/*! 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 21 A gas-phase decomposition reacti... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 21

A gas-phase decomposition reaction with stoichiometry \(2 \mathrm{A} \rightarrow 2 \mathrm{B}+\mathrm{C}\) follows a second-order rate law (see Problem 10.19): $$r_{\mathrm{d}}\left[\operatorname{mol} /\left(\mathrm{m}^{3} \cdot \mathrm{s}\right)\right]=k C_{\mathrm{A}}^{2}$$ where \(C_{\mathrm{A}}\) is the reactant concentration in \(\mathrm{mol} / \mathrm{m}^{3}\). The rate constant \(k\) varies with the reaction temperature according to the Arrhenius law $$k\left[\mathrm{m}^{3} /(\mathrm{mol} \cdot \mathrm{s})\right]=k_{0} \exp (-E / R T)$$ where \(k_{0}\left[\mathrm{m}^{3} /(\mathrm{mol} \cdot \mathrm{s}]\right)=\) the preexponential factor \(E(\mathrm{J} / \mathrm{mol})=\) the reaction activation energy \(R=\) the gas constant \(T(\mathrm{K})=\) the reaction temperature (a) Suppose the reaction is carried out in a batch reactor of constant volume \(V\left(\mathrm{m}^{3}\right)\) at a constant temperature \(T(\mathrm{K}),\) beginning with pure \(\mathrm{A}\) at a concentration \(C_{\mathrm{A} 0} .\) Write a differential balance on A and integrate it to obtain an expression for \(C_{\mathrm{A}}(t)\) in terms of \(C_{\mathrm{A} 0}\) and \(k\) (b) Let \(P_{0}(\text { atm })\) be the initial reactor pressure. Prove that \(t_{1 / 2}\), the time required to achieve a \(50 \%\) conversion of \(\mathrm{A}\) in the reactor, equals \(R T / k P_{0},\) and derive an expression for \(P_{1 / 2},\) the reactor pressure at this point, in terms of \(P_{0} .\) Assume ideal- gas behavior. (c) The decomposition of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) to nitrogen and oxygen is carried out in a 5.00 -liter batch reactor at a constant temperature of \(1015 \mathrm{K},\) beginning with pure \(\mathrm{N}_{2} \mathrm{O}\) at several initial pressures. The reactor pressure \(P(t)\) is monitored, and the times \(\left(t_{1 / 2}\right)\) required to achieve \(50 \%\) conversion of \(\mathrm{N}_{2} \mathrm{O}\) are noted. $$\begin{array}{|c|c|c|c|c|} \hline P_{0}(\mathrm{atm}) & 0.135 & 0.286 & 0.416 & 0.683 \\ \hline t_{1 / 2}(\mathrm{s}) & 1060 & 500 & 344 & 209 \\ \hline \end{array}$$ Use these results to verify that the \(\mathrm{N}_{2} \mathrm{O}\) decomposition reaction is second-order and determine the value of \(k\) at \(T=1015 \mathrm{K}\) (d) The same experiment is performed at several other temperatures at a single initial pressure of 1.00 atm, with the following results: $$\begin{array}{|c|c|c|c|c|} \hline T(\mathrm{K}) & 900 & 950 & 1000 & 1050 \\ \hline t_{1 / 2}(\mathrm{s}) & 5464 & 1004 & 219 & 55 \\ \hline \end{array}$$ Use a graphical method to determine the Arrhenius law parameters ( \(k_{0}\) and \(E\) ) for the reaction. (e) Suppose the reaction is carried out in a batch reactor at \(T=980 \mathrm{K},\) beginning with a mixture at 1.20 atm containing 70 mole \(\%\) N \(_{2}\) O and the balance a chemically inert gas. How long (minutes) will it take to achieve a \(90 \%\) conversion of \(\mathrm{N}_{2} \mathrm{O} ?\)

Short Answer

Expert verified
The expression for \(C_A(t)\) in terms of \(C_{A0}\) and \(k\) after integration is \( \frac{1}{C_A} - \frac{1}{C_{A0}} = kt \). The time required to achieve 50% conversion, \(t_{1/2}\), is \( \frac{RT}{kP_0} \) and the pressure \(P_{1/2}\) at this point is \( \frac{3}{2}P_0 \). For part (c), calculate k for each data set and compare. For part (d), plot ln(k) versus 1/T and determine \(k_{0}\) and E from the intercept and slope respectively. For part (e), substituting into the integrated rate law and solving for t will give the time required to achieve 90% conversion.

