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Chapter 6: Problem 26

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 $$ \begin{aligned} 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{NO}_{2}(g) \\ \Delta H &=-114.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Short Answer

Expert verified
The total amount of heat evolved when \(1.26 \times 10^{4} g\) of \(NO_{2}\) is produced is given by the result of the calculation in Step 3.

Step by step solution

01

Convert grams of \(NO_{2}\) to moles

Using the molar mass of \(NO_{2}\) (46.01 g/mol), number of moles of \(NO_{2}\) can be calculated as \(1.26 \times 10^{4}~ g \,NO_{2} \times \frac{1 \,mol\, NO_{2}}{46.01 \,g\, NO_{2}}\).
02

Apply the stoichiometry of the reaction and enthalpy change

From the balanced chemical reaction, it can be seen that \(2 \,mol \,of \, NO_{2}\) are produced for every \(-114.6 \,kJ\) of heat evolved. Therefore, the heat evolution for the calculated number of moles of \(NO_{2}\) can be calculated as \( \text{Moles} \,of\, NO_{2} \times \frac{-114.6 \,kJ}{2 \,mol\, NO_{2}}\).
03

Calculation

Calculate the total amount of heat evolved by multiplying the moles of\(NO_{2}\) (from Step 1) and the heat release per mole of \(NO_{2}\) (determined in Step 2).

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Most popular questions from this chapter

Calculate the work done in joules when 1.0 mole of water vaporizes at 1.0 atm and \(100^{\circ} \mathrm{C}\). Assume that the volume of liquid water is negligible compared with that of steam at \(100^{\circ} \mathrm{C}\), and ideal gas behavior.

From a thermochemical point of view, explain why a carbon dioxide fire extinguisher or water should not be used on a magnesium fire.

From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$

From these data, $$ \begin{aligned} \mathrm{S}(\text { rhombic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-296.06 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{S}(\text { monoclinic })+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{SO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-296.36 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate the enthalpy change for the transformation $$ S \text { (rhombic) } \longrightarrow \mathrm{S} \text { (monoclinic) } $$ (Monoclinic and rhombic are different allotropic forms of elemental sulfur.)

You are given the following data: $$ \begin{array}{c} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \quad \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \quad \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \\ \Delta H^{\circ}=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}(g)+\operatorname{Br}(g) \longrightarrow \operatorname{HBr}(g) $$
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