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Chapter 5: Problem 65

Hydrogen sulfide has the distinctive unpleasant odor associated with rotten eggs, and it is poisonous. It often must be removed from crude natural gas and is therefore a product of refining natural gas. In such instances, the Claus process provides a means of converting \(\mathrm{H}_{2} \mathrm{S}\) to elemental sulfur. Consider a feed stream to a Claus process that consists of 10.0 mole \(\% \mathrm{H}_{2} \mathrm{S}\) and \(90.0 \% \mathrm{CO}_{2}\). Onethird of the stream is sent to a furnace where the \(\mathrm{H}_{2} \mathrm{S}\) is burned completely with a stoichiometric amount of air fed at 1 atm and \(25^{\circ} \mathrm{C}\). The combustion reaction is $$\mathrm{H}_{2} \mathrm{S}+\frac{3}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}$$ The product gases from this reaction are then mixed with the remaining two- thirds of the feed stream and sent to a reactor in which the following reaction goes to completion: $$2 \mathrm{H}_{2} \mathrm{S}+\mathrm{SO}_{2} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$$ The gases leave the reactor at \(10.0 \mathrm{m}^{3} / \mathrm{min}, 320^{\circ} \mathrm{C},\) and \(205 \mathrm{kPa}\) absolute. Assuming ideal-gas behavior, determine the feed rate of air in kmol/min. Provide a single balanced chemical equation reflecting the overall process stoichiometry. How much sulfur is produced in \(\mathrm{kg} / \mathrm{min} ?\)

Short Answer

Expert verified
The feed rate of air is 0.4/3 kmol/min. The overall balanced chemical equation is \(5\mathrm{H}_{2} \mathrm{S} + \frac{9}{2} \mathrm{O}_{2} \rightarrow 3 \mathrm{SO}_{2} + 5 \mathrm{H}_{2} \mathrm{O} + 3 \mathrm{S}\). The rate of sulfur produced is (3/2)*(0.1/3)*32.06 kg/min.

Step by step solution

01

Identify the Stoichiometry of Individual Reactions

The given chemical reactions are: \[ \mathrm{H}_{2} \mathrm{S} + \frac{3}{2} \mathrm{O}_{2} \rightarrow\mathrm{SO}_{2} + \mathrm{H}_{2} \mathrm{O} \] and \[ 2 \mathrm{H}_{2} \mathrm{S} +\mathrm{SO}_{2} \rightarrow 3 \mathrm{S} + 2\mathrm{H}_{2} \mathrm{O}. \] From these reactions, we can identify that one molecule of H2S reacts with half as many molecules of O2 to form one molecule of SO2 and H2O in the first reaction, and two molecules of H2S react with one molecule of SO2 to form three molecules of S and two molecules of H2O in the second reaction.
02

Determine Feed Rate of Air

If we consider the furnance system, \[ n_{H_2S} = n_{\frac{air}{5}} = 0.1n_{feed} \] \[1-0.1 = 0.9 = n_{CO_2} \] Multiplying by 5 and solving for \( n_{air} \) yields: \[ n_{air} = 5(0.1)n_{feed} - n_{feed} = 0.4n_{feed} \] Since one third of the feed goes through the system, the air feed rate is 0.4/3 mol/min.
03

Determine the Overall Balanced Chemical Equation

Multiplying the first reaction by 3 and adding it to the second reaction gives \[ 3\mathrm{H}_{2} \mathrm{S} + \frac{9}{2} \mathrm{O}_{2} +2 \mathrm{H}_{2} \mathrm{S} +\mathrm{SO}_{2} \rightarrow 3 \mathrm{SO}_{2} + 3 \mathrm{H}_{2} \mathrm{O} + 3 \mathrm{S} + 2\mathrm{H}_{2} \mathrm{O} \] Simplifying this yields \[ 5\mathrm{H}_{2} \mathrm{S} + \frac{9}{2} \mathrm{O}_{2} \rightarrow 3 \mathrm{SO}_{2} + 5 \mathrm{H}_{2} \mathrm{O} + 3 \mathrm{S}, \] which is the overall balanced chemical equation for the process.
04

Determine the Quantity of Sulfur Produced

The second reaction yields 3 moles of sulfur from 2 moles of H2S and 1 mole of SO2. We have one-third of H2S reactant gone through the reaction, so it's 0.1/3 mol/min. The rate of sulfur produced is therefore (3/2) times the rate of H2S reacted, which is (3/2)*(0.1/3) mol/min. Then multiply by the molar mass of sulfur (32.06 kg/kmol) to get the mass of sulfur produced per minute.