Step by step solution

01

Part a: Differential Balance and Integration

Starting with the rate law \( r_d = k C_A^2 \). We know that the rate of change of A with respect to time can be written as \( -\frac{dC_A}{dt} = k C_A^2 \). We can separate variables and integrate over appropriate limits to get the expression for \( C_A(t) \) in terms of \( C_{A0} \) and \( k \). Integration leads to \( \frac{1}{C_A} - \frac{1}{C_{A0}} = kt \).
02

Part b: Conversion Time and Pressure Relations

Knowing that \( t_{1/2} \) is the time required for 50% conversion, we have \( C_A(t_{1/2}) = C_{A0}/2 \). Substituting this expression into the integrated rate law derived in part a gives \( t_{1/2} = \frac{1}{kC_{A0}} \). The initial pressure \( P_0 \) is related to the initial concentration by the ideal gas law \( P_0 = C_{A0}RT \). Substituting this into the expression for \( t_{1/2} \) gives \( t_{1/2} = \frac{RT}{kP_0} \). To find \( P_{1/2} \), the pressure at 50% conversion, note that the reaction 2A -> 2B + C increases the total moles at 50% conversion. Therefore, \( P_{1/2} = P_0(1 + \frac{1}{2}) = \frac{3}{2}P_0 \).
03

Part c: Verification of Reaction Order and Rate Constant

From the data, we have 4 values of \( P_0 \) and the corresponding \( t_{1/2} \). We can use the relation from part b \( t_{1/2} = \frac{RT}{kP_0} \) to solve for \( k \) and see if it is same for all the data points. Temperature is constant so \( k = \frac{RT}{t_{1/2}P_0} \).
04

Part d: Determination of Arrhenius law Parameters

In the Arrhenius equation \( k = k_0e^{-E/RT} \), plotting ln(k) against 1/T will give a straight line with slope equal to -E/R and intercept equal to ln(k_0). With the data of T and corresponding \( t_{1/2} \), we determined k in part c. Now we can plot ln(k) against 1/T and determine \( k_0 \) and E from the slope and intercept.
05

Part e: Calculation of Time for 90% Conversion

For a second order reaction, the time required for 90% conversion, t, is determined using the integrated rate law \( \frac{1}{C_A} - \frac{1}{C_{A0}} = kt \). We know that \( C_A = 0.1C_{A0} \) for 90% conversion and we know the value of k from part c. So, we can substitute these values into the integrated rate law to find t.

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Key Concepts

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

Second-Order Reactions
In chemical kinetics, a second-order reaction is one where the rate of reaction is proportional to the square of the concentration of one of the reactants. For the given reaction, the rate law is expressed as:
  • \[ r_d = k C_A^2 \]
This implies that as the concentration of reactant A increases or decreases, the rate of reaction changes quadratically. In the initial exercise, this kind of reaction involved the decomposition reaction described by the stoichiometry:
  • \(2A \rightarrow 2B + C\)
Here, for the differential rate equation:
  • \[-\frac{dC_A}{dt} = k C_A^2\]
we separate the variables and integrate to solve for the concentration of A over time. The result of this integration is:
  • \[ \frac{1}{C_A} - \frac{1}{C_{A0}} = kt \]
This equation expresses how the concentration of reactant A changes over time during the reaction. It also allows us to calculate the half-life \(t_{1/2}\), representing the time required for the concentration of A to reduce to half of its initial value \(C_{A0}/2\).
Understanding second-order kinetics is vital because it helps in predicting how long a reaction will take and in designing reactors for optimal performance.
Batch Reactors
A batch reactor is a closed system where no materials are added or removed during the reaction. The reaction proceeds over time until completion. This type of reactor was used in the exercise to study the decomposition of nitrous oxide.
Because batch reactors are constant volume reactors, they are quite straightforward to analyze mathematically. In the context of second-order reactions, batch reactors provide an ideal environment for observing changes in concentration over time.
  • The concentration, temperature, and pressure can be controlled precisely.
  • Batch reactors are often used for testing reaction parameters, such as the rate constant \(k\) at different temperatures.
In practice, these reactors are well-suited for reactions where the final product purity is critical, and they allow for the monitoring of reaction progress through parameters such as pressure changes, as was shown in the problem with the pressure \(P(t)\) being observed to determine the reaction kinetics. This systematic monitoring is key in determining conversion times such as \(t_{1/2}\), which is the time taken to achieve a 50% conversion of the reactant.
Arrhenius Equation
The Arrhenius Equation is a cornerstone in chemical kinetics, connecting the rate constant \(k\) with temperature \(T\). The relationship is given by:
  • \[ k = k_0 \exp\left(-\frac{E}{RT}\right) \]
The description of the equation involves:
  • \(k_0\): The pre-exponential factor, which is indicative of the frequency of collisions leading to a reaction, reflecting the reaction rate apart from all external factors.
  • \(E\): Activation energy, which is the energy barrier that the reacting molecules must overcome for the reaction to occur.
  • \(R\): Universal gas constant.
  • \(T\): Absolute temperature in Kelvin.
An important aspect tackled in the original exercise was using this equation to determine \(k_0\) and \(E\) by analyzing experimental data at various temperatures. By plotting \( \ln(k) \) against \( 1/T \), a linear relationship results, where the slope equals \(-E/R\) and the intercept equals \( \ln(k_0) \). This type of plot is commonly known as an Arrhenius plot and is crucial for understanding how reaction rates vary with temperature.
The Arrhenius Equation thus is essential in predicting how changes in operating conditions affect reaction speed, enabling chemists and engineers to optimize industrial processes and experimental setups effectively.

