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Chapter 10: Problem 35

The following chemical reactions take place in a liquid-phase batch reactor of constant volume \(V\). $$\begin{aligned} &\mathrm{A} \rightarrow 2 \mathrm{B} \quad r_{1}[\mathrm{mol} \mathrm{A} \text { consumed } /(\mathrm{L} \cdot \mathrm{s})]=0.100 C_{\mathrm{A}}\\\ &\mathbf{B} \rightarrow \mathbf{C} \quad r_{2}[\mathrm{mol} \mathbf{C} \text { generated } /(\mathbf{L} \cdot \mathbf{s})]=0.200 C_{\mathrm{B}}^{2} \end{aligned}$$ where the concentrations \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) are in mol/L. The reactor is initially charged with pure \(\mathrm{A}\) at a concentration of 1.00 mol/L. (a) Write expressions for ( \(i\) ) the rate of generation of \(\mathrm{B}\) in the first reaction and (ii) the rate of consumption of \(\mathrm{B}\) in the second reaction. (If this takes you more than about 10 seconds, you're missing the point.) (b) Write mole balances on A, B, and C, convert them into expressions for \(d C_{\mathrm{A}} / d t, d C_{\mathrm{B}} / d t\), and \(d C_{\mathrm{C}} / d t,\) and provide boundary conditions. (c) Without doing any calculations, sketch on a single graph the plots you would expect to obtain of \(C_{\mathrm{A}}\) versus \(t, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t .\) Clearly show the function values at \(t=0\) and \(t \rightarrow \infty\) and the curvature (concave up, concave down, or linear) in the vicinity of \(t=0 .\) Briefly explain your reasoning. (d) Solve the equations derived in Part (b) using a differential equation- solving program. On a single graph, show plots of \(C_{\mathrm{A} \text { versust }}, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t\) from \(t=0\) to \(t=50\) s. Verify that your predictions in Part (c) were correct. If they were not, change them and revise your explanation.

Short Answer

Expert verified
The rates of the reactions are: Generation of \(B: 2*r_1 = 2* 0.100 C_A = 0.200 C_A \) Consumption of \( B: r_2 = 0.200 C_B^2 \).Mole Balances result in differential equations: \(\frac{dC_A}{dt} = -r_1 = -0.100 C_A\), \(\frac{dC_B}{dt} = 2*r_1 - r_2 = 0.200 C_A - 0.200 C_B^2\), \(\frac{dC_C}{dt} = r_2 = 0.200 C_B^2\).Boundary Conditions: Initially, \(C_A = 1mol/L\), \(C_B = 0mol/L\) and \(C_C = 0mol/L\). Predicted trends: concentration of \(A\) will decrease, concentration of \(B\) will first increase then decrease, \(C\) will keep increasing. The resulting numerical solution will provide detailed insight.

Step by step solution

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01

Writing the Rate of Reactions

The question provides the rate of reactions \(r_1\) and \(r_2\). For reaction A to B, the rate of generation of B is equal to twice the rate of consumption of A. Hence, for part (i), the rate of generation of B is \(2r_1 = 2* 0.100 C_A = 0.200 C_A \), since \(r_1\) is multiplied by 2 as one molecule of A generates 2 molecules of B. For reaction B to C, the rate of consumption of B is equal to the rate of generation of C. Hence, for part (ii), the rate of consumption of B is \(r_2 = 0.200 C_B^2 \).
02

Mole Balances, Differential Expressions & Boundary Conditions

For the second part (b), we write the mole balances. Note that reactant A is consumed in one reaction and reactant B is both generated and consumed, while substance C is only generated. Hence for A, B and C, the rates of change of concentration with time are governed by the reactions. These are given as follows: \(\frac{dC_A}{dt} = -r_1 = -0.100 C_A\), \(\frac{dC_B}{dt} = 2*r_1 - r_2 = 0.200 C_A - 0.200 C_B^2\), \(\frac{dC_C}{dt} = r_2 = 0.200 C_B^2\). The boundary condition is that initially, we only have A in the reactor, hence \(C_A = 1mol/L\), \(C_B = 0mol/L\) and \(C_C = 0mol/L\).
03

Predicting Concentrations Over Time

For part (c), to predict the change of concentrations over time, it can be considered that as time proceeds, the concentration of A will decrease (as it is consumed), the concentration of B will first increase as it is produced in the first reaction until all of A is consumed and then decrease as B is consumed in the second reaction, and the concentration of C will keep increasing as B is turned into C.
04

Solving the Differential Equations

The final part (d) is quite intricate and would typically require the use of a discrete differential equation software to solve the equations given in step 2. The graphs of these results would then validate or disprove the predictions made in step 3 and provide numerical insight into the processes underway.

