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The reaction of \(1 \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\) and \(1 \mathrm{~mol} \mathrm{~N}_{2}(\mathrm{~g})\) to yield \(2 \mathrm{~mol} \mathrm{NO}(\mathrm{g})\) is endothermic, with \(\Delta H=+181.8 \mathrm{~kJ} .\) Cal- culate the enthalpy change observed when \(2.20 \mathrm{~g} \mathrm{~N}_{2}(\mathrm{~g})\) reacts with an excess of oxygen.

Short Answer

Expert verified
The enthalpy change for the reaction is 14.27 kJ.

Step by step solution

01

Understand Given Data and Required Calculation

We have the enthalpy change for the reaction of 1 mol of \( \mathrm{O}_2 \) with 1 mol of \( \mathrm{N}_2 \) resulting in 2 mol of \( \mathrm{NO} \). The reaction is: \[ \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) \] and is endothermic with \( \Delta H = +181.8 \mathrm{\, kJ} \). We need to calculate the enthalpy change when 2.20 g of \( \mathrm{N}_2 \) reacts.
02

Convert Mass to Moles

First, convert 2.20 g of \( \mathrm{N}_2 \) to moles using the molar mass.\( \mathrm{N}_2 \) has a molar mass ≈ 28.02 g/mol. Calculate moles of \( \mathrm{N}_2 \):\[ \text{Moles of } \mathrm{N}_2 = \frac{2.20 \text{ g}}{28.02 \text{ g/mol}} \approx 0.0785 \text{ mol} \]
03

Calculate Enthalpy Change for Given Moles of \( \mathrm{N}_2 \)

The reaction's enthalpy change (+181.8 kJ) is for the reaction involving 1 mol of \( \mathrm{N}_2 \). To find the enthalpy change for 0.0785 mol of \( \mathrm{N}_2 \), scale \( \Delta H \) by the mole ratio:\[ \Delta H' = 0.0785 \text{ mol} \times \frac{181.8 \text{ kJ}}{1 \text{ mol}} = 14.27 \text{ kJ} \] Thus, the enthalpy change for the reaction of 2.20 g \( \mathrm{N}_2 \) is 14.27 kJ.

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Key Concepts

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

Endothermic Reaction
An endothermic reaction is a chemical process that absorbs energy, specifically in the form of heat, from its surroundings. In this particular reaction between nitrogen gas (\(\mathrm{N}_2\)) and oxygen gas (\(\mathrm{O}_2\)), heat energy is required to break the strong bonds present in the diatomic molecules, forming nitrogen monoxide (\(\mathrm{NO}\)).
This absorption of heat is what makes the reaction endothermic. As a result, the reaction feels cooler to the touch, because it is drawing in heat from the environment. The enthalpy change, denoted as \(\Delta H\), for this reaction is \(+181.8 \, \mathrm{kJ}\), indicating that the products have a higher energy than the reactants. This positive \(\Delta H\) signifies that the reaction is absorbing energy. When describing reactions, the sign of \(\Delta H\) is crucial:
  • Positive \(\Delta H\): Endothermic, heat absorbed.
  • Negative \(\Delta H\): Exothermic, heat released.
Understanding whether a reaction is endothermic or exothermic helps predict how it will behave under different conditions.
Moles Calculation
The concept of moles is central to understanding chemical reactions, as it allows chemists to quantify substances. In this exercise, we needed to know how many moles of \( \mathrm{N}_2 \) are present when given a mass of 2.20 grams. To calculate the number of moles, you use the formula:
\[\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}} \]
Given that the molar mass of \( \mathrm{N}_2 \) is approximately 28.02 g/mol, the calculation becomes:
\[\text{Moles of } \mathrm{N}_2 = \frac{2.20 \, \text{g}}{28.02 \, \text{g/mol}} \approx 0.0785 \mathrm{mol}\]
This step is crucial because chemical reactions are depicted in terms of moles. Understanding moles allows us to relate the mass of a substance to the amount (in moles) required or produced in a chemical reaction, which further enables calculations of energy changes, like enthalpy changes in this example.
Reaction Stoichiometry
Reaction stoichiometry refers to the quantitative relationships between reactants and products in a chemical reaction. It is based on the balanced chemical equation, which in this exercise is:
\( \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) \)
This equation tells us that one mole of nitrogen gas reacts with one mole of oxygen gas to produce two moles of nitrogen monoxide. The stoichiometric coefficients (the numbers in front of the molecules) indicate the proportions in which the substances react and form.
Stoichiometry is essential when calculating enthalpy changes. In this scenario, the enthalpy change (\(\Delta H = +181.8 \, \mathrm{kJ}\)) is based on one mole of \(\mathrm{N}_2\) reacting. However, if you have a different amount of \(\mathrm{N}_2\), you must scale \(\Delta H\), using: \[\Delta H' = \text{Moles of } \mathrm{N}_2 \times \frac{\Delta H}{1 \, \text{mol}}\] This ensures that the calculated energy change corresponds correctly to the amount of substance involved in your reaction. Overall, understanding stoichiometry is vital for accurately predicting and calculating chemical reaction outcomes.

