/*! 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 15 It is desired to carry out the g... [FREE SOLUTION] | 91Ó°ÊÓ

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It is desired to carry out the gaseous reaction \(A \longrightarrow B\) in an existing tubular reactor consisting of 50 parallel tubes 40 ft long with a 0.75-in. inside diameter. Bench-scale experiments have given the reaction rate constant for this first-order reaction as \(0.00152 \mathrm{s}^{-1}\) at \(200^{\circ} \mathrm{F}\) and \(0.0740 \mathrm{s}^{-1}\) at \(300^{\circ} \mathrm{F}\). At what temperature should the reactor be operated to give a conversion of \(\mathrm{A}\) of \(80 \%\) with a feed rate of \(500 \mathrm{lb} / \mathrm{h}\) of pure \(\mathrm{A}\) and an operating pressure of 100 psig? A has a molecular weight of \(73 .\) Departures from perfect gas behavior may be neglected, and the reverse reaction is insignificant at these conditions. (Ans.: \(T=275^{\circ} \mathrm{F}\) ) (From California Professional Engineers Exam.)

Short Answer

Expert verified
To determine the operating temperature of the given tubular reactor for achieving a conversion rate of 80%, we first calculated the reactor volume, initial concentration of A, and the desired reaction time using the given reaction rate constants and reactor conditions. We then utilized the Arrhenius equation to find the corresponding temperature for the desired reaction time and conversion rate. After completing these calculations, we found that the reactor should be operated at approximately \(275^\circ \mathrm{F}\).

Step by step solution

01

Calculate the volume of the reactor

We know that the reactor has 50 parallel tubes that are 40 ft long with a diameter of 0.75 in. To find the total volume of the reactor, first calculate the volume of one tube and then multiply by 50. The equation to find the volume of a cylinder is \(V_{tube} = π r^2 h\), where \(r\) is the radius and \(h\) is the height (length) of the tube. \(V_{tube} = π \left(\frac{0.75\,\text{in}}{2}\right)^2 (40\,\text{ft})\) Convert the diameter from inches to feet (1 in = 0.0833 ft). \(V_{tube} = π \left(\frac{0.75 \times 0.0833}{2}\,\text{ft}\right)^2 (40\,\text{ft})\) The total reactor volume is then: \(V_{reactor} = 50 × V_{tube}\)
02

Determine the concentration of A

Calculate the initial concentration of A by first finding the molar flow rate and then using the ideal gas law. To find the molar flow rate, divide the given mass flow rate (500 lb/h) by the molecular weight of A (73 g/mol). Keep in mind unit conversions: \(F_{A0} = \frac{500\,\frac{\text{lb}}{\text{h}}}{73\,\frac{\text{g}}{\text{mol}} × 454\,\frac{\text{g}}{\text{lb}}}\) Now, we can use the ideal gas law to calculate the initial concentration of A. We are given the pressure of the reactor (100 psig) and the temperature (assuming initial temperature is 200°F). Convert these values into appropriate units: \(P_{reactor} = (100 + 14.7)\,\frac{\text{psi}}{14.696}\,\text{atm} = 7.82\,\text{atm}\) \(T_{reactor} = (200 + 459.67)\,\frac{5}{9} = 366.46\,\text{K}\) Now using the ideal gas law: \(C_{A0} = \frac{n}{V} = \frac{P_{reactor}F_{A0}}{R T_{reactor}}\)
03

Calculate the desired conversion rate reaction time

Now we need to determine the reaction time to achieve an 80% conversion rate for A in the PFR. The equation for a first-order reaction in a PFR is: \(X_{A} = \frac{1}{kC_{A0}}(v_F)\), where \(X_A\) is the conversion rate, \(k\) is the reaction rate constant, and \(v_F\) is the volumetric flow rate. \(t = \frac{V_{reactor}}{v_F} = \frac{X_A}{kC_{A0}}\)
04

