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Chapter 4: Problem 52

Acetylene is hydrogenated to form ethane. The feed to the reactor contains \(1.50 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}\) (a) Calculate the stoichiometric reactant ratio (mol \(\mathrm{H}_{2}\) react/mol \(\mathrm{C}_{2} \mathrm{H}_{2}\) react) and the yield ratio (kmol \(\mathbf{C}_{2} \mathbf{H}_{6}\) formed/kmol \(\mathbf{H}_{2}\) react (b) Determine the limiting reactant and calculate the percentage by which the other reactant is in excess. (c) Calculate the mass feed rate of hydrogen ( \(\mathrm{kg} / \mathrm{s}\) ) required to produce \(4 \times 10^{6}\) metric tons of ethane per year, assuming that the reaction goes to completion and that the process operates for 24 hours a day, 300 days a year. (d) There is a definite drawback to running with one reactant in excess rather than feeding the reactants in stoichiometric proportion. What is it? [Hint: In the process of Part (c), what does the reactor effluent consist of and what will probably have to be done before the product ethane can be sold or used?]

Short Answer

Expert verified
Stoichiometric reactant ratio is \(2.00 \, \text{mol H}_2/\text{mol C}_2\text{H}_2\), yield ratio is \(1.00 \, \text{kmol C}_2\text{H}_6/\text{kmol H}_2\). \(\text{H}_2\) is the limiting reactant, and \(\text{C}_2\text{H}_2\) is in excess by 25%. Feed rate of hydrogen required to meet production goal can be calculated but depends on molecular weights and time. Potential drawbacks of running the process with one reactant in excess include the need for expensive and energy-intensive separation processes.

Step by step solution

01

Calculate the stoichiometric reactant ratio and the yield ratio

We know that the reaction is \(C_2H_2 + 2H_2 鈫� C_2H_6\). This tells us that for every mole of acetylene, we need two moles of hydrogen gas, and we form 1 mole of ethane. Therefore, the stoichiometric reactant ratio is \(2.00 \, \text{mol H}_2/\text{mol C}_2\text{H}_2\) and the yield ratio is \(1.00 \, \text{kmol C}_2\text{H}_6/\text{kmol H}_2\).
02

Determine the limiting reactant and the percentage by which the other reactant is in excess

Since the feed to the reactor contains only \(1.50 \, \text{mol H}_2/\text{mol C}_2\text{H}_2\), there is less hydrogen than needed for every mole of acetylene, making \(H_2\) the limiting reactant. The reactant \(C_2H_2\) is in excess by \((2.00 - 1.50)/2.00 * 100 = 25%\).
03

Calculate the mass feed rate of hydrogen required to produce the desired amount of ethane

We know that 1 mole of \(H_2\) is needed to produce 1 mole of \(C_2H_6\). With a molecular weight of \(2.016 \, \text{kg/kmol}\) for \(H_2\) and \(30.069 \, \text{kg/kmol}\) for \(C_2H_6\), the yearly required feed rate of \(H_2\) to produce \(4 * 10^9 \, \text{kg}\) of \(C_2H_6\) is \((4 * 10^9 \, \text{kg C}_2\text{H}_6)(\text{kmol C}_2\text{H}_6/30.069 \, \text{kg C}_2\text{H}_6)(2.016 \, \text{kg H}_2/\text{kmol H}_2)\). Dividing this by the number of seconds in a year ( \(24 * 60 * 60 * 300 = 25,920,000\) s) gives the required mass feed rate of \(H_2\).
04

Discuss the drawbacks of running the process with one reactant in excess

Feeding the reactor with an excess of acetylene means that unreacted acetylene will be present in the effluent from the reactor. This unreacted acetylene needs to be separated from the product ethane before the ethane can be sold or used. This separation could be costly and energy-intensive, reducing the overall efficiency of the process.

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Key Concepts

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

Stoichiometric Reactant Ratio
In any chemical reaction, understanding the stoichiometric reactant ratio is crucial. This ratio tells us the exact amount of reactants needed to perfectly balance a chemical equation, ensuring that all reactants are used up to form the intended products without anything left over. This is important for efficient chemical process principles.

