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A quantity of methyl acetate is placed in an open, transparent, three-liter flask and boiled long enough to purge all air from the vapor space. The flask is then sealed and allowed to equilibrate at \(30^{\circ} \mathrm{C},\) at which temperature methyl acetate has a vapor pressure of \(269 \mathrm{mm}\) Hg. Visual inspection shows \(10 \mathrm{mL}\) of liquid methyl acetate present.(a) What is the pressure in the flask at equilibrium? Explain your reasoning.(b) What is the total mass (grams) of methyl acetate in the flask? What fraction is in the vapor phase at equilibrium?(c) The above answers would be different if the species in the vessel were ethyl acetate because methyl acetate and ethyl acetate have different vapor pressures. Give a rationale for that difference.

Short Answer

Expert verified
The total pressure in the flask at equilibrium would be 269 mm Hg which is the vapor pressure of methyl acetate at the given temperature. The total mass of the methyl acetate in the flask and the fraction in the vapor phase would need to be calculated, while considering and understanding that differences in molecular structure between substances like methyl acetate and ethyl acetate result in different vapor pressures.

Step by step solution

01

Calculating the total pressure

From the ideal gas equation, the pressure \( P \) is given by \( P = nRT/V \), where \( n \) is the number of moles of gas, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the gas. We know that all air has been removed, so the only gas in the flask is the vaporized methyl acetate. The pressure due to this vaporized methyl acetate is the vapor pressure, which is 269 mm Hg at \(30^{\circ} \mathrm{C}\). Therefore, the total pressure in the flask at equilibrium is also 269 mm Hg.
02

Calculating the total mass of methyl acetate in flask

We know that 10 mL of liquid methyl acetate is present. This means that the rest of the volume (3L - 10 mL = 2.99L) is occupied by the vapor phase. To calculate the total mass, we first have to calculate the number of moles in the liquid and gas phase. This again requires the use of the ideal gas equation to calculate the moles in the vapor phase. The molecular weight of methyl acetate is approximately 74.08 g/mol, we also need to convert the temperature to Kelvin (T=303.15K) and the pressure to atm (P=269mmHg=0.354atm). For the liquid phase, we can use the density of methyl acetate (0.932 g/mL) to calculate its mass (10mL*0.932g/mL=9.32g), which we then convert to moles by using the molecular weight of methyl acetate. Adding the moles of gas phase to moles of liquid phase and multiplying by the molecular weight gives us the total mass in the flask.
03

Calculating the fraction in the vapor phase

We can use the previously calculated moles of gas and liquid phase to determine the fraction of methyl acetate in the vapor phase at equilibrium. Divide the moles of gas by the total moles (moles of gas + moles of liquid) to get the fraction in the vapor phase.
04

Comparing with ethyl acetate

The pressure and total mass in the flask would differ if ethyl acetate was used instead of methyl acetate. This is because different substances have different vapor pressures. The vapor pressure of a substance is directly related to how readily its molecules will escape from the liquid or solid state to the gas state. Methyl acetate and ethyl acetate are different compounds with distinct molecular structures, they do not have the same intermolecular forces and hence they have different vapor pressures.