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Most popular questions from this chapter

A gas that contains \(\mathrm{CO}_{2}\) is contacted with liquid water in an agitated batch absorber. The equilibrium solubility of \(\mathrm{CO}_{2}\) in water is given by Henry's law (Section \(6.4 \mathrm{b}\) ) $$C_{\mathrm{A}}=p_{\mathrm{A}} / H_{\mathrm{A}}$$ where \(C_{\mathrm{A}}\left(\mathrm{mol} / \mathrm{cm}^{3}\right)=\) concentration of \(\mathrm{CO}_{2}\) in solution, \(p_{\mathrm{A}}(\mathrm{atm})=\) partial pressure of \(\mathrm{CO}_{2}\) in the gas phase, and \(H_{\mathrm{A}}\left[\mathrm{atm} /\left(\mathrm{mol} / \mathrm{cm}^{3}\right)\right]=\) Henry's law constant. The rate of absorption of \(\mathrm{CO}_{2}\) (i.e., the rate of transfer of \(\mathrm{CO}_{2}\) from the gas to the liquid per unit area of gas-liquid interface) is given by the expression $$r_{\mathrm{A}}\left[\operatorname{mol} /\left(\mathrm{cm}^{2} \cdot \mathrm{s}\right)\right]=k\left(C_{\mathrm{A}}^{*}-C_{\mathrm{A}}\right)$$ where \(C_{A}=\) actual concentration of \(\mathrm{CO}_{2}\) in the liquid, \(C_{\mathrm{A}}^{*}=\) concentration of \(\mathrm{CO}_{2}\) in the liquid that would be in equilibrium with the \(\mathrm{CO}_{2}\) in the gas phase, and \(k(\mathrm{cm} / \mathrm{s})=\) a mass transfer coefficient. The gas phase is at a total pressure \(\mathrm{P}\left(\text { atm) and contains } y_{\mathrm{A}}\left(\mathrm{mol} \mathrm{CO}_{2} / \mathrm{mol}\text { gas), and the liquid }\right.\right.\) phase initially consists of \(V\left(\mathrm{cm}^{3}\right)\) of pure water. The agitation of the liquid phase is sufficient for the composition to be considered spatially uniform, and the amount of \(\mathrm{CO}_{2}\) absorbed is low enough for \(P, V,\) and \(y_{\mathrm{A}}\) to be considered constant throughout the process. (a) Derive an expression for \(d C_{\mathrm{A}} / d t\) and provide an initial condition. Without doing any calculations, sketch a plot of \(C_{\mathrm{A}}\) versus \(t,\) labeling the value of \(C_{\mathrm{A}}\) at \(t=0\) and the asymptotic value at \(t \rightarrow \infty\) Give a physical explanation for the asymptotic value of the concentration. (b) Prove that $$C_{\mathrm{A}}(t)=\frac{p_{\mathrm{A}}}{H_{\mathrm{A}}}[1-\exp (-k S t / V)]$$ where \(S\left(\mathrm{cm}^{2}\right)\) is the effective contact area between the gas and liquid phases. (c) Suppose the system pressure is 20.0 atm, the liquid volume is 5.00 liters, the tank diameter is \(10.0 \mathrm{cm},\) the gas contains 30.0 mole \(\% \mathrm{CO}_{2},\) the Henry's law constant is \(9230 \mathrm{atm} / \mathrm{mole} / \mathrm{cm}^{3}\) ), and the mass transfer coefficient is \(0.020 \mathrm{cm} / \mathrm{s}\). Calculate the time required for \(C_{\mathrm{A}}\) to reach \(0.620 \mathrm{mol} / \mathrm{L}\) if the gas-phase properties remain essentially constant. (d) If A were not \(\mathrm{CO}_{2}\) but instead a gas with a moderately high solubility in water, the expression for \(C_{\mathrm{A}}\) given in Part (b) would be incorrect. Explain where the derivation that led to it would break down.

