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Chapter 5: Problem 54

Propylene is hydrogenated in a batch reactor: $$\mathrm{C}_{3} \mathrm{H}_{6}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})$$ Equimolar amounts of propylene and hydrogen are fed into the reactor at \(25^{\circ} \mathrm{C}\) and a total absolute pressure of 32.0 atm. The reactor temperature is raised to \(235^{\circ} \mathrm{C}\) and held constant thereafter until the reaction is complete. The propylene conversion at the beginning of the isothermal period is \(53.2 \%\) You may assume ideal-gas behavior for this problem, although at the high pressures involved this assumption constitutes an approximation at best. (a) What is the final reactor pressure? (b) What is the percentage conversion of propylene when \(P=35.1\) atm? (c) Construct a graph of pressure versus fractional conversion of propylene covering the isothermal period of operation. Use the graph to confirm the results in Parts (a) and (b). (Suggestion: Use a spreadsheet.)

Short Answer

Expert verified
a) The final reactor pressure should be determined using the Ideal Gas Law and the known conditions given at the point where the reaction is complete. b) The conversion of Propylene when the pressure is \(35.1\) atm should be calculated from the derived relationship between mole fractions and pressure. c) The pressure-conversion graph should reflect the calculated results from part (a) and (b) as well as show the change in pressure with varying conversion percentages.

Step by step solution

01

Analyze the Reaction

The reaction involves the conversion of Propylene (\(C_3H_6\)) and Hydrogen (\(H_2\)) into Propane (\(C_3H_8\)). This implies that for each mole of Propylene that reacts, we lose one mole of Hydrogen but gain one mole of Propane. Use this principle to calculate the initial and final moles of each component, taking into account the initial conversion of \(53.2\%\).
02

Use the Ideal Gas Law

The Ideal Gas Law equation is \(PV = nRT\), where P=pressure, V=volume, n=mole, R=universal gas constant, and T=temperature. Assume volume is constant and all other variables are known or can be calculated. So, the relationship between pressure and the mole fraction can be obtained. Calculate the final pressure using the conditions given when the reaction is complete.
03

Calculate the Conversion

Determine the conversion of propylene for a pressure of \(35.1\) atm using the relationship between pressure and the mole fraction derived in the previous step.
04

Graphing Pressure vs Conversion

Plot a graph of pressure against propylene conversion using the relationship derived in Step 2. This must reflect the change in pressure with varying amounts of conversion. This graph should answer part (c) of the problem and confirm the results obtained in part (a) and (b).

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Key Concepts

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

Batch Reactors
Batch reactors are a type of chemical reactor where reactants are loaded into the reactor, possibly along with a solvent, and the reaction occurs over time with no additional reactants being added or products removed during the process. This type of reactor is particularly useful for small to moderate scale productions and for reactions that require precise control over reaction time, temperature, and pressure.

In the context of propylene hydrogenation, the batch reactor allows the hydrogen and propylene to react completely within a specified temperature range. As the reactor temperature is raised to 235掳C and kept constant, the conditions are optimized for the reaction and completion at a reliable rate. The batch system is essential for reactions needing strict control to achieve the desired conversion rates, such as the 53.2% conversion at the start of the isothermal period mentioned in the solution.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemical engineering and chemistry. It relates the pressure, volume, temperature, and the number of moles of a gas, expressed as: \[ PV = nRT \] Where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance (in moles), \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. This equation assumes that the gases behave ideally, meaning they have no interactions between molecules and occupy no volume themselves.

In the propylene hydrogenation problem, the Ideal Gas Law is used to relate the pressure with the conversion of propylene. Even though high pressures could deviate from ideal behavior, it provides a suitable approximation. By holding the volume constant and knowing the temperature, we can solve for changes in pressure corresponding to changes in the number of moles as propylene converts to propane.
Propylene Hydrogenation
Propylene hydrogenation is a chemical reaction where propylene \((C_3H_6)\) is converted into propane \((C_3H_8)\) using hydrogen \((H_2)\) under the influence of a catalyst, often at elevated temperatures and pressures. This reaction is significant in the petrochemical industry for producing propane, which is an important fuel and chemical feedstock.

