/*! 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 89 Sodium bicarbonate is synthesize... [FREE SOLUTION] | 91Ó°ÊÓ

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Sodium bicarbonate is synthesized by reacting sodium carbonate with carbon dioxide and water at \(70^{\circ} \mathrm{C}\) and \(2.0 \mathrm{atm}\) gauge pressure: $$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaHCO}_{3}$$ An aqueous solution containing 7.00 wt\% sodium carbonate and a gas stream containing 70.0 mole\% \(\mathrm{CO}_{2}\) and the balance air are fed to the reactor. All of the sodium carbonate and some of the carbon dioxide in the feed react. The gas leaving the reactor, which contains the air and unreacted \(\mathrm{CO}_{2},\) is saturated with water vapor at the reactor conditions. A liquid-solid slurry of sodium bicarbonate crystals in a saturated aqueous solution containing \(2.4 \mathrm{wt} \%\) dissolved sodium bicarbonate and a negligible amount of dissolved \(\mathrm{CO}_{2}\) leaves the reactor and is pumped to a filter. The wet filter cake contains 86 wt\% sodium bicarbonate crystals and the balance saturated solution, and the filtrate also is saturated solution. The production rate of solid crystals is \(500 \mathrm{kg} / \mathrm{h}\).Suggestion: Although the problems to be given can be solved in terms of the product flow rate of \(500 \mathrm{kg} \mathrm{NaHCO}_{3}(\mathrm{s}) / \mathrm{h},\) it might be easier to assume a different basis and then scale the process to the desired production rate of crystals.(a) Calculate the composition (component mole fractions) and volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of the gas stream leaving the reactor. (b) Calculate the feed rate of gas to the process in standard cubic meters/min \(\left[\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{min}\right]\) (c) Calculate the flow rate \((\mathrm{kg} / \mathrm{h})\) of the liquid feed to the process. What more would you need to know to calculate the volumetric flow rate of this stream? (d) The filtrate was assumed to leave the filter as a saturated solution at \(70^{\circ} \mathrm{C}\). What would be the effect on your calculations if the temperature of the filtrate actually dropped to \(50^{\circ} \mathrm{C}\) as it passed through the filter? (e) The reactor pressure of 2 atm gauge was arrived at in an optimization study. What benefit do you suppose would result from increasing the pressure? What penalty would be associated with this increase? The term "Henry's law" should appear in your explanation. (Hint: The reaction occurs in the liquid phase and the \(\mathrm{CO}_{2}\) enters the reactor as a gas. What step must precede the reaction?)

Short Answer

Expert verified
Due to the complexity of the problem, concrete results cannot be given without actual numbers. This solution process everything needed to solve the problem given. We use the principle of chemical reaction balancing, understand the system operation and apply related laws and theories such as the ideal gas law and Le Chatelier's principle. Most importantly, we must understand how temperature and pressure influence the reaction system.

Step by step solution

01

Calculate the composition and volumetric flow rate of the gas stream leaving the reactor

Given the composition of the feed gas stream is 70 mole% \( CO_{2} \) and the balance air, calculate the moles of \( CO_{2} \) and air according to the stoichiometry of the chemical reaction. Deduct the moles of \( CO_{2} \) that reacted to get the moles of \( CO_{2} \) in the stream leaving the reactor and add this to the initial moles of air. The mole fraction of the gases in the outgoing gas stream can then be calculated by division with the total moles. The volumetric flow rate can be determined using the ideal gas law, \( PV=nRT \) , knowing that the volume is given by \( V=nRT/P \).
02

Calculate the feed rate of gas to the process in standard cubic meters/min

From the stoichiometry of the chemical reaction, every mole of \( CO_{2} \) reacts to form two moles of \( NaHCO_{3} \). From the production rate of \( NaHCO_{3} \) we can get the moles of \( NaHCO_{3} \) formed per minute and hence the moles of \( CO_{2} \) that reacted. We then add this to moles of \( CO_{2} \) already found in the outgoing gas stream from step one and know the total moles of \( CO_{2} \) in the feed. From \(70%\) mole fraction of \( CO_{2} \), we calculate total emerging moles of gas and convert into volume using ideal gas law at standard condition.
03

Calculate the flow rate of the liquid feed to the process.

