/*! 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 26 (Modified \(P 8-9_{B}\) ) The ga... [FREE SOLUTION] | 91Ó°ÊÓ

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(Modified \(P 8-9_{B}\) ) The gas-phase exothermic elementary reaction $$A \stackrel{A}{\longrightarrow} B+C$$ is carried out in a moving-bed reactor. $$k=0.33 \exp \left[\frac{E_{r}}{R}\left(\frac{1}{450}-\frac{1}{T}\right)\right] \mathbf{s}^{-1} \text {. with } \frac{E_{r}}{R}=3777 \mathrm{K}$$ Heat is removed by a heat exchanger jacketing the reactor. $$\frac{U a}{\rho_{b}}=\frac{0.8 \mathrm{J}}{\mathrm{s} \cdot \mathrm{kg} \mathrm{cat} \cdot \mathrm{K}}$$ The flow rate of the coolant in the exchanger is sufficiently high that the ambient temperature is constant at \(50^{\circ} \mathrm{C}\). Pure \(\mathrm{A}\) enters the reactor at a rate of 5.42 mol/s at a concentration of \(0.27 \mathrm{mol} / \mathrm{dm}^{3}\). Both the solid catalyst and the reactant enter the reactor at a temperature of \(450 \mathrm{K}\), and the heat transfer coefficient between the catalyst and gas is virtually infinite. The heat capacity of the solid catalyst is \(100 \mathrm{J} / \mathrm{kg}\) cat/K.

Short Answer

Expert verified
In order to analyze the gas-phase exothermic elementary reaction in a moving-bed reactor and heat exchanger system, we can follow these steps: 1. Identify the given variables and parameters, including the rate constant, heat removal term, and flow rates. 2. Analyze the heat exchange process to determine the heat balance between the heat removed by the heat exchanger and the heat generated in the reactor. 3. Calculate the equilibrium temperature using an iterative method or numerical methods, such as the Newton-Raphson method, considering the given rate constant and heat balance equations. 4. Determine the reactor performance by calculating the conversion of reactant A, the reactant and product concentrations, and reactor throughput based on the calculated temperature. By following these steps, we can analyze the system and evaluate the reactor's performance, such as efficiency, conversion, and selectivity towards desired products.

Step by step solution

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1. Identify the variables and parameters#gta_content#First, identify and list the given parameters and variables in the problem. - Elementary reaction: \(A \stackrel{A}{\longrightarrow} B+C\) - Rate Constant (k): \(0.33 \exp{\left[\frac{E_r}{R}\left(\frac{1}{450}-\frac{1}{T}\right)\right]} \, s^{-1}\) with \(\frac{E_r}{R} = 3777 \, K\) - Heat Removal Term (\(\frac{Ua}{\rho_b}\)): \(\frac{0.8\, J}{s \cdot kg_{\text{cat}} \cdot K}\) - Coolant flow rate: high enough to maintain ambient temperature at \(50^{\circ}C\) - Reactant flow rate: \(5.42 \, mol/s\) - Reactant concentration: \(0.27 \, mol/dm^{3}\) - Initial temperature of solid catalyst and reactant: \(450 \, K\) - Heat capacity of the solid catalyst: \(100 \, J/(kg_{\text{cat}} \cdot K)\)

2. Analyze the heat exchange process#gta_content#In this step, we have to determine whether the heat exchanger provides enough heat to maintain a constant reaction temperature. To do this, we need to analyze the heat balance between the heat removed through the heat exchanger and the heat generated in the reactor. The heat removed by the heat exchanger (Q) can be calculated using the equation: \[Q = \frac{Ua}{\rho_b}(T_0-T)\] where \(T\) is the reaction temperature and \(T_0\) is the ambient temperature (set by the coolant). The heat generated in the reactor (H) is the product of the rate of reaction (R) and the heat of reaction (\(\Delta H_{r}\)): \[H = R \cdot \Delta H_{r}\]
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3. Calculate the equilibrium temperature#gta_content#We must find the reaction temperature T that balances the heat generated and removed. We can use the given rate constant k, the heat balance equations, and the heat removal term to solve for T. First, find the rate of reaction R, using the rate constant k in the form of Arrhenius equation: \[R = kA = 0.33 \exp{\left[\frac{E_r}{R}\left(\frac{1}{450}-\frac{1}{T}\right)\right]} A\] Now, use the heat balance equations from Step 2 to set up an equation to solve for the equilibrium temperature T: \[\frac{Ua}{\rho_b}(T_0-T) = R \cdot \Delta H_{r}\] Plugging in the rate of reaction R and the given parameters, we can solve for the equilibrium temperature T using an iterative method or numerical methods, such as the Newton-Raphson method.

