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Chapter 4: Problem 56

The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product stream is analyzed and found to contain 51.7 mole \(\% \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\) and 17.3\% HBr. The feed to the reactor contains only ethylene and hydrogen bromide. Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is \(165 \mathrm{mol} / \mathrm{s}\), what is the extent of reaction?

Short Answer

Expert verified
The fractional conversion of the limiting reactant, HBr, is approximately 64.2%. The reactant C2H4 is in excess by about 35.8%. The extent of the reaction is 51.15 mol/s.

Step by step solution

01

Determining the limiting reactant

Since the reaction between ethylene (\(C_{2}H_{4}\)) and hydrogen bromide (\(HBr\)) leads to the formation of ethyl bromide(\(C_{2}H_{5}Br\)), the stoichiometric reaction is as follows: \(C_{2}H_{4} + HBr \rightarrow C_{2}H_{5}Br\). We can determine the limiting reactant by comparing the mole fractions from the analysis of the product stream and the initial feed. Since the product stream contains \(17.3\% HBr\), it would mean that \(HBr\) is the limiting reactant because it has not been completely consumed.
02

Calculating fractional conversion of limiting reactant

The fractional conversion of the limiting reactant can be calculated using the equation: \[conversion = \frac{initial\ moles - final\ moles}{initial\ moles}\] Here, as \(HBr\) is the limiting reactant, the initial moles of \(HBr\) would have been equal to the total moles minus moles of \(C_{2}H_{4}\). Therefore, the conversion = \(1 - \frac{0.173}{0.483}\) = \(0.642\). Thus, approximately \(64.2\%\) of the limiting reactant \(HBr\) is converted to the product.
03

Determining the excess of the other reactant

The percentage by which the other reactant is in excess is given by the equation: \[ excess = \frac{initial\ moles - converted\ moles}{initial\ moles} * 100\% \] Here, \(C_{2}H_{4}\) is the other reactant. Using the stoichiometry of the reaction, we should observe that the moles of \(C_{2}H_{4}\) should be same as the consumed moles of \(HBr\). Hence, the excess of \(C_{2}H_{4}\) = \(1 - conversion\) = \(1 - 0.642\) = \(35.8\%\) Therefore, approximately \(35.8\%\) of \(C_{2}H_{4}\) is in excess.
04

Determining the extent of reaction

The extent of the reaction can be determined by using the equation: \[extent\ of\ reaction = \frac{\Delta n}{\nu}\] where \(\Delta n\) is the change in mole numbers and \(\nu\) is the stoichiometric coefficient, which is equal to 1 for both reactants in our case. Here, the mole numbers change by the number of moles of the limiting reactant initially present - number of moles of limiting reactant finally present = \(0.483*165 - 0.173*165\) = \(51.15\) mol/s. Hence, the extent of reaction = \(\frac{51.15}{1}\) = \(51.15\) mol/s.

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Key Concepts

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

Limiting Reactant
Understanding the concept of a limiting reactant is crucial in chemical reactions, as it dictates the maximum amount of product that can be formed. This reactant is the one that will be completely used up first, and thus limits the progression of the reaction. In a chemical equation, the mole ratio of reactants is defined by the balanced equation, and the reactant present in the lesser amount, according to this ratio, is the limiting one.

Consider a simple reaction where ingredient A reacts with ingredient B to form product C. If you run out of A before B, then A is the limiting reactant. Similarly, in our given exercise, hydrogen bromide (HBr) was identified as the limiting reactant because the product stream contained unreacted HBr, suggesting all the ethylene (C2H4) could not react as there wasn't enough HBr available.
Fractional Conversion Calculation
The fractional conversion calculation quantifies the proportion of a reactant that has been used up in the reaction; it is a measure of the reaction's progress. The calculation is straightforward: it's the initial amount of the reactant minus the amount that remains, divided by the initial amount. Expressed as a fraction (hence, 'fractional'), it can also be converted into a percentage.

In our example, to calculate the fractional conversion of hydrogen bromide, you subtract the final moles of HBr from the initial moles (which can be found using the mole fraction of HBr in the feed) and then divide by the initial moles. Mathematically, this is represented by the formula \[\text{Conversion} = \frac{\text{initial moles} - \text{final moles}}{\text{initial moles}}\] When successfully calculated, this gives you direct insight into how much of your limiting reactant was consumed during the chemical reaction.
Excess Reactant Percentage
While the limiting reactant is completely used up, the excess reactant is the one left over after the reaction goes to completion. Knowing the percentage of the excess reactant is helpful for understanding the reaction efficiency and planning for resource management in industrial processes.

