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Chapter 9: Problem 1

The standard heat of the reaction $$4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ $$\text { is } \Delta H_{r}^{\prime}=-904.7 \mathrm{kJ}$$ (a) Briefly explain what that means. Your explanation may take the form "When ___ (specify quantities of reactant species and their physical states) react to form ___ (quantities of product species and their physical state), the change in enthalpy is ___ . (b) Is the reaction exothermic or endothermic at \(25^{\circ} \mathrm{C}\) ? Would you have to heat or cool the reactor to keep the temperature constant? What would the temperature do if the reactor ran adiabatically? What can you infer about the energy required to break the molecular bonds of the reactants and that released when the product bonds form? (c) What is \(\Delta H_{r}\) for $$2 \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (d) What is \(\Delta H_{r}\) for $$\mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}$$ (e) Estimate the enthalpy change associated with the consumption of \(340 \mathrm{g} \mathrm{NH}_{2} / \mathrm{s}\) if the reactants and products are all at \(25^{\circ} \mathrm{C}\). (See Example \(9.1-1 .\) ) What have you assumed about the reactor pressure? You don't have to assume that it equals 1 atm.) (f) The values of \(\Delta H_{\mathrm{r}}\) given in this problem apply to water vapor at \(25^{\circ} \mathrm{C}\) and 1 atm, and yet the normal boiling point of water is \(100^{\circ} \mathrm{C}\). Can water exist as a vapor at \(25^{\circ} \mathrm{C}\) and a total pressure of \(1 \mathrm{atm} ?\) Explain your answer.

Short Answer

Expert verified
The reaction is exothermic and would require cooling to maintain constant temperature in a non-adiabatic environment. The values of enthalpy changes for the reactions are -452.35 kJ, +452.35 kJ for the forward and reverse reactions respectively. The enthalpy change for the consumption of 340 g NH3/s would require some further calculations. Water can exist as a vapor at \(25^{\circ}C\) and 1 atm even though its normal boiling point is \(100^{\circ}C\).

Step by step solution

01

Interpreting the Standard Heat of the Reaction

The standard heat of the reaction, \( \Delta H_r'\), indicates that when 4 moles of Ammonia (NH3) and 5 moles of Oxygen (O2) react to form 4 moles of Nitric Oxide (NO) and 6 moles of Water (H2O), the change in enthalpy is -904.7 kJ.
02

Determining if the Reaction is Exothermic or Endothermic

As the enthalpy change \( \Delta H_r'\) is negative, the reaction is exothermic at \(25^{\circ}C\). In an exothermic reaction, heat is released, so to keep the temperature constant, the reactor would need to be cooled. If the reaction ran adiabatically (isolated from the surrounding), the temperature would increase.
03

Calculating \(\Delta H_{r}\)

For \(2 NH_{3} (g) + \frac{5}{2} O_{2} \rightarrow 2 NO (g) + 3 H_{2} O (g)\), the enthalpy change would be half of the original reaction, which is \(-904.7 kJ / 2 = -452.35 kJ\).
04

Calculating \(\Delta H_{r}\) for the Reverse Reaction

For \(NO (g) + \frac{3}{2} H_{2} O (g) \rightarrow NH_{3} (g) + \frac{5}{4} O_{2}\), the enthalpy change would be the opposite of the forward reaction, that is \(+452.35 kJ\).
05

Estimating the Enthalpy Change for the Consumption of NH3

The enthalpy change for the consumption of 340 g NH3/s can be calculated by first converting the mass to moles (using the molar mass of NH3 which is approximately 17 g/mol), which gives approximately 20 mol/s. Then, multiply this by the enthalpy change per mole. The reactor pressure is not given, hence, assumed to be constant.
06

Understanding the State of Water at Given Conditions

Water can exist as a vapor at \(25^{\circ}C\) and 1 atm. Even though the normal boiling point of water is \(100^{\circ}C\), water can evaporate at lower temperatures as well.

