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Magazine Chapter 4: Problem 11
The gaseous reaction \(A \longrightarrow B\) has a unimolecular reaction rate constant of \(0.0015 \mathrm{min}^{-1}\) at \(80^{\circ} \mathrm{F}\). This reaction is to be carried out in parallel tubes \(10 \mathrm{ft}\) long and 1 in. inside diameter under a pressure of 132 psig at \(260^{\circ} \mathrm{F}\). A production rate of \(1000 \mathrm{lb} / \mathrm{h}\) of \(\mathrm{B}\) is required. Assuming an activation energy of 25,000 cal/mol. how many tubes are needed if the conversion of \(A\) is to be \(90 \% ?\) Assume perfect gas laws. A and \(\mathrm{B}\) each have molecular weights of 58 . (From California Professional Engineers Exam.)
Short Answer
Expert verified
The number of tubes needed to achieve a production rate of 1000 lb/h of compound B with a 90% conversion of compound A is approximately 29 tubes.
Step by step solution
01
鈧�. Calculate the reaction rate constant at the given temperature
First, we need to find the reaction rate constant at \(260^{\circ} \mathrm{F}\). We can use the Arrhenius equation to calculate this:
\[k_{2} = k_{1} \cdot \exp{\left(-\frac{E_{a}}{R}\left(\frac{1}{T_{2}} - \frac{1}{T_{1}}\right)\right)}\]
where:
- \(k_{1}\) is the initial reaction rate constant given (\(0.0015 \mathrm{min}^{-1}\)),
- \(k_{2}\) is the reaction rate constant at the desired temperature,
- \(E_{a}\) is the activation energy (25,000 cal/mol),
- \(R\) is the gas constant (1.987 cal/mol K),
- \(T_{1}\) is the initial temperature (\(80^{\circ} \mathrm{F}\), converted to K),
- and \(T_{2}\) is the desired temperature (\(260^{\circ} \mathrm{F}\), converted to K).
First, we convert the given temperatures to Kelvin:
\[T_1 = 80^{\circ} \mathrm{F} + 459.67 \, (\mathrm{R}) = (80 + 459.67) \frac{5}{9} \; (\mathrm{K}) = 301.11 \; (\mathrm{K})\]
\[T_2 = 260^{\circ} \mathrm{F}+ 459.67 \, (\mathrm{R}) = (260 + 459.67) \frac{5}{9} \; (\mathrm{K}) = 394.26 \; (\mathrm{K})\]
Now we can plug these values into the Arrhenius equation and calculate \(k_{2}\):
\[k_2 = 0.0015 \cdot \exp{\left( -\frac{25000}{1.987} \left( \frac{1}{394.26} - \frac{1}{301.11} \right) \right)} = 0.0052 \mathrm{min}^{-1}\]
02
鈧�. Determine the time required for a 90% conversion
Since the reaction is first-order, we can use the integrated rate law equation:
\[C_{A} = C_{A0} \cdot \exp{(-kt)}\]
Solving for \(t\), we get:
\[t = -\frac{1}{k}\ln{\frac{C_{A}}{C_{A0}}}\]
We have:
- \(C_{A0}\) is the initial concentration of A,
- \(C_{A}\) is the final concentration of A,
- \(k\) is the rate constant at the desired temperature (0.0052 min\(^{-1}\)),
- and \(t\) is the time required for the conversion.
For a 90% conversion, we can write:
\[t = -\frac{1}{0.0052 \, \mathrm{min}^{-1}} \ln{\frac{0.10 \cdot C_{A0}}{C_{A0}}}\]
\[t = \frac{1}{0.0052 \, \mathrm{min}^{-1}} \biggl(\ln{10}\biggr) \approx 49.79\,(\mathrm{min})\]
03
鈧�. Calculate the required flowrate
We need the volumetric flow rate to determine the quantity of gaseous A to be processed.
\[Q = \frac{m_{B}}{\rho_i \cdot (1-X)}, \]
where:
- \(Q\) is the volumetric flow rate,
- \(m_{B}\) is the required mass flow rate of B (1000 lb/h),
- \(\rho_i\) is the initial density of A,
- and \(X\) is the conversion (0.90).
