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Chapter 9: Problem 66

Explain why the bond enthalpy of a molecule is usually defined in terms of a gas-phase reaction. Why are bond-breaking processes always endothermic and bond-forming processes always exothermic?

Short Answer

Expert verified
Bond enthalpy is defined in terms of a gas-phase reaction because gases demonstrate predictable behavior according to kinetic theory, and their homogenous, isotropic nature reduces interparticle interactions, simplifying calculations. Bond-breaking processes are endothermic, as they require energy to overcome the forces binding atoms together. Conversely, bond-forming processes are exothermic because they release energy as the system stabilizes.

Step by step solution

01

Understanding Bond Enthalpy & Gas-Phase Reaction

Bond enthalpy is defined as the amount of energy required to break one mole of a bond in a diatomic molecule while the substance is in its gaseous state. This definition is based on gaseous reactions because gases are homogenous and isotropic, and their behaviour can be defined properly by the kinetic theory of gases. There is also minimum interaction between the particles, which simplifies the calculations.
02

Explaining Endothermic Bond-Breaking Process

An endothermic process absorbs heat from its surroundings, meaning it requires energy to proceed. When bonds are broken, energy is needed to overcome the forces holding the atoms together, thus bond-breaking is an endothermic process.
03

Expounding Exothermic Bond-Forming Process

On the other hand, an exothermic process releases heat into its surroundings. When a bond is formed, the system loses energy as it's released into the environment, hence bond-forming is an exothermic process.

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Most popular questions from this chapter

Explain what an ionic bond is.

Classify these bonds as ionic, polar covalent, or \(\mathrm{co}-\) valent, and give your reasons: (a) the CC bond in \(\mathrm{H}_{3} \mathrm{CCH}_{3},\) (b) the KI bond in \(\mathrm{KI},(\mathrm{c})\) the \(\mathrm{NB}\) bond in \(\mathrm{H}_{3} \mathrm{NBCl}_{3},\) (d) the \(\mathrm{ClO}\) bond in \(\mathrm{ClO}_{2}\).

Write Lewis structures for these molecules: (a) ICl, (b) \(\mathrm{PH}_{3}\) (c) \(\mathrm{P}_{4}\) (each \(\mathrm{P}\) is bonded to three other P atoms), (d) \(\mathrm{H}_{2} \mathrm{~S},\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}\) (f) \(\mathrm{HClO}_{3},(\mathrm{~g}) \mathrm{COBr}_{2}\) (C is bonded to \(\mathrm{O}\) and \(\mathrm{Br}\) atoms ).

Give three examples of compounds that do not satisfy the octet rule. Write a Lewis structure for each.

Using this information: $$ \begin{array}{rr} \mathrm{C}(s) \longrightarrow \mathrm{C}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=716 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{H}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=872.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ and the fact that the average \(\mathrm{C}-\mathrm{H}\) bond enthalpy is \(414 \mathrm{~kJ} / \mathrm{mol}\), estimate the standard enthalpy of formation of methane \(\left(\mathrm{CH}_{4}\right)\).
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