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Chapter 9: Problem 52

A 2.00 mole \(\%\) sulfuric acid solution is neutralized with a 5.00 mole\% sodium hydroxide solution in a continuous reactor. All reactants enter at \(25^{\circ} \mathrm{C}\). The standard heat of solution of sodium sulfate is \(-1.17 \mathrm{kJ} / \mathrm{mol} \mathrm{Na}_{2} \mathrm{SO}_{4},\) and the heat capacities of all solutions may be taken to be that of pure liquid water [4.184 kJ/(kg.'C)]. (a) How much heat (kJ/kg acid solution fed) must be transferred to or from the reactor contents (state which it is) if the product solution emerges at \(40^{\circ} \mathrm{C} ?\) (b) Estimate the product solution temperature if the reactor is adiabatic, neglecting heat transferred between the reactor contents and the reactor wall.

Short Answer

Expert verified
The detailed solutions will depend on the specific amounts of the acid and alkali solutions. However, in general, if the heat released by the reaction exceeds the heat required to heat the product solution to 40掳C, heat must be transferred out of the reactor. Conversely, if less heat is released, heat must be added. In an adiabatic reactor, the final product temperature would increase due to the heat of the reaction, calculated using the formula in step 5.

Step by step solution

01

Determining amount of reactants

First, calculate the amount of sodium hydroxide needed to neutralize sulfuric acid. From the molecular formula \(H_{2}SO_{4} + 2NaOH \to Na_{2}SO_{4} + 2H_{2}O\), we see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. So for every 2 moles of sulfuric acid, we need 4 moles of sodium hydroxide to neutralize it.
02

Calculate the heat released by the reaction

The reaction releases heat when forming sodium sulfate. Calculate the amount of heat released per kg of acid solution fed using the formula \(q_{rxn} = n_{Na_2SO_4} \times \Delta H_{f{Na_2SO_4}}\) where \(n_{Na_2SO_4}\) is the number of moles of sodium sulfate that is formed after neutralization and \(\Delta H_{f{Na_2SO_4}}\) is the standard heat of formation of sodium sulfate given as -1.17 kJ/mol.
03

Calculate the heat required to heat the product solution to 40掳C

Then calculate the amount of heat required to increase the product solution temperature to 40掳C using the formula \(q_{heating} = m_{water} \times C_{p_{water}} \times \Delta T\) where \(m_{water}\) is the mass of the product solution, \(C_{p_{water}}\) is the heat capacity of the solution which is equal to that of water, and \(\Delta T\) is the change in temperature.
04

Determine whether heat is transferred to or from the reactor

Now, compare the heat released by the reaction \(q_{rxn}\) with the heat required to heat the solution \(q_{heating}\). If \(q_{heating} > q_{rxn}\), heat must be transferred into the reactor. If \(q_{rxn} > q_{heating}\), heat must be transferred out of the reactor.
05

Estimate solution temperature in adiabatic reactor

Next, for an adiabatic reactor, the heat of reaction would be equal to the heat required to increase the temperature of the solution since heat is not being transferred to or from the system. By rearranging the formula in step 3, we can estimate the product solution temperature in the adiabatic reactor: \(T_{final} = T_{initial} + \frac{q_{rxn}}{m_{water} \times C_{p_{water}}}\)

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Key Concepts

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

Thermochemistry
Thermochemistry is a branch of thermodynamics focused on the study of energy changes during chemical reactions, particularly the exchange of heat. Heat, in this context, is a form of energy transfer between systems and their surroundings caused by a temperature difference.
Understanding thermochemistry involves looking at the heat of reactions, specific heat capacities of materials, and how temperature changes influence the reactants and products in a chemical process.

For instance, when a chemical reaction occurs at constant pressure, the heat exchanged with the surroundings is known as enthalpy change, \( \Delta H \). This value can be either negative, indicating an exothermic process where heat is released, or positive for an endothermic process where heat is absorbed from the surroundings.
Heat of Reaction
The heat of reaction, also known as the enthalpy change of reaction, represents the amount of heat absorbed or released during a chemical reaction. It is denoted by \( \Delta H \) and is measured in joules per mole (J/mol) or kilojoules per mole (kJ/mol).

In the given exercise, the standard heat of solution of sodium sulfate, \( \Delta H_{f\text{Na}_2\text{SO}_4} \), is a form of the heat of reaction, indicating the heat released when one mole of sodium sulfate forms. Since energy is released, this specific reaction has a negative heat of reaction, signifying that it is exothermic. Calculating the total heat of reaction requires knowing the amount of reactant used and how it translates to energy per kilogram of the solution.
Neutralization Reaction
A neutralization reaction usually occurs between an acid and a base, resulting in the formation of water and a salt. In the exercise, sulfuric acid (\(H_2SO_4\)) reacts with sodium hydroxide (\(NaOH\)) to produce sodium sulfate (\(Na_2SO_4\)) and water (\(H_2O\)).