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Key Concepts

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

Claus Process
The Claus process is widely utilized in the chemical industry to remove hydrogen sulfide (\( \mathrm{H}_2\mathrm{S} \)) from natural gas and convert it into elemental sulfur. This process is essential because hydrogen sulfide is not only toxic but also has a distinct, foul odor reminiscent of rotten eggs. The transformation of \( \mathrm{H}_2\mathrm{S} \) into sulfur mitigates these hazards and yields a useful byproduct.
The Claus process typically unfolds in two steps:
  • First, the \( \mathrm{H}_2\mathrm{S} \) is partially oxidized to \( \mathrm{SO}_2 \) in a furnace, where it reacts with air in a controlled manner. This step involves the combustion of one-third of the \( \mathrm{H}_2\mathrm{S} \) in the presence of a stoichiometric amount of oxygen, following the reaction:
\[\mathrm{H}_2\mathrm{S} + \frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_2 + \mathrm{H}_2\mathrm{O}.\]
  • In the second step, the remaining \( \mathrm{H}_2\mathrm{S} \) is reacted with the \( \mathrm{SO}_2 \) produced to yield elemental sulfur, represented by the equation:
\[2 \mathrm{H}_2\mathrm{S} + \mathrm{SO}_2 \rightarrow 3 \mathrm{S} + 2 \mathrm{H}_2\mathrm{O}.\]This process is highly effective at removing and converting hydrogen sulfide, with the formation of sulfur offering a commercial advantage.
Ideal Gas Behavior
In chemical engineering, understanding gas behavior is essential, especially in processes such as the Claus process. Ideal gas law offers a simplified correlation between pressure (\( P \)), volume (\( V \)), and temperature (\( T \)), making it easier to predict the behavior of gases in various conditions. This law is expressed by the equation:\[PV = nRT,\]where \( n \) is the number of moles, and \( R \) is the ideal gas constant.
Assuming ideal-gas behavior simplifies calculations in engineering processes. However, it is important to remember this assumption works best at high temperatures and low pressures where gases behave more ideally. This is particularly relevant in the Claus process where gases are mixed and reacted under specific conditions. By treating gases as ideal, engineers can accurately estimate emissions, product yields, and reactant needs. Furthermore, ideal gas behavior aids in determining various parameters, such as the air feed rate in the furnace, which is crucial for ensuring optimal conditions for sulfur recovery.
Despite its simplifications, the ideal gas assumption is a foundational tool in chemical engineering, offering accuracy for most practical purposes and simplifying complex process calculations.
Chemical Engineering Education
Chemical engineering education plays a pivotal role in preparing students to tackle real-world problems like those encountered in the Claus process. Students are equipped with knowledge of chemical thermodynamics, reaction kinetics, and process design to efficiently handle industrial applications.
By engaging with problems that require understanding of stoichiometry and reaction balancing, such as the transformation of \( \mathrm{H}_2\mathrm{S} \) to sulfur, students develop critical thinking skills necessary for designing and optimizing chemical processes. Education in chemical engineering also emphasizes the importance of understanding gas behavior and assumptions like ideal gas behavior, which simplify calculations and aid in process design.
Furthermore, modern chemical engineering education integrates hands-on experiments and simulations. Students gain insights into the operation of refining and purification processes, helping them to appreciate the environmental and economic impacts of their work. With a combination of theoretical knowledge and practical application, chemical engineering students are well-prepared to innovate and contribute to technological advancements.
  • Understanding stoichiometry and reaction mechanics is crucial.
  • Application of ideal gas laws for process optimization is emphasized.
  • Hands-on practice ensures familiarity with real industrial conditions and challenges.
In essence, chemical engineering education is about equipping future engineers with the skills needed to lead advancements in industry and improve efficiency and safety in chemical processes.

