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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

The standard heat of combustion \(\left(\Delta \hat{H}_{c}\right)\) of liquid 2,3,3 -trimethylpentane \(\left[\mathrm{C}_{8} \mathrm{H}_{18}\right]\) is reported in a table of physical properties to be \(-4850 \mathrm{kJ} / \mathrm{mol} .\) A footnote indicates that the reference temperature for the reported value is \(25^{\circ} \mathrm{C}\) and the presumed combustion products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). (a) In your own words, briefly explain what all that means. (b) There is some question about the accuracy of the reported value, and you have been asked to determine the heat of combustion experimentally. You burn 2.010 grams of the hydrocarbon with pure oxygen in a constant-volume calorimeter and find that the net heat released when the reactants and products \(\left[\mathrm{CO}_{2}(\mathrm{g}) \text { and } \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\right]\) are all at \(25^{\circ} \mathrm{C}\) is sufficient to raise the temperature of \(1.00 \mathrm{kg}\) of liquid water by \(21.34^{\circ} \mathrm{C}\). Write an energy balance to show that the heat released in the calorimeter equals \(n_{\mathrm{C}_{3} \mathrm{H}_{18}} \Delta \hat{U}_{\mathrm{c}}^{\mathrm{S}},\) and calculate \(\Delta \tilde{U}_{\mathrm{c}}^{\mathrm{o}}(\mathrm{kJ} / \mathrm{mol}) .\) Then calculate \(\Delta \hat{H}_{c}^{c}\) (See Example 9.1-2.) By what percentage of the measured value does the tabulated value differ from the measured one? (c) Use the result of Part (b) to estimate \(\Delta \hat{H}_{f}\) for 2,3,3 -trimethylpentane. Why would the heat of formation of 2,3,3 -trimethylpentane probably be determined this way rather than directly from the formation reaction?