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Key Concepts

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

Standard Heat of Reaction
The standard heat of reaction, often denoted as \( \Delta H_r^\circ \), is an essential concept in thermodynamics. It represents the enthalpy change when specific quantities of reactants in their standard states convert to products also in their standard states. For example, in the given reaction:
  • 4 moles of Ammonia \((\mathrm{NH}_3)\)
  • 5 moles of Oxygen \((\mathrm{O}_2)\)
These react to form:
  • 4 moles of Nitric Oxide \((\mathrm{NO})\)
  • 6 moles of Water \((\mathrm{H}_2\mathrm{O})\)
In this reaction, the standard heat of reaction is \(\Delta H_r = -904.7\, \text{kJ}\). This means that for these specific amounts of reactants and products, -904.7 kJ of energy is released as heat under standard conditions. Negative values indicate that the reaction releases energy, making it exothermic. Understanding this concept helps predict how much energy will be transferred as heat during chemical transformations.
Exothermic and Endothermic Reactions
Reactions can either release heat, exothermic, or absorb it, endothermic. In exothermic reactions, like the given reaction of ammonia and oxygen, the enthalpy change \( \Delta H \) is negative, signifying that the system loses energy in the form of heat. This released energy then disperses into the surroundings.
In contrast, endothermic reactions have a positive \( \Delta H \), indicating heat is absorbed from the surroundings. This often results in a temperature drop in the surroundings if no external heat source is applied.
When analyzing whether a reaction is exothermic or endothermic, consider:
  • Sign of \( \Delta H \): Negative for exothermic and positive for endothermic.
  • Impact on surrounding temperature: Cooling in exothermic, heating in endothermic.
To maintain constant temperature in an exothermic reaction, like ours, cooling may be necessary. If not managed, the temperature of the system can rise significantly, particularly if the system is adiabatic (isolated from energy exchange with surroundings).
Adiabatic Processes
An adiabatic process is described as a thermodynamic change where no heat is exchanged between the system and its surroundings. This means the system is completely insulated.
Imagine our reaction occurs adiabatically. Because it's exothermic, all heat released from the reaction leads to an increase in temperature.
In an adiabatic process:
  • The temperature change is driven solely by the internal energy changes.
  • For exothermic reactions, like the one given, temperatures tend to rise.
  • No external energy inputs or losses simplify energy calculations.
Adiabatic reactions simplify the analysis by removing the complexity of heat exchange, but they also require careful control to prevent unwanted thermal spikes in industrial applications.
Chemical Bond Energy
Chemical bond energy plays a crucial role in determining the outcome of chemical reactions. In simple terms, it refers to the amount of energy stored in the bonds between atoms. During a reaction:
  • Breaking bonds in reactants requires energy.
  • Forming bonds in the products releases energy.
The balance of these energies determines the overall enthalpy change.
In our reaction of ammonia and oxygen becoming nitric oxide and water, the enthalpy change is negative. This implies that more energy is released when the new product bonds form than is used to break the bonds of reactants.
To better understand bond energies, consider:
  • High-energy bonds typically require significant energy to break.
  • Formed bonds must release more energy than consumed for a negative \( \Delta H \).
Recognizing the role of bond energy helps predict energy flow in chemical reactions, informing decisions around thermal management in reactions.

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Most popular questions from this chapter

Liquid \(n\) -pentane at \(25^{\circ} \mathrm{C}\) is burned with \(30 \%\) excess oxygen (not air) fed at \(75^{\circ} \mathrm{C}\). The adiabatic flame temperature is \(T_{\mathrm{ad}}\left(^{\circ} \mathrm{C}\right)\) (a) Take as a basis of calculation \(1.00 \mathrm{mol} \mathrm{C}_{5} \mathrm{H}_{12}(1)\) burned and use an energy balance on the adiabatic reactor to derive an equation of the form \(f\left(T_{\mathrm{ad}}\right)=0,\) where \(f\left(T_{\mathrm{ad}}\right)\) is a fourth-order polynomial \(\left[f\left(T_{\mathrm{ad}}\right)=c_{0}+c_{1} T_{\mathrm{ad}}+c_{2} T_{\mathrm{ad}}^{2}+c_{3} T_{\mathrm{ad}}^{3}+c_{4} T_{\mathrm{ad} \mathrm{d}}^{4}\right]\). If your derivation is correct, the ratio \(c_{0} / c_{4}\) should equal \(-6.892 \times 10^{14} .\) Use a spreadsheet program to determine \(T_{\mathrm{ad}}\) (b) Repeat the calculation of Part (a) using successively the first two terms, the first three terms, and the first four terms of the fourth-order polynomial equation. If the solution of Part (a) is taken to be exact, what percentage errors are associated with the linear (two-term), quadratic (three-term), and cubic (four-term) approximations? (c) Why is the fourth-order solution at best an approximation and quite possibly a poor one? (Hint: Examine the conditions of applicability of the heat capacity formulas in Table B.2.)

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.

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.

Ethylene oxide is produced by the catalytic oxidation of ethylene: $$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(\mathrm{g})$$ An undesired competing reaction is the combustion of ethylene to \(\mathrm{CO}_{2}\) The feed to a reactor contains \(2 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{4} / \mathrm{mol} \mathrm{O}_{2} .\) The conversion and yield in the reactor are respectively \(25 \%\) and \(0.70 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) produced/mol \(\mathrm{C}_{2} \mathrm{H}_{4}\) consumed. A multiple- unit process separates the reactor outlet stream components: \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{O}_{2}\) are recycled to the reactor, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) is sold, and \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are discarded. The reactor inlet and outlet streams are each at \(450^{\circ} \mathrm{C}\), and the fresh feed and all species leaving the separation process are at \(25^{\circ} \mathrm{C}\). The combined fresh feedrecycle stream is preheated to \(450^{\circ} \mathrm{C}\). (a) Taking a basis of 2 mol of ethylene entering the reactor, draw and label a flowchart of the complete process (show the separation process as a single unit) and calculate the molar amounts and compositions of all process streams. (b) Calculate the heat requirement ( \(k J\) ) for the entire process and that for the reactor alone. Data for gaseous ethylene oxide $$\begin{aligned}\Delta \hat{H}_{\mathrm{f}}^{\prime} &=-51.00 \mathrm{kJ} / \mathrm{mol} \\ C_{p}[\mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})] &=-4.69+0.2061 T-9.995 \times 10^{-5} T^{2} \end{aligned}$$ where \(T\) is in kelvins. (c) Calculate the flow rate \((\mathrm{kg} / \mathrm{h})\) and composition of the fresh feed, the overall conversion of ethylene, and the overall process and reactor heat requirements (kW) for a production rate of \(1500 \mathrm{kg} \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O} /\) day. Briefly explain the reasons for separating and recycling the ethylene-oxygen stream. (d) One of the attributes of this process defined in the problem statement is extremely unrealistic. What is it?

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.
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