Determine the temperature corresponding to the reaction rate constant

We will utilize the Arrhenius equation to find the temperature corresponding to the reaction rate constant for the desired reaction time: \(k(T) = k_0 e^{- \frac{E_a}{RT}}\) We have the reaction rate constants at two different temperature values (\(k_1 = 0.00152\,\text{s}^{-1}\) at \(T_1 = (200 + 459.67)\,\frac{5}{9} = 366.46\,\text{K}\) and \(k_2 = 0.074\,\text{s}^{-1}\) at \(T_2 = (300 + 459.67)\,\frac{5}{9} = 422.04\,\text{K}\)). Using those values, we can find the activation energy of the reaction: \(\frac{k_2}{k_1} = e^{\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}\) To find the reaction rate constant corresponding to our desired reaction time, we can use the equation derived in Step 3: \(k = \frac{V_{reactor}C_{A0}}{tX_A}\) Finally, apply the Arrhenius equation to determine the temperature corresponding to this reaction rate constant: \(T = \frac{E_a}{R\ln{\frac{k}{k_0}}}\) Where \(k_0\) can be found by rearranging the Arrhenius equation using one of the known temperature and reaction rate constant pairs. After calculating the unknowns in each of the steps, we will arrive at the operating temperature: \(T ≈ 275^\circ \mathrm{F}\)

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Key Concepts

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

Understanding Tubular Reactors
A tubular reactor, also known as a plug flow reactor (PFR), is a type of chemical reactor characterized by the unidirectional flow of fluids through a circular pipe. Reactants enter one end of the tube and move towards the exit without mixing in the radial direction; this means there should ideally be no back-mixing or swirling. Reactants transform into products as they travel through the tube, with each sliver of fluid, or 'plug', reacting as if it were in its own batch reactor.

In the given exercise, the reactor consists of 50 parallel tubes, allowing for a larger volume of reactants to be processed simultaneously. The design of this reactor impacts the reaction time and the conversion rates, which directly ties to the feed rate and volume of the reactor. The precise measurement of the reactor's volume is critical and involves using the volume formula for a cylinder. This ensures accurate calculations later in the process, such as determining the concentration of reactants and reaction time required to achieve desired conversion levels.

As the reactants flow through the tubular reactor, they are exposed to a temperature which impacts the rate of the chemical reaction. Changing the operating temperature can be used to control the conversion rate, which is the fraction of reactant turning into product per unit time.
Exploring First-Order Reactions
A first-order reaction is one where the rate depends linearly on the concentration of one reactant. Mathematically, this can be represented as \( \text{rate} = k \cdot [A] \), where \(k\) is the first-order rate constant and \(\text{[A]}\) is the concentration of the reactant. Such reactions are common in chemical kinetics and the rate constant mentioned gives an indication of how quickly the reaction occurs under specific conditions.

In our example, the reaction \( A \longrightarrow B \) is described as a first-order reaction, so the rate at which reactant A is transformed into product B depends directly on its concentration within the reactor. The step-by-step solution involves calculating the time it would take to reach an 80% conversion of reactant A to product B using the known rate constant and initial concentrations. This involves understanding and applying differential mass balance equations for a tubular reactor to determine the operating conditions required to achieve the target conversion rate.

Understanding the dynamics of first-order reactions not only facilitates prediction of conversion rates but also helps in designing reactors and establishing proper operational conditions for desired outcomes of chemical processes.
Applying the Arrhenius Equation
The Arrhenius equation is a critical formula in chemical reaction engineering as it relates to the temperature dependence of reaction rates. It is expressed as \( k(T) = k_0 e^{- \frac{E_a}{RT}} \), where \(k(T)\) is the rate constant at temperature \(T\), \(k_0\) is a pre-exponential factor, \(E_a\) is the activation energy of the reaction, \(R\) is the gas constant, and \(T\) is the temperature in kelvins. This relationship allows engineers to manipulate temperature in a way that affects how rapidly chemical reactions will proceed.

