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Chapter 5: Problem 28

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

Short Answer

Expert verified
About \(-15686.58\,\text{kJ}\) of heat is released.

Step by step solution

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01

Determine the Molar Mass of NO2

To find how much energy is released, first calculate the molar mass of \(\mathrm{NO}_2\). It is calculated by adding the atomic masses: \(14.01\) (for nitrogen) + \(2 \times 16.00\) (for two oxygens) which equals \(46.01\, \text{g/mol}\).
02

Calculate Moles of NO2 Produced

Next, use the weight of \(\mathrm{NO}_2\) produced, which is \(1.26 \times 10^4 \text{ g}\), to find the moles. Use the formula: \(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\). Thus, \[\text{Moles of } \mathrm{NO}_2 = \frac{1.26 \times 10^4 \text{ g}}{46.01 \text{ g/mol}} \approx 273.87 \text{ mol}\].
03

Calculate Total Energy Change

Since the reaction gives off \(-114.6\,\text{kJ/mol}\) for the formation of \(2\) moles of \(\mathrm{NO}_2\), calculate the total energy released. Thus, for \(273.87\,\text{mol}\) of \(\mathrm{NO}_2\), the energy released is \[\Delta H = \frac{-114.6\,\text{kJ}}{2 \text{ mol}} \times 273.87 \text{ mol} \approx -15686.58 \text{ kJ}\].
04

Provide Final Answer with Sign Indicating Exothermic Reaction

The energy released when \(1.26 \times 10^4 \text{ g}\) of \(\mathrm{NO}_2\) is produced is approximately \(-15686.58\,\text{kJ}\). The negative sign indicates the reaction is exothermic.

Key Concepts

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

Molar Mass
The molar mass of a compound is a fundamental concept when working with chemical equations. Essentially, it tells us how much one mole of a substance weighs in grams. This value is crucial for converting between grams and moles, which are the units most chemical reactions are written in.
To find the molar mass of a compound like nitrogen dioxide (\(\mathrm{NO}_2\), you add up the atomic masses of its constituent atoms. Here, nitrogen contributes an atomic mass of 14.01 grams per mole, and each oxygen atom contributes 16.00 grams per mole. Since there are two oxygen atoms, we calculate the molar mass as:
  • Nitrogen: 14.01 g/mol
  • Oxygen: 2 x 16.00 = 32.00 g/mol
Adding these together gives us a molar mass for \(NO_2\) of 46.01 g/mol. Knowing this allows us to then proceed to calculate the number of moles of \(NO_2\) produced in a given reaction.
Exothermic Reaction
An exothermic reaction is a chemical reaction that releases energy to its surroundings. This release generally occurs in the form of heat, making the surroundings feel warmer. The energy released during such reactions is represented by a negative enthalpy change, denoted as \(\Delta H\).
In our reaction: \[ 2 \mathrm{NO}(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}_2(g) \]The enthalpy change \(\Delta H\) is -114.6 kJ/mol. The negative sign here signifies that energy is released when nitrogen monoxide (NO) reacts with oxygen (\(O_2\)) to form nitrogen dioxide (\(NO_2\)). The energy is released because the products of the reaction (\(\mathrm{NO}_2\)) are at a lower energy state than the reactants (\(\mathrm{NO}\) and \(\mathrm{O}_2\)).
Overall, understanding whether a reaction is exothermic or endothermic (absorbing energy) is essential for predicting energy changes and determining the reaction's impact on its surroundings.
Energy Calculation
Calculating the energy change in a reaction allows us to understand the extent of energy absorbed or released. When you know the moles of a substance involved, and the energy change per mole, you can compute the total energy released or absorbed.
In this exercise, the reaction between NO and \(\mathrm{O}_2\) to form \(\mathrm{NO}_2\) involves an enthalpy change of \(-114.6\) kJ for every two moles of \(\mathrm{NO}_2\) produced.
To find out the total energy change when \(273.87\) moles of \(\mathrm{NO}_2\) are produced, we use the relation:\[\Delta H = \frac{-114.6\,\text{kJ}}{2 \text{ mol}} \times 273.87 \text{ mol}\]This calculation results in a total energy release of \(-15686.58\,\text{kJ}\). The negative sign indicates the process is exothermic. Such calculations help chemists and engineers predict energy changes in reactions, which is crucial for designing and scaling up chemical processes.

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Most popular questions from this chapter

A man ate 0.50 pound of cheese (an energy intake of \(4 \times 10^{3} \mathrm{~kJ}\) ). Suppose that none of the energy was stored in his body. What mass (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes \(44.0 \mathrm{~kJ}\) to vaporize 1 mole of water.)

Calculate the work done (in joules) when \(1.0 \mathrm{~mole}\) of water is frozen at \(0^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\). The volumes of 1 mole of water and ice at \(0^{\circ} \mathrm{C}\) are 0.0180 and \(0.0196 \mathrm{~L},\) respectively. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .)\)

Calculate the standard enthalpy of formation for diamond, given that $$ \begin{aligned} \text { C(graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}(\text { diamond })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \\ \Delta H^{\circ} &=-395.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

A sample of nitrogen gas expands in volume from 1.6 to \(5.4 \mathrm{~L}\) at constant temperature. Calculate the work done in joules if the gas expands (a) against a vacuum, (b) against a constant pressure of 0.80 atm, and (c) against a constant pressure of 3.7 atm. (See Equation 5.4. \((1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J})\).

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: \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)_{\Delta H^{\circ}}=-65.2 \mathrm{~kJ} / \mathrm{mol}\) The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime \(\left[0.946 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right]\) and slaked lime \(\left[1.20 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right]\) 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 500.0 -g sample of water reacts with an equimolar amount of \(\mathrm{CaO}\) (both at an initial temperature of \(\left.25^{\circ} \mathrm{C}\right)\), what is the final temperature of the product, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) Assume that the product absorbs all 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 and \(-285.8 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the standard enthalpy of formation of \(\mathrm{Ca}(\mathrm{OH})_{2}\).
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