Key Concepts

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

Batch Reactor
A batch reactor is a closed system where the reaction mixture is sealed inside a reactor vessel. There is no continuous input or output of material in a batch reactor.
This means that the concentrations of reactants and products change over time as the reaction progresses. Since the volume remains constant, it simplifies the calculations and analysis.
  • No inflow or outflow: Everything happens in a closed vessel.
  • Constant volume: Easy to calculate concentration changes.
  • Time-dependent reactions: Ideal for detailed kinetic studies.
Mole Balance
Mole balance is a crucial concept in chemical reaction engineering that helps keep track of the moles of reactants and products in a reaction. In a batch reactor, mole balances are used to derive equations that describe how concentrations of substances change over time.
Mole balances are applied to every species in the reaction:
  • Conservation of mass: Total mass is constant, only distribution changes.
  • Mole balance equations: Derived from reaction kinetics.
  • Account for generation and consumption: Ensures accurate modeling.
Rate of Reaction
The rate of reaction is an expression that tells us how fast a reaction is proceeding. It depends on the concentration of reactants and specific rate constants provided in the exercise.
For example, in the given reactions:
  • From A to B: The rate of generation of B, \( r_1 \), is directly proportional to the concentration of A, given by \( 0.100 C_A \).
  • From B to C: The rate of consumption of B, \( r_2 \), depends on the square of B's concentration, \( 0.200 C_B^2 \).
Understanding these rates is essential to predict how concentrations evolve over time.
Differential Equations
Differential equations are mathematical expressions that describe how a quantity changes over time. In chemical kinetics, these equations help us model the rate of concentration change for each reactant and product.
The equations derived for concentration changes are:
  • For A: \( \frac{dC_A}{dt} = -0.100 C_A \)
  • For B: \( \frac{dC_B}{dt} = 0.200 C_A - 0.200 C_B^2 \)
  • For C: \( \frac{dC_C}{dt} = 0.200 C_B^2 \)
These differential equations, along with initial conditions, allow us to solve for concentrations as a function of time using numerical methods.

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Most popular questions from this chapter

In an enzyme-catalyzed reaction with stoichiometry \(\mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}\) is consumed at a rate given by an expression of the Michaelis-Menten form: $$r_{\mathrm{A}}[\operatorname{mol} /(\mathrm{L} \cdot \mathrm{s})]=\frac{k_{1} C_{\mathrm{A}}}{1+k_{2} C_{\mathrm{A}}}$$ where \(C_{\mathrm{A}}(\operatorname{mol} / \mathrm{L})\) is the reactant concentration, and \(k_{1}\) and \(k_{2}\) depend only on temperature. (a) The reaction is carried out in an isothermal batch reactor with constant reaction mixture volume \(V\) (liters), beginning with pure \(A\) at a concentration \(C_{\mathrm{A} 0}\). Derive an expression for \(d C_{\mathrm{A}} / d t\), and provide an initial condition. Sketch a plot of \(C_{\mathrm{A}}\) versus \(t,\) labeling the value of \(C_{\mathrm{A}}\) at \(t=0\) and the asymptotic value as \(t \rightarrow \infty\) (b) Solve the differential equation of Part (a) to obtain an expression for the time required to achieve a specified concentration \(C_{\mathrm{A}}\) (c) Use the expression of Part (b) to devise a graphical method of determining \(k_{1}\) and \(k_{2}\) from data for In versus the pour plot should involve fitting a straight line and determining the two parameters \(C_{\mathrm{A}}\) (int the parting of the partating and and the conting are a contation a conting from the slope and intercept of the line. (There are several possible solutions.) Then apply your method to determine \(k_{1}\) and \(k_{2}\) for the following data taken in a 2.00 -liter reactor, beginning with A at a concentration \(C_{\mathrm{A} 0}=5.00 \mathrm{mol} / \mathrm{L}\) $$\begin{array}{|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 60.0 & 120.0 & 180.0 & 240.0 & 480.0 \\ \hline C_{\mathrm{A}}(\mathrm{mol} / \mathrm{L}) & 4.484 & 4.005 & 3.561 & 3.154 & 1.866 \\ \hline \end{array}$$

A solution containing hydrogen peroxide with a mass fraction \(x_{\mathrm{p} 0}\) \(\left(\mathrm{kg} \mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{kg} \text { solution }\right)\) is added to a storage tank at a steady rate \(\dot{m}_{0}(\mathrm{kg} / \mathrm{h})\). During this process, the liquid level reaches a corroded spot in the tank wall and a leak develops. As the filling continues, the leak rate \(\dot{m}_{1}(\mathrm{kg} / \mathrm{h})\) becomes progressively worse. Moreover, once it is in the tank the peroxide begins to decompose at a rate $$r_{\mathrm{d}}(\mathrm{kg} / \mathrm{h})=k M_{\mathrm{p}}$$ where \(M_{\mathrm{p}}(\mathrm{kg})\) is the mass of peroxide in the tank. The tank contents are well mixed, so that the peroxide concentration is the same at all positions. At a time \(t=0\) the liquid level reaches the corroded spot. Let \(M_{0}\) and \(M_{\mathrm{p} 0}\) be the total liquid mass and mass of peroxide, respectively, in the tank at \(t=0,\) and let \(M(t)\) be the total mass of liquid in the tank at any time thereafter. (a) Show that the leakage rate of hydrogen peroxide at any time is \(\dot{m}_{1} M_{\mathrm{p}} / M\) (b) Write differential balances on the total tank contents and on the peroxide in the tank, and provide initial conditions. Your solution should involve only the quantities \(\dot{m}_{0}, \dot{m}_{1}, x_{\mathrm{p} 0}, k, M, M_{0}, M_{\mathrm{p}}\) \(M_{\mathrm{p} 0},\) and \(t\)

