/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 The solubility coefficient of a ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
a) The Henry's Law constant will be calculated as per the conversion explained above. b) To determine the amount of \(CO_2\) dissolved, moles of \(CO_2\) will be calculated using Henry's law and then will be converted into mass with the molar mass. c) Finally, the volume that the dissolved \(CO_2\) would occupy if it were released, is to be calculated using the ideal gas law.

Step by step solution

01

Convert Solubility Coefficient to Henry's Constant

To convert the solubility coefficient \(0.0901 \, \mathrm{cm}^{3} \mathrm{CO}_{2} \, (\mathrm{STP}) / \mathrm{cm}^{3} \mathrm{H}_{2} \mathrm{O} (\mathrm{l})\) to Henry's constant in atm/mole fraction we first need to convert the volume of CO2 to moles at STP. At STP, 1 mole of a gas occupies 22.4 liters. \nSo, \(1 \, \mathrm{cm}^{3}=10^{-3} \, \mathrm{liters}\), therefore, \(0.0901 \, \mathrm{cm}^{3} = 0.0901 \times 10^{-3} \, \mathrm{moles}\) \nThen, Henry's law constant \(K_{H} = \frac{P}{X}\), where P is the partial pressure of the gas and X is the mole fraction. Here P is 1 atm and hence the mole fraction X becomes \(X= \frac{n_{\mathrm{CO2}}}{n_{\mathrm{CO2}} + n_{\mathrm{H2O}}}\) . Since, n_{\mathrm{H2O}} >> n_{\mathrm{CO2}} , X ~= n_{\mathrm{CO2}} and thus, \(K_{H} = 1 \, \mathrm{atm} / 0.0901 \times 10^{-3} \, \mathrm{moles}\)
02

Calculate the amount of dissolved CO2

For a 12-ounce bottle, the volume of water = 12/33.8 liters. At 2.5 atm, we can calculate the number of moles of CO2. Since the Henry's Law states that the amount of dissolved gas is directly proportional to its partial pressure in the gas phase, the amount of CO2 dissolved would be: \n\(n_{\mathrm{CO2}} = P/K_{H} = 2.5 \, \mathrm{atm} / K_{H}\) and the mass of CO2 would be \(m_{\mathrm{CO2}} = n_{\mathrm{CO2}} \times M_{\mathrm{CO2}}\) where \(M_{\mathrm{CO2}} = 44 \, \mathrm{g}\, / \, \mathrm{mole}\)
03

Calculate the volume of released CO2

To find the volume that the CO2 would occupy when released, we’d use the Ideal Gas Law equation \(PV = nRT\). We need to convert the temperature to Kelvin by adding 273 to the Celsius temperature. Thus, the volume is : \(V = nRT / P = n_{\mathrm{CO2}} \times R \times (37 + 273) \, K / 1 \, atm\), where R = 0.08206 L-atm/mol-K.

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Key Concepts

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

Solubility Coefficient
The solubility coefficient tells us how much gas can dissolve in a liquid at a specific pressure. It represents the volume of gas (in cubic centimeters, STP) that dissolves in one cubic centimeter of solvent at a partial pressure of 1 atm. In simpler terms, it measures how easily a gas mixes with a liquid. For example, at 20°C, the solubility coefficient of CO₂ in water is 0.0901 cm³ of CO₂ (STP) per cm³ of water.
Understanding this concept is important because it helps us predict the behavior of gases in liquids. When a gas has a high solubility coefficient, it means the gas can dissolve significantly in the liquid. This concept is used frequently in fields like chemistry and environmental science.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates pressure, volume, temperature, and moles of a gas. It is expressed as: \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant (0.08206 L-atm/mol-K), and \(T\) is the temperature in Kelvin.
This equation helps predict how a gas will behave under different conditions or how it will change when any of these variables are altered. For instance, when calculating the volume that dissolved COâ‚‚ would occupy if released, the Ideal Gas Law allows us to account for the new conditions at body temperature by adjusting these variables to reflect the properties of gases.
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 moles of one component to the total moles in the mixture. For a gas such as COâ‚‚ dissolving in water, the mole fraction \(X\) can be calculated using the formula: \(X = \frac{n_{\mathrm{CO2}}}{n_{\mathrm{CO2}} + n_{\mathrm{H2O}}}\).
Here, \(n_{\mathrm{CO2}}\) represents the moles of CO₂ and \(n_{\mathrm{H2O}}\) represents the moles of water. In most cases, because the moles of water are much larger compared to the moles of dissolved gas, the mole fraction simplifies to \(X \approx n_{\mathrm{CO2}}\). Understanding the mole fraction is crucial in calculating Henry’s law constant and predicting how much gas will dissolve under certain conditions.

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Most popular questions from this chapter

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.

Nitric acid is a chemical intermediate primarily used in the synthesis of ammonium nitrate, which is used in the manufacture of fertilizers. The acid also is important in the production of other nitrates and in the separation of metals from ores. Nitric acid may be produced by oxidizing ammonia to nitric oxide over a platinum-rhodium catalyst, then oxidizing the nitric oxide to nitrogen dioxide in a separate unit where it is absorbed in water to form an aqueous solution of nitric acid.The reaction sequence is as follows:$$\begin{aligned} 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \\\4 \mathrm{NO}+2 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}_{2} \\\4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2} & \rightarrow 4 \mathrm{HNO}_{3}(\mathrm{aq}) \end{aligned}$$.Ammonia vapor produced by vaporizing pure liquid ammonia at 820 kPa absolute is mixed with air, and the combined stream enters the ammonia oxidation unit. Air at \(30^{\circ} \mathrm{C}, 1\) atm absolute, and \(50 \%\) relative humidity is compressed and fed to the process. A fraction of the air is sent to the cooling and hydration units, while the remainder is passed through a heat exchanger and mixed with the ammonia. The total oxygen fed to the process is the amount stoichiometrically required to convert all of the ammonia to HNO \(_{3},\) while the fraction sent to the ammonia oxidizer corresponds to the stoichiometric amount required to convert ammonia to NO.The ammonia reacts completely in the oxidizer, with \(97 \%\) forming NO and the rest forming \(\mathrm{N}_{2}\). Only a negligible amount of \(\mathrm{NO}_{2}\) is formed in the oxidizer. However, the gas leaving the oxidizer is subjected to a series of cooling and hydration steps in which the NO is completely oxidized to \(\mathrm{NO}_{2}\) which in turn combines with water (some of which is present in the gas from the oxidizer and the rest is added) to form a 55 wt\% aqueous solution of nitric acid. The product gas from the process may be taken to contain only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). (a) Taking a basis of \(100 \mathrm{kmol}\) of ammonia fed to the process, calculate (i) the volumes \(\left(\mathrm{m}^{3}\right)\) of the ammonia vapor and air fed to the process using the compressibility-factor equation of state; (ii) the amount (kmol) and composition (in mole fractions) of the gas leaving the oxidation unit; (iii) the required volume of liquid water \(\left(\mathrm{m}^{3}\right)\) that must be fed to the cooling and hydration units; and (iv) the fraction of the air fed to the ammonia oxidizer. (b) Scale the results from Part (a) to a new basis of 100 metric tons per hour of 55\% nitric acid solution.(c) Nitrogen oxides (collectively referred to as \(\mathrm{NO}_{x}\) ) are a category of pollutants that are formed in many ways, including processes like that described in this problem. List the annual emission rates of the three largest sources of \(\mathrm{NO}_{x}\) emissions in your home region. What are the effects of exposure to excessive concentrations of \(\mathrm{NO}_{x} ?\) (d) A platinum-rhodium catalyst is used in ammonia oxidation. Fxplain the function of the catalyst, describe its structure, and explain the relationship of the structure to the function.

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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. 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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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