In a batch reactor setup, equimolar amounts of propylene and hydrogen are allowed to react. The importance of understanding this process lies in efficiently driving the reaction forward to maximize conversion rates, such as the 53.2% conversion discussed earlier and calculated more deeply during ideal gas law applications. Adjustments in temperature, pressure, and possibly catalysts can improve the reaction efficiency and outcome.
Pressure-Conversion Relationship
The relationship between pressure and conversion during a chemical reaction is a vital concept in reaction engineering. As the reaction proceeds, the number of moles of gas changes, thereby affecting the pressure when volume is held constant. For the propylene hydrogenation reaction, as conversion increases, the pressure initially increases because reactants are in higher mole quantities than the products.

Visualizing this pressure-conversion relationship is central to analyzing reactor performance and is done by plotting graphs that show how pressure changes with varying conversions. By understanding this graph, engineers can predict the pressure at certain conversions, such as finding at what conversion the pressure reaches 35.1 atm and confirming it with experimental or calculated values, as the problem solution suggests for parts (a) and (b).

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Most popular questions from this chapter

Spray drying is a process in which a liquid containing dissolved or suspended solids is injected into a chamber through a spray nozzle or centrifugal disk atomizer. The resulting mist is contacted with hot air, which evaporates most or all of the liquid, leaving the dried solids to fall to a conveyor belt at the bottom of the chamber. Powdered milk is produced in a spray dryer \(6 \mathrm{m}\) in diameter by \(6 \mathrm{m}\) high. Air enters at \(167^{\circ} \mathrm{C}\) and \(-40 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\). The milk fed to the atomizer contains \(70 \%\) water by mass, all of which evaporates. The outlet gas contains 12 mole \(\%\) water and leaves the chamber at \(83^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) (absolute) at a rate of \(311 \mathrm{m}^{3} / \mathrm{min}\). (a) Calculate the production rate of dried milk and the volumetric flow rate of the inlet air. Estimate the upward velocity of air (m/s) at the bottom of the dryer. (b) Engineers often face the challenge of what to do to a process when demand for a product increases (or decreases). Suppose in the present case production must be doubled. (i) Why is it unlikely that the flow rates of feed and air can simply be increased to achieve the new production rate? (ii) An obvious option is to buy another dryer like the existing one and operate the two in parallel. Give two advantages and two disadvantages of this option. (iii) Still another possibility is to buy a larger dryer to replace the original unit. Give two advantages and two disadvantages of doing so. Estimate the approximate dimensions of the larger unit.

In a metered-dose inhaler (MDI), such as those used for asthma medication, medicine is delivered by a compressed-gas propellant. (The device is similar in concept to a can of spray paint.) When the inhaler is activated, a fixed amount of the medicine suspended in the propellant is expelled from the mouthpiece and inhaled. In the past, chlorofluorocarbons (CFCs) were used as propellants; however, because of their reactivity with the Earth's ozone layer, they have been replaced by hydrofluorocarbons (HFCs), which do not react with ozone. In one brand of inhalers, the original CFC propellant has been replaced by HFC 227 ea \(\left(\mathrm{C}_{3} \mathrm{HF}_{7},\right.\) heptafluoropropane). The volume of the inhaler propellant reservoir is \(1.00 \times 10^{2} \mathrm{mL}\), and the propellant is charged into the reservoir to a gauge pressure of 4.443 atm at \(23^{\circ} \mathrm{C}\). An online search for properties of HFC 227ea yields the information that the critical temperature and pressure of the substance are \(374.83 \mathrm{K}\) and 28.74 atm, and the acentric factor is \(\omega=0.180\). (a) Assuming ideal-gas behavior, estimate the mass(g) of propellant in the fully charged inhaler. (b) Someone in the manufacturer's Quality Control Division has raised a concern that assuming ideal-gas behavior might be inaccurate at the charging pressure. Use the SRK equation of state to recalculate the moles of propellant at the specified conditions. What percentage error resulted from using the ideal-gas assumption?