4. Determine the reactor performance#gta_content#Once the equilibrium temperature T is found, we can evaluate the reactor's performance. We can calculate the conversion of reactant A, the reactant, and product concentrations, and reactor throughput. Using the calculated temperature T, we can find the rate of reaction R: \[R = 0.33 \exp{\left[\frac{E_r}{R}\left(\frac{1}{450}-\frac{1}{T}\right)\right]} A\] With R and the reactant flow rate (\(5.42 \, mol/s\)), we can find the conversion of reactant A: \[\text{Conversion} = \frac{R}{\text{Flow Rate}}\] Using the conversion and the initial reactant concentration (\(0.27 \, mol/dm^{3}\)), we can calculate the concentrations of reactants and products at the equilibrium temperature: \[[A] = (1 - \text{Conversion}) \cdot 0.27\] \[[B] = [C] = \text{Conversion} \cdot 0.27\] We can now analyze the reactor performance based on the calculated values, e.g., the efficiency, conversion, and selectivity towards desired products.

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Key Concepts

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

Elementary Reaction
In the context of chemical kinetics, an elementary reaction refers to a single-step chemical reaction where reactants transform into products in one single, indivisible step without any intermediate species. This simplification is particularly useful in engineering and chemistry because it allows us to describe the rate of the reaction with a rate law based on stoichiometric coefficients directly from the balanced equation.

For instance, in the exercise, the reaction given is \(A \longrightarrow B + C\) indicating that one molecule of substance A decomposes into molecules of substances B and C directly. Since this reaction is elementary, its rate can be directly correlated with the reactant concentration to the power of its stoichiometric coefficient, which, in this case, is one since the reaction involves the decomposing of a single A molecule. Understanding the nature of an elementary reaction is pivotal to predicting the reaction rate and designing the appropriate reactor system.
Rate Constant
The rate constant (\(k\)) in chemical kinetics is a proportionality factor that connects the reaction rate to the reactant concentrations in the rate law expression. Essentially, it provides the speed at which the reaction proceeds under certain temperature conditions, and it highly depends on the reaction's activation energy and the ambient temperature. The Arrhenius equation embodies this temperature dependency and is often used to calculate the rate constant.

In the given exercise, the expression for the rate constant is \(k=0.33 \exp{\left[\frac{E_r}{R}\left(\frac{1}{450}-\frac{1}{T}\right)\right]}\) where \(E_r/R\) represents the temperature-independent term of the activation energy divided by the gas constant. Knowledge of the rate constant at various temperatures is critical for predicting reactor behavior and calculating the rate of the elementary reaction.
Heat Exchanger
A heat exchanger is a system used to transfer heat between two or more fluids without mixing them. In the context of chemical reactors, such as the moving-bed reactor addressed in the exercise, heat exchangers are integral for controlling the reactor temperature, especially for exothermic reactions that release heat.

The exercise highlights a situation where a heat exchanger jackets the reactor, removing excess heat to maintain a steady reaction temperature. This is achieved by circulating a coolant with a high flow rate to keep the ambient temperature constant at \(50^\circ\text{C}\). Having an effective heat exchanger system like this enables the reactor to operate safely and efficiently. The parameter \(\frac{Ua}{\rho_b}\) provided in the exercise relates to the heat transfer capacity of the system, essential for determining the heat removal rate and the reactor’s ability to reach its equilibrium temperature.
Equilibrium Temperature
Equilibrium temperature is the steady state temperature where the heat generated by the chemical reaction balances with the heat removed by the cooling system, in this case, the heat exchanger. It represents a point of thermal balance and is crucial for the reactor operation since it can dictate the reaction rate and thereby the overall reactor performance.