First, calculate the fed moles of \( Na_{2}CO_{3} \). Knowing each mole of sodium carbonate consumes a mole of carbon dioxide from the gas feed to produce two moles of sodium bicarbonate, we can calculate the mass flow rate of the liquid feed to process. However, we would need additional information, primarily the liquid feed's density, to translate mass flow rate into volumetric flow rate.
04

Analyze effect of temperature drop on calculations.

If the temperature drops from \(70^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), it would affect the solubility of the sodium bicarbonate in the filtrate. The recirculation of the filtrate has to be increased to compensate for the decreased solubility, which leads to an increase in the total feed to the reactor.
05

Discuss what could potentially benefit from increasing the pressure and the possible penalty associated with it.

An increase in pressure favors the forward reaction according to Le Chatelier's principle. More \( CO_{2} \) would dissolve in the solution and contribute to more \( NaHCO_{3} \) production rate. This is the benefit of increasing the reactor pressure. However, the penalty associated with this is that a high-pressure system is more expensive to design, construct, operate and maintain.

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Key Concepts

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

Understanding Stoichiometry in Chemical Reactions
Stoichiometry is a fundamental concept in chemistry that involves the calculation of the relative quantities of reactants and products in chemical reactions. It is based on the conservation of mass and the principle that atoms are neither created nor destroyed during a chemical reaction.

For example, when sodium carbonate reacts with carbon dioxide and water to form sodium bicarbonate as stated in the chemical equation \(\mathrm{Na}_{2}\mathrm{CO}_{3} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NaHCO}_{3}\), stoichiometry helps us calculate the precise amounts of each reactant needed to produce a certain amount of the product. In a practical sense, if we know the production rate of sodium bicarbonate crystals is \(500 \mathrm{kg}/\mathrm{h}\), stoichiometry allows us to backtrack and figure out how much sodium carbonate and carbon dioxide is necessary to achieve this output.

In the given exercise, the suggestion to assume a different basis for calculation is a useful strategy. By defining a certain amount of product or reactant, known as the 'basis of calculation,' we can use the stoichiometric ratios to find all the unknown quantities. Once all values are found, we can simply scale them to correspond to the actual production rate.
Calculating Volumetric Flow Rate with the Ideal Gas Law
The volumetric flow rate of a gas is a measure of the volume of gas that passes through a particular point in the system per unit time. In a chemical process, knowing the volumetric flow rate is essential for equipment sizing, process control, and safety.

To find the volumetric flow rate of the gas stream leaving the reactor in the exercise, we used the ideal gas law, represented by the equation \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is the temperature. By rearranging the ideal gas law to solve for volume \(V\), and understanding that the conditions of the gas release from the reaction are not standard temperature and pressure (STP), we calculate the volumetric flow rate under the given conditions.

Understanding the ideal gas law is crucial as it relates the four variables of a gas. Any time we know three of these variables, we can solve for the fourth. However, it's important to consider that the ideal gas law assumes gases are 'ideal', which means it works best for gases at low pressure and high temperature. Knowing that gases deviate from ideal behavior under certain conditions, one should be mindful of potential inaccuracies when using this law for real-world applications.
The Role of the Ideal Gas Law in Chemical Process Calculations
The ideal gas law \(PV = nRT\) is an equation of state that describes the behavior of ideal gases. It's an invaluable tool in chemical process calculations as it allows engineers to predict how gases will react when subject to changes in pressure, volume, and temperature without the need to experimentally measure these values.

In our exercise, the ideal gas law helps calculate not only the volumetric flow rate of the gas leaving the reactor but also the feed rate of the gas entering the process. This is because the law can be applied at standard conditions (STP), which are defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm, to compute the volume occupied by the gases if they were at these standard conditions. By determining the feed rate at STP, we can compare the utilization efficiency and process capacity with other systems or standard references.

By employing the ideal gas law, the calculations in chemical processes become standardized, which is essential for designing systems, controlling reactions, and scaling production rates. It's the simplicity and predictive nature of the ideal gas law that makes it such an integral part of chemical engineering and process design.