The excess reactant percentage is found by the formula \[ \text{Excess} = \frac{\text{initial moles} - \text{converted moles}}{\text{initial moles}} \times 100\% \] In this formula, 'converted moles' is the number of moles of the excess reactant that actually participated in the reaction, which can be deduced from the stoichiometry of the chemical equation. In our reaction, the ethylene was in excess; thus, we could determine how much ethylene did not react and remained in the reactor.
Extent of Reaction
Calculating the extent of reaction gives us another piece of important information, it represents the quantity in moles of a product formed or reactant consumed in a given reaction. The extent is calculated by determining the change in moles (\(\text{螖n}\)) of reactants or products, and dividing by the stoichiometric coefficient (\(u\)), which is the 'balanced equation coefficient' for each substance in the reaction.

For the reaction between ethylene and hydrogen bromide, the stoichiometric coefficients for both reactants are equal to 1 (as per the balanced equation), simplifying the calculation. The extent of reaction tells us how far the reaction has gone in terms of molar consumption and is directly related to the flow rate of the product stream in continuous reactors, like in the case of our exercise.

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Most popular questions from this chapter

A liquid mixture containing 30.0 mole \(\%\) benzene \((\mathrm{B}), 25.0 \%\) toluene \((\mathrm{T}),\) and the balance xylene \((\mathrm{X})\) is fed to a distillation column. The bottoms product contains 98.0 mole \(\% \mathrm{X}\) and no \(\mathrm{B},\) and \(96.0 \%\) of the \(\mathrm{X}\) in the feed is recovered in this stream. The overhead product is fed to a second column. The overhead product from the second column contains \(97.0 \%\) of the \(\mathrm{B}\) in the feed to this column. The composition of this stream is 94.0 mole\% B and the balance T. (a) Draw and label a flowchart of this process and do the degree-of-freedom analysis to prove that for an assumed basis of calculation, molar flow rates and compositions of all process streams can be calculated from the given information. Write in order the equations you would solve to calculate unknown process variables. In each equation (or pair of simultaneous equations), circle the variable(s) for which you would solve. Do not do the calculations. (b) Calculate (i) the percentage of the benzene in the process feed (i.e., the feed to the first column) that emerges in the overhead product from the second column and (ii) the percentage of toluene in the process feed that emerges in the bottom product from the second column.

Gas streams containing hydrogen and nitrogen in different proportions are produced on request by blending gases from two feed tanks: Tank A (hydrogen mole fraction \(=x_{\mathrm{A}}\) ) and Tank \(\mathrm{B}\) (hydrogen mole fraction \(=x_{\mathrm{B}}\) ). The requests specify the desired hydrogen mole fraction, \(x_{\mathrm{p}}\), and mass flow rate of the product stream, \(\dot{m}_{\mathrm{P}}(\mathrm{kg} / \mathrm{h})\) (a) Suppose the feed tank compositions are \(x_{\mathrm{A}}=0.10 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol}\) and \(x_{\mathrm{B}}=0.50 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol},\) and the desired blend-stream mole fraction and mass flow rate are \(x_{\mathrm{P}}=0.20 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol}\) and \(\dot{m}_{\mathrm{P}}=100 \mathrm{kg} / \mathrm{h} .\) Draw and label a flowchart and calculate the required molar flow rates of the feed mixtures, \(\dot{n}_{\mathrm{A}}(\mathrm{kmol} / \mathrm{h})\) and \(\dot{n}_{\mathrm{B}}(\mathrm{kmol} / \mathrm{h})\) (b) Derive a series of formulas for \(\dot{n}_{\mathrm{A}}\) and \(\dot{n}_{\mathrm{B}}\) in terms of \(x_{\mathrm{A}}, x_{\mathrm{B}}, x_{\mathrm{P}},\) and \(\dot{m}_{\mathrm{P}} .\) Test them using the values in Part (a). (c) Write a spreadsheet that has column headings \(x_{\mathrm{A}}, x_{\mathrm{B}}, x_{\mathrm{P}}, \dot{m}_{\mathrm{P}}, \dot{n}_{\mathrm{A}},\) and \(\dot{n}_{\mathrm{B}}\). The spreadsheet should calculate entries in the last two columns corresponding to data in the first four. In the first six data rows of the spreadsheet, do the calculations for \(x_{\mathrm{A}}=0.10, x_{\mathrm{B}}=0.50,\) and \(x_{\mathrm{P}}=\) \(0.10,0.20,0.30,0.40,0.50,\) and \(0.60,\) all for \(\dot{m}_{\mathrm{P}}=100 \mathrm{kg} / \mathrm{h} .\) Then in the next six rows repeat the calculations for the same values of \(x_{\mathrm{A}}, x_{\mathrm{B}},\) and \(x_{\mathrm{p}}\) for \(\dot{m}_{\mathrm{p}}=250 \mathrm{kg} / \mathrm{h} .\) Explain any of your results that appear strange or impossible. (d) Enter the formulas of Part (b) into an equation-solving program. Run the program to determine \(\dot{n}_{\mathrm{A}}\) and \(\dot{n}_{\mathrm{B}}\) for the 12 sets of input variable values given in Part (c) and explain any physically impossible results.