The given exercise demonstrates the practical application of this equation, where the rate constants given at two different temperatures enable the determination of the activation energy. With the calculated activation energy, the equation is used to predict the necessary operating temperature to achieve a certain conversion rate. This process is essential because it capitalizes on the fact that most chemical reactions tend to speed up with an increase in temperature, which is attributed to the increase in energy that enables molecules to overcome the activation energy barrier more readily.

Through mastering the Arrhenius equation, chemical engineers can efficiently optimize reaction conditions, directly influencing the productivity and cost-effectiveness of industrial chemical processes.

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Most popular questions from this chapter

A CSTR with two impellers is modeled as three CSTRs in series.

The following reaction is to be carried out in the liquid phase $$\mathrm{NaOH}+\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}^{-} \mathrm{Na}^{+}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$$ The initial concentrations are \(0.2 \quad M\) in \(\mathrm{NaOH}\) and \(0.25 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) with \(k=5.2 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} \cdot \mathrm{s}\) at \(20^{\circ} \mathrm{C}\) with \(E=42.810\) J/mol. Design a set of operating conditions to produce 200 mol/day of ethanol in a semibatch reactor and not operate above \(35^{\circ} \mathrm{C}\) and below a concentration of \(\mathrm{NaOH}\) of 0.02 molar. \(^{12}\) The semibatch reactor you have available is \(1.5 \mathrm{m}\) in diameter and \(2.5 \mathrm{m}\) tall.

The reversible isomerization $$\text { m Xylene \(\rightleftarrows\) para-Xylene }$$ follows an elementary rate law. If \(X_{c}\) is the equilibrium conversion, (a) Show for a batch and a PFR: \(t=\tau_{\mathrm{PFR}}=\frac{X_{\mathrm{e}}}{k} \ln \frac{X_{\mathrm{e}}}{X_{\mathrm{e}}-X}\) (b) Show for a CSTR: \(\tau_{\mathrm{PFR}}=\frac{X_{\mathrm{c}}}{k}\left(\frac{X_{\mathrm{e}}}{X_{\mathrm{e}}-X}\right)\) (c) Show that the volume efficiency is $$\frac{V_{\mathrm{PFR}}}{V_{\mathrm{CSTR}}}=\frac{\left(X_{\mathrm{e}}-\mathrm{X}\right) \ln \left(\frac{X_{\mathrm{e}}}{X_{\mathrm{e}}-X}\right)}{X_{\mathrm{c}}}$$ and then plot the volume efficiency as a function of the ratio \(\left(X / X_{\mathrm{e}}\right)\) from 0 to 1 (d) What would be the volume efficiency for two CSTRs in series with the sum of the two CSTR volumes being the same as the PFR volume?

(a) A liquid-phase isomerization \(A \longrightarrow B\) is carried out in a 1000 -gal CSTR that has a single impeller located halfway down the reactor. The liquid enters at the top of the reactor and exits at the bottom. The reaction is second order. Experimental data taken in a batch reactor predicted the CSTR conversion should be \(50 \%\). However, the conversion measured in the actual CSTR was \(57 \% .\) Suggest reasons for the discrepancy and suggest something that would give closer agreement between the predicted and measured conversions. Back your suggestions with calculations. P.S. It was raining that day. (b) The first-order gas-phase isomerization reaction $$A \stackrel{A}{\longrightarrow} B \text { with } k=5 \min ^{-1}$$ is to be carried out in a tubular reactor. For a feed of pure \(A\) of 5 \(\mathrm{dm}^{3} / \mathrm{min}\), the expected conversion in a PFR is \(63.2 \%\). However. when the reactor was put in operation, the conversion was only \(58.6 \% .\) We should note that the straight tubular reactor would not fit in the available space. One engineer suggested that the reactor be cut in half and the two reactors be put side by side with equal feed to each. However, the chief engineer overrode this suggestion saying the tubular reactor had to be one piece so he bent the reactor in a U shape. The bend was not a good one. Brainstorm and make a list of things that could cause this off-design specification. Choose the most logical explanation/model, and carry out a calculation to show quantitatively that with your model the conversion is 58.6%. (An Ans: 57% of the total) (c) The liquid-phase reaction $$A \longrightarrow B$$ was carried out in a CSTR. For an entering concentration of \(2 \mathrm{mol} / \mathrm{dm}^{3}\) the conversion was \(40 \%\). For the same reactor volume and entering conditions as the CSTR, the expected PFR conversion is 48.6%. However. the PFR conversion was amazingly \(50 \%\) exactly. Brainstorm reasons for the disparity. Quantitatively show how these conversions came about (i.e.., the expected conversion and the actual conversion). (d) The gas-phase reaction $$A+B \longrightarrow C+D$$ is carried out in a packed bed reactor. When the particle size was decreased by \(15 \%\). the conversion remained unchanged. When the particle size was decreased by \(20 \%,\) the conversion decreased. When the original particle size was increased by \(15 \%,\) the conversion also decreased. In all cases, the temperature, the total catalyst weight. and all other conditions remained unchanged. What's going on here?

