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Chapter 6: Problem 45

Sulfur trioxide (SO \(_{3}\) ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) The vapor in equilibrium with the solution contains both \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). If enough \(\mathrm{SO}_{3}\) is added, all of the water reacts and the solution becomes pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If still more \(\mathrm{SO}_{3}\) is added, it dissolves to form a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure \(\mathrm{SO}_{3}\). Twenty percent oleum by definition contains \(20 \mathrm{kg}\) of dissolved \(\mathrm{SO}_{3}\) and \(80 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as \(\% \mathrm{SO}_{3}\) by mass, with the constituents of the oleum considered to be \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Prove that a \(15.0 \%\) oleum contains \(84.4 \% \mathrm{SO}_{3}\) (b) Suppose a gas stream at \(40^{\circ} \mathrm{C}\) and 1.2 atm containing 90 mole \(\% \mathrm{SO}_{3}\) and \(10 \% \mathrm{N}_{2}\) contacts a liquid stream of 98 wt\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (aq), producing \(15 \%\) oleum. Tabulated equilibrium data indicate that the partial pressure of \(S O_{3}\) in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of \(S O_{3}\) in the outlet gas if this gas is in equilibrium with the liquid product at \(40^{\circ} \mathrm{C}\) and 1 atm, and (ii) the ratio ( \(\mathrm{m}^{3}\) gas feed) \(/\) (kg liquid feed).

Short Answer

Expert verified
a) 84.286% (by mass), i) 0.00151 (mole fraction) ii) 237.76 (m^3 gas feed) / (kg liquid feed)

Step by step solution

01

Calculation of mass percentage of SO3 in 15.0% oleum

First, let's assume that the total weight of oleum is 100 g, so according to the definition, it consists of 15 g SO3 and 85 g of H2SO4. However, the weight of H2SO4 can be represented as a combination of SO3 and H2O. Since the molecular weight of H2O is 18 and the molecular weight of H2SO4 is 98, the weight of water in H2SO4 is (18/98) * 85 g = 15.714 g. Therefore, the weight of SO3 present in H2SO4 is 85 - 15.714 g = 69.286 g. So, the total weight of SO3 in 100 g of 15% oleum is 15g + 69.286 g = 84.286 g. Therefore, the mass percentage of SO3 is (84.286/100) * 100 = 84.286%.
02

Calculation of mole fraction of SO3 in the outlet gas

Next, we鈥檒l find the mole fraction of SO3 in the outlet gas. Using Henry's law, we know that at equilibrium, the ratio of the partial pressure of a gas above a solution to the concentration of the gas in the solution is constant. Given that the partial pressure of SO3 is 1.15 mm Hg, we convert this to atm by dividing by 760 (because 1 atm = 760 mm Hg), we get 1.15/760 = 0.00151 atm. As a result, the mole fraction of SO3 in the outlet gas, if this gas is in equilibrium with the liquid product at 40 degree Celsius and 1 atm, is 0.00151.
03

Calculation of the ratio (m^3 gas feed) / (kg liquid feed)

For this ratio, we need to first determine the molar flow rates of SO3 and N2 in the gas feed. Given that the gas stream consists of 90 mol % SO3 and 10 % N2 and its total pressure is 1.2 atm, the partial pressure of SO3 is 0.9*1.2 = 1.08 atm. We can find the molar flow rate of SO3 using the ideal gas law, PV = nRT. Rearranging, n = PV/RT = (1.08 atm * V) / (0.0821 * 313 K), where V is the volume of the gas feed in m^3, R is the ideal gas constant 0.0821 atm*m^3/(mol*K), and T is the temperature in K (273 + 40 = 313K). Similarly, we can find the mass flow rate of the liquid feed knowing that it is 98 wt % H2SO4, so the mass of H2SO4 in 1 kg of the liquid feed is 0.98 kg. Using the molar mass of H2SO4 is 98 g/mol, we find the molar flow rate = 0.98 kg * 1000/98 = 10 mol. The desired ratio is (m^3 gas feed) / (kg liquid feed) = V/1 = 0.0821 * 313K * 10 / 1.08 atm = 237.76 m^3/kg.