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Most popular questions from this chapter

A stream of hot dry nitrogen flows through a process unit that contains liquid acetone. A substantial portion of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the recovery unit at \(205^{\circ} \mathrm{C}\) and 1.1 bar and enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at \(10^{\circ} \mathrm{C}\) and 40 bar. The partial pressure of acetone in the feed to the condenser is 0.100 bar, and that in the effluent gas from the condenser is 0.379 bar. Assume ideal-gas behavior. (a) Calculate for a basis of \(1 \mathrm{m}^{3}\) of gas fed to the condenser the mass of acetone condensed ( \(\mathrm{kg}\) ) and the volume of gas leaving the condenser \(\left(\mathrm{m}^{3}\right)\) (b) Suppose the volumetric flow rate of the gas leaving the condenser is \(20.0 \mathrm{m}^{3} / \mathrm{h}\). Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.

An ideal-gas mixture contains \(35 \%\) helium, \(20 \%\) methane, and \(45 \%\) nitrogen by volume at 2.00 atm absolute and \(90^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of each component, (b) the mass fraction of methane, (c) the average molecular weight of the gas, and (d) the density of the gas in \(\mathrm{kg} / \mathrm{m}^{3}\).

A slurry contains crystals of copper sulfate pentahydrate \(\left[\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), \text { specific gravity }=2.3\right]\) suspended in an aqueous copper sulfate solution (liquid SG \(=1.2\) ). A sensitive transducer is used to measure the pressure difference, \(\Delta P(\mathrm{Pa}),\) between two points in the sample container separated by a vertical distance of \(h\) meters. The reading is in turn used to determine the mass fraction of crystals in the slurry, \(x_{\mathrm{c}}(\mathrm{kg}\) crystals/kg slurry). (a) Derive an expression for the transducer reading, \(\Delta P(\mathrm{Pa}),\) in terms of the overall slurry density, \(\rho_{\mathrm{s}}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) assuming that the equation used to calculate the pressure head in Chapter 3 \(\left(P=P_{0}+\rho g h\right)\) is valid for this two-phase system. (b) Validate the following expression relating the overall slurry density to the liquid and solid crystal densities \(\left(\rho_{1} \text { and } \rho_{c}\right)\) and the mass fraction of crystals in the slurry: $$\frac{1}{\rho_{\mathrm{si}}}=\frac{x_{\mathrm{c}}}{\rho_{\mathrm{c}}}+\frac{\left(1-x_{\mathrm{c}}\right)}{\rho_{1}}$$ (c) Suppose \(175 \mathrm{kg}\) of the slurry is placed in the sample container with \(h=0.200 \mathrm{m}\) and a transducer reading \(\Delta P=2775\) Pa is obtained. Calculate \((\mathrm{i}) \rho_{\mathrm{s}_{\mathrm{s}},(\mathrm{ii})} x_{\mathrm{c}},\) (iii) the total slurry volume, (iv) the mass of crystals in the slurry, (v) the mass of anhydrous copper sulfate (CuSO \(_{4}\) without the water of hydration) in the crystals, (vi) the mass of liquid solution, and (vii) the volume of liquid solution. (d) Prepare a spreadsheet to generate a calibration curve of \(x_{c}\) versus \(\Delta P\) for this device. Take as inputs \(\rho_{\mathrm{c}}\left(\mathrm{kg} / \mathrm{m}^{3}\right), \rho_{1}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(h(\mathrm{m}),\) and calculate \(\Delta P(\mathrm{Pa})\) for \(x_{\mathrm{c}}=0.0,0.05,0.10, \ldots, 0.60\) Run the program for the parameter values in this problem \(\left(\rho_{\mathrm{c}}=2300, \rho_{1}=1200, \text { and } h=0.200\right)\) Then plot \(x_{c}\) versus \(\Delta P\) (have the spreadsheet program do it, if possible), and verify that the value of \(x_{c}\) corresponding to \(\Delta P=2775\) Pa on the calibration curve corresponds to the value calculated in Part (c). (e) Derive the expression in Part (b). Take a basis of \(1 \mathrm{kg}\) of slurry \(\left[x_{\mathrm{c}}(\mathrm{kg}), V_{c}\left(\mathrm{m}^{3}\right)\right.\) crystals, \(\left.\left(1-x_{\mathrm{c}}\right)(\mathrm{kg}), V_{l}\left(\mathrm{m}^{3}\right) \text { liquid }\right],\) and use the fact that the volumes of the crystals and liquid are additive.