Finding the equilibrium temperature involves setting the heat generated by the reaction equal to the heat removed by the heat exchanger, as seen in the exercise with the equation \(\frac{Ua}{\rho_b}(T_0-T) = R \cdot \Delta H_{r}\). Being able to calculate the equilibrium temperature allows engineers to design reactors that optimize reaction conditions for maximum efficiency and yield.
Reactor Performance
Reactor performance encompasses various indicators of how effectively a chemical reactor operates, such as conversion rate, selectivity, yield, and throughput. These performance metrics are vital for engineers to evaluate and optimize reactor designs for industrial chemical processes.

In our exercise, we assess the performance of a moving-bed reactor by calculating the conversion of reactant A using the established rate law and observable parameters such as reactant flow and concentration. By analyzing factors like conversion rates and maintaining proper equilibrium temperature, we can infer the efficiency of the reactor and make adjustments to optimize production. For students tackling similar problems, focusing on these core metrics will aid in understanding the implications of reactor design and operation on the overall production process.

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Most popular questions from this chapter

The elementary irreversible gas-phase catalytic reaction $$A \stackrel{k_{1}}{\longrightarrow} B$$ is carried out isothermically in a batch reactor. The catalyst deactivation follows a first-order decay law and is independent of the concentrations of both A and B. (a) Determine a general expression for catalyst activity as a function of time. (b) Make a qualitative sketch of catalyst activity as a function of time. Does \(a(t)\) ever equal zero for a first-order decay law? (c) Write out the general algorithm and derive an expression for conversion as a function of time, the reactor parameters, and the catalyst parameters. Fill in the following algorithm Mole balance Rate law Decay law Stoichiometry Combine Solve 1\. Separate 2\. Integrate $$\left[\text {Ans.}: X=1-\exp \left[-\frac{k_{1} W}{k_{d} V_{0}}\left(1-\exp \left(-k_{d} t\right)\right)\right]\right]$$ (d) Calculate the conversion and catalyst activity in the reactor after \(10 \mathrm{min}\) utes at \(300 \mathrm{K}\). (e) How would you expect your results in parts (b) and (d) to change if the reaction were run at \(400 \mathrm{K}\) ? Briefly describe the trends qualitatively. (f) Calculate the conversion and catalyst activity in the reactor after 10 minutes if the reaction were run at \(400 \mathrm{K}\) instead of \(300 \mathrm{K}\). Do your results match the predictions in part (e)? Additional information: \(C_{\mathrm{A} 0}=1 \mathrm{mol} / \mathrm{dm}^{3}\) \(V_{0}=1 \mathrm{dm}^{3}\) \(W=1 \mathrm{kg}\) \(k_{d}=0.1 \min ^{-1}\) at \(300 \mathrm{K}\) \(E_{d} / R=2000 \mathrm{K}\) \(k_{1}=0.2 \mathrm{dm}^{3} /(\mathrm{kg} \text { cat } \cdot \min )\) at \(300 \mathrm{K}\) \(E_{A} / R=500 \mathrm{K}\).