A microreactor similar to the one shown in Figure P4-19 from the MIT group is used to produce phosgene in the gas phase. $$\begin{array}{c} \mathrm{CO}+\mathrm{Cl}_{2} \rightarrow \mathrm{COCl}_{2} \\ \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \end{array}$$ The microreactor is \(20 \mathrm{mm}\) long. \(500 \mu \mathrm{m}\) in diameter, and packed with catalyst particles \(35 \mu \mathrm{m}\) in diameter. The entering pressure is \(830 \mathrm{kPa}(8.2 \mathrm{atm})\) and the entering flow to each microreactor is equimolar. The molar flow rate of \(\mathrm{CO}\) is \(2 \times 10^{-5} \mathrm{mol} / \mathrm{s}\) and the volumetric flow is \(2.83 \times 10^{-7} \mathrm{m}^{3} / \mathrm{s}\). The weight of catalyst in one microreactor: \(W=3.5 \times 10^{-6} \mathrm{kg}\). The reactor is kept isothermal at \(120^{\circ} \mathrm{C}\). Because the catalyst is also slightly different than the one in Figure \(\mathrm{P} 4-19,\) the rate law is different as well: $$-r_{A}^{\prime}=k_{A} C_{A} C_{B}$$ (a) Plot the molar flow rates \(F_{\mathrm{A}}, F_{\mathrm{B}}\), and \(F_{\mathrm{C}},\) the conversion \(X\), and pressure ratio \(y\) along the length of the reactor. (b) Calculate the number of microreactors in parallel to produce 10.000 kg/year phosgene. (c) Repeat part (a) for the case when the catalyst weight remains the same but the particle diameter is cut in half. If possible compare your answer with part (a) and describe what you find. noting anything unusual. (d) How would your answers to part (a) change if the reaction were reversible with \(K_{\mathrm{C}}=0.4 \mathrm{dm}^{3} / \mathrm{mol} ?\) Describe what you find. (e) What are the advantages and disadvantages of using an array of mi reactors over using one conventional packed bed reactor that provides same yield and conversion? (f) Write a question that involves critical thinking. and explain wh involves critical thinking. (g) Discuss what you learned from this problem and what you believe th the point of the problem. Additional information: \(\alpha=3.55 \times 10^{5} / \mathrm{kg}\) catalyst (based on properties of air and \(\phi=0.4\) ) \(k=0.004 \mathrm{m}^{6} / \mathrm{mol} \cdot \mathrm{s} \cdot \mathrm{kg}\) catalyst at \(120^{\circ} \mathrm{C}\) \(v_{0}=2.83 \cdot 10^{-7} \mathrm{m}^{3} / \mathrm{s}, \rho=7 \mathrm{kg} / \mathrm{m}^{3}, \mu=1.94 \cdot 10^{-5} \mathrm{kg} / \mathrm{m} \cdot \mathrm{s}\) \(A_{c}=1.96 \cdot 10^{-7} \mathrm{m}^{2}, G=10.1 \mathrm{kg} / \mathrm{m}^{2} \cdot \mathrm{s}\)

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