This kind of reaction is typically exothermic, releasing heat. Knowing the stoichiometry of the reaction allows us to calculate the amount of heat liberated when the acid is neutralized. The molarity of the reactants helps us determine the amount of product formed during the reaction, thus facilitating the heat calculation based on the heat of reaction for sodium sulfate formation.
Adiabatic Process
In thermodynamics, an adiabatic process is one where no heat is exchanged with the surroundings. In other words, the system is perfectly insulated, and all the heat generated by the reaction is used to change the internal energy of the system, which often results in a temperature change of its contents.

In the case of the continuous reactor mentioned in the exercise, estimating the final temperature of an adiabatic reaction requires us to assume that all the heat of the reaction is utilized to raise the temperature of the product solution. To find this temperature, we need to know the mass and specific heat capacity of the solution along with the amount of heat produced by the chemical reaction.

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Most popular questions from this chapter

Methane at \(25^{\circ} \mathrm{C}\) is burned in a boiler furnace with \(10.0 \%\) excess air preheated to \(100^{\circ} \mathrm{C}\). Ninety percent of the methane fed is consumed, the product gas contains \(10.0 \mathrm{mol} \mathrm{CO}_{2} / \mathrm{mol} \mathrm{CO},\) and the combustion products leave the furnace at \(400^{\circ} \mathrm{C}\). (a) Calculate the heat transferred from the furnace, \(-\dot{Q}(\mathrm{kW}),\) for a basis of \(100 \mathrm{mol} \mathrm{CH}_{4}\) fed/s. (The greater the value of \(-\dot{Q}\), the more steam is produced in the boiler.) (b) Would the following changes increase or decrease the rate of steam production? (Assume the fuel feed rate and fractional conversion of methane remain constant.) Briefly explain your answers. (i) Increasing the temperature of the inlet air; (ii) increasing the percent excess air for a given stack gas temperature; (iii) increasing the selcctivity of \(\mathrm{CO}_{2}\) to \(\mathrm{CO}\) formation in the furnace; and (iv) increasing the stack gas temperature.

A gaseous fuel containing methane and ethane is burned with excess air. The fuel enters the furnace at \(25^{\circ} \mathrm{C}\) and 1 atm, and the air enters at \(200^{\circ} \mathrm{C}\) and 1 atm. The stack gas leaves the furnace at \(800^{\circ} \mathrm{C}\) and 1 atm and contains 5.32 mole\% \(\mathrm{CO}_{2}, 1.60 \%\) CO, \(7.32 \%\) O \(_{2}, 12.24 \% \mathrm{H}_{2} \mathrm{O}\), and the balance \(\mathrm{N}_{2}\). (a) Calculate the molar percentages of methane and ethane in the fuel gas and the percentage excess air fed to the reactor. (b) Calculate the heat (kJ) transferred from the reactor per cubic meter of fuel gas fed. (c) A proposal has been made to lower the feed rate of air to the furnace. State advantages and a drawback of doing so.

In a coal gasification process, carbon (the primary constituent of coal) reacts with steam to produce carbon monoxide and hydrogen (synthesis gas). The gas may either be burned or subjected to further processing to produce any of a variety of chemicals. A coal contains 10.5 wt\% moisture (water) and 22.6 wt\% noncombustible ash. The remaining fraction of the coal contains 81.2 wife \(\mathrm{C}, 13.4 \%\) O, and \(5.4 \%\) H. A coal slurry containing \(2.00 \mathrm{kg}\) coal/kg water is fed at \(25^{\circ} \mathrm{C}\) to an adiabatic gasification reactor along with a stream of pure oxygen at the same temperature. The following reactions take place in the reactor: $$\begin{array}{l}\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+131.3 \mathrm{kJ} \\\\\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-393.5 \mathrm{kJ} \\ 2 \mathrm{H}(\mathrm{in} \mathrm{coal})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ} \approx-242 \mathrm{kJ}\end{array}$$ Gas and slag (molten ash) leave the reactor at \(2500^{\circ} \mathrm{C}\). The gas contains \(\mathrm{CO}, \mathrm{H}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}^{14}\) (a) Feeding oxygen to the reactor lowers the yield of synthesis gas, but no gasifier ever operates without supplementary oxygen. Why does the oxygen lower the yield? Why it is nevertheless always supplied. (Hint: All the necessary information is contained in the first two stoichiometric equations and associated heats of reaction shown above.) (b) Suppose the oxygen gas fed to the reactor and the oxygen in the coal combine with all the hydrogen in the coal (Reaction 3) and with some of the carbon (Reaction 2), and the remainder of the carbon is consumed in Reaction 1. Taking a basis of 1.00 kg coal fed to the reactor and letting \(n_{0}\) equal the moles of \(\mathrm{O}_{2}\) fed, draw and label a flowchart. Then derive expressions for the molar flow rates of the four outlet gas species in terms of \(n_{0}\). [Partial solution: \(n_{\mathrm{H}_{2}}=\left(51.3-n_{0}\right)\) mol \(\mathrm{H}_{2} . \mathrm{J}\) (c) The standard heat of combustion of the coal has been determined to be -21,400 kJ/kg, taking \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) to be the combustion products. Use this value and the given clemental composition of the coal to prove that the standard heat of formation of the coal is \(-1510 \mathrm{kJ} / \mathrm{kg}\). Then use an energy balance to calculate \(n_{0},\) using the following approximate heat capacities in your calculation: Take the heat of fusion of ash (the heat required to convert ash to slag) to be \(710 \mathrm{kJ} / \mathrm{kg}\).