A kettle containing 3.00 liters of water at a temperature of \(18^{\circ} \mathrm{C}\) is placed on an electric stove and begins to boil in three minutes. (a) Write an energy balance on the water and determine an expression for \(d T / d t,\) neglecting evaporation of water before the boiling point is reached, and provide an initial condition. Sketch a plot of \(T\) versus \(t\) from \(t=0\) to \(t=4\) minutes. (b) Calculate the average rate (W) at which heat is being added to the water. Then calculate the rate (g/s) at which water vaporizes once boiling begins. (c) The rate of heat output from the stove element differs significantly from the heating rate calculated in Part (b). In which direction, and why?

An iron bar \(2.00 \mathrm{cm} \times 3.00 \mathrm{cm} \times 10.0 \mathrm{cm}\) at a temperature of \(95^{\circ} \mathrm{C}\) is dropped into a barrel of water at \(25^{\circ} \mathrm{C} .\) The barrel is large enough so that the water temperature rises negligibly as the bar cools. The rate at which heat is transferred from the bar to the water is given by the expression $$\dot{Q}(\mathrm{J} / \mathrm{min})=U A\left(T_{\mathrm{b}}-T_{\mathrm{w}}\right)$$ where \(U\left[=0.050 \mathrm{J} /\left(\mathrm{min} \cdot \mathrm{cm}^{2} \cdot^{\circ} \mathrm{C}\right)\right]\) is a heat transfer coefficient, \(A\left(\mathrm{cm}^{2}\right)\) is the exposed surface area of the bar, and \(T_{\mathrm{b}}\left(^{\circ} \mathrm{C}\right)\) and \(T_{\mathrm{w}}\left(^{\circ} \mathrm{C}\right)\) are the surface temperature of the bar and the water temperature, respectively. The heat capacity of the bar is \(0.460 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right) .\) Heat conduction in iron is rapid enough for the temperature \(T_{\mathrm{b}}(t)\) to be considered uniform throughout the bar. (a) Write an energy balance on the bar, assuming that all six sides are exposed. Your result should be an expression for \(d T_{\mathrm{b}} / d t\) and an initial condition. (b) Without integrating the equation, sketch the expected plot of \(T_{\mathrm{b}}\) versus \(t\), labeling the values of \(T_{\mathrm{b}}\) at \(t=0\) and \(t \rightarrow \infty\) (c) Derive an expression for \(T_{\mathrm{b}}(t)\) and check it three ways. How long will it take for the bar to cool to \(30^{\circ} \mathrm{C} ?\)

A 2000 -liter tank initially contains 400 liters of pure water. Beginning at \(t=0\), an aqueous solution containing \(1.00 \mathrm{g} / \mathrm{L}\) of potassium chloride flows into the tank at a rate of \(8.00 \mathrm{L} / \mathrm{s}\) and an outlet stream simultaneously starts flowing at a rate of \(4.00 \mathrm{L} / \mathrm{s}\). The contents of the tank are perfectly mixed, and the densities of the feed stream and of the tank solution, \(\rho(g / L),\) may be considered equal and constant. Let \(V(t)(\mathrm{L})\) denote the volume of the tank contents and \(C(t)(\mathrm{g} / \mathrm{L})\) the concentration of potassium chloride in the tank contents and outlet stream. (a) Write a balance on total mass of the tank contents, convert it to an equation for \(d V / d t\), and provide an initial condition. Then write a potassium chloride balance, show that it reduces to $$\frac{d C}{d t}=\frac{8-8 C}{V}$$ and provide an initial condition. (Hint: You will need to use the mass balance expression in your derivation.) (b) Without solving either equation, sketch the plots you expect to obtain for \(V\) versus \(t\) and \(C\) versus \(t\) If the plot of \(C\) versus \(t\) has an asymptotic limit as \(t \rightarrow \infty,\) determine what it is and explain why it makes sense. (c) Solve the mass balance to obtain an expression for \(V(t)\). Then substitute for \(V\) in the potassium chloride balance and solve for \(C(t)\) up to the point when the tank overflows. Calculate the \(\mathrm{KCl}\) concentration in the tank at that point.
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