/*! 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 51 A dilute aqueous solution of sul... [FREE SOLUTION] | 91Ó°ÊÓ

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A dilute aqueous solution of sulfuric acid at \(25^{\circ} \mathrm{C}\) is used to absorb ammonia in a continuous reactor, thereby producing ammonium sulfate, a fertilizer: $$2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})$$ (a) If the ammonia enters the absorber at \(75^{\circ} \mathrm{C}\), the sulfuric acid enters at \(25^{\circ} \mathrm{C}\), and the product solution emerges at \(25^{\circ} \mathrm{C}\), how much heat must be withdrawn from the unit per mol of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) produced? (All needed physical property data may be found in Appendix B.) (b) Estimate the final temperature if the reactor of Part (a) is adiabatic and the product of the solution contains 1.00 mole \(\%\) ammonium sulfate. Take the heat capacity of the solution to be that of pure liquid water [4.184 kJ/(kg.'C)]. (c) In a real (imperfectly insulated) reactor, would the final solution temperature be less than, equal to, or greater than the value calculated in Part (b), or is there no way to tell without more information? Briefly explain your answer.

Short Answer

Expert verified
In Part a, the heat to be withdrawn per mol of product was calculated using the enthalpy of reaction and heat capacities of substances. In Part b, the final temperature of the adiabatic reactor was calculated based on energy produced by the reaction and heat capacity of the solution. For Part c, it was inferred that a real (imperfectly insulated) reactor would have lower final solution temperature than the adiabatic due to heat loss.

Step by step solution

01

Part a: Calculation of heat withdrawn

To determine the heat to be withdrawn, differential heat balances will be performed. During the absorption process, 1 mol of sulfuric acid reacts with 2 mol of ammonia to form 1 mol of ammonium sulfate. From Appendix B, we get values for enthalpy of reaction, heat capacities of all substances. We use \(\Delta H = \Delta H_{rxn} + \Delta H_{cooling}\), where \(\Delta H_{rxn}\) is heat due to reaction and \(\Delta H_{cooling}\) is heat of cooling to bring ammonia from 75°C to 25°C. Heat of cooling \(\Delta H_{cooling} = 2 * Cp_{NH_3} * (T_{initial} - T_{final})\). Plug in values to get the amount of heat withdrawn.
02

Part b: Final Temperature Calculation

An adiabatic reactor does not exchange heat with its surroundings. So, the heat produced in the reactor because of the reaction must be accounted for the increase in temperature. Initial temperature will be taken as 25°C (temperature of the entering sulfuric acid). Use the equation \(q_{absorption} = q_{reaction} + q_{heating}\) where \(q_{absorption}\) is enthalpy absorbed by absorber and \(q_{heating}\) is energy required to increase temperature from 25°C to T_final. As this is adiabatic conditions, \(q_{absorption} = 0\), so \(-\Delta H_{reaction} = Cp_{solution} * (T_{final} - T_{initial})\). Here \(\Delta H_{reaction}\) is enthalpy change due to reaction and \(Cp_{solution}\) is heat capacity of solution. Solve this to find \(T_{final}\).
03

Part c: Final solution temperature in real reactor

In an imperfectly insulated reactor, some heat will escape due to heat loss. As a result, not all heat generated by a reaction will be available to increase the temperature of the solution. Therefore, the temperature of the final solution in a real reactor would be lower than in an adiabatic reactor.

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Key Concepts

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

Heat Balances
When dealing with chemical processes, it's essential to consider the heat balances of reactions. This involves calculating the heat produced or absorbed during a chemical process. In the context of sulfuric acid absorbing ammonia to form ammonium sulfate, we perform a heat balance to find out how much heat needs to be withdrawn to maintain the product temperature at a desired level.

The heat balance equation often breaks down as follows:
  • \(\Delta H = \Delta H_{rxn} + \Delta H_{cooling}\)
Here, \(\Delta H_{rxn}\) is the enthalpy change from the chemical reaction, and \(\Delta H_{cooling}\) is the heat removed to cool the substances to a desired final temperature. In this problem, to ensure that the final product temperature is 25°C, we subtract the heat generated in the reaction and cool the entering ammonia from 75°C to 25°C.

Understanding heat balances helps maintain control over reactions, both for safety and efficiency reasons.
Adiabatic Reactors
Adiabatic reactors are a type of reactor where no heat is exchanged with the environment. In these reactors, all heat generated by the reaction is used to change the temperature of the system.

In the case of our sulfuric acid and ammonia reaction, an adiabatic process means that the heat produced by the conversion to ammonium sulfate directly increases the temperature of the reaction mixture. We use the equation:
  • \(-\Delta H_{reaction} = Cp_{solution} \times (T_{final} - T_{initial})\)
This equation implies that any heat from the reaction raises the temperature of the solution from the initial 25°C. Understanding adiabatic reactions is crucial because they give insight into how much the temperature will change without outside heat transfer.

Such knowledge is useful for designing reactors and predicting behavior under various conditions.
Enthalpy of Reaction
The enthalpy of reaction, \(\Delta H_{rxn}\), is a measure of the heat absorbed or released during a chemical reaction at constant pressure. In our exercise, the enthalpy of reaction tells us how much heat is generated or consumed when ammonia reacts with sulfuric acid.

This value is crucial for calculating overall heat balance. For example, if \(\Delta H_{rxn}\) is negative, the reaction is exothermic, releasing heat. Conversely, if it's positive, the reaction is endothermic, absorbing heat. Knowing \(\Delta H_{rxn}\) helps determine how much external heat or cooling is required to achieve desired product conditions.