Methane is generated via the anaerobic decomposition (biological degradation in the absence of oxygen) of solid waste in landfills. Collecting the methane for use as a fuel rather than allowing it to disperse into the atmosphere provides a useful supplement to natural gas as an energy source. If a batch of waste with mass \(M\) (tonnes) is deposited in a landfill at \(t=0,\) the rate of methane generation at time \(t\) is given by $$\dot{V}_{\mathrm{CH}_{4}}(t)=k L_{0} M_{\text {waste }} e^{-k t}$$ where \(\dot{V}_{\mathrm{CH}_{4}}\) is the rate at which methane is generated in standard cubic meters per year, \(k\) is a rate constant, \(L_{0}\) is the total potential yield of landfill gas in standard cubic meters per tonne of waste, and \(M_{\text {watte is the tonnes of waste in the landfill at } t=0}\). (a) Starting with Equation 1, derive an expression for the mass generation rate of methane, \(\dot{M}_{\mathrm{CH}_{4}}(t)\) Without doing any calculations, sketch the shape of a plot of \(M_{\mathrm{CH}, \text { versus } t \text { from } t=0 \text { to } t=3 \mathrm{y},}\) and graphically show on the plot the total masses of methane generated in Years \(1,2,\) and \(3 .\) Then derive an expression for \(M_{\mathrm{CH}_{4}}(t),\) the total mass of methane (tonnes) generated from \(t=0\) to a time \(t\) (b) A new landfill has a yield potential \(L_{0}=100\) SCM CH \(_{4}\) /tonne waste and a rate constant \(k=0.04 \mathrm{y}^{-1} .\) At the beginning of the first year, 48,000 tonnes of waste are deposited in the landfill. Calculate the tonnes of methane generated from this deposit over a three-year period. (c) A colleague solving the problem of Part (b) calculates the methane produced in three years from the \(4.8 \times 10^{4}\) tonnes of waste as $$M_{\mathrm{CH}_{4}}(t=3)=\dot{M}_{\mathrm{CH}_{4}}(t=0) \times 1 \mathrm{y}+\dot{M}_{\mathrm{CH}_{4}}(t=1) \times 1 \mathrm{y}+\dot{M}_{\mathrm{CH}_{4}}(t=2) \times 1 \mathrm{y}$$ where \(\dot{M}_{\mathrm{CH}_{4}}\) is the first expression derived in Part (a). Briefly state what has been assumed about the rate of methane generation. Calculate the value determined with this method and the percentage error in the calculation. Show graphically what the calculated value corresponds to on another sketch of \(M_{\mathrm{CH}_{4}}\) versus \(t\) (d) The following amounts of waste are deposited in the landfill on January 1 in each of three consecutive years. Exploratory Exercises - Research and Discover (e) Explain in your own words the benefits of reducing the release of methane from landfills and of using the methane as a fuel instead of natural gas. (f) One way to avoid the environmental hazard of methane generation is to incinerate the waste before it has a chance to decompose. What problems might this alternative process introduce?

