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Ch.8 - Basic Concepts of Chemical Bonding
All textbooksBrown 14th EditionCh.8 - Basic Concepts of Chemical BondingProblem 105d
Chapter 8, Problem 105d

Acetylene (C2H2) and nitrogen (N2) both contain a triple bond, but they differ greatly in their chemical properties. (d) Calculate the enthalpy of oxidation per mole for N2 and for C2H2 (the enthalpy of formation of N2O5(g) is 11.30 kJ/mol).

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Identify the chemical reactions for the oxidation of N2 and C2H2. For N2, the oxidation reaction can be represented as: \( \text{N}_2 + \frac{5}{2} \text{O}_2 \rightarrow \text{N}_2\text{O}_5 \). For C2H2, the complete combustion reaction is: \( \text{C}_2\text{H}_2 + \frac{5}{2} \text{O}_2 \rightarrow 2 \text{CO}_2 + \text{H}_2\text{O} \).
Use the enthalpy of formation values to calculate the enthalpy change for each reaction. The enthalpy of formation for N2O5(g) is given as 11.30 kJ/mol. Assume standard enthalpy of formation for N2 and O2 is zero since they are in their elemental form.
For the oxidation of N2, calculate the enthalpy change using the formula: \( \Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \). Substitute the known values to find \( \Delta H \) for the formation of N2O5.
For the oxidation of C2H2, use the enthalpy of formation values for CO2 and H2O. The standard enthalpy of formation for CO2(g) is -393.5 kJ/mol and for H2O(l) is -285.8 kJ/mol. Calculate the enthalpy change using the same formula: \( \Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \).
Compare the calculated enthalpy changes for the oxidation of N2 and C2H2 to understand the difference in their chemical properties and energy release during oxidation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy of Oxidation

Enthalpy of oxidation refers to the heat change that occurs when a substance is completely oxidized. This value is crucial for understanding the energy changes in chemical reactions, particularly combustion. For hydrocarbons like acetylene, the enthalpy of oxidation can be significantly different from that of diatomic molecules like nitrogen, due to their distinct molecular structures and bonding.
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Triple Bonds

A triple bond is a type of chemical bond where three pairs of electrons are shared between two atoms, resulting in a very strong bond. In the case of nitrogen (N2), the triple bond contributes to its stability and inertness, while in acetylene (C2H2), the triple bond affects its reactivity and combustion properties. Understanding the nature of these bonds is essential for predicting the behavior of these molecules during oxidation.
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Enthalpy of Formation

The enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It is a key concept in thermodynamics and is used to calculate the enthalpy changes in reactions. In this question, the enthalpy of formation of N2O5 is provided, which can be used in conjunction with the enthalpy of oxidation to determine the overall energy changes during the oxidation of nitrogen and acetylene.
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Related Practice
Textbook Question

Acetylene (C2H2) and nitrogen (N2) both contain a triple bond, but they differ greatly in their chemical properties. (b) By referring to Appendix C, look up the enthalpies of formation of acetylene and nitrogen. Which compound is more stable?

Textbook Question

Acetylene (C2H2) and nitrogen (N2) both contain a triple bond, but they differ greatly in their chemical properties. (c) Write balanced chemical equations for the complete oxidation of N2 to form N2O5(g) and of acetylene to form CO2(g) and H2O(g). Write a balanced chemical equation for the complete oxidation of acetylene to form CO2(g) and H2O(g).

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Textbook Question

Acetylene (C2H2) and nitrogen (N2) both contain a triple bond, but they differ greatly in their chemical properties. (e) Both N2 and C2H2 possess triple bonds with quite high bond enthalpies (Table 8.3). Calculate the enthalpy of hydrogenation per mole for both compounds: acetylene plus H2 to make methane, CH4; nitrogen plus H2 to make ammonia, NH3.

Textbook Question

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of 69.6% S and 30.4% N. Measurements of its molecular mass yield a value of 184.3 g/mol. The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance.

Textbook Question

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of 69.6% S and 30.4% N. Measurements of its molecular mass yield a value of 184.3 g/mol. The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.)