In the production of ammonia $$\mathrm{NO}+\frac{5}{2} \mathrm{H}_{2} \rightleftarrows \mathrm{H}_{2} \mathrm{O}+\mathrm{NH}_{3}$$ the following side reaction occurs: $$\mathrm{NO}+\mathrm{H}_{2} \rightleftarrows \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{N}_{2}$$ Ayen and Peters [Ind. Eng. Chem. Process Des. Dev., 1.204(1962) ] studied catalytic reaction of nitric oxide with Girdler \(G-50\) catalyst in a differential reactor at atmospheric pressure. Table \(\mathrm{P} 10-16\) shows the reaction rate of the side reaction as a function of \(P_{\mathrm{H}_{2}}\) and \(P_{\mathrm{NO}}\) at a temperature of \(375^{\circ} \mathrm{C}\) $$\begin{array}{lcc} & \text { TABLE PIO-16. } & \text { FoRMATION OF WATER } \\ \hline & & \text {Reaction Rate} \\ & & r_{\mathrm{H}_{2}} \times 10^{5}(\mathrm{g} \mathrm{mol} / \mathrm{min} \cdot \mathrm{g} \text { cat }) \\ P_{\mathrm{H}_{2}}(\mathrm{atm}) & P_{\mathrm{NO}}(\mathrm{atm}) & T=375^{\circ} \mathrm{C}, W=2.39 \mathrm{g} \\ \hline 0.00922 & 0.0500 & 1.60 \\\0.0136 & 0.0500 & 2.56 \\\0.0197 & 0.0500 & 3.27 \\\0.0280 & 0.0500 & 3.64 \\ 0.0291 & 0.0500 & 3.48 \\\0.0389 & 0.0500 & 4.46 \\\0.0485 & 0.0500 & 4.75 \\\0.0500 & 0.00918 & 1.47 \\\0.0500 & 0.0184 & 2.48 \\ 0.0500 & 0.0298 & 3.45 \\\0.0500 & 0.0378 & 4.06 \\\0.0500 & 0.0491 & 4.75 \\\\\hline\end{array}$$ The following rate laws for side reaction (2), based on various catalytic mechanisms, were suggested: $$\begin{array}{l}r_{H, 0}=\frac{k K_{\mathrm{N} 0} P_{\mathrm{NO}} P_{\mathrm{H}_{2}}}{1+K_{\mathrm{N} 0} P_{\mathrm{NO}}+K_{\mathrm{H}_{2}} P_{\mathrm{H}_{2}}} \\ r_{\mathrm{H}_{2} \mathrm{O}}=\frac{k K_{\mathrm{H}_{2}} K_{\mathrm{NO}} P_{\mathrm{NO}}}{1+K_{\mathrm{NO}} P_{\mathrm{NO}}+K_{\mathrm{H}_{2}} P_{\mathrm{H}_{2}}} \\ r_{\mathrm{H}_{2} \mathrm{O}}=\frac{k_{1} K_{\mathrm{H}_{2}} K_{\mathrm{NO}} P_{\mathrm{NO}} P_{\mathrm{H}_{2}}}{\left(1+K_{\mathrm{NO}} P_{\mathrm{KO}}+K_{\mathrm{H}_{2}} P_{\mathrm{H}_{2}}\right)^{2}} \end{array}$$ Find the parameter values of the different rate laws and determine which rate law best represents the experimental data

With the increasing demand for xylene in the petrochemical industry, the production of xylene from toluene disproportionation has gained attention in recent years [lnd. Eng. Chem. Res., 26. 1854 (1987)]. This reaction, 2 Toluene \(\longrightarrow\) Benzene \(+\) Xylene $$2 \mathrm{T} \stackrel{\text { coulyst }}{\longrightarrow} \mathrm{B}+\mathrm{X}$$ was studied over a hydrogen mordenite catalyst that decays with time. As a first approximation, assume that the catalyst follows second-order decay. $$r_{d}=k_{d} a^{2}$$and the rate law for low conversions is$$ -r_{\mathrm{T}}^{\prime}=k_{\mathrm{T}} P_{\mathrm{T}} a$$ with \(k_{\mathrm{T}}=20 \mathrm{g} \mathrm{mol} / \mathrm{h} \cdot \mathrm{kg}\) cat \(\cdot \mathrm{atm}\) and \(k_{d}=1.6 \mathrm{h}^{-1}\) at \(735 \mathrm{K}\) (a) Compare the conversion time curves in a batch reactor containing \(5 \mathrm{kg}\) cat at different initial partial pressures ( 1 atm, 10 atm. etc.). The reaction volume containing pure toluene initially is \(1 \mathrm{dm}^{3}\) and the temperature is \(735 \mathrm{K}\).