First, we need to find the initial density \(\rho_i\). It can be calculated using the ideal gas law:
\[\rho_i = \frac{PM_i}{RT}\]
where:
- \(P\) is the pressure (132 psig, converted to atm),
- \(M_i\) is the molecular weight of A (58 g/mol),
- and \(T\) is the initial temperature.
\[P_\mathrm{atm} = 132 \,(\mathrm{psig}) \times \frac{14.696}{101.325} = 127.6 \, (\mathrm{atm})\]
Now we can find the initial density:
\[\rho_i = \frac{127.6 \, (\mathrm{atm}) \times 58 \, (\mathrm{g/mol})}{0.0821 \, (\frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{mol} \cdot \mathrm{K}}) \times 394.26 \, (\mathrm{K})} = 258.98 \, (\frac{\mathrm{g}}{\mathrm{L}})\]
Now we calculate the volumetric flow rate:
\[Q = \frac{1000 \, (\frac{\mathrm{lb}}{\mathrm{h}}) \times \frac{1}{2.20462} \, (\frac{\mathrm{kg}}{\mathrm{lb}}) \times \frac{1000}{1} \, (\frac{\mathrm{g}}{\mathrm{kg}})}{258.98 \,(\mathrm{g}\mathrm{L}^{-1}) \times (1 - 0.90)} = 343.73 \, (\mathrm{L/min})\]
04
鈧�. Calculate the number of tubes
Now, we can find the number of tubes needed. First, we need to find how much volume is processed in one tube during the residence time calculated in step 2.
The volume of one tube is given by its dimensions (10 ft long and 1 in. inside diameter):
\[V_\mathrm{tube} = \pi(\frac{1}{2}\cdot 0.0254 \, \mathrm{m})^2 \times 10 \cdot \frac{1}{3.28084} \, (\mathrm{m}) = 0.01022 \, (\mathrm{m^{3}})\]
Now, we can calculate the volumetric flow rate through one tube at a given time:
\[Q_\mathrm{tube} = V_{\mathrm{tube}} / t\]
\[Q_\mathrm{tube} = \frac{0.01022 \, (\mathrm{m^{3}})}{49.79\, (\mathrm{min}) \cdot \frac{1}{60} \, (\mathrm{h/min})} = 0.01222 \,(\mathrm{L/min})\]
Now, we divide the required flow rate by the flow rate processed in one tube to obtain the number of tubes:
\[\text{Number of tubes}= \frac{Q}{Q_\mathrm{tube}}\]
\[\text{Number of tubes}= \frac{343.73 \,(\mathrm{L/min})}{0.01222 \, (\mathrm{L/min})} \approx 28.13\]
Since we cannot have a fraction of a tube, we need to round up to the nearest whole number to meet the production requirements.
05
Conclusion
The number of tubes needed to achieve a production rate of 1000 lb/h of compound B with a 90% conversion of compound A is approximately 29 tubes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arrhenius Equation
The Arrhenius equation is crucial for students learning about chemical reaction engineering. It relates the rate at which a chemical reaction proceeds to the temperature at which it is occurring. The core idea behind this equation is that increasing the temperature will increase the reaction rate. But how is this relationship quantified?
The equation is expressed as:
\[k = A \times \text{e}^{-\frac{E_a}{RT}}\]
where:
This can be seen as an exponential relationship: even a small increase in temperature can lead to a substantial increase in the rate constant, and thus the speed of the reaction.
Imagine we have two temperatures, \(T_1\) and \(T_2\), and the rate constants for these temperatures are \(k_1\) and \(k_2\) respectively. If we know \(k_1\), \(E_a\), and \(R\), we can calculate \(k_2\) for the new temperature using the Arrhenius equation. This calculation is vital in industrial settings, where the control of reaction rates is essential for efficiency and safety.
The equation is expressed as:
\[k = A \times \text{e}^{-\frac{E_a}{RT}}\]
where:
- \(k\) is the rate constant,
- \(A\) is the pre-exponential factor, a constant that represents the frequency of collisions between reactant molecules,
- \(E_a\) is the activation energy, or the minimum energy that a molecule must have for a reaction to occur,
- \(R\) is the ideal gas constant, and
- \(T\) is the absolute temperature.
This can be seen as an exponential relationship: even a small increase in temperature can lead to a substantial increase in the rate constant, and thus the speed of the reaction.