In our given problem involving the hydrogenation of acetylene (\[C_2H_2 + 2H_2 \rightarrow C_2H_6\]), the stoichiometric reactant ratio informs us that for every mole of acetylene (\(C_2H_2\)), we require two moles of hydrogen gas (\(H_2\)). Thus, the stoichiometric reactant ratio is \(2 \text{ mol H}_2/\text{mol C}_2\text{H}_2\). This means to convert acetylene to ethane completely, the ideal proportion of hydrogen to acetylene must be maintained at this ratio. Deviating from this ratio can lead to inefficiencies, such as having excess reactants.
Limiting Reactant
The concept of a limiting reactant is fundamental in chemical reactions because it dictates how much product can be formed. The limiting reactant is the reactant that gets completely used up first, thus determining the maximum amount of product formed.

In our scenario, the feed has only \(1.50 \text{ mol H}_2/\text{mol C}_2\text{H}_2\), which is less than the stoichiometric ratio of \(2.00 \text{ mol H}_2/\text{mol C}_2\text{H}_2\). Therefore, hydrogen (\(H_2\)) is the limiting reactant, as it runs out first, halting the reaction despite having unreacted acetylene.

The acetylene (\(C_2H_2\)) is in excess by \([\,(2.00 - 1.50)/2.00 \,] \times 100 = 25\%\). Identifying the limiting reactant helps in calculating the reaction's yield and understanding the reaction's efficiency.
Mass Feed Rate Calculation
Mass feed rate calculation is critical in ensuring that a chemical process is economically feasible and efficient. It determines the amount of reactant that needs to be supplied to achieve a desired production level.

In the problem, we are tasked to calculate the mass feed rate of hydrogen required to produce \(4 \times 10^6\) metric tons of ethane annually. Given ethane's molecular weight \( (30.069 \, \text{kg/kmol})\) and hydrogen's molecular weight \( (2.016 \, \text{kg/kmol})\), the mass of hydrogen needed is derived through a series of conversions:
  • First, calculate the moles of ethane required: \[\,(4 \times 10^9 \, \text{kg C}_2\text{H}_6) \, (\text{kmol C}_2\text{H}_6/30.069 \, \text{kg C}_2\text{H}_6)\,\]
  • Convert it based on the reaction ratio to find required moles of hydrogen.
  • Finally, divide the total mass of hydrogen by the total operational seconds in a year \((24 \times 60 \times 60 \times 300)\) to determine the continuous feed rate in \(\text{kg/s}\)
By tackling these calculations, you ensure the process operates smoothly, continuously, and economically viable for the given production requirements.

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Most popular questions from this chapter

Two streams flow into a 500 -gallon tank. The first stream is 10.0 wt\% ethanol and \(90.0 \%\) hexane (the mixture density, \(\rho_{1},\) is \(0.68 \mathrm{g} / \mathrm{cm}^{3}\) ) and the second is \(90.0 \mathrm{wt} \%\) ethanol, \(10.0 \%\) hexane \(\left(\rho_{2}=0.78 \mathrm{g} / \mathrm{cm}^{3}\right) .\) After the tank has been filled, which takes 22 \(\mathrm{min}\), an analysis of its contents determines that the mixture is 60.0 wt\% ethanol, \(40.0 \%\) hexane. You wish to estimate the density of the final mixture and the mass and volumetric flow rates of the two feed streams. (a) Draw and label a flowchart of the mixing process and do the degree-of- freedom analysis. (b) Perform the calculations and state what you assumed.

A liquid-phase chemical reaction \(\mathrm{A} \rightarrow \mathrm{B}\) takes place in a well-stirred tank. The concentration of \(\mathrm{A}\) in the feed is \(C_{\mathrm{A} 0}\left(\operatorname{mol} / \mathrm{m}^{3}\right),\) and that in the tank and outlet stream is \(C_{\mathrm{A}}\left(\mathrm{mol} / \mathrm{m}^{3}\right) .\) Neither concentration varies with time. The volume of the tank contents is \(V\left(\mathrm{m}^{3}\right)\) and the volumetric flow rate of the inlet and outlet streams is \(\dot{V}\left(\mathrm{m}^{3} / \mathrm{s}\right)\). The reaction rate (the rate at which \(\mathrm{A}\) is consumed by reaction in the tank) is given by the expression $$r(\text { mol } A \text { consumed } / \mathrm{s})=k V C_{\mathrm{A}}$$ (a) Is this process continuous, batch, or semibatch? Is it transient or steady-state? (b) What would you expect the reactant concentration \(C_{\mathrm{A}}\) to equal if \(k=0\) (no reaction)? What should it approach if \(k \rightarrow \infty\) (infinitely rapid reaction)? (c) Write a differential balance on \(A,\) stating which terms in the general balance equation (accumulation = input + generation - output - consumption) you discarded and why you discarded them. Use the balance to derive the following relation between the inlet and outlet reactant concentrations: $$C_{\mathrm{A}}=\frac{C_{\mathrm{A} 0}}{1+k V / \dot{V}}$$ Verify that this relation predicts the results in Part (b).

