The amount of energy required to break a bond is same as the amount of energy released…
Chemistry · JEE Advanced · NTA Exams — Alkyl and Aryl Halides
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below
\[\begin{array}{l} \mathrm{H}_{3} \mathrm{C}-\mathrm{H}(\mathrm{~g}) \longrightarrow \mathrm{H}_{3} \mathrm{C}^{*}(\mathrm{~g})+\mathrm{H}^{*}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=105 \mathrm{kcal} \mathrm{~mol}^{-1} \\ \mathrm{Cl}-\mathrm{Cl}(\mathrm{~g}) \longrightarrow \mathrm{Cl}^{*}(\mathrm{~g})+\mathrm{Cl}^{*}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=58 \mathrm{kcal} \mathrm{~mol}^{-1} \\ \mathrm{H}_{3} \mathrm{C}-\mathrm{Cl}_{(\mathrm{g})} \longrightarrow \mathrm{H}_{3} \mathrm{C}^{*}(\mathrm{~g})+\mathrm{Cl}^{*}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=85 \mathrm{kcal} \mathrm{~mol}^{-1} \\ \mathrm{H}-\mathrm{Cl}_{(\mathrm{g})} \longrightarrow \mathrm{H}^{*}(\mathrm{~g})+\mathrm{Cl}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=103 \mathrm{kcal} \mathrm{~mol}^{-1} \end{array}\]
For the following reaction
\(\Delta \mathrm{H}^{\mathrm{o}}=-58 \mathrm{kcal} \mathrm{~mol}^{-1}\)
The correct statement is
\[\begin{array}{l} \mathrm{H}_{3} \mathrm{C}-\mathrm{H}(\mathrm{~g}) \longrightarrow \mathrm{H}_{3} \mathrm{C}^{*}(\mathrm{~g})+\mathrm{H}^{*}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=105 \mathrm{kcal} \mathrm{~mol}^{-1} \\ \mathrm{Cl}-\mathrm{Cl}(\mathrm{~g}) \longrightarrow \mathrm{Cl}^{*}(\mathrm{~g})+\mathrm{Cl}^{*}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=58 \mathrm{kcal} \mathrm{~mol}^{-1} \\ \mathrm{H}_{3} \mathrm{C}-\mathrm{Cl}_{(\mathrm{g})} \longrightarrow \mathrm{H}_{3} \mathrm{C}^{*}(\mathrm{~g})+\mathrm{Cl}^{*}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=85 \mathrm{kcal} \mathrm{~mol}^{-1} \\ \mathrm{H}-\mathrm{Cl}_{(\mathrm{g})} \longrightarrow \mathrm{H}^{*}(\mathrm{~g})+\mathrm{Cl}(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=103 \mathrm{kcal} \mathrm{~mol}^{-1} \end{array}\]
For the following reaction
\(\Delta \mathrm{H}^{\mathrm{o}}=-58 \mathrm{kcal} \mathrm{~mol}^{-1}\)
The correct statement is
- Initiation step is exothermic with \({ }^{\cdot} \mathrm{CH}_{3}\)
- Propagation step involving \(\Delta \mathrm{H}^{\mathrm{o}}=-2 \mathrm{kcal} \mathrm{~mol}^{-1}\)formation is exothermic with \(\mathrm{CH}_{3} \mathrm{Cl}\)
- Propagation step involving \(\Delta \mathrm{H}^{\circ}=+27 \mathrm{kcal} \mathrm{~mol}^{-1}\)formation is endothermic with \(\Delta \mathrm{H}^{\mathrm{o}}=-25 \mathrm{kcal} \mathrm{~mol}^{-1}\)
- The reaction is exothermic with \(\mathrm{Cl}_{2} \rightarrow 2 \mathrm{Cl}^{*} \text { (Initiation step) } \Delta \mathrm{H}=58 \mathrm{kcal} / \mathrm{mol}\)
Answer
(D) The reaction is exothermic with Cl _ 2 2 Cl ^ * (Initiation step) H =58 kcal / mol
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