The wastewater treatment plant at the Ossabaw Paper Company paper mill generates about 24 tonnes of sludge per day. The consistency of the sludge is \(35 \%,\) meaning that the sludge contains 35 wt\% solids and the balance liquids. The mill currently spends \(\$ 40\) /tonne to dispose of the sludge in a landfill. The plant environmental cngincer has determined that if the sludge consistency could be increased to \(75 \%\) the sludge could be incinerated (burned) to generate useful energy and to eliminate the environmental problems associated with landfill disposal. A flowchart for a preliminary design of the proposed sludge-treatment process follows. For simplicity, we will assume that the liquid in the sludge is just water. Process description: The sludge from the wastewater treatment plant (Stream ? passes through a dryer where a portion of the water in the sludge is vaporized. The heat required for the vaporization comes from condensing saturated steam at 4.00 bar (Stream ?. The steam fed to the dryer is produced in the plant's oil-fired boiler from feedwater at \(20^{\circ} \mathrm{C}\) (Stream ? The heat required to produce the steam is transferred from the boiler furnace, where fuel oil (Stream ? from the boiler furnace, where fuel oil (Stream ? .The concentrated sludge coming from the dryer (Stream ? which has a consistency of \(75 \%,\) is fed to an incinerator. The heating value of the sludge is insufficient to keep the incinerator temperature high enough for complete combustion, so natural gas (Stream ? is used as a supplementary fuel. A stream of outside air at \(25^{\circ} \mathrm{C}\) (Stream ? Is heated to \(110^{\circ} \mathrm{C}\) and fed to the incinerator along with the concentrated sludge and natural gas. The waste gas from the incinerator is discharged to the atmosphere. Fuel oil: The oil is a low-sulfur No. 6 fuel oil. Its ultimate (elemental) analysis on a weight basis is \(87 \% \mathrm{C}, 10 \% \mathrm{H}, 0.84 \% \mathrm{S},\) and the balance oxygen, nitrogen, and nonvolatile ash. The higher heating value of the oil is \(3.75 \times 10^{4} \mathrm{kJ} / \mathrm{kg}\) and the heat capacity is \(C_{p}=1.8 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) Boiler: The boiler has an efficiency of \(62 \%,\) meaning that \(62 \%\) of the heating value of the fuel oil burned is used to produce saturated steam at 4.00 bar from boiler feedwater at 20^{\circ } \mathrm { C } \text { . Fuel oil at } 6 5 ^ { \circ } \mathrm { C } and dry air at \(125^{\circ} \mathrm{C}\) are fed to the boiler furnace. The air feed rate is \(25 \%\) in excess of the amount theoretically required for complete consumption of the fuel. Sludge: The sludge from the wastewater treatment plant contains \(35 \%\) w/w solids (S) and the balance liquids (which for the purposes of this problem may be treated as only water) and enters the dryer at \(22^{\circ} \mathrm{C} .\) The sludge includes a number of volatile organic species, some of which may be toxic, and has a terrible odor. The heat capacity of the solids is approximately constant at \(2.5 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\). Dryer: The dryer has an efficiency of \(55 \%,\) meaning that the heat transferred to the sludge, \(Q_{2},\) is \(55 \%\) of the total heat lost by the condensing steam, and the remainder, \(Q_{3}\), is lost to the surroundings. The dryer operates at 1 atm, and the water vapor and concentrated sludge emerge at the corresponding saturation temperature. The steam condensate leaves the dryer as a liquid saturated at 4.00 bar. Incinerator: The concentrated sludge has a heating value of \(19,000 \mathrm{kJ} / \mathrm{kg}\) dry solids. For a feed sludge of \(75 \%\) consistency, the incinerator requires 195 SCM natural gas/tonne wet sludge \(\left[1 \mathrm{SCM}=1 \mathrm{m}^{3}(\mathrm{STP})\right]\) The theoretical air requirement for the sludge is 2.5 SCM air/ 10,000 kJ of heating value. Air is fed in \(100 \%\) excess of the amount theoretically required to burn the sludge and the natural gas. (a) Use material and energy balances to calculate the mass flow rates (tonnes/day) of Streams ? ? ? ? ? ? and ? and heat flows \(\dot{Q}_{0}, \dot{Q}_{1}, \ldots, \dot{Q}_{4}(\mathrm{kJ} / \text { day }) .\) Take the molecular weight of air to be \(29.0 .\) (Caution: Before you start doing lengthy and unnecessary energy balance calculations on the boiler furnace, remember the given furnace efficiency.) Exploratory Exercises - Research and Discover (b) The money saved by implementing this process will be the current cost of disposing of the wastewater plant sludge in a landfill. Two major costs of implementing the process are the installed costs of the new dryer and incinerator. What other costs must be taken into account when determining the economic feasibility of the process? Why might management decide to go ahead with the project even if it proves to be unprofitable? (c) What opportunities exist for improving the energy economy of the process? (Hint: Think about the need to preheat the fuel oil and the boiler and incinerator air streams and consider heat exchange possibilities.) (d) The driving force for the introduction of this process is to eliminate the environmental cost of sludge disposal. What is that cost- -that is, what environmental penalties and risks are associated with using landfills for hazardous waste disposal? What environmental problems might incineration introduce?