The indicator-dilution method is a technique used to determine flow rates of fluids in channels for which devices like rotameters and orifice meters cannot be used (e.g., rivers, blood vessels, and largediameter pipelines). A stream of an easily measured substance (the tracer) is injected into the channel at a known rate, and the tracer concentration is measured at a point far enough downstream of the injection point for the tracer to be completely mixed with the flowing fluid. The larger the flow rate of the fluid, the lower the tracer concentration at the measurement point. A gas stream that contains 1.50 mole \(\% \mathrm{CO}_{2}\) flows through a pipeline. Twenty (20.0) kilograms of \(\mathrm{CO}_{2}\) per minute is injected into the line. A sample of the gas is drawn from a point in the line 150 meters (a) Estimate the gas flow rate (kmol/min) upstream of the injection point. (b) Eighteen seconds elapse from the instant the additional \(\mathrm{CO}_{2}\) is first injected to the time the \(\mathrm{CO}_{2}\) concentration at the measurement point begins to rise. Assuming that the tracer travels at the average velocity of the gas in the pipeline (i.e., neglecting diffusion of \(\mathrm{CO}_{2}\) ), estimate the average velocity (m/s). If the molar gas density is \(0.123 \mathrm{kmol} / \mathrm{m}^{3}\), what is the pipe diameter?

Ammonia is oxidized to nitric oxide in the following reaction: $$4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$$ (a) Calculate the ratio (lb-mole \(\mathrm{O}_{2}\) react/lb-mole NO formed). (b) If ammonia is fed to a continuous reactor at a rate of \(100.0 \mathrm{kmol} \mathrm{NH}_{3} / \mathrm{h}\), what oxygen feed rate (kmol/h) would correspond to 40.0\% excess O_? (c) If \(50.0 \mathrm{kg}\) of ammonia and \(100.0 \mathrm{kg}\) of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

Methanol is synthesized from carbon monoxide and hydrogen in a catalytic reactor. The fresh feed to the process contains 32.0 mole \(\%\) CO, \(64.0 \%\) H \(_{2}\), and \(4.0 \%\) Ne. This stream is mixed with a recycle stream in a ratio 5 mol recycle/ 1 mol fresh feed to produce the feed to the reactor, which contains 13.0 mole\% \(\mathrm{N}_{2}\). A low single-pass conversion is attained in the reactor. The reactor effluent goes to a condenser from which two streams emerge: a liquid product stream containing essentially all the methanol formed in the reactor, and a gas stream containing all the \(\mathrm{CO}, \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) leaving the reactor. The gas stream is split into two fractions: one is removed from the process as a purge stream, and the other is the recycle stream that combines with the fresh feed to the reactor. (a) Assume a methanol production rate of \(100 \mathrm{kmol} / \mathrm{h}\). Perform the DOF for the overall system and all subsystems to prove that there is insufficient information to solve for all unknowns. (b) Briefly explain in your own words the reasons for including (i) the recycle stream and (ii) the purge stream in the process design.
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