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Chapter 6: Problem 26

Determine the amount of heat (in \(\mathrm{kJ}\) ) given off when \(1.26 \times 10^{4} \mathrm{~g}\) of \(\mathrm{NO}_{2}\) are produced according to the equation $$ \begin{aligned} 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{NO}_{2}(g) \\ \Delta H &=-114.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Short Answer

Expert verified
15705.2 kJ of heat is released.

Step by step solution

01

Determine Molar Mass of NO2

Calculate the molar mass of \( NO_2 \): Nitrogen (N) has an atomic mass of about \( 14.01 \, \text{g/mol} \), and Oxygen (O) has an atomic mass of \( 16.00 \, \text{g/mol} \). The formula for \( NO_2 \) includes one Nitrogen atom and two Oxygen atoms, so the molar mass is \( 14.01 + 2 \times 16.00 = 46.01 \, \text{g/mol} \).
02

Convert Grams of NO2 to Moles

To find out how many moles \( 1.26 \times 10^4 \, \text{g} \) of \( NO_2 \) is, use the molar mass: \[ \text{moles of } NO_2 = \frac{1.26 \times 10^4 \, \text{g}}{46.01 \, \text{g/mol}} \approx 274 \text{ mol} \].
03

Calculate Heat Released

According to the reaction, the formation of \( 2 \text{ mol} \) of \( NO_2 \) releases \( 114.6 \, \text{kJ} \). Therefore, \( 1 \text{ mol} \) releases \( \frac{114.6}{2} = 57.3 \, \text{kJ} \). To find the heat released by 274 mol, calculate \( 274 \times 57.3 \approx 15705.2 \, \text{kJ} \).

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Key Concepts

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

Stoichiometry
Stoichiometry is like a recipe for a chemical reaction. It uses the balanced chemical equation to determine the proportions in which chemicals react. This is vital to predict how much of each product is formed. In our case, the balanced equation provided is:
  • \( 2 \text{NO(g)} + \mathrm{O}_2\text{(g)} \rightarrow 2 \text{NO}_2\text{(g)} \)
From this equation, we learn that two moles of nitrogen monoxide \( \text{NO} \) react with one mole of oxygen \( \text{O}_2 \) to produce two moles of nitrogen dioxide \( \text{NO}_2 \).
Stoichiometry allows us to use these ratios to convert amounts of reactants to products. For instance, if we know the amount of \( \text{NO}_2 \) produced, we can figure out how much heat will be released based on stoichiometric ratios from the reaction's enthalpy change \( \Delta H \). Thus, mastering stoichiometry enables you to precisely predict reaction quantities.
Molar Mass
Molar mass is a key concept that tells us the mass of one mole of a substance in grams. For any compound like \( \text{NO}_2 \), you can calculate molar mass by adding together the atomic masses of all the atoms in a molecule. This calculation is important because it lets us convert grams of a substance to moles, and vice versa.
To calculate the molar mass of \( \text{NO}_2 \), you need to sum the mass of one nitrogen atom (\( 14.01 \text{ g/mol} \)) with two oxygen atoms (\( 2 \times 16.00 \text{ g/mol} \)), giving a total of \( 46.01 \text{ g/mol} \).
Knowing the molar mass allows us to take a specific mass of \( \text{NO}_2 \) and determine how many moles it corresponds to, which is essential for further calculations involving reaction heats or other chemical properties.
Exothermic Reactions
Exothermic reactions are reactions that release energy in the form of heat. This energy release occurs because the total energy of the products is lower than that of the reactants. The enthalpy change \( \Delta H \) is negative, indicating that energy is being released.
In our example, the chemical reaction for the formation of \( \text{NO}_2 \) releases \( 114.6 \text{ kJ/mol} \) for every 2 moles of \( \text{NO}_2 \) produced according to the given equation. This means, when \( 1 \text{ mol} \) of \( \text{NO}_2 \) forms, \( 57.3 \text{ kJ} \) of energy is released.
Understanding exothermic reactions helps in predicting the heat output in large-scale applications. When you compute for a significantly larger amount of a product, like the \( 1.26 \times 10^4 \text{ g} \) of \( \text{NO}_2 \) in this exercise, you realize that a significant amount of energy is released, emphasizing the practical importance of thermodynamics in chemical reactions.

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Most popular questions from this chapter

For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}=\) \(\Delta H_{\mathrm{f}}^{\infty} ?\) (a) \(\mathrm{H}_{2}(g)+\mathrm{S}\) (rhombic) \(\longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(\mathrm{C}\) (diamond) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{CuO}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cu}(s)\) (d) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\)

Calculate the standard enthalpy change for the reaction $$ 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ given that $$ \begin{aligned} 2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ}=&-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-822.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Lime is a term that includes calcium oxide \((\mathrm{CaO},\) also called quicklime) and calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2},\right.\) also called slaked lime]. It is used in the steel industry to remove acidic impurities, in air-pollution control to remove acidic oxides such as \(\mathrm{SO}_{2}\), and in water treatment. Quicklime is made industrially by heating limestone \(\left(\mathrm{CaCO}_{3}\right)\) above \(2000^{\circ} \mathrm{C}\) : $$ \begin{aligned} \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \\ \Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Slaked lime is produced by treating quicklime with water: $$ \begin{aligned} \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) \\ \Delta H^{\circ}=-65.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime \(\left(0.946 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) and slaked lime \(\left(1.20 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a 500 -g sample of water reacts with an equimolar amount of \(\mathrm{CaO}\) (both at an initial temperature of \(25^{\circ} \mathrm{C}\) ), what is the final temperature of the product, \(\mathrm{Ca}(\mathrm{OH})_{2}\) ? Assume that the product absorbs all of the heat released in the reaction. (b) Given that the standard enthalpies of formation of \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are \(-635.6 \mathrm{~kJ} / \mathrm{mol}\) and \(-285.8 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the standard enthalpy of formation of \(\mathrm{Ca}(\mathrm{OH})_{2}\)

The average temperature in deserts is high during the day but quite cool at night, whereas that in regions along the coastline is more moderate. Explain.

Consider these changes. (a) \(\mathrm{Hg}(l) \longrightarrow \mathrm{Hg}(g)\) (b) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{3}(g)\) (c) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) At constant pressure, in which of the reactions is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?
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