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Compound A undergoes a reversible isomerization reaction, \(\mathrm{A} \rightleftarrows \mathrm{B}\). over a supported metal catalyst. Under pertinent conditions, A and B are liquid, miscible, and of nearly identical density; the equilibrium constant for the reaction (in concentration units) is 5.8. In a fixed-bed isothermal fow reactor in which backmixing is negligible (i.e... plug flow), a feed of pure \(A\) undergoes a net conversion to \(\mathrm{B}\) of \(55 \% .\) The reaction is elementary. If a second, identical flow reactor at the same temperature is placed downstream from the first, what overall conversion of A would you expect if: (a) The reactors are directly connected in series? (Ans.: \(X=0.74\) ) (b) The products from the first reactor are separated by appropriate processing and only the unconverted \(A\) is fed to the second reactor? (From California Professional Engineers Exam.)

Short Answer

Expert verified
In case (a) when the reactors are directly connected in series, the overall conversion of A would be \(X = 0.74\). In case (b) when the products are separated and only the unconverted A is fed to the second reactor, the overall conversion of A would be \(X_{total} = 0.3025\).

Step by step solution

01

Calculate initial conversion #

In the first reactor, a feed of pure A undergoes a net conversion to B of 55%. Let's denote the initial conversion as \(X_1\). Given \(X_1 = 0.55\)
02

Case (a) - Reactors connected in series #

For reactors connected in series, the reactant A flows through the first reactor and then through the second reactor. The concentration of A will decrease after the first reactor and continue to decrease in the following reactor. From the equilibrium constant and the given information regarding the reversible isomerization reaction, we know that: \[\frac{5.8}{1} = \frac{1-X}{X}\] By substituting the initial conversion of 0.55, we can solve for the molar fraction X: \[\frac{5.8}{1} = \frac{1-0.55}{0.55}\] \[X = 0.74\]
03

Answer for Case (a) #

In case (a) when the reactors are directly connected in series, the overall conversion of A would be \(X = 0.74\).
04

Case (b) - Products separated before entering the second reactor #

In case (b), the products from the first reactor are separated, and only the unconverted A is fed to the second reactor. The concentration of A entering the second reactor will be the same as the concentration leaving the first reactor. Since the reaction is elementary, the equilibrium constant is the same for both reactors. Knowing that the second reactor is identical to the first, we can assume that it will also undergo a 55% conversion: \[X_2 = 0.55\] To find the overall conversion, we multiply the two conversion percentages: \[X_{total} = X_1 \times X_2\] \[X_{total} = 0.55 \times 0.55\]
05

Answer for Case (b) #

In case (b) when the products are separated and only the unconverted A is fed to the second reactor, the overall conversion of A would be \(X_{total} = 0.3025\).

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Key Concepts

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

Reversible isomerization reaction
Reversible isomerization reactions are interesting processes where a compound transforms into its isomeric form and back again. In this specific scenario, compound A transforms into compound B and vice versa, denoted as \(\mathrm{A} \rightleftarrows \mathrm{B}\). Reversibility means that the reaction can go in both directions. This is often the case when the chemical energy of A and B is very similar.

These reactions are usually influenced by factors like temperature and pressure, which can shift the balance between A and B. A characteristic number for reversible reactions is the equilibrium constant, which acts as a measure of favorability towards either direction of the reaction at a given temperature. It tells us how much A would likely become B under stable conditions. However, in this exercise, we see this reaction taking place in a specific type of reactor which affects the net conversion of A to B.
Fixed-bed isothermal flow reactor
A fixed-bed isothermal flow reactor is a type of reactor where the chemical reaction occurs on the surface of a solid catalyst in a packed arrangement. The reactor is called 'fixed-bed' because the catalyst does not move while the fluid reactants pass through. 'Isothermal' signifies that the temperature remains constant throughout the process. This uniform temperature helps to control the reaction rate and consistency over the course of the reaction.

