_{3}\) at \(60^{\circ} \mathrm{C}\) is cooled and comes to equilibrium at \(30^{\circ} \mathrm{C},\) what percentage of the dissolved salt crystallizes?

The solubility coefficient of a gas may be defined as the number of cubic centimeters (STP) of the gas that dissolves in \(1 \mathrm{cm}^{3}\) of a solvent under a partial pressure of 1 atm. The solubility coefficient of \(\mathrm{CO}_{2}\) in water at \(20^{\circ} \mathrm{C}\) is \(0.0901 \mathrm{cm}^{3} \mathrm{CO}_{2}(\mathrm{STP}) / \mathrm{cm}^{3} \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) Calculate the Henry's law constant in atm/mole fraction for \(\mathrm{CO}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) from the given solubility coefficient. (b) How many grams of \(\mathrm{CO}_{2}\) can be dissolved in a \(12-\mathrm{oz}\) bottle of soda at \(20^{\circ} \mathrm{C}\) if the gas above the soda is pure \(\mathrm{CO}_{2}\) at a gauge pressure of 2.5 atm ( 1 liter \(=33.8\) fluid ounces)? Assume the liquid properties are those of water. (c) What volume would the dissolved \(C O_{2}\) occupy if it were released from solution at body temperature and pressure \(-37^{\circ} \mathrm{C}\) and 1 atm?

A fuel cell is an electrochemical device in which hydrogen reacts with oxygen to produce water and DC electricity. A 1-watt proton-exchange membrane fuel cell (PEMFC) could be used for portable applications such as cellular telephones, and a \(100-\mathrm{kW}\) PEMFC could be used to power an automobile. The following reactions occur inside the PEMFC:Anode: \(\quad \mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) Cathode: \(\quad \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}\) Overall: \(\quad \overline{\mathrm{H}}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}\) A flowchart of a single cell of a PEMFC is shown below. The complete cell would consist of a stack of such cells in series, such as the one shown in Problem 9.19.The cell consists of two gas channels separated by a membrane sandwiched between two flat carbonpaper electrodes- -the anode and the cathode- -that contain imbedded platinum particles. Hydrogen flows into the anode chamber and contacts the anode, where \(\mathrm{H}_{2}\) molecules are catalyzed by the platinum to dissociate and ionize to form hydrogen ions (protons) and electrons. The electrons are conducted throughthe carbon fibers of the anode to an extemal circuit, where they pass to the cathode of the next cell in the stack. The hydrogen ions permeate from the anode through the membrane to the cathode.Humid air is fed into the cathode chamber, and at the cathode \(\mathrm{O}_{2}\) molecules are catalytically split to form oxygen atoms, which combine with the hydrogen ions coming through the membrane and electrons coming from the external circuit to form water. The water desorbs into the cathode gas and is carried out of the cell. The membrane material is a hydrophilic polymer that absorbs water molecules and facilitates the transport of the hydrogen ions from the anode to the cathode. Electrons come from the anode of the cell at one end of the stack and flow through an extemal circuit to drive the device that the fuel cell is powering, while the electrons coming from the device flow back to the cathode at the opposite end of the stack to complete the circuit. is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. On the other hand, if there is not enough water in the cathode gas (less than \(85 \%\) relative humidity), the membrane dries out and cannot transport hydrogen efficiently, which also leads to reduced performance. 400-sell 300-yolt PEMFS anerates at stady state witha nonwer outnul of 36 k W, The air fod to It is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. On the other hand, if there is not enough water in the cathode gas (less than \(85 \%\) relative humidity), the membrane dries out and cannot transport hydrogen efficiently, which also leads to reduced performance.A 400-cell 300-volt PEMFC operates at steady state with a power output of 36 kW. The air fed to the cathode side is at \(20.0^{\circ} \mathrm{C}\) and roughly 1.0 atm (absolute) with a relative humidity of \(70.0 \%\) and a volumetric flow rate of \(4.00 \times 10^{3}\) SLPM (standard liters per minute). The gas exits at \(60^{\circ} \mathrm{C}\). (a) Explain in your own words what happens in a single cell of a PEMFC. (b) The stoichiometric hydrogen requirement for a PEMFC is given by \(\left(n_{\mathrm{Hz}}\right)_{\text {conanmad }}=I N / 2 F,\) where \(I\) is the current in amperes (coulomb/s), \(N\) is the number of single cells in the fuel cell stack, and \(F\) is the Faraday constant, 96,485 coulombs of charge per mol of electrons. Derive this expression. (Hint: Recall that since the cells are stacked in series the same current flows through each one, and the same quantity of hydrogen must be consumed in each single cell to produce that current at each anode.) (c) Use the expression of Part (b) to determine the molar rates of oxygen consumed and water generated in the unit with the given specifications, both in units of mol/min. (Remember that power = voltage \(\times\) current.) Then determine the relative humidity of the cathode exit stream, \(h_{\mathrm{r} \text { rout. }}\) (d) Determine the minimum cathode inlet flow rate in SLPM to prevent the fuel cell from flooding ( \(h_{\mathrm{r}, \text { out }}=100 \%\) ) and the maximum flow rate to prevent it from drying \(\left(h_{\mathrm{r}, \text { out }}=85 \%\right)\) .

An adult inhales approximately 12 times per minute, taking in about 500 mL of air with each inhalation. Oxygen and carbon dioxide are exchanged in the lungs, but there is essentially no exchange of nitrogen. The exhaled air has a mole fraction of nitrogen of 0.75 and is saturated with water vapor at body temperature, \(37^{\circ} \mathrm{C}\). If ambient conditions are \(25^{\circ} \mathrm{C}, 1\) atm, and \(50 \%\) relative humidity, what volume of liquid water (mL) would have to be consumed over a two-hour period to replace the water loss from breathing? How much would have to be consumed if the person is on an airplane where the temperature, pressure, and relative humidity are respectively \(25^{\circ} \mathrm{C}, 1 \mathrm{atm},\) and \(10 \% ?\)

Acetone is to be extracted with \(n\) -hexane from a \(40.0 \mathrm{wt} \%\) acetone- \(60.0 \mathrm{wt} \%\) water mixture at \(25^{\circ} \mathrm{C} .\) The acetone distribution coefficient (mass fraction acetone in the hexane-rich phase/mass fraction acetone in the water-rich phase) is \(0.34 .^{18}\) Water and hexane may be considered immiscible. Three different processing alternatives are to be considered: a two-stage process and two single-stage processes.(a) In the first stage of the proposed two-stage process, equal masses of the feed mixture and pure hexane are blended vigorously and then allowed to settle. The organic phase is withdrawn and the aqueous phase is mixed with \(75 \%\) of the amount of hexane added in the first stage. The mixture is allowed to settle and the two phases are separated. What percentage of the acetone in the original feed solution remains in the water at the end of the process?(b) Suppose all of the hexane added in the two-stage process of Part (a) is instead added to the feed mixture and the process is carried out in a single equilibrium stage. What percentage of the acetone in the feed solution remains in the water at the end of the process?(c) Finally, suppose a single-stage process is used but it is desired to reduce the acetone content of the water to the final value of Part (a). How much hexane must be added to the feed solution?(d) Under what circumstances would each of the three processes be the most cost-effective? What additional information would you need to make the choice?

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