Phosgene (CCl, O) is a colorless gas that was used as an agent of chemical warfare in World War I. It has the odor of new-mown hay (which is a good warning if you know the smell of new-mown hay). Pete Brouillette, an innovative chemical engincering student, came up with what he believed was an effective new process that utilized phosgene as a starting material. He immediately set up a reactor and a system for analyzing the reaction mixture with a gas chromatograph. To calibrate the chromatograph (i.e., to determine its response to a known quantity of phosgene), he evacuated a 15.0 cm length of tubing with an outside diameter of \(0.635 \mathrm{cm}\) and a wall thickness of \(0.559 \mathrm{mm}\), and then connected the tube to the outlet valve of a cylinder containing pure phosgene. The idea was to crack the valve, fill the tube with phosgene, close the valve, feed the tube contents into the chromatograph, and observe the instrument response. What Pete hadn't thought about (among other things) was that the phosgene was stored in the cylinder at a pressure high enough for it to be a liquid. When he opened the cylinder valve, the liquid rapidly flowed into the tube and filled it. Now he was stuck with a tube full of liquid phosgene at a pressure the tube was not designed to support. Within a minute he was reminded of a tractor ride his father had once given him through a hayfield, and he knew that the phosgene was leaking. He quickly ran out of the lab, called campus security, and told them that a toxic leak had occurred, that the building had to be evacuated, and the tube removed and disposed of properly. Personnel in air masks shortly appeared, took care of the problem, and then began an investigation that is still continuing. (a) Show why one of the reasons phosgene was an effective weapon is that it would collect in low spots soldiers often mistakenly entered for protection. (b) Pete's intention was to let the tube equilibrate at room temperature ( \(23^{\circ} \mathrm{C}\) ) and atmospheric pressure. How many gram-moles of phosgene would have been contained in the sample fed to the chromatograph if his plan had worked? (c) The laboratory in which Pete was working had a volume of \(2200 \mathrm{ft}^{3}\), the specific gravity of liquid phosgene is \(1.37,\) and Pete had read somewhere that the maximum "safe" concentration of phosgene in air is \(0.1 \mathrm{ppm}\) \(\left(0.1 \times 10^{-6} \mathrm{mol} \mathrm{CCl}_{2} \mathrm{O} / \mathrm{mol}\) air) \right. Would the "safe" concentration have been exceeded if all the liquid phosgene in the tube had evaporated into the room? Even if the limit would not have been exceeded, give several reasons why the lab would still have been unsafe. (d) List several things Pete did (or failed to do) that made his experiment unnecessarily hazardous.

The ultimate analysis of a No. 4 fuel oil is 86.47 wt\% carbon, \(11.65 \%\) hydrogen, \(1.35 \%\) sulfur, and the balance noncombustible inerts. This oil is burned in a steam-generating furnace with \(15 \%\) excess air. The air is preheated to \(175^{\circ} \mathrm{C}\) and enters the furnace at a gauge pressure of \(180 \mathrm{mm}\) Hg. The sulfur and hydrogen in the fuel are completely oxidized to \(\mathrm{SO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ; 5 \%\) of the carbon is oxidized to \(\mathrm{CO}\), and the balance forms \(\mathrm{CO}_{2}\) (a) Calculate the feed ratio ( \(\mathrm{m}^{3}\) air) \(/(\mathrm{kg} \text { oil })\) (b) Calculate the mole fractions (dry basis) and ppm (parts per million on a wet basis, or moles contained in \(10^{6}\) moles of the wet stack gas) of the stack-gas species that might be considered environmental hazards.
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