Methyl ethyl ketone (MEK) is an important industrial solvent that can be produced from the dehydrogenation of butan- 2 -ol (Bu) over a zinc oxide catalyst [Ind. Eng. Chem. Res.. 27, 2050 (1988)]: $$\mathrm{Bu} \rightarrow \mathrm{MEK}+\mathrm{H}_{2}$$ The following data giving the reaction rate for MEK were obtained in a differential reactor at \(490^{\circ} \mathrm{C}\) $$\begin{array}{lllllll}\hline P_{\mathrm{Bu}}(\mathrm{atm}) & 2 & 0.1 & 0.5 & 1 & 2 & 1 \\ P_{\mathrm{MEK}}(\mathrm{atm}) & 5 & 0 & 2 & 1 & 0 & 0 \\ P_{\mathrm{H}_{2}}(\mathrm{atm}) & 0 & 0 & 1 & 1 & 0 & 10 \\ r_{\mathrm{MEK}}^{\prime}(\mathrm{mol} / \mathrm{h} \cdot \mathrm{g} \text { cat. }) & 0.044 & 0.040 & 0.069 & 0.060 & 0.043 & 0.059 \\\\\hline\end{array}$$ (a) Suggest a rate law consistent with the experimental data. (b) Suggest a reaction mechanism and rate-limiting step consistent with the rate law. (Hint: Some spccies might be weakly adsorbed.) (c) What do you believe to he the point of this problem? (d) Plot conversion (up to \(90^{\circ}\) is and reaction rate as a function of catalyst weight for an entering molar flow rate of pure butan-2-ol of \(10 \mathrm{mol} / \mathrm{min}\) and an entering pressure \(P_{0}=10\) atm. \(W_{\max }=23 \mathrm{kg}\) (e) Write a question that requires critical thinking and then explain why your question requires critical thinking. [Hint: See Preface Section B.2.] (f) Repeat part (d) accounting for pressure drop and \(\alpha=0.03 \mathrm{kg}^{-1} .\) Plot \(y\) and \(X\) as a function of catalyst weight down the reactor.

Titanium dioxide is a wide-bandgap semiconductor that is showing promise as an insulating dielectric in VLSI capacitors and for use in solar cells. Thin films of \(\mathrm{Ti} \mathrm{O}_{2}\) are to be prepared by chemical vapor deposition from gaseous titanium tetraisopropoxide (TTIP). The overall reaction is $$\mathrm{Ti}\left(\mathrm{OC}_{3} \mathrm{H}_{7}\right)_{4} \longrightarrow \mathrm{TiO}_{2}+4 \mathrm{C}_{3} \mathrm{H}_{6}+2 \mathrm{H}_{2} \mathrm{O}$$ The reaction mechanism in a CVD reactor is believed to be [K. L. Siefering and G. L. Griffin, \(J\). Electrochem. Soc.. 137,814 (1990)] $$\begin{array}{c}\mathrm{TTIP}(\mathrm{g})+\operatorname{TTIP}(\mathrm{g}) \rightleftarrows 1+\mathrm{P}_{1} \\\\\mathrm{I}+\mathrm{S} \rightleftarrows \mathrm{I} \cdot \mathrm{S} \\ \mathrm{I} \cdot \mathrm{S} \longrightarrow \mathrm{Ti} \mathrm{O}_{2}+\mathrm{P}_{2} \end{array}$$ where I is an active intermediate and \(P_{1}\) is one set of reaction products (e.g.. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{C}_{3} \mathrm{H}_{6}\) ) and \(\mathrm{P}_{2}\) is another set. Assuming the homogeneous gas-phase reaction for TTIP is in equilibrium. derive a rate law for the deposition of \(\mathrm{TiO}_{2}\) The experimental results show that at \(200^{\circ} \mathrm{C}\) the reaction is second order at low partial pressures of TTIP and zero order at high partial pressures, while at \(300^{\circ} \mathrm{C}\) the reaction is second order in TTIP over the entire pressure range. Discuss these results in light of the rate law vou derived.

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