Carbon disulfide, a key component in the manufacture of rayon fibers, is produced in the reaction between methane and sulfur vapor over a metal oxide catalyst: $$\begin{array}{c}\mathrm{CH}_{4}(\mathrm{g})+4 \mathrm{S}(\mathrm{v}) \rightarrow \mathrm{CS}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \\ \Delta H_{\mathrm{r}}\left(700^{\circ} \mathrm{C}\right)=-274 \mathrm{kJ} \end{array}$$ Methane and molten sulfur, each at \(150^{\circ} \mathrm{C}\), are fed to a heat exchanger in stoichiometric proportion. Heat is exchanged between the reactor feed and product streams, and the sulfur in the feed is vaporized. The gascous methane and sulfur leave the exchanger and pass through a second preheater in which they are heated to \(700^{\circ} \mathrm{C}\), the temperature at which they enter the reactor. Heat is transferred from the reactor at a rate of \(41.0 \mathrm{kJ} / \mathrm{mol}\) of feed. The reaction products emerge from the reactor at \(800^{\circ} \mathrm{C}\), pass through the heat exchanger, and emerge at \(200^{\circ} \mathrm{C}\) with sulfur as a liquid. Use the following heat capacity data to perform the requested calculations: \(C_{p}\left[J /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right] \approx 29.4\) for \(\mathrm{S}(1), 36.4\) for \(\mathrm{S}(\mathrm{v}), 71.4\) for \(\mathrm{CH}_{4}(\mathrm{g}), 31.8\) for \(\mathrm{CS}_{2}(\mathrm{v}),\) and 44.8 for \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (a) Estimate the fractional conversion achieved in the reactor. In enthalpy calculations, take the feed and product species at \(700^{\circ} \mathrm{C}\) as references. (b) Suppose the heat of reaction at \(700^{\circ} \mathrm{C}\) had not been given. What would be different in your solution to Part (a)? (Be thorough in your explanation.) Sketch the process paths from the feed to the products built into both the calculation of Part (a) and your alternative calculation. Explain why the result would be the same regardless of which method you used. (c) Suggest a method to improve the energy economy of the process.

A mixture of air and a fine spray of gasoline at ambient (outside air) temperature is fed to a set of pistonfitted cylinders in an automobile engine. Sparks ignite the combustible mixtures in one cylinder after another, and the consequent rapid increase in temperature in the cylinders causes the combustion products to expand and drive the pistons. The back-and-forth motion of the pistons is converted to rotary motion of a crank shaft, motion that in turn is transmitted through a system of shafts and gears to propel the car. Consider a car driving on a day when the ambient temperature is 298 K and suppose that the rate of heat loss from the engine to the outside air is given by the formula $$-\dot{Q}_{1}\left(\frac{\mathrm{kJ}}{\mathrm{h}}\right) \approx \frac{15 \times 10^{6}}{T_{\mathrm{a}}(\mathrm{K})}$$ where \(T_{\mathrm{a}}\) is the ambient temperature. (a) Take gasoline to be a liquid with a specific gravity of 0.70 and a higher heating value of \(49.0 \mathrm{kJ} / \mathrm{g}\), assume complete combustion and that the combustion products leaving the engine are at \(298 \mathrm{K}\), and calculate the minimum feed rate of gasoline (gal/h) required to produce 100 hp of shaft work. (b) If the exhaust gases are well above \(298 \mathrm{K}\) (which they are), is the work delivered by the pistons more or less than the value determined in Part (a)? Explain. (c) If the ambicnt temperature is much lower than \(298 \mathrm{K}\), the work delivered by the pistons would decrease. Give two reasons.

Sulfur dioxide is oxidized to sulfur trioxide in a small pilot-plant reactor. SO \(_{2}\) and \(100 \%\) excess air are fed to the reactor at \(450^{\circ} \mathrm{C}\). The reaction proceeds to a \(65 \% \mathrm{SO}_{2}\) conversion, and the products emerge from the reactor at \(550^{\circ} \mathrm{C}\). The production rate of \(\mathrm{SO}_{3}\) is \(1.00 \times 10^{2} \mathrm{kg} / \mathrm{min}\). The reactor is surrounded by a water jacket into which water at \(25^{\circ} \mathrm{C}\) is fed. (a) Calculate the feed rates (standard cubic meters per second) of the \(\mathrm{SO}_{2}\) and air feed streams and the extent of reaction, \(\xi\) (b) Calculate the standard heat of the SO_ oxidation reaction, \(\Delta H_{\mathrm{t}}^{\mathrm{r}}(\mathrm{kJ}) .\) Then, taking molecular species at \(25^{\circ} \mathrm{C}\) as references, prepare and fill in an inlet-outlet enthalpy table and write an energy balance to calculate the necessary rate of heat transfer ( \(\mathrm{kW}\) ) from the reactor to the cooling water. (c) Calculate the minimum flow rate of the cooling water if its temperature rise is to be kept below \(15^{\circ} \mathrm{C}\) (d) Briefly state what would have been different in your calculations and results if you had taken elemental species as references in Part (b).
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