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The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Use the values of \(\Delta H_{\mathrm{f}}^{\circ}\) in Appendix 4 to calculate the value of \(\Delta H^{\circ}\) for each of the preceding reactions. b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Is the overall reaction exothermic or endothermic?

Short Answer

Expert verified
The overall equation for the Ostwald process is: $$4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) + \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{HNO}_3(aq) + 6 \mathrm{H}_2 \mathrm{O}(g)$$ To determine if the reaction is exothermic or endothermic, use the calculated values of ΔH° for each reaction and sum them up: ΔH_total° = ΔH_1° + ΔH_2° + ΔH_3°. If ΔH_total° is negative, the reaction is exothermic; if positive, it is endothermic.

Step by step solution

01

Calculate ΔH° for each reaction

To find the value of ΔH° for each reaction, we will use the equation: ΔH° = Σ (ΔHf° of products) - Σ (ΔHf° of reactants) For Reaction 1: $$ΔH_1° = [4ΔHf°(NO) + 6ΔHf°(H_2O_{(g)})] - [4ΔHf°(NH_3) + 5ΔHf°(O_2)]$$ For Reaction 2: $$ΔH_2° = [2ΔHf°(NO_2)] - [2ΔHf°(NO) + ΔHf°(O_2)]$$ For Reaction 3: $$ΔH_3° = [2ΔHf°(HNO_3) + ΔHf°(NO)] - [3ΔHf°(NO_2)+ΔHf°(H_2O_{(l)})]$$ Insert the corresponding ΔHf° values from Appendix 4 (not provided) and calculate ΔH° for each reaction.
02

Combine the equations to write the overall equation

Now, we need to add the three given equations to form the overall equation for the production of nitric acid and water. Starting with the first reaction, and then adding the others, eliminate and combine the reactants and products to obtain the overall equation: $$ 4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_2 \mathrm{O}(g) $$ $$ 2 \mathrm{NO}(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}_2(g) $$ $$ 3 \mathrm{NO}_2(g) + \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{HNO}_3(aq) + \mathrm{NO}(g) $$ The overall equation will be: $$ 4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) + \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{HNO}_3(aq) + 6 \mathrm{H}_2 \mathrm{O}(g) $$
03

Determine if the reaction is exothermic or endothermic

To determine if the overall reaction is exothermic or endothermic, we need to calculate the total ΔH° for the overall reaction: $$ΔH_{total}° = ΔH_1° + ΔH_2° + ΔH_3°$$ If the total ΔH° is negative, the reaction is exothermic. If the total ΔH° is positive, the reaction is endothermic. Once you calculate the total ΔH° using the values determined in Step 1, you can identify the nature of the overall reaction (exothermic or endothermic).

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Key Concepts

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

Enthalpy Change
Enthalpy change, represented as \( \Delta H \), is a term in chemistry that refers to the heat absorbed or released during a chemical reaction under constant pressure. This concept is essential for understanding the energy dynamics of reactions in the Ostwald process.

When a reaction occurs, the bonds in the reactants must be broken and new bonds are formed to create the products. Breaking bonds requires energy, while forming bonds releases energy. The difference between the energy consumed in breaking bonds and the energy released in forming new bonds determines whether the reaction is endothermic (absorbs heat) or exothermic (releases heat).

In the Ostwald process, we calculate the enthalpy change for each step by summing the standard enthalpies of formation (\( \Delta H_{\mathrm{f}}^{\circ} \) of the products and subtracting the sum of the standard enthalpies of formation of the reactants. The standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its elements in their standard states. By doing these calculations, we can analyze the energetics of nitric acid production.
Nitric Acid Production
The production of nitric acid is a significant industrial process, which is primarily achieved through the Ostwald process. The process is a series of chemical reactions that converts ammonia (\( \mathrm{NH}_3 \) into a dilute nitric acid (\( \mathrm{HNO}_3 \) solution.

The Ostwald process starts with the catalytic oxidation of ammonia to produce nitrogen monoxide (NO), followed by the oxidation of NO to nitrogen dioxide (\( \mathrm{NO}_2 \) and finally the absorption of \( \mathrm{NO}_2 \) in water to produce nitric acid. It is important for students to grasp not only the sequence of reactions but also the reasons behind each step, such as catalyst use and temperature control, which affect the yield and concentration of the produced nitric acid.
Chemical Reactions
Chemical reactions are processes that involve the rearrangement of atoms to form new substances. Understanding the types of reactions and how substances interact is fundamental to grasping the Ostwald process. The Ostwald process involves several reaction types:
  • Oxidation: Ammonia is oxidized to form nitrogen monoxide, and then further oxidized to nitrogen dioxide.
  • Combination: Nitrogen dioxide and water combine to produce nitric acid.

Each step has its balance, which can be illustrated and understood through the stoichiometry of the involved chemicals. It's important to note that not all reactions proceed in a single direction; some may reach a state of dynamic equilibrium, affecting the yield of the desired product.
Thermochemistry
Thermochemistry is the study of the heat and energy changes associated with chemical reactions. It provides insights into the energy requirements and changes during the different stages of the Ostwald process. By calculating the enthalpy change for each reaction step, we can determine aspects such as the amount of heat released or absorbed, which informs us about the stability and spontaneity of the reactions.

To evaluate whether the overall Ostwald process for producing nitric acid is energy-efficient, it is crucial to understand the net enthalpy change. As mentioned, if the total enthalpy change over the entire process is negative, the reaction is exothermic, releasing energy to the surroundings, and is typically more favorable for industrial purposes. Thermochemistry, thus, plays a vital role in optimizing industrial production processes like the Ostwald process.

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Most popular questions from this chapter

In a coffee-cup calorimeter, \(1.60 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is mixed with \(75.0 \mathrm{~g}\) water at an initial temperature of \(25.00^{\circ} \mathrm{C}\). After dissolution of the salt, the final temperature of the calorimeter contents is \(23.34^{\circ} \mathrm{C}\). Assuming the solution has a heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of \(\mathrm{kJ} / \mathrm{mol}\).

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm}\). In the process, there is a heat gain by the system of \(350 . \mathrm{J} .\) c. A piston expands against \(1.00\) atm of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

Use the values of \(\Delta H_{\mathrm{f}}^{\circ}\) in Appendix 4 to calculate \(\Delta H^{\circ}\) for the following reactions. b. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow 3 \mathrm{CaSO}_{4}(s)+2 \mathrm{H}_{3} \mathrm{PO}_{4}(l)\) c. \(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Consider the substances in Table 6.1. Which substance requires the largest amount of energy to raise the temperature of \(25.0 \mathrm{~g}\) of the substance from \(15.0^{\circ} \mathrm{C}\) to \(37.0^{\circ} \mathrm{C} ?\) Calculate the energy. Which substance in Table \(6.1\) has the largest temperature change when \(550 . \mathrm{g}\) of the substance absorbs \(10.7 \mathrm{~kJ}\) of energy? Calculate the temperature change.

Consider the following reaction: \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-891 \mathrm{~kJ}\) Calculate the enthalpy change for each of the following cases: a. \(1.00 \mathrm{~g}\) methane is burned in excess oxygen. b. \(1.00 \times 10^{3} \mathrm{~L}\) methane gas at 740 . torr and \(25^{\circ} \mathrm{C}\) are burned in excess oxygen.

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