A \(100 \mathrm{kmol} / \mathrm{h}\) stream that is 97 mole \(\%\) carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and \(3 \%\) carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) is to be recovered from the bottom of a distillation column. The feed to the column is 16 mole \(\% \mathrm{CS}_{2}\) and \(84 \% \mathrm{CCl}_{4},\) and \(2 \%\) of the \(\mathrm{CCl}_{4}\) entering the column is contained in the overhead stream leaving the top of the column. (a) Draw and label a flowchart of the process and do the degree-of-freedom analysis. (b) Calculate the mass and mole fractions of \(\mathrm{CCl}_{4}\) in the overhead stream, and determine the molar flow rates of \(\mathrm{CCl}_{4}\) and \(\mathrm{CS}_{2}\) in the overhead and feed streams. (c) Suppose the overhead stream is analyzed and the mole fraction of \(\mathrm{CS}_{2}\) is found to be significantly lower than the value calculated in Part (b). List as many reasons as you can for the discrepancy, including possible violations of assumptions made in Part (b).

An evaporation-crystallization process of the type described in Example \(4.5-2\) is used to obtain solid potassium sulfate from an aqueous solution of this salt. The fresh feed to the process contains 19.6 wt\% \(\mathrm{K}_{2} \mathrm{SO}_{4}\). The wet filter cake consists of solid \(\mathrm{K}_{2} \mathrm{SO}_{4}\) crystals and a \(40.0 \mathrm{wt} \% \mathrm{K}_{2} \mathrm{SO}_{4}\) solution, in a ratio \(10 \mathrm{kg}\) crystals/kg solution. The filtrate, also a \(40.0 \%\) solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0\% is evaporated. The evaporator has a maximum capacity of 175 kg water evaporated/s. (a) Assume the process is operating at maximum capacity. Draw and label a flowchart and do the degree-of-freedom analysis for the overall system, the recycle-fresh feed mixing point, the evaporator, and the crystallizer. Then write in an efficient order (minimizing simultaneous equations) the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don't do the calculations. (b) Calculate the maximum production rate of solid \(\mathrm{K}_{2} \mathrm{SO}_{4}\), the rate at which fresh feed must be supplied to achieve this production rate, and the ratio kg recycle/kg fresh feed. (c) Calculate the composition and feed rate of the stream entering the crystallizer if the process is scaled to 75\% of its maximum capacity. (d) The wet filter cake is subjected to another operation after leaving the filter. Suggest what it might be. Also, list what you think the principal operating costs for this process might be. (e) Use an equation-solving computer program to solve the equations derived in Part (a). Verify that you get the same solutions determined in Part (b).

In the production of soybean oil, dried and flaked soybeans are brought into contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil. (a) Draw a flowchart of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract). (b) The soybeans contain 18.5 wt\% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to \(2.0 \mathrm{kg}\) hexane per \(\mathrm{kg}\) beans. The residual solids leaving the extraction unit contain 35.0 wt\% hexane, all of the non-oil solids that entered with the beans, and \(1.0 \%\) of the oil that entered with the beans. For a feed rate of \(1000 \mathrm{kg} / \mathrm{h}\) of dried flaked soybeans, calculate the mass flow rates of the extract and residual solids, and the composition of the extract. (c) The product soybean oil must now be separated from the extract. Sketch a flowchart with two units, the extraction unit from Parts (a) and (b) and the unit separating soybean oil from hexane. Propose a use for the recovered hexane.
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