Imagine we have two temperatures, \(T_1\) and \(T_2\), and the rate constants for these temperatures are \(k_1\) and \(k_2\) respectively. If we know \(k_1\), \(E_a\), and \(R\), we can calculate \(k_2\) for the new temperature using the Arrhenius equation. This calculation is vital in industrial settings, where the control of reaction rates is essential for efficiency and safety.
Rate Constant Calculation
Understanding the rate constant calculation is critical for mastering chemical kinetics and reaction engineering. The rate constant, \(k\), denotes the speed at which a reaction takes place, and it is influenced by various factors, including temperature, as discussed in the Arrhenius equation. Let's get practical and see how you calculate \(k\) using that equation.
For instance, in the given exercise, we're dealing with a reaction at two different temperatures. The initial rate constant \(k_1\) at \(T_1\) is provided, and we aim to find the new rate constant \(k_2\) at \(T_2\). Converting temperatures to Kelvin avoids the complication of negative values since absolute zero in Kelvin is the lowest possible temperature. The resulting temperatures are then used with the given activation energy and gas constant to resolve the equation for \(k_2\).
A concrete example from our exercise is calculating \(k_2\) at \(260^{\text{o}} F\) using \(k_1\) given at \(80^{\text{o}} F\). The conversion of temperatures to Kelvin and the usage of \(E_a\) (activation energy in cal/mol) allows us to extract the value of \(k_2\). This step is paramount because the rate constant at the operating temperature dictates how fast the reaction will proceed and, in practical applications, influences the design and scale of the reactors we need.
For instance, in the given exercise, we're dealing with a reaction at two different temperatures. The initial rate constant \(k_1\) at \(T_1\) is provided, and we aim to find the new rate constant \(k_2\) at \(T_2\). Converting temperatures to Kelvin avoids the complication of negative values since absolute zero in Kelvin is the lowest possible temperature. The resulting temperatures are then used with the given activation energy and gas constant to resolve the equation for \(k_2\).
A concrete example from our exercise is calculating \(k_2\) at \(260^{\text{o}} F\) using \(k_1\) given at \(80^{\text{o}} F\). The conversion of temperatures to Kelvin and the usage of \(E_a\) (activation energy in cal/mol) allows us to extract the value of \(k_2\). This step is paramount because the rate constant at the operating temperature dictates how fast the reaction will proceed and, in practical applications, influences the design and scale of the reactors we need.
Conversion Calculation
The goal of many chemical reactions in industry is to convert reactants into products efficiently. The extent to which this occurs is called conversion, usually expressed as a percentage. It has direct implications on the economy of the process, the design of equipment, and operational parameters.
In reaction engineering, calculating conversion typically begins with understanding stoichiometry and the rate laws governing the reaction. The rate law indicates how the rate depends on the concentration of reactants. For a first-order reaction, the integrated rate law is used, showing how concentration changes with time:
\[C = C_0 \text{e}^{-kt}\]
where \(C_0\) is the initial concentration, \(C\) is the concentration after time \(t\), and \(k\) is the rate constant.
The conversion, \(X\), is usually defined as:\[X = \frac{C_0 - C}{C_0}\]
In the context of our exercise, we're aiming for a 90% conversion鈥攎eaning that 90% of reactant A has turned into product B. To calculate the time needed for this conversion, we rearrange the integrated rate law to solve for \(t\), using the known rate constant at the operating temperature and the desired conversion. These calculations are essential when you need to predict how long your reaction must run to achieve the desired product yield.
In reaction engineering, calculating conversion typically begins with understanding stoichiometry and the rate laws governing the reaction. The rate law indicates how the rate depends on the concentration of reactants. For a first-order reaction, the integrated rate law is used, showing how concentration changes with time:
\[C = C_0 \text{e}^{-kt}\]
where \(C_0\) is the initial concentration, \(C\) is the concentration after time \(t\), and \(k\) is the rate constant.
The conversion, \(X\), is usually defined as:\[X = \frac{C_0 - C}{C_0}\]
In the context of our exercise, we're aiming for a 90% conversion鈥攎eaning that 90% of reactant A has turned into product B. To calculate the time needed for this conversion, we rearrange the integrated rate law to solve for \(t\), using the known rate constant at the operating temperature and the desired conversion. These calculations are essential when you need to predict how long your reaction must run to achieve the desired product yield.