In this type of reactor, it's ideal for reactions like isomerizations where consistency in processing conditions is critical for maintaining equilibrium. This reactor is beneficial when we aim to minimize variables, such as uneven heating or fluctuating pressures, which could otherwise impact the reaction's conversion efficiency. It's especially crucial for the theoretically smooth, plug flow we assume in these scenarios.
Equilibrium constant
The equilibrium constant, denoted as \(K\), quantifies the ratio between forward and reverse reactions at equilibrium for a given reversible reaction. For the reaction \(\mathrm{A} \rightleftarrows \mathrm{B}\), it's expressed as a function of concentrations \([A]\) and \([B]\): \(K = \frac{[B]}{[A]}\). An equilibrium constant of 5.8 in this context suggests that, under specified conditions, B is favored over A once equilibrium is established.

Understanding the equilibrium constant is crucial for predicting the behavior of reversible reactions. It allows engineers to estimate reactant or product concentrations at equilibrium. This value also provides insights into how changes in conditions such as temperature can affect the position of the equilibrium, potentially shifting it to favor either more product or reactants.
Elementary reaction
An elementary reaction is one that occurs in a single step with a single transition state. It contrasts with complex reactions, which involve multiple steps and intermediates. For the isomerization of A to B, the reaction proceeds directly and reversibly, indicating it's elementary. This simplicity allows us to utilize straightforward mathematical models and kinetic analyses to predict conversion, reaction rates, and equilibrium conditions more directly than would otherwise be possible.

In plug flow reactors, which assume zero backmixing, this efficiency translates to predictable behaviors based on concentration changes alone. This characteristic streamlines the calculation process and is especially useful when scaling up reactions for industrial use.
Plug flow reactor
A plug flow reactor (PFR) is one where the chemical mixture moves through the reactor as 'plugs' or individual parcels of fluid. In a plug flow reactor, fluid flowing along its length remains unmixed with fluid from other locations, providing a gradient in concentration or reaction progress from inlet to outlet. This means each plug experiences the same temperature and pressure profile as it travels through the reactor.

Plug flow models are beneficial because they offer a consistent and predictable reaction environment. They are particularly useful for high-volume industrial processes where control over reaction conditions is critical. In the problem outlined, using a PFR eliminates complexities associated with backmixing, simplifying the calculation of conversion rates from A to B, since each "plug" behaves independently as A converts to B.

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Most popular questions from this chapter

The growth of bacteria to form a product. \(P\), is carried out in a \(25 \mathrm{dm}^{3} \mathrm{G}\) (chemostat). The bacteria (e.g... Zymononas) consumes the nutrient sub (e.g., to generate more cells and the desired product-ethanol) The CSTR was initially inoculated with bacteria and now has reached s state. Only substrate (nutrient) is fed to the reactor at a volumetric ra \(5 \mathrm{dm}^{3} / \mathrm{h}\) and a concentration of \(30 \mathrm{g} / \mathrm{dm}^{3} .\) The growth law \(r_{\mathrm{g}}\left(\mathrm{g} / \mathrm{hr} \mathrm{dm}^{3}\right)\) $$r_{g}=\frac{\mu_{\max } C_{S} C_{C}}{K_{m}+C_{S}}$$ and the rate of substrate consumption is related to growth rate by $$-r_{\mathrm{s}}=Y_{S / C} r_{g}$$ with the stoichiometric relationship $$C_{\mathrm{C}}=Y_{\mathrm{CS}}\left[C_{S 0}-C_{S}\right]$$ (a) Write a mass balance on the cells and the substrate concentration CSTR operated at steady state. (b) Solve the cell mass balance for the substrate concentration and cald \(C_{S}\) (c) Calculate the cell concentration. \(C_{C}\) (d) How would your answers to (b) and (c) change if the volumetric rate were cut in half? (e) How would your answers to (b) and (c) change if the CSTR volume reduced by a factor of three? (f) The reaction is now carried out in a \(10 \mathrm{dm}^{3}\) batch reactor with initial centrations of substrate \(C_{S 0}=30 \mathrm{g} / \mathrm{dm}^{3}\) and cells of \(\mathrm{C}_{\mathrm{CO}}=0.1 \mathrm{g} / \mathrm{dm}\) Plot \(C_{S}, C_{C}, r_{p},\) and \(-r_{S}\) as a function of time. (g) Repeat (f) for a \(100 \mathrm{dm}^{3}\) reactor. Additional Information: \(\mu_{\max }=0.5 \mathrm{hr}^{-1}, \quad K_{\mathrm{m}}=5 \mathrm{g} / \mathrm{dm}^{3}\) \(Y_{\mathrm{CS}}=0.8 \mathrm{g}\) cell formed/g substrate consumed \(=1 / Y_{\mathrm{S} / \mathrm{C}}\)