An adult takes about 12 breaths per minute, inhaling roughly \(500 \mathrm{mL}\) of air with each breath. The molar compositions of the inspired and expired gases are as follows: $$\begin{array}{lcc} \hline \text { Species } & \text { Inspired Gas (\%) } & \text { Expired Gas (\%) } \\ \hline \mathrm{O}_{2} & 20.6 & 15.1 \\ \mathrm{CO}_{2} & 0.0 & 3.7 \\ \mathrm{N}_{2} & 77.4 & 75.0 \\ \mathrm{H}_{2} \mathrm{O} & 2.0 & 6.2 \\ \hline \end{array}$$ The inspired gas is at \(24^{\circ} \mathrm{C}\) and 1 atm, and the expired gas is at body temperature and pressure \(\left(37^{\circ} \mathrm{C}\right.\) and 1 atm). Nitrogen is not transported into or out of the blood in the lungs, so that \(\left(\mathrm{N}_{2}\right)_{\text {in }}=\left(\mathrm{N}_{2}\right)_{\text {out }}\) (a) Calculate the masses of \(\mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. (b) Calculate the volume of air exhaled per milliliter inhaled. (c) At what rate (g/min) is this individual losing weight by merely breathing? (d) The rate at which oxygen is transferred from the air in the lungs to the blood is roughly proportional to \(\left[\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{air}}-\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\right],\) where \(\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\) is a quantity related to the concentration of oxygen in the blood. Compared to regions where atmospheric pressure is 14.7 psia, what effect does the atmospheric pressure in Denver, which is approximately 12.1 psi, have on the transport rate and breathing rate? How does the body adjust to address this condition?

The volume of a dry box (a closed chamber with dry nitrogen flowing through it) is \(2.0 \mathrm{m}^{3}\). The dry box is maintained at a slight positive gauge pressure of \(10 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\) and room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) If the contents of the box are to be replaced every five minutes, calculate the required mass flow rate of nitrogen in \(g / \min\) by (a) direct solution of the ideal-gas equation of state and (b) conversion from standard conditions. You may assume the gas in the dry box is well mixed.

The lower flammability limit (LFL) and the upper flammability limit (UFL) of propane in air at 1 atm are, respectively, 2.3 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) and 9.5 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8} .^{17}\) If the mole percent of propane in a propane-air mixture is between \(2.3 \%\) and \(9.5 \%,\) the gas mixture will burn explosively if exposed to a flame or spark; if the percentage is outside these limits, the mixture is safe-a match may burn in it but the flame will not spread. If the percentage of propane is below the LFL, the mixture is said to be too lean to ignite; if it is above the UFL, the mixture is too rich to ignite. (a) Which would be safer to release into the atmosphere- -a fuel-air mixture that is too lean or too rich to ignite? Explain. (b) A mixture of propane in air containing 4.03 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) is fed to a combustion furnace. If there is a problem in the furnace, the mixture is diluted with a stream of pure air to make sure that it cannot accidentally ignite. If propane enters the furnace at a rate of \(150 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8} / \mathrm{s}\) in the original fuel- air mixture, what is the minimum molar flow rate of the diluting air? (c) The actual diluting air molar flow rate is specified to be \(130 \%\) of the minimum value. Assuming the fuel mixture (4.03 mole\% \(\mathrm{C}_{3} \mathrm{H}_{8}\) ) enters the furnace at the same rate as in Part (b) at \(125^{\circ} \mathrm{C}\) and 131 kPa and the diluting air enters at \(25^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\), calculate the ratio \(\left(\mathrm{m}^{3} \text { diluting air) } /\right.\) (m \(^{3}\) fuel gas) and the mole percent of propane in the diluted mixture. (d) Give several possible reasons for feeding air at a value greater than the calculated minimum rate.
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