Understanding this concept allows engineers and scientists to safely and effectively design and operate chemical processes, ensuring the desired outcomes and maintaining safety standards.

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Most popular questions from this chapter

Formaldehyde is produced commercially by the catalytic oxidation of methanol. In a side reaction, methanol is oxidized to \(\mathrm{CO}_{2}\) $$\begin{array}{l}\mathrm{CH}_{3} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CH}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\\\\mathrm{CH}_{3} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\end{array}$$ A mixture containing 55.6 mole \(\%\) methanol and the balance oxygen enters a reactor at \(350^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) at a rate of \(4.60 \times 10^{4} \mathrm{L} / \mathrm{s}\). The reaction products emerge at the same temperature and pressure at a rate of \(6.26 \times 10^{4} \mathrm{L} / \mathrm{s} .\) An analysis of the products yields a molar composition of \(36.7 \% \mathrm{CH}_{2} \mathrm{O}, 4.1 \% \mathrm{CO}_{2}\) \(14.3 \% \mathrm{O}_{2},\) and \(44.9 \% \mathrm{H}_{2} \mathrm{O} .\) The required reactor cooling rate is calculated to be \(1.05 \times 10^{5} \mathrm{kW}\) (a) Is the calculated cooling rate correct for the given stream data? (b) The stream data cannot be correct. Prove it.

A culture of the fungus aspergillus niger is used industrially in the manufacture of citric acid and other organic species. Cells of the fungus have an ultimate analysis of \(\mathrm{CH}_{1,79} \mathrm{N}_{0.2} \mathrm{O}_{0.5}\), and the heat of formation of this species is necessary to approximate the heat duty for the bioreactor in which citric acid is to be produced. You collect a dried sample of the fungus and determine its heat of combustion to be \(-550 \mathrm{kJ} / \mathrm{mol} .\) Estimate the heat of formation \((\mathrm{kJ} / \mathrm{mol})\) of the dried fungus cells.

An ultimate analysis of a coal is a series of operations that yields the percentages by mass of carbon, hydrogen, nitrogen, oxygen, and sulfur in the coal. The heating value of a coal is best determined in a calorimeter, but it may be estimated with reasonable accuracy from the ultimate analysis using the Dulong formula: $$H H V(\mathrm{k} J / \mathrm{kg})=33,801(\mathrm{C})+144,158[(\mathrm{H})-0.125(\mathrm{O})]+9413(\mathrm{S})$$ where (C), (H), (O), and (S) are the mass fractions of the corresponding elements. The 0.125(O) term accounts for the hydrogen bound in the water contained in the coal. (a) Derive an expression for the higher heating value ( \(H H V\) ) of a coal in terms of \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{S},\) and compare your result with the Dulong formula. Suggest a reason for the difference. (b) A coal with an ultimate analysis of \(75.8 \mathrm{wt} \% \mathrm{C}, 5.1 \% \mathrm{H}, 8.2 \% \mathrm{O}, 1.5 \% \mathrm{N}, 1.6 \% \mathrm{S},\) and \(7.8 \%\) ash (noncombustible) is burned in a power-plant boiler fumace. All of the sulfur in the coal forms \(\mathrm{SO}_{2}\) The gas leaving the furnace is fed through a tall stack and discharged to the atmosphere. The ratio \(\phi\) (\(\mathrm{kg} \mathrm{SO}_{2}\) in the stack gas/kJ heating value of the fuel) must be below a specified value for the power plant to be in compliance with Environmental Protection Agency regulations regarding sulfur emissions. Estimate \(\phi\), using the Dulong formula for the heating value of the coal. (c) An earlier version of the EPA regulation specified that the mole fraction of \(\mathrm{SO}_{2}\) in the stack gas must be less than a specified amount to avoid a costly fine and the required installation of an expensive stack gas scrubbing unit. When this regulation was in force, a few unethical plant operators blew clear air into the base of the stack while the furnace was operating. Briefly explain why they did so and why they stopped this practice when the new regulation was introduced.

Hydrogen is produced in the steam reforming of propane: $$\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow 3 \mathrm{CO}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g})$$ The water-gas shift reaction also takes place in the reactor, leading to the formation of additional hydrogen: $$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ The reaction is carried out over a nickel catalyst in the tubes of a shell- and-tube reactor. The feed to the reactor contains steam and propane in a 6: 1 molar ratio at \(125^{\circ} \mathrm{C}\), and the products emerge at \(800^{\circ} \mathrm{C}\). The excess steam in the feed assures essentially complete consumption of the propane. Heat is added to the reaction mixture by passing the exhaust gas from a nearby boiler over the outside of the tubes that contain the catalyst. The gas is fed at \(4.94 \mathrm{m}^{3} / \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8}\), entering the unit at \(1400^{\circ} \mathrm{C}\) and 1 atm and leaving at \(900^{\circ} \mathrm{C} .\) The unit may be considered adiabatic. (a) Calculate the molar composition of the product gas, assuming that the heat capacity of the heating gas is \(0.040 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\cdot} \mathrm{C}\right)\) (b) Is the reaction process exothermic or endothermic? Explain how you know. Then explain how running the reaction in a reactor-heat exchanger improves the process economy.

Use Hess's law to calculate the standard heat of the water-gas shift reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ from each of the two sets of data given here. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+1226 \mathrm{Btu}\) $$\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta \hat{H}_{\mathrm{v}}=+18,935 \mathrm{Btu} / \mathrm{lb}-\mathrm{mole}$$ $$\begin{aligned}&\text { (b) } \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-121,740 \mathrm{Btu}\\\&\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ}=-104,040 \mathrm{Btu} \end{aligned}$$

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