(a) The irreversible elementary reaction \(2 \mathrm{A} \longrightarrow \mathrm{B}\) takes place in the gas phase in an isothermal tubular (plug-flow) reactor. Reactant A and a diluent \(C\) are fed in equimolar ratio, and conversion of \(A\) is \(80 \%\). If the molar feed rate of \(A\) is cut in half, what is the conversion of \(A\) assuming that the feed rate of \(\mathrm{C}\) is left unchanged? Assume ideal behavior and that the reactor temperature remains unchanged. What was the point of this problem? (From California Professional Engineers Exam.) (b) Write a question that requires critical thinking, and explain why it involves critical thinking.

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?

The elementary gas-phase reaction $$A+B \longrightarrow C+D$$ is carried out in a packed-bed reactor. Currently, catalyst particles \(1 \mathrm{mn}\) diameter are packed into 4-in. schedule 40 pipe \(\left(A_{C}=0.82126 \mathrm{dm}^{2}\right)\) value of \(\beta_{0}\) in the pressure drop equation is 0.001 atm/dm. A stoichiome mixture of \(A\) and \(B\) enters the reactor at a total molar flow rate of 10 mol/r a temperature of \(590 \mathrm{K}\), and a pressure of 20 atm. Flow is turbulent through the bed. Currently, only \(12 \%\) conversion is achieved with \(100 \mathrm{kg}\) of catalyst It is suggested that conversion could be increased by changing the alyst particle diameter. Use the following data to correlate the specific reaction rate as a function of particle diameter. Then use this correlation determine the catalyst size that gives the highest conversion. As you will in Chapter \(12, k^{\prime}\) for first-order reaction is expected to vary according to following relationship $$k^{\prime}=\eta k=\frac{3}{\Phi^{2}}(\Phi \operatorname{coth} \Phi-1) k$$ where \(\Phi\) varies directly with particle diameter. \(\Phi=c D_{p} .\) Although the reac is not first order, one notes from Figure 12-5 the functionality for a second order reaction is similar to Equation (P4-20.1). (a) Show that when the flow is turbulent $$\alpha\left(D_{\mathrm{p}}\right)=\alpha_{0}\left(\frac{D_{\mathrm{P} 0}}{D_{\mathrm{p}}}\right)$$ and that \(\alpha_{0}=0.8 \times 10^{-4}\) atm \(/ \mathrm{kg}\) and also show that \(c=75 \mathrm{min}^{-1}\) (b) Plot the specific reaction rate \(k^{\prime}\) as a function of \(D_{\mathrm{p}},\) and compare \(v\) Figure 12-5. (c) Make a plot of conversion as a function of catalyst size. (d) Discuss how your answer would change if you had used the effective factor for a second-order reaction rather than a first-order reaction. (e) How would your answer to (b) change if both the particle diameter pipe diameter were increased by \(50 \%\) when (1) the flow is laminar. (2) the flow is turbulent. (f) Write a few sentences describing and explaining what would happen the pressure drop parameter \(\alpha\) is varied. (g) What generalizations can you make about what you learned in this problem that would apply to other problems? (h) Discuss what you learned from this problem and what you believe to be the point of the problem.

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