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Ch.5 - Thermochemistry
All textbooksBrown 14th EditionCh.5 - ThermochemistryProblem 86
Chapter 5, Problem 86

(a) The nitrogen atoms in an N2 molecule are held together by a triple bond; use enthalpies of formation in Appendix C to estimate the enthalpy of this bond, D(N‚N). (b) Consider the reaction between hydrazine and hydrogen to produce ammonia, N2H41g2 + H21g2¡2 NH31g2. Use enthalpies of formation and bond enthalpies to estimate the enthalpy of the nitrogen– nitrogen bond in N2H4. (c) Based on your answers to parts (a) and (b), would you predict that the nitrogen–nitrogen bond in hydrazine is weaker than, similar to, or stronger than the bond in N2 ?

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(a) To estimate the enthalpy of the N≡N bond in N₂, use the enthalpy of formation data from Appendix C. The enthalpy of formation for N₂ is zero because it is in its elemental form. Use the bond enthalpy values to calculate the energy required to break the N≡N bond.
(b) For the reaction N₂H₄(g) + H₂(g) → 2 NH₃(g), use the enthalpies of formation for N₂H₄, H₂, and NH₃ from Appendix C. Calculate the enthalpy change of the reaction (ΔH_rxn) using the formula: ΔH_rxn = ΣΔH_f(products) - ΣΔH_f(reactants).
(b) Use bond enthalpies to estimate the enthalpy of the N-N bond in N₂H₄. Consider the bonds broken and formed during the reaction. Calculate the total energy required to break the bonds in the reactants and the energy released when forming the bonds in the products.
(b) The difference between the total energy required to break the bonds and the energy released when forming the bonds gives the enthalpy change of the reaction. Use this information to estimate the N-N bond enthalpy in N₂H₄.
(c) Compare the estimated N-N bond enthalpy in N₂H₄ with the N≡N bond enthalpy in N₂. Determine if the N-N bond in hydrazine is weaker, similar, or stronger than the triple bond in N₂ based on the calculated values.

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Key Concepts

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

Enthalpy of Formation

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 crucial concept in thermodynamics, allowing chemists to calculate the energy changes associated with chemical reactions. By using standard enthalpies of formation, one can estimate the overall enthalpy change for a reaction, which is essential for understanding bond energies and stability.
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Bond Enthalpy

Bond enthalpy, or bond dissociation energy, is the energy required to break one mole of a specific type of bond in a gaseous molecule. It provides insight into the strength of chemical bonds; higher bond enthalpies indicate stronger bonds. In the context of the question, comparing the bond enthalpy of the nitrogen-nitrogen bond in N2 and N2H4 helps assess the relative stability and reactivity of these molecules.
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Triple Bond Characteristics

A triple bond consists of one sigma bond and two pi bonds, resulting in a strong interaction between two atoms, as seen in N2. This type of bond is characterized by its high bond enthalpy, making it one of the strongest types of covalent bonds. Understanding the nature of triple bonds is essential for predicting the behavior of nitrogen-containing compounds and their reactivity in chemical reactions.
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Chemical Bonds Example 2
Related Practice
Textbook Question

Use bond enthalpies in Table 5.4 to estimate H for each of the following reactions: (a)

Textbook Question

Consider the reaction 2 H2(g) + O2(g) → 2 H2O(l). (a) Use the bond enthalpies in Table 5.4 to estimate H for this reaction, ignoring the fact that water is in the liquid state.