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Most popular questions from this chapter

Why is it unnecessary to include the enthalpies of formation of elements, such as \(\mathrm{P}_{4}(\mathrm{~s}), \mathrm{H}_{2}(\mathrm{~g})\), or \(\mathrm{C}\) (graphite), in a table of standard enthalpies of formation?

State in words the meaning of the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \\\ \Delta H=-1411 \mathrm{~kJ} \end{aligned} $$

In the 1880 s, Frederick Trouton noted that the enthalpy of vaporization of \(1 \mathrm{~mol}\) pure liquid is approximately 88 times the boiling point, \(T_{\mathrm{b}}\), of the liquid on the Kelvin scale. This relationship is called Trouton's rule and is represented by the thermochemical equation liquid \(\rightarrow\) gas \(\Delta H=88 \cdot T_{\mathrm{b}}\) joules Combined with an empirical formula from chemical analysis, Trouton's rule can be used to find the molecular formula of a compound, as illustrated here. A compound that contains only carbon and hydrogen is \(85.6 \% \mathrm{C}\) and \(14.4 \% \mathrm{H}\). Its enthalpy of vaporization is \(389 \mathrm{~J} / \mathrm{g}\), and it boils at a temperature of \(322 \mathrm{~K}\). (a) What is the empirical formula of this compound? (b) Use Trouton's rule to calculate the approximate enthalpy of vaporization of one mole of the compound. Combine the enthalpy of vaporization per mole with that same quantity per gram to obtain an approximate molar mass of the compound. (c) Use the results of parts (a) and (b) to find the molecular formula of this compound. Remember that the molecular mass must be exactly a whole-number multiple of the empirical formula mass, so considerable rounding may be needed.

Ammonia is produced commercially by the direct reaction of the elements. The formation of \(5.00 \mathrm{~g}\) gaseous \(\mathrm{NH}_{3}\) by this reaction releases \(13.56 \mathrm{~kJ}\) of heat. $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ (a) What is the sign of the enthalpy change for this reaction? (b) Calculate \(\Delta H\) for the reaction, assuming molar amounts of reactants and products.

When \(50.0 \mathrm{~g}\) water at \(41.6{ }^{\circ} \mathrm{C}\) was added to \(50.0 \mathrm{~g}\) water at \(24.3^{\circ} \mathrm{C}\) in a calorimeter, the temperature increased to \(32.7^{\circ} \mathrm{C}\). When \(4.82 \mathrm{~g} \mathrm{KClO}_{3}(\mathrm{~s})\) was added to \(100.0 \mathrm{~g}\) water in the calorimeter \(\left(\right.\) at \(\left.24.3^{\circ} \mathrm{C}\right),\) the temperature decreased to \(20.6^{\circ} \mathrm{C}\). (a) What is the heat capacity of this calorimeter? (b) What is the enthalpy of solution of \(\mathrm{KClO}_{3}(\mathrm{~s})\) in \(\mathrm{kJ} / \mathrm{mol} ?\)

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