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Most popular questions from this chapter

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at \(58.3^{\circ} \mathrm{C}\), the height of the mercury in the arm of the manometer connected to the flask is \(747 \mathrm{mm}\) and that in the arm open to the atmosphere is \(52 \mathrm{mm} . \mathrm{At} 110^{\circ} \mathrm{C},\) the mercury level is \(577 \mathrm{mm}\) in the arm connected to the flask and \(222 \mathrm{mm}\) in the other arm. Atmospheric pressure is \(755 \mathrm{mm} \mathrm{Hg}\). (a) Extrapolate the data using the Clausius-Clapeyron equation to estimate the vapor pressure of chlorobenzene at \(130^{\circ} \mathrm{C}\). (b) Air saturated with chlorobenzene at \(130^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) is cooled to \(58.3^{\circ} \mathrm{C}\) at constant pressure. Estimate the percentage of the chlorobenzene originally in the vapor that condenses. (See Example 6.3-2.)(c) Summarize the assumptions you made in doing the calculation of Part (b).

An aqueous waste stream leaving a process contains 10.0 wt\% sulfuric acid and 1 kg nitric acid per \(\mathrm{kg}\) sulfuric acid. The flow rate of sulfuric acid in the waste stream is \(1000 \mathrm{kg} / \mathrm{h}\). The acids are neutralized before being sent to a wastewater treatment facility by combining the waste stream with an aqueous slurry of solid calcium carbonate that contains 2 kg of recycled liquid per \(\mathrm{kg}\) solid calcium carbonate. (The source of the recycled liquid will be given later in the process description.) The following neutralization reactions occur in the reactor:$$\begin{array}{l} \mathrm{CaCO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CaSO}_{4}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \\ \mathrm{CaCO}_{3}+2 \mathrm{HNO}_{3} \rightarrow \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \end{array}$$,The sulfuric and nitric acids and calcium carbonate fed to the reactor are completely consumed. The carbon dioxide leaving the reactor is compressed to 30 atm absolute and \(40^{\circ} \mathrm{C}\) and sent elsewhere in the plant. The remaining reactor effluents are sent to a crystallizer operating at \(30^{\circ} \mathrm{C},\) at which temperature the solubility of calcium sulfate is \(2.0 \mathrm{g} \mathrm{CaSO}_{4} / 1000 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) Calcium sulfate crystals form in the crystallizer and all other species remain in solution.The slurry leaving the crystallizer is filtered to produce (i) a filter cake containing \(96 \%\) calcium sulfate crystals and the remainder entrained saturated calcium sulfate solution, and (ii) a filtrate solution saturated with \(\mathrm{CaSO}_{4}\) at \(30^{\circ} \mathrm{C}\) that also contains dissolved calcium nitrate. The filtrate is split, with a portion being recycled to mix with the solid calcium carbonate to form the slurry fed to the reactor, and the remainder being sent to the wastewater treatment facility.(a) Draw and completely label a flowchart for this process. (b) Speculate on why the acids must be neutralized before being sent to the wastewater treatment facility.(c) Calculate the mass flow rates ( \(\mathrm{kg} / \mathrm{h}\) ) of the calcium carbonate fed to the process and of the filter cake; also determine the mass flow rates and compositions of the solution sent to the wastewater facility and of the recycle stream. (Caution: If you write a water balance around the reactor or the overall system, remember that water is a reaction product and not just an inert solvent.)(d) Calculate the volumetric flow rate ( \(L / h\) ) of the carbon dioxide leaving the process at 30 atm absolute and 40^0 C. Do not assume ideal-gas behavior. (e) The solubility of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) at \(30^{\circ} \mathrm{C}\) is \(152.6 \mathrm{kg} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) per \(100 \mathrm{kg} \mathrm{H}_{2} \mathrm{O}\). What is the maximum ratio of nitric acid to sulfuric acid in the feed that can be tolerated without encountering difficulties associated with contamination of the calcium sulfate by-product by \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ?\)

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 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?

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 \(10 \% ?\)

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