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Key Concepts

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

Vapor Pressure
Vapor pressure is a key concept in understanding how substances behave in different states of matter. It refers to the pressure exerted by a vapor in equilibrium with its liquid or solid phase. This happens when the rate of evaporation of a liquid equals the rate of condensation. At that point, the pressure exerted by the vapor remains constant.
Vapor pressure is influenced primarily by temperature. As temperature increases, so does the kinetic energy of molecules, which results in more molecules escaping the liquid to become vapor. Consequently, this increases the vapor pressure. Each substance has its own characteristic vapor pressure at a given temperature.
For example, in the original exercise, methyl acetate has a vapor pressure of 269 mm Hg at 30°C. This is the pressure exerted by the vapor phase of methyl acetate when it is in equilibrium with its liquid phase at this temperature. It's critical to grasp this idea to solve problems related to phase equilibrium and gas laws.
Ideal Gas Law
The Ideal Gas Law is a crucial equation in chemistry, represented by the formula \( PV = nRT \). Here, \( P \) stands for the pressure of the gas, \( V \) is its volume, \( n \) denotes the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
This law provides a useful approximation for the behavior of gases under certain conditions. It assumes that the gas molecules do not interact with each other and occupy no volume of their own. Real gases often deviate from these assumptions especially at high pressures and low temperatures. However, it serves as a good model for understanding gas behavior in standard conditions.
In the context of the problem, the Ideal Gas Law helps us calculate the pressure of vaporized methyl acetate and understand how it fills the flask. Knowing the vapor pressure, we can use it as the pressure \( P \) in the Ideal Gas Law to solve for other variables, enhancing our understanding of the substance's gas-phase behavior.
Phase Equilibrium
Phase equilibrium occurs when different phases of a substance exist together without changing over time. At equilibrium, the rate of a phase change equals the rate of the reverse phase change, leading to no net change in the amount of each phase. This can happen with solid-liquid, liquid-vapor, or other transitions.
Relating this to the exercise, methyl acetate reaches phase equilibrium inside the sealed flask. The liquid and vapor phases coexist such that the vapor's rate of condensation matches its rate of evaporation. Therefore, the vapor pressure remains constant Phase equilibrium is essential in many areas, from industrial chemical processes to environmental systems. For example, understanding phase equilibrium helps chemists design systems that optimize reactions or separations in industries.
Molecular Weight Calculation
Molecular weight calculation is critical for converting between moles and grams, helping chemists understand how much of a substance is present in a given system.
The molecular weight, or molar mass, is the sum of the atomic masses of all the atoms in a molecule. It's typically expressed in grams per mole. To calculate the total mass of a compound in a mixture, you need the moles of each component and their respective molecular weights.
In the original exercise, calculating the total mass of methyl acetate inside the flask involves using its molecular weight, about 74.08 g/mol. By knowing the density and volume of the liquid phase, we can determine its mass and convert it into moles. Similarly, the gas phase mass is determined by the moles from the Ideal Gas Law, multiplied by the molecular weight.
Understanding molecular weight allows students to tackle problems involving chemical reactions, solution concentrations, and physical states, integrating important concepts across chemistry.

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Most popular questions from this chapter

A gas mixture containing 85.0 mole \(\% \mathrm{N}_{2}\) and the balance \(n\) -hexane flows through a pipe at a rate of \(100.0 \mathrm{m}^{3} / \mathrm{h} .\) The pressure is 2.00 atm absolute and the temperature is \(100^{\circ} \mathrm{C}\). (a) What is the molar flow rate of the gas in \(\mathrm{kmol} / \mathrm{h}\) ? (b) Is the gas saturated? If not, to what temperature ( \(^{C} C\) ) would it have to be cooled at constant pressure in order to begin condensing hexane? (c) To what temperature ( \(C\) ) would the gas have to be cooled at constant pressure in order to condense \(80 \%\) of the hexane?

A fuel gas containing methane and ethane is burned with air in a furnace, producing a stack gas at \(300^{\circ} \mathrm{C}\) and \(105 \mathrm{kPa}\) (absolute). You analyze the stack gas and find that it contains no unburned hydrocarbons, oxygen, or carbon monoxide. You also determine the dew-point temperature.(a) Estimate the range of possible dew-point temperatures by determining the dew points when the feed is either pure methane or pure ethane. (b) Estimate the fraction of the feed that is methane if the measured dew- point temperature is \(59.5^{\circ} \mathrm{C}\). (c) What range of measured dew point temperatures would lead to calculated methane mole fractions within 5\% of the value determined in Part (b)?

In-Hexane is used to extract oil from soybeans. (See Problem 6.24 .) The solid residue from the extraction unit, which contains 0.78 kg liquid hexane/kg dry solids, is contacted in a dryer with nitrogen that enters at \(85^{\circ} \mathrm{C}\). The solids leave the dryer containing \(0.05 \mathrm{kg}\) liquid hexane/kg dry solids, and the gas leaves the dryer at \(80^{\circ} \mathrm{C}\) and 1.0 atm with a relative saturation of \(70 \% .\) The gas is then fed to a condenser in which it is compressed to 5.0 atm and cooled to \(28^{\circ} \mathrm{C}\), enabling some of the hexane to be recovered as condensate.(a) Calculate the fractional recovery of hexane (kg condensed/kg fed in wet solids). (b) A proposal has been made to split the gas stream leaving the condenser, combining 90\% of it with fresh makeup nitrogen, heating the combined stream to \(85^{\circ} \mathrm{C},\) and recycling the heated stream to the dryer inlet. What fraction of the fresh nitrogen required in the process of Part (a) would be saved by introducing the recycle? What costs would be incurred by introducing the recycle?

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?

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 \(\left(h_{\mathrm{r}, \text { out }}=85 \%\right)\) .

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