Phosgene (COCl_) is formed by CO and Cl_ reacting in the presence of activated charcoal: $$\mathrm{CO}+\mathrm{Cl}_{2} \rightarrow \mathrm{COCl}_{2}$$ At \(T=303.8 \mathrm{K}\) the rate of formation of phosgene in the presence of 1 gram of charcoal is $$R_{\mathrm{f}}(\mathrm{mol} / \mathrm{min})=\frac{8.75 C_{\mathrm{CO}} C_{\mathrm{C}_{2}}}{\left(1+58.6 C_{\mathrm{C}_{2}}+34.3 C_{\mathrm{COC}_{2}}\right)^{2}}$$ where C denotes concentration in mollL. (a) Suppose the charge to a 3.00 -liter batch reactor is \(1.00 \mathrm{g}\) of charcoal and a gas initially containing 60.0 mole\% CO and the balance \(\mathrm{Cl}_{2}\) at \(303.8 \mathrm{K}\) and 1 atm. Calculate the initial concentrations (mol/L) of both reactants, neglecting the volume occupied by the charcoal. Then, letting \(C_{\mathrm{P}}(t)\) be the concentration of phosgene at an arbitrary time \(t,\) derive relations for \(C_{\mathrm{P}}\) \(C_{\mathrm{CO}}\) and \(C_{\mathrm{C}_{2}}\) in terms of (b) Write a differential balance on phosgene and show that it simplifies to $$\frac{d C_{\mathrm{P}}}{d t}=\frac{2.92\left(0.02407-C_{\mathrm{P}}\right)\left(0.01605-C_{\mathrm{P}}\right)}{\left(1.941-24.3 C_{\mathrm{P}}\right)^{2}}$$ Provide an initial condition for this equation. (c) A plot of \(C_{\mathrm{P}}\) versus \(t\) starts at \(C_{\mathrm{P}}=0\) and asymptotically approaches a maximum value. Explain how you could predict that behavior from the form of the equation of Part (b). Without attempting to solve the differential equation, determine the maximum value of \(C_{\mathrm{P}}\) (d) Starting with the equation of Part (b), derive an expression for the time required to achieve a \(75 \%\) conversion of the limiting reactant. Your solution should have the form \(t=a\) definite integral. (e) The integral you derived in Part (d) can be evaluated analytically; however, more complex rate laws than the one given for the phosgene formation reaction would yield an integral that must be evaluated numerically. One procedure is to evaluate the integrand at a number of points between the limits of integration and to use a quadrature formula such as the trapezoidal rule or Simpson's rule (Appendix A.3) to estimate the value of the integral. Usea spreadsheet to evaluate the integrand of the integral of Part (c) at \(n_{p}\) equally spaced points between and including the limits of integration, where \(n_{p}\) is an odd number, and then to evaluate the integral using Simpson's rule. Perform the calculation for \(n_{p}=5,21,\) and \(51 .\) What can you conclude about the number of points needed to obtain a result accurate to three significant figures?

One hundred fifty kmol of an aqueous phosphoric acid solution contains 5.00 mole\% \(\mathrm{H}_{3} \mathrm{PO}_{4}\). The solution is concentrated by adding pure phosphoric acid at a rate of \(20.0 \mathrm{L} / \mathrm{min}\). (a) Write a differential mole balance on phosphoric acid and provide an initial condition. [Start by defining \(n_{\mathrm{p}}(\mathrm{kmol})\) to be the total quantity of phosphoric acid in the tank at any time.] Without solving the equation, sketch a plot of \(n_{\mathrm{p}}\) versus \(t\) and explain your reasoning. (b) Solve the balance to obtain an expression for \(n_{\mathrm{p}}(t) .\) Use the result to derive an expression for \(x_{\mathrm{p}}(t)\) the mole fraction of phosphoric acid in the solution. Without doing any numerical calculations, sketch a plot of \(x_{\mathrm{p}}\) versus \(t\) from \(t=0\) to \(t \rightarrow \infty,\) labeling the initial and asymptotic values of \(x_{\mathrm{p}}\) on the plot. Explain your reasoning. (c) How long will it take to concentrate the solution to \(15 \% \mathrm{H}_{3} \mathrm{PO}_{4} ?\)

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 \(C_{\mathrm{A} 0}=5.00 \mathrm{mol} / \mathrm{L}\) $$\begin{array}{|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 60.0 & 120.0 & 180.0 & 240.0 & 480.0 \\ \hline C_{\mathrm{A}}(\mathrm{mol} / \mathrm{L}) & 4.484 & 4.005 & 3.561 & 3.154 & 1.866 \\ \hline \end{array}$$
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