Formaldehyde is produced by decomposing methanol over a silver catalyst: $$\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{HCHO}+\mathrm{H}_{2}$$ To provide heat for this endothermic reaction, some oxygen is included in the feed to the reactor, leading to the partial combustion of the hydrogen produced in the methanol decomposition. The feed to an adiabatic formaldehyde production reactor is obtained by bubbling a stream of air at 1 atm through liquid methanol. The air leaves the vaporizer saturated with methanol and contains \(42 \%\)methanol by volume. The stream then passes through a heater in which its temperature is raised to \(145^{\circ} \mathrm{C} .\) To avoid deactivating the catalyst, the maximum temperature attained in the reactor must be limited to \(600^{\circ} \mathrm{C}\). For this purpose, saturated steam at \(145^{\circ} \mathrm{C}\) is metered into the air-methanol stream, and the combined stream cnters the reactor. A fractional methanol conversion of \(70.0 \%\) is achicved in the reactor, and the product gas contains 5.00 mole\% hydrogen. The product gas is cooled to \(145^{\circ} \mathrm{C}\) in a waste heat boiler in which saturated steam at 3.1 bar is generated from liquid water at \(30^{\circ} \mathrm{C}\). Several absorption and distillation units follow the waste heat boiler, and formaldehyde is ultimately recovered in an aqueous solution containing 37.0 wt\% HCHO. The plant is designed to produce 36 metric kilotons of this solution per year, operating 350 days/yr. (a) Draw the process flowchart and label it completely. Show the absorption/distillation train as a single unit with the reactor product gas and additional water entering and the formaldehyde solution and a gas stream containing methanol, oxygen, nitrogen, and hydrogen leaving. (b) Calculate the operating temperature of the methanol vaporizer. (c) Calculate the required feed rate of steam to the reactor \((\mathrm{kg} / \mathrm{h})\) and the molar flow rate and composition of the product gas. (d) Calculate the rate ( \(\mathrm{kg} / \mathrm{h}\) ) at which steam is generated in the waste heat boiler. (e) Enough saturated steam was added to the feed to the reactor to keep the reactor outlet temperature at \(600^{\circ} \mathrm{C}\). Explain in your own words (i) why adding steam lowers the outlet temperature, and (ii) the cconomic drawbacks of higher and lower outlet temperatures.

Calcium chloride is a salt used in a number of food and medicinal applications and in brine for refrigeration systems. Its most distinctive property is its affinity for water. in its anhydrous form it efficiently absorbs water vapor from gases, and from aqueous liquid solutions it can form (at different conditions) calcium chloride hydrate \(\left(\mathrm{CaCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\right)\) dihydrate \(\left(\mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)\) tetrahydrate \(\left(\mathrm{CaCl}_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}\right),\) and hexahydrate \(\left(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right)\) You have been given the task of determining the standard heat of the reaction in which calcium chloride hexahydrate is formed from anhydrous calcium chloride: $$\mathrm{CaCl}_{2}(\mathrm{s})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}): \quad \Delta H_{\mathrm{r}}^{\circ}(\mathrm{k} \mathrm{J})=?$$ By definition, the desired quantity is the heat of hydration of calcium chloride hexahydrate. You cannot carry out the hydration reaction directly, so you resort to an indirect method. You first dissolve 1.00 mol of anhydrous \(\mathrm{CaCl}_{2}\) in \(10.0 \mathrm{mol}\) of water in a calorimeter and determine that \(64.85 \mathrm{kJ}\) of heat must be transferred away from the calorimeter to keep the solution temperature at \(25^{\circ} \mathrm{C}\). You next dissolve 1.00 mol of the hexahydrate salt in 4.00 mol of water and find that 32.1 kJ of heat must be transferred to the calorimeter to keep the temperature at \(25^{\circ} \mathrm{C}\). (a) Use these results to calculate the desired heat of reaction. (Suggestion: Begin by writing out the stoichiometric equations for the two dissolution processes.) (b) Calculate the standard heat of reaction in \(\mathrm{kJ}\) for \(\mathrm{Ca}(\mathrm{s}), \mathrm{Cl}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}\) reacting to form \(\mathrm{CaCl}_{2}\) (aq, \(r=10\) ). (c) Speculate about why the standard heat of reaction in forming calcium chloride hexahydrate cannot be measured directly by reacting the anhydrous salt with water in a calorimeter.
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