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Key Concepts

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

Sulfuric Acid
Sulfuric acid (H extsubscript{2}SO extsubscript{4}) is one of the most important industrial chemicals. It's widely used in various industries like fertilizer production, oil refining, and wastewater processing. This powerful acid is known for its strong dehydrating and oxidizing properties.
Understanding its chemical makeup is essential: a molecule of sulfuric acid consists of two hydrogen atoms (H), one sulfur atom (S), and four oxygen atoms (O).
  • Sulfuric acid is highly corrosive and should be handled with care.
  • It can lead to severe chemical burns upon contact.
  • Proper safety equipment, including gloves and goggles, is crucial when working with it.
In aqueous solutions, sulfuric acid behaves as a strong acid, fully dissociating into its ions: \[H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}\]This dissociation provides the solution with high acidity.
The process of dissolving SO extsubscript{3} in H extsubscript{2}O to form H extsubscript{2}SO extsubscript{4} is exothermic, meaning it releases heat. This reaction is represented by:\[SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\]Understanding sulfuric acid's properties and safe handling practices is key to its effective and secure use.
Vapor Equilibrium
Vapor equilibrium refers to the balance between a liquid or solution and its vapor phase. In the context of sulfuric acid and oleum, it helps in understanding how SO extsubscript{3} and water vapor reach a stable state above the liquid mixture.
When a solution like sulfuric acid is prepared, some of its components will evaporate until the rate at which they enter the vapor phase equals the rate at which they return to the liquid. This balance constitutes vapor equilibrium.
  • In a mixture of SO extsubscript{3} and H extsubscript{2}SO extsubscript{4}, equilibrium is reached when the concentration of SO extsubscript{3} in the vapor equals its concentration in the liquid phase.
  • This equilibrium impacts the behavior of the solution and its concentration over time.
  • Changing the temperature or pressure can shift the equilibrium, altering the vapor and liquid concentrations.
When dealing with vapor equilibrium, it's critical to consider factors such as temperature, which affects the rate of evaporation and hence the equilibrium state.
For example, in oleum, which is a mixture of sulfuric acid and excess SO extsubscript{3}, the vapor in equilibrium is primarily SO extsubscript{3}. Understanding these principles is fundamental for industries that use these chemicals to produce consistent results efficiently.
Oleum Composition
Oleum, also known as fuming sulfuric acid, is a solution of sulfur trioxide (SO extsubscript{3}) dissolved in sulfuric acid (H extsubscript{2}SO extsubscript{4}). It is more powerful than sulfuric acid alone, especially because of its ability to release SO extsubscript{3}.
In the context of oleum, its composition is often measured by the percentage of free SO extsubscript{3} present in the mixture. For example, a 15% oleum means that for every 100 units of the solution, 15 units are free SO extsubscript{3}.
  • This composition can be represented in two ways: percentage of SO extsubscript{3} in terms of total solution or as part of its mixture with H extsubscript{2}SO extsubscript{4} and water.
  • In a 15% oleum, the effective percentage of pure SO extsubscript{3} when considering H extsubscript{2}O is 84.4% SO extsubscript{3}.
  • Oleum's high reactivity makes it critical for specific industrial applications, especially in explosives and synthesis reactions.
The presence of free SO extsubscript{3} provides oleum with stronger dehydrating capabilities than sulfuric acid alone. It鈥檚 often used in manufacturing processes where stronger acidic conditions or more vigorous reactions are needed.
Henry's Law
Henry's Law is a key principle in chemistry that describes the behavior of gases in contact with liquids. It states that at constant temperature, the amount of a given gas that dissolves in a liquid is directly proportional to its partial pressure above the liquid.
This relationship is typically represented as:\[C = k_H \cdot P\]where:
  • \(C\) is the concentration of the gas in the liquid,
  • \(k_H\) is Henry's law constant, and
  • \(P\) is the partial pressure of the gas.
This principle is crucial in processes involving gas-liquid equilibria, such as those in the production of oleum from sulfuric acid. In such systems, understanding how SO extsubscript{3} behaves when in equilibrium with its liquid phase under varying conditions helps in predicting and controlling the amount of SO extsubscript{3} that is dissolved.For example, in the specified exercise, using Henry's Law allows for calculating the mole fraction of SO extsubscript{3} in the outlet gas. It connects the partial pressure of SO extsubscript{3} to its concentration in the solution, providing insights into how much of the gas will be dissolved under different atmospheric conditions.Overall, mastering Henry's Law is essential for anyone dealing with chemical processes involving gases and liquids. It helps predict how changes in pressure or temperature could affect the solubility and behavior of gases.

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Most popular questions from this chapter

The sulfur dioxide content of a stack gas is monitored by passing a sample stream of the gas through an SO_ analyzer. The analyzer reading is \(1000 \mathrm{ppm} \mathrm{SO}_{2}\) (parts per million on a molar basis). The sample gas leaves the analyzer at a rate of \(1.50 \mathrm{L} / \mathrm{min}\) at \(30^{\circ} \mathrm{C}\) and \(10.0 \mathrm{mm}\) Hg gauge and is bubbled through a tank containing 140 liters of initially pure water. In the bubbler, \(S O_{2}\) is absorbed and water evaporates. The gas leaving the bubbler is in equilibrium with the liquid in the bubbler at \(30^{\circ} \mathrm{C}\) and 1 atm absolute. The \(\mathrm{SO}_{2}\) content of the gas leaving the bubbler is periodically monitored with the \(\mathrm{SO}_{2}\) analyzer, and when it reaches \(100 \mathrm{ppm} \mathrm{SO}_{2}\) the water in the bubbler is replaced with 140 liters of fresh water.(a) Speculate on why the sample gas is not just discharged directly into the atmosphere after leaving the analyzer. Assuming that the equilibrium between \(S O_{2}\) in the gas and dissolved \(S O_{2}\) is described by Henry's law, explain why the SO_ content of the gas leaving the bubbler increases with time. What value would it approach if the water were never replaced? Explain. (The word "solubility" should appear in your explanation.)(b) Use the following data for aqueous solutions of \(\mathrm{SO}_{2}\) at \(30^{\circ} \mathrm{C}^{14}\) to estimate the Henry's law constant in units of \(\mathrm{mm}\) Hg/mole fraction:$$\begin{array}{|l|c|c|c|c|c|}\hline \mathrm{g} \mathrm{SO}_{2} \text { dissolved/ } 100 \mathrm{g}\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & 0.0 & 0.5 & 1.0 & 1.5 & 2.0 \\\\\hline p_{\mathrm{SO}_{2}}(\mathrm{mm} \mathrm{Hg}) & 0.0 & 37.1 & 83.7 &132 & 183 \\\\\hline\end{array}.$$(c) Estimate the SO_concentration of the bubbler solution (mol SO_/liter), the total moles of SO_ dissolved, and the molar composition of the gas leaving the bubbler (mole fractions of air, \(\mathrm{SO}_{2}\), and water vapor) at the moment when the bubbler solution must be changed. Make the following assumptions: \bullet. The feed and outlet streams behave as ideal gases. \bullet Dissolved SO_ is uniformly distributed throughout the liquid. ? The liquid volume remains essentially constant at 140 liters. \- The water lost by evaporation is small enough for the total moles of water in the tank to be considered constant. \- The distribution of SO_ between the exiting gas and the liquid in the vessel at any instant of time is governed by Henry's law, and the distribution of water is governed by Raoult's law (assume \(\left.x_{\mathrm{H}_{2} \mathrm{O}} \approx 1\right)\).(d) Suggest changes in both scrubbing conditions and the scrubbing solution that might lead to an increased removal of \(\mathrm{SO}_{2}\) from the feed gas.

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.

The pressure in a vessel containing methane and water at \(70^{\circ} \mathrm{C}\) is 10 atm. At the given temperature, the Henry's law constant for methane is \(6.66 \times 10^{4}\) atm/mole fraction. Estimate the mole fraction of methane in the liquid.

The vapor pressure of ethylene glycol at several temperatures is given below:$$\begin{array}{|l|r|r|r|r|r|r|}\hline T\left(^{\circ} \mathrm{C}\right) & 79.7 & 105.8 & 120.0 & 141.8 & 178.5 & 197.3 \\\\\hline p^{*}(\mathrm{mm} \mathrm{Hg}) & 5.0 & 20.0 & 40.0 & 100.0 & 400.0 & 760.0 \\\\\hline\end{array}$$ a semilog plot e vapor-pressure data and determine a linear expression for \(\ln p^{*}\) function of \(1 / T(\mathrm{K}) .\) Use the results to estimate the heat of vaporization \((\mathrm{kJ} / \mathrm{mol})\) of ethylene glycol, and then use that value in the Clausius-Clapeyron equation to estimate the vapor pressures at each of the temperatures given in the table.(b) Repeat Part (a) using the Slope and Intercept functions of APEx to obtain the expression for \(\ln p^{*}\) vs. \(1 / T(\mathrm{K})\).(c) Use the results from Part (b) to estimate vapor pressures of ethylene glycol at \(50^{\circ} \mathrm{C}, 80^{\circ} \mathrm{C},\) and \(110^{\circ} \mathrm{C} .\) Also estimate the boiling point of this substance at system pressures of \(760 \mathrm{mm} \mathrm{Hg}\) and 2000 mm Hg. Compare all five results with those obtained directly using APEx functions. In which of the estimates at the given temperatures and pressures would you have the least